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^,*r4g*sftE3
. Revision Exercises
- Mid-Year/Final Specimen Papers t
. Answers with Contplete Worke{
yTspi(5
. S'lid-Yexrr$imal Speeimen Fap*rs
I Mensuration of Pyramids, Cones and Spheres... . " 1
Tutorial 8.................. . .... . .. .... . 14
Thinking Skills Corner......... . ...... . .... ..... ... 41
9 Graphs of Linear Equations in Two Unknowns . 44
Tutorial 9.................. . .. . 55
Thinking Skills Corner.... . . ..... . .. .. " "" " " 72
Revision Exercise 4 73
10 Graphs of Ouadratic Equations. .. . .. .
13 Probabi1ity...................
84
97Tutorial 10.........
Thinking SkillsCorner...... . .. .. . ..
Set Language and Notaiion.. .. .. ..
Revision Exercise 5
12 Statistics
109
l1 111
124
154
156
240
255
275
277
168
194
Thinking Skills Corner.............. . .. .
Revision Exercise 6
Final Examination Specimen Paper 1 ..... .
Final Examination Specimen Paper 2 .
Answers with Complete Worked Solutions. . .. . . '
296
321
Mensuration of Pyramids,
Gones and Spheres
/EF Pvramids
I ptramia is a solid with a polygon-shaped base and the remaining triangular
faces (lateral faces) meeting at a vertex. A pynmid is named accordinpito the ;hape
lriar]gle-based
pyramid
(also called a
tetrahedfon)
2. In a dght pyramid, the vertex is directly above the centre of tle base. Otherwise. lr
is called an oblique Dvramid.
right pyramid oblique pyramids
square-based
pyramlcl
rectangle-based
pyranid
hexagon based
pyramid
will be used
4.
In a pyra.mid.
. rhe Frlrendicular height of height is the peeendicular distance from the
vertex to the base of the pyramid.
. the slanl height is the height of each of the triangular faces.
. fte slant edge is the edge where the ftiangular faces meet each other.
Volume of p)'ramid = ; Base area \ Heighl
Total surface area of pyramid = Total area of all its faces
= Base area + Area of all lateral faces
(t> WORKED EXAMPLE 1:
The diagram shows a pyramid with a horizontal rectan-
gular base ABCD. yN is perpendicular to ARCD. Given
that yN= 12 cm,.4-B = 8 cm and BC = 5 cm, calculaie the
volume of the pyramid.
SOLUTION:
Volume of pyramid = I x Base area x Height
]r1s,.s;,.rz
160 cml
:i
(D WORKED E){AMPLE 2:
The diagram shows a tetrahedron VABC.
Given that,4-8 = 15 m, CD = 10 m and the volume of the
letraledron is 225 m3, find rhe heighr of the pymmid.
soLuTtoN:
Let the height of the p)Tamid be h m.
Volume of pyramid = 225 m3 (civen)
X Base area X Height = 225l
3
1
5
x[],.rs'ro) ,.r
h
225x3x2
15x10
. . The height of the pyramid is 9 m.
1D WORKED EXAMPLE 3:
The diagram shows a pyramid with a square base of sides l0
and height 12 cm.
(a) Find the slant height of the pFamid.
(b) Draw a net of the pyramid.
(c) Find the total surface area of the p'.ramid.
soLufloN:
(a) AB=10+2=5cm
U.ing Pyrhagora\' theorem on Ay,4-8.
VB'=12'+5'
= 169
VB = ,\a
=13cm
.. The slant hei8hr of the pyramid ir ll cm.
cbaprer 8: Mensuradon oflymd! CltE od 5.6-.: :
(c) Area of sqlare base
=t0x l0
= 100 cnf
Area of each triangular face
l
=;XBasexHeight
l
=;, WORKED EXAMPLE 2:
The volume of a cone with a circular base is 216t mm3. If the base ndius of the cone is
9 mm, find the height of the cone.
SOLUTIONT
Volume of cone = 216r mmr (Given)
t-
- nlh = 2l6n
I
t^
- . n 9' / h =216n
2\@x3
'' ){ 9'
-8mm
._. The heighl of rhe cone is 8 mm
(b)
:
i
I
(D WORKED EXAMPLE 3:
The diagram shows sector AOB of a circle cut from a piece of cadboard. The mms i-e
sector is 21 cm and ZAOB is 150'. The edges OA and.OB are joined together ro male.
hollow cone.
Find
(a) the base mdius,
(b) the height,
(c) the volume
of the cone formed.
22_
SOLUTION:
(a)
t5n" ,,)
Lenpth of arcAB = := \ 2 \ 
= 
\ 2l- Jrr(r' I
=55cm
Let r cm be the base mdius of the hollow cone formed.
2x22
2x+x
(b)
. . The base radius of the cone formed is 8.75 cm.
Let the height of the cone formed be l, cm.
Using Pythagoras' theorem,
h'+8.75'=21'
= 19.09
* l9.l cm (correct to 3 sig. fig.)
. . The height of the cone formed is 19.1 cm.
t_(cl Volume of cone formed = ,-ay'lr
t22=; " ; " 8.75, x le.oe
- 1530 cmr (correct ro 3 sig. fig.)
h" = 21" -8.75"
h =.lzt, - t.ts,
Chdpter 8: Mensuntion of Py.*tal 
"on". 
-U "00".", 
fl
(D WORKED EXAMPLE 4:
A solid cone has a circular base of radius 10.5 m and a total surface area of 940 m'?. Find
the slant height of the cone.
[Take z= 3.142.]
soLuT|oN:
Total surface arca = 940 m? (Given)
tul + ,rf
3.142 x 10.5 x I + 3.142 x lO.5'
32.9911 + 346.4055
32.9911
I
= 940
= 940
= 940
= 593.5945
- 18.0 m (conect to 3 sig. fig.)
. . The slant height of the cone is 18.0 m.
(E> WORKED EXAMPLE 5:
A solid cone of height 12 cm is placed into a cyiinder so that their bases and heights are the
same. 2750 cm3 of water is then poued into the cylinder such that the waterjust covers the
vertex of the corle,
(a) Calculate the radius of the cone.
(b) The cone is then removed from the cylinder Find the drop in the water level.
SOLUTION:
12cm
(a) Volume of cylinder Volume of cone = Volume of water
^t^n'h -nr'h = 2'750
,)
-nr'h = 2750
2"
axnxr'> WORKED EXAMPLE 1:
Find
(a) the volume,
(b) the surface area
of a sphere ofradius 21 cm.
77
ITake ,r = J.lI-
SOLUTION:
4-(a) Volume of sphere = aff'
=ar..]'zt'
= 38 808 cm3
(b) Surface area of spherc = 4nr2
-a'!"1'
= 5544 cm,
Volume of sphere = Jzr' A
Surface area of sphere = 4rr'?
Volume of hemisohere = I -zr //q"t '.da'e' J -ffi-
Sudace area of hemisphere = Curved surface + Flat surface
= 2nr2 + ar2
= 3fir
fr) **..,u
(-> WORKED EXAMPLE 2:
(a) The volume of a spherc is 2800 nt' Find its radius.
(b) The total surface area of a hemisphere is 65 mm:. Find its radius
SOLUTION:
(a) Volume of spherc = 2800 mi (Given)
-rr.r'= 2800
j 2800 x 3
- 8.74 m (corect to 3 sig. fig.)
. . The radius of the spherc is 8.74 m.
Total surface arca of hemisphere'u
:lt,
dtrt
..s$o "t-f'dJ",
"K
The equation of the line of symmehy of the cufle y = x' + pt + 2l is ,r = ,1. Find the value of
The diagram shows the curve with equation 1 = .rl + ,t + c. Find the values of d, , and c.
chaprer l0: Graphs of Quadra,'. **,t"", @
3. The table gives the n and )-coordinates of some points which lie on a cufle.
4.
(c)
(a) Using a scale of2 cm to reprcsent 1 unit on each nxis, plot these poinls and dftw a smooth
curve through them.
(b) Write down thc equation of the line of symmetry.
(c) The points (4.2, ,) and (d, &) lie on the culve. Use your graph to find the values of.r and
b.
(d) The vatues ofr and ), are related by fhe equation J = Pr'+ 0r Use two points on the
curve to obtain a pair of simultaneous equations. Hence calculate the value ofP and the
value of q.
A zookeeper wants to build an enclosure having two similar rectangular stalls to keep
animals. He consrucm a bdck wall on one side and uses 150 metrcs of wire fencing to
construct the olher sides of the enclosure. Given that the width of each recta-ngle^ is .)
metres. show that the lotal area, A of the enclosurc is given by A = (1501 3x') m'.
Using a scale of I cm to represent I unit on the .r- axis and 2 cm to rcpresent 10 units on
rhc A a\i\. drau lhe gfaph ofA - l50r h lor20'y'.to.
Use your graph to find the maximum value ofthe total enclosed aJea and Ihe corresponding
dimensions of each stall.
(b)
@ tt*..",n, ,*- tu
oolh
I the
I the
rg Ia
is.r
rs o:
rdin:
Set Language and
Notation
@ s"t Notation
2.
-t_
l. A set is a collcction of lvcll-defined objects-
Each object in the set is callcd an element or a member of the sel.
We use bnces, { }, to enclosc the clements of t sel. We use capital leltels. e g. A, B.
C, ... to label a sct and sma1l leuers. e.3. d, r, c. ... to dcnote the elements in a set.
E.g. I The set of vowels in 1he English aiphnbel can be wrrtten as:
A = Id, e, i, (). ul
E.g- 2 The set of pime numbers smaller than l0 can be wdttcn as:
B = {2.3.5.7}
E.g. 3 The set of the days of the week can be written as:
C = {Morday, Tuesday, Wedncsday, ThuNday, Friday, Saturday. Sund.ly}
cr,rorcr '. ser ri,a,,xrc.N! *. . ' fii'\-
I
To defme a set, we can
5.
@
@
@
E.g. I A={2,4,6,8}
2 is an element of A. We write 2 e A.
4 is an element ofA. We write 4 € A.
6 is an element ofA. We write 6 € A.
8 is an element ofA. We wriie 8 € A.
5 i\ nol an elemenl ofA. We wrile 5 e A.
F,.8.2 B = {'I : ,r is a rcot of (r lXx - 3) = 0l
(r-lx.v-3)=0
..,v l=0 or .r 3=0
t=I or x=3
List the elements
Describe the elements
Use set builder notation
If r is an el€m€nt of set A, we write .t € A
If r is not an €lem€nt of set A, we write r e A.
e is an element of
is not an element of
1.-R 1G t2 )dR -dt2
A= [2,3,5,11 A is the set of prime
numbers smaller than 10.
A={:r:xisaprime
number WORKED EXAMPLE 2.
{x : r is the square of natulal numbers and -! YlgPKgP EXAMPLE 3:
P= Ia.bl Q=Ia,b,dl R=tb,al s= Ib,eJ
Use e, g, C, q or = to describe the rclation between the following sets.
(a) PandQ (b) PandR (c) Sands
soLuTroN:
(^) PCQ
(b) Since P e R andR e P, wehaveP=R.
(c) sgo
i
I
6i) ""*"*o*^..r"
WORKED EXAMPLE 4r
(a) Ljst the subsets of
(i) {p}, (n) {l,21.
Ia, bj [a, b, cj
(b) Find the number of subsers of {d, ,, c}.
soLuTtoN:
(a) (i) The subsets of {Z} are O and {pl.
(ii) The subsets of { 1, 2l are A, {1}, {2 } and {1,2}.
(D,) a
13. The Universalset, denoted by €is the set that contains all elements being considered
in a given discussion.
Ibj lb, cj
tcl
.. {a, b, c} has 8 subsets.
€ | uni\enal set
E.g. t A = {Letters in the word 'mathematics'}
B = {Irtters in the word 'statistics'}
... A = lm, a, t, h, e, i, c, sj
B = Is, t, t1, i, cl
The universal set of the sets A and B could be the set of lette$ of the English
alphabet.
. . t = \a, b, c, d. e, f, g, h, i, j, k, I, n, n, o, p. q, r., r, r. L, r, w, .x, r. ..1
E.g.3 The unive.sal set ofA = { apples, pexrs, orarges} could be E = {All fruits }.
E.g. 4 The universal set of B = {3, 6, 9. 12J could be € = {.r : i is a multiple of 3 }.
choDFr L: s.r Lrnsuase and 
^"r,,." 
Lr)- \- _/
@ Venn olagrams and Complement of a Set
2.
t. We can also use a Venn diagram to express a set.
In a Venn diagram,
. a large rectangle is used to represent the universalset, €.
. circles or ovals are drawn inside the rectangle to represent the subsets of €.
(@ WORKED EXAMPLE l:
GiventhatE={1,2,3,4,5,6.7,81,A=12,4.5landB={1,2,5,71,&awaVenndiagrao
to represent the sets,
soLuTtoN:
Sets C and D in the example above are called disjoint sets.
Iftwo sets have Ilo elements in common, then ihe two sets are called disjoint sets.
E.g. I IfA = { 1, 3, 5, 7l and B = {2, 4, 6, 81, tlen,4 and B are disjoint sets.
F,.9.2 If H = {Months of the year starting with the lefter '"f I and
tr = { Months of the year starting with the letter 'M' L then
11 and tra are disjoint sets.
(@ WORKED EXAMPLE 2:
Given that €= {p,4, r, s,l, rrl, C
the sets.
soLuTtoN:
3.
= {p, t, &l and D = { r", s}, draw a Venn diagram to represe(
D
t'.)
@ ",*"'"0* 
,"'. ,"
WORKED EXAMPLE 3:
Giventhat E= {5, 10, 15,20,25 J, P = {5, 10} and Q = { 5, r0, 15,201, draw a Venn diagram
to represent the sets.
SOLUTIONT
4. The complement of set A, written A' is the set of all elements in the univemal set €
that are not ir,4. We read A' as 'A complement' or A prime'
A'= {x:i e € ardreAl
E.g l {3, 4, 5, 7, 9l and A = {4, 7, 9}, find A'
A,=t3,5i.-jw
chapter rr, set Lansuaee and *"." 
@
@
lrl W(n|G[' EIAIPLE 4:
$e=Lrr,12,13,74,15;76,11,18], A = {12, 14, 16, 18} and E = { ll, 13, 15, l7},
(a) draw a Venn diagam to reprcsent the sets,
(b, list the elemenr of 4'and B.
soLuTtoN:
(AJ €
(b) A' = { 11, 13, 15, t7l
B' = 112, t4, t6, tqj
and B = i l, 2, 5, 71, find
SOLUTION:
(a) ,{uB= U,2,4,5,'7J,
(b) n(AuB)=s
@ unioti anil Intersection of Sets
t. The union of two sets, ,4 and B is the
A and B. It is denoted by A u B.
AUB= {r:r€A orr € B}
u I the union of
(a) Ar,B, (b) ,?(A u B).
WOBKED EXAMPLE 1:
tf € = {t,2,3,4, 5,6,'1, 8}, A = {1, 4, 7l
set of elements which are in A or in , or in both
6i) 
'"r"."ri", 
ruo. zn
WORKED EXAMPLE 2:
Shade the following regions in the Venn diagram.
(c) Aw B'
(b) (A u B)'
n
Y
il
-x
A*#
(a) A uB
(d) A'uB
SOLUTION:
(a) AUB
(c) Ae B'
(b) (A \J B)'
(e) A'e B'
Teacher's fips:
Step @: Shade the region A.
Slep @: Shade the regioD B'.
The required region A u B' includes
any shaded parts.
Chaoter I l: Set Lanauase tud N",".". f;)- \_/
A'eB
A'w B'
Step @: Shade the region A'.
Step @: Shade the region B.
The required region A' Lr B includes
any shaded par1s.
Step @: Shade the region A'.
Step @: Shade the region B'.
The required rcgion A' u B' includes
any shaded Parts.
i
t (D ",,**o*'*-'u
WORKED EXAi'PLE 3:
lf e = [a, b, c, d, e,f, s\, A = lb, e, dl at\d. B = Ia, d, e, gl, ftnd
(c) (A u B)',(n\ A',
SOLUTION:
(b) B" (d) A'u B'.
(a)
o)
(c)
B'
A'=[a,c,f,stffi
= [b, c, fl
,ff'
(Aw B)' = lc,fl
A^B={x:r€Aandt€Bl
^ 
the intersection of
(d) e
A'w B, = Ia, b, c,f, gl
Alt€rnative method:
A' e B'= la, c.f, gl \J lb, c,fl
= Ia, b, c,f, gl
The intersection of two sets, A and B is the set of elements which are common to
both A and B. It is denoted by A 
^ 
B.
Chapter ll: Sel Language and|.I",",t." 
@
WORKED EXAMPLE 4:
If € = {2, 4, 6, 8, 10, l2l, A = 14,8, loJ and B = {2, 6, 81, find
(b) ,(A 
^ 
B).
10
(a) A^4,
soLuTtoN:
(a) Ar-\B= {8}
(b) n(A r-\ B) = 1
WORKED EXAMPLE 5:
Given that € = {r : r is a natuml number smaller than 9}, find A n B in each of
following.
(a) A = {1, 2, 3} and B = {4, 5J.
(b) A = {5,7} and B = {1,5,7J.
I
SOLUTION:
e= {1,2,3,4,5,6,7,8l
(a) At.'B=Q
(b) A^B= {5,7J
/-]\
(+ 
)
\_-/
@
I
234
(,t] **-** r*- *
WORKED EXAMPLE 6:
If€= {8,9, 10, 11, 12, 13, 14, 15, 16}, A = {8, r0, 12, 14,
C = {9, 11, 13, 15}, find
(a) At.'B, (b) n(A^B), (c) A^C,
(e\ A' ^8, 
(D A^B', (s) A'u B,
SOLUTION:
(a) AaB=|2,141
(b\ n(A n B\ =2
(c\ A. C=O
(d) n(A^C)=0
(e) A'^B= {9, 11, 13, 15} ^ {11, 12, 13, l4J
= {11, 13i
(, A .\ B' = {8, 10, 12, 14, 16} ^ {8, 9, 10, 15, 16}
= {8, 10, 161
(g) A'v B= {9, lt, 13, 15} u {11, 12, 13, 14}
= {9, 11, 12, 13, 14, 15}
(h) A'uB'={9, 11, 13,15}u {8,9, 10, 15, 16}
= {8,9, 10, 1r, 13, 15, 161
(A' o B')' = {12, t4l
161, B = {11, 12, 13, 14} and
(d) /r(A . C),
(h) (A' 0 B)'.
WORKED EXAMPLE 7:
If e = la, b, c, d., e. f. gj, A = la, d, fL znd. B = lb, d, e, fL rtnd
(a) (A^B)', (b) A 
^B',
soLuT|0N:
(a) (A 
^ 
B)' = la, b, c, e, sl €
(c) A' 
^ 
B, (d) A' 
^ 
B'.
(b) A^B'={a}
Chapr€r 1l: Sel Language and *r*" 
@
(c) A'. A = {l', e}
A'. A'= 1..8]
3. The table bel(nv is e sunmluy of set language and notalion.
€
e is not an element of
C is a subset of
g is not a subset of
C is e proper subset of
q is not a proper subset of
number oi elements in set A
€ universal set
Aotll emply set of null set
complement of sel A
A\JB union of ,4 md ,B
A.B intenection of A md I
1. The list of somc rcal numbers are given below.
Integ€rs: ..., 3, 2, 1,0, 1,2, 3,...
Negative integers: -1, 2. 3,1, 5, ...
Positive integ€rs: 1, 2, 3, ,t, 5, ...
Whofe numbe.s: 0, 1,2,3, 4, 5, ...
Natural numb€rsr 1, 2. 3, 4. 5, ...
A prime numbe. is a natural number that has exactly 2 different factors, 1 and itself.
Prime numbers: 2, 3, 5, 7. I l, 13, l7, 19,23, ...
A conposite number is a natural number that has more than 2 diflerenl factors.
Composite numbers: 4, 6, 8, 9, 10, 12, ...
A rational number can be written in the fornr
l'tl
E.s. -l -. --.8.9- are ratronal n mbcrs.
@ ,o*.-.'* r,-.u
f, where a and b are intcgers and , + 0.
The shaded regions represent each of the following sets.
(A U B),
A' 'J B
Y
N- n
x
A
A'^B
(A 
^ 
B)'
A^B'
A'r B' A'^ B'
Chalter ll: Set Languge and *"*t 
@
Tutorial
1. (D
(ii)
List the elements ofA.
Find r(A).
A = {Letters in the word 'strategy'li (a) tol {Days of the week stafting wilh
the letter 'T'l
(d) A = {x : .{ is a positive odd number
smaller than 101
(e) A = {.r : jr is a prime number
smaller than 20]
(f) A= {n : r is a factor of I 8 }
\x : r = 2k + 1, i is a positjve
integerrrr i lrr.:. sr
Chapter I I : set Lansnage and r*,,t G,)
9. Fill in the boxes with e or C to make each statement ffue'
at ,1, -1,o,r, 
o, ul ro t,rrl' jt", r,.r
@ ,1,, . ){apple, 
pear, oranse} (d) {s, 18} L ., lftt, tf
10. If A = ic, d, t], state whether each of the following is True (T) or False (F)'
i_ - I--'.
I I (b) {a'rl cA i
(d) {c}cA i ,
(l AeA I
itr -.-, l
---
@) lai=o I j
G) [c,a,t]c tr,r,.t I I
I I . If B = { Letters in the word 'statistics' I ' 
state whether each of the following is True (T) or False
(F).
n16y = 10
I
i' l
il
(a) c€A
G) OEA
(^) la, ci C B
@, ACB
(e) {d, t, t, c, si C B
(b)
(d)
@ ""*.-",", '*- ,"
P= {.rr: r is arootofl I =0}
o= \rJ
R= {0, l,2}
s= t 1, 1l
Use C, C, e or G to describe the relation between the following sets.
(a) PndQ
(c) O and R
A = {x : .I is a positive odd integer10, ll, 12 }, R = {.r : r is a factor of 18}andS= {.x:5t 1 > 34},
trDd
(a) R, (c) F r-\ S, (d) R u s',
(e) n(R'. S'), (0
s,
/r(R' \J Y).
Chaple.ll: Serl, suge dd *.rt"" 
@
€ = {Letters in the word 'lrigonomehy'}
A = {Letters in the word 'geometry }
B = {Letters in the word 'time'}
(a) List the elements of
ti, rA ' Br'. rii,4'r8. riii)
(b) Find
(i.) ,r(A' 
^ 
B'), (iD
numberl.
List the elements of
(a) A', (b) B"
(e) {1, 3, 7} r-] B.
(A \r B)', (iv) AUB'.
n(A' 'J B').
42. €= {t:-risapositive integerandtI
rre ) axi-. Jrcw the gruph of \ - r, lrr'\ 'tl lor '1 
3}
(i) List the elements ofM r') M
(ii) Filtd n(-M).
RdkronErerc$e5 
@
31. (a) On the Venn diagram, add the set R which is such that R C M and R 
^ 
N = A.
(b) € = {r is an integer, I2:
The dot diagram below shows the time laken, in minutes, by each of a group of students fo
complete a task.
(b)
10 25 30
cr."rt". rz,sr"ri.ri". @
(b)
(a) How many students were there in the Broup?
(b.) Find the percentage of students who took more thnn 25 minutes to complete the ta+
(c) Comment bdefly on the data.
SOLUTION:
(a) Number of students in the group = 20
Nrmber of students who took more than 25 minutes to complete the task = 5
Requjred percentage
= _ r00Ea
The data vary between 5 and 30. The shofiest time taken was 5 min and the longes
time taken was 30 min. The most common time taken was 25 min. The data clust*
arcund 25.
(c)
,7
I
0
8
3
2
4
2
The numbers on the left of the ve(ical line are the stems. Here tle
stems are the tens digits of the ages.
The Dumbers on the right of the vertical line are the leaves. Here the
leaves are the units digits.
@ st"- and Leaf Diagrams
2
l
4
5
6
In the above example, 2 7 represents 27.
Frcm the stem and leaf diagmm, we can see that
the youngest employee is 27 years old
- the oldest employee is 63 years old.
- the most common age is 42 years,
L We can use a st€m and leaf diagram to organise data so that if is easy to spot patterlls
and make comparisons. No original values arc lost in a stem and leaf diagram_
2. In a stem and leaf diagram, each value is splir into two parts, the st€m and the leat
E.g. The stem and leaf diagmm below shows the ages of 20 employees of a banl
Stem Leaf
@ ***.",",'r*- ru
WORKED E)(AMPLE 1:
The masses, measured to the nearest kiloBram, of 20 boys are given below
Represent the data in a stem and leaf diagram-
Srcp @: Determine the stem and teaf units. Shce the data arc 2-digit numbers, we take
the tens digit to be tle stem and the units digit as the leaf-
E.g. For the value 57, 5 is the stem and 7 is the leaf.
For the vatue 49, 4 is the stem and 9 is the leaf.
The masses mnge frcm 40s to 80s. Here, we take the ten digits of the number
as the stems. Place the stem digirs, i.e. 4, 5, 6, 7 and 8 in a column ftom the least
to the grcatest on the left side of the vertical line.
Step @:
Stem Leaf
4
5
6
'7
8
Cr'"lr- rZ,St"ti.rr", @
Step O: The leaves are the units digits. Write each leaf to the right of each stem i
ascending order starting fiom tbe lcft
E.g.4l is split into its slem digit '4' and its leafdigil 'l
Stem Leaf
I
0
0
6
t
2
I
0
5
6
1
cI+ WORKED EXAMPLE 2:
The data below rcpresents the time' in mjnutes, taken by a group of 30 students to sohe -:
mrthemttics question.
Represent the data set in a stcm and leaf diagram
SOLUTION:
Stem Leaf
17 899
Tte above stem and lcaf diagram groups lhe data into 3 class intervals When eroupirr
data. try to group then into 5 to ll class intcrvals'
Tbe stem ancl leaf diagram on the fbllowing pag e \ho$\ cach stcm dr\phled t$ice ie one
stem for the leaves 0-4, and the other stem lor ler\es 5 a The re\ultrngdLiSrurnrscalled
a stem and l€af diagram wjth split stems-
045668
011222146
000133579
L
2
3
@ "'n.."..,*-,u
Stem Leal
I
1
2
2
3
3
04
56
01
6',7
00
5',7
68
l2
78
0l
9
22
89
33
41
9
WORKED EXAMPLE 3:
The stem and leaf diagram below shows the heights, in cm, of a $oup of children.
Stem Leaf
13
t4
15
17
5788
011246
0000134579
3345889
156
n.) rJ: represenG L' cm r
Find the number of children in the group.
Find the height of the tallest and shofiest child.
Find the percentage of children who are taller than 150 cm.
Descdbe the shape of the distribution.
(a)
(b)
(d)
(b)
soLufloN:
(a) Number of children in the group
-4+6+10+'7 +3
=30
Height of tallest child = 176 cm
Height of sho(est child = 135 cm
Number of childrcn who arc taller than 150 cm = 16
Required percentage
= _ 100'70
Cha!'terlr:sramdcs @
(d) The distdbution is synrmeffical with a peak at the middle. Most of the children
heights between 150 cm and 159 cm. There are about the same number of
with heights above and below the middle values.
(E> WORKED EXAMPLE 4:
The stem and leaf diagram below shows the masses, in kilograms, of 40 parcels
Stem Leaf
22
22
23
23
24
033
000r
6671
ot23
6788
7'7
12
88
4
34
88
2223
8
1i",,,.139'-",,r', 1o G I
(a) Wdte down the mos[ common mass.
(b) 507, of the parcels have a mass of below t kg each. Find the value ofr'
SOLUTION:
(a) The most common mass = 23.2 kg
(b) 50Vo of 40
== 40
=20
.. 20 parcel. have a mas. of below 2J.3 Lg.
... x = 23.3 kc.
3. We use a back-to-back stem and leaf diagram to comparc two sets ol rclated data
To consffuct a back-to-back stem and leaf diagram, we dmw a stem and leaf diagm!
with a colnmon stem in the middle and leaves of each data on both sides of the sten
@ ".*-"0*.,.,'u
WORKED EXAMPLE 5:
The ages of the teachers of two different schools are recorded below.
40 61 52
s8 50 37
65 64 54
58 52 41
63
56
43 59 38
50 63 s3
30 5l
11 29
24
40
37
25
28
30
25
28
30 26
33
35
2:7
2',7
40
(a) Represent the data in a back to-back stem and leaf diagram.
(b) Which school had the oldest Ieacher?
(c) Which school had the youngest teacher?
(d) Compare the distribution of the ages of the teachem of these two schools.
soLuTtoN:
(a) Leaves for School A Stem Leaves for School B
988643
5
87
310
2200
4331
7 8 89455667
000357
001
1
2
l
5
o
(b) School A had the oldest teacher (65 years old)
School B had the youngest teacher (24 years old)
The ages of ihe teachers in School A clustef around 50 to 60 years old. The ages of
the teachers in School -8 cluster arcund 20 to 30 yean old.
Thus, the average age of the teacheN in School A is more than those in School B.
(c)
(d)
@
@t uoa"
1.
2.
The mean, median and mode are measures of central tendency' A measwe df
""nJ 
t"na"n"V i, 
^ 
tingle value that descdbes where the data :re centred' i-e- ia'
avemge value.
The most common value in a set of data is called the mode
3. ln some distributions' no value appears more than once So there ts no mode'
ln other distributions, therc may be morc th:n one mode'
(@ WORKED EXAMPLE 1:
Find the mode(s) of tle following sets of numbe$'
(a) 54, 69' 70, 72, 80
(b) 24,26' 2E ' 28' 28 ' 29,29
(c) 5,6,'7,'7' 9' 12' 12, t5
SOLUTION:
(a) 5,+, 69, 70' 72. 80
There is no mode.
(b) 24, 26, 28, 28, 28, 29 ' 29
Mode = 28
@>
(c) 5. 6,1 ,7, 9. 12,12. 15
Mode=7and12
WORKED EXAMPLE 2:
la) The heights of 9 bols. in cm are:
'53. 1ro, 164. 156. 175. 156 178 175 156
Find fte mode.
(b) If the heights of 3 more boys. 178 cm' 175 cm and 180 cm are added to the heigtB
of the 9 boys. find the mode'
SOLUTION:
(a) 156, 156, 156, 158, 175, 175, 178. l7q
(b) 156, 156, 156, 158, 164, 1?5, 175' 175'
Mode = 156 cm
Mode = 156 cm and 175 cm
r @ "o*"-",n, 
,'- ,"
I78, 178, 17q. I80
WORKED EXAMPLE 3:
Find the mode of each of the following.
(a)
7 8 910 11 t2 13
Lensth (cm)
Leaf
3
4
5
6
7
(b)
134
00?
022
455
033
7
888
26'7
556
36
(d)
40
30
20
t0
0
Red Blue G€en
Colour oi ca6
Sllver
soLufloN:
(a) Mode = 1l cm
(b) Mode = 48 and 65
(c) Mode = $21
(d) Mode = Silver
Chaeter12:stathlics @
@ u"aiutt
1
2.
(a)
The value exactly in the middle of a set of ordered numbers (ascending or descenG
ing) is the median.
Steps to find the m€dian of a s€t of tr data:
(l@
SOLUTION:
Arrange the numbels in ascending order first.
Middle position
(a) 2, 3,5, 6 , 8,9, 10
Median = 6
Middle posillon
(b) 12, 14, 15, 16, 18, 19
l5 + l6
Median = -:-
= 15.5
(@ WORKED EXAMPLE 2:
Find the median of each of the ibllowing.
WORKED EXAMPLE 1:
Find fhe median of the following sets of numbers.
(a) 3, 6, 2, 8, 10, 9, s (b) 12, 16, 19, 15, 18, 14
rhr
---#
5678910
nme (seconds)
-#21 22 23 24 25 26
Speed (kn/h)
€) Arrange the numbers in ascending order, i.e. from the leasf to the greatesl-
@ If11 is odd, the median is the middle value.
If n is ev€n. the median is the mean of the two middle values-
@ m*..*n. t*-.u
(d)
2
3
4
5
1 
r,l.rjo 6;;;,,r. l
12
46
-; ''-:: ^ -
SOLUTION:
(a) Total number of data = 12
1) + 1
Middle polihon = ----= 6 5th Position
. . Median = Mean of 6th and 7th values
2
= 24.5 ktn4t
0123315
24558
013478889
0223399
003
1225
0466',1
3455889
0013
chapd12:srarinrs 6]
(b) Total number of data = 15
15+1
Middle position = ---
= 8th Position
..Median=7s
Total number of data ='7 + 5 + 9 + 7 = 28
,R+l
Middle position = --
= 14.5th Position
Median = Mean of 14th and 15th values
41 +13
2
(d) Total number of dat^ = 3 + 4 + 5 + 1 + 4 = 23
23+l
Middle Position = -- ^
= 12th Positlon
Median = 44.7 mm
(e) Total number of apples = 3 + 2 + 4+ 3 + 6 = 18
li{ + l
Middle position = )-
= 9.51h Position
Median = Mean of 9th and 10th values
84+86
=2
=8sg
(0 Total numb€r of cals = 9 + 17 +
51 +1
Middle position = 2-
= 26th Position
Median = 2 people
12+8+4+1=51
@ "".*",", t*- r"
The Mean
I - The mean of a set of data is obtained by dividing the sum of all the dafa by the total
number of dcta.
Sumofdata
Number of data
2. The mean of a set of r data, .rr, .rr, -ri, .... .v,,, denoted by t (read jr bar) is given by
rt +r2 +:r? +... +"!,
8+9+4+5+7+12(a) Mean =
WORKED EXAMPLE 1:
(a) Find the mean of the following set of numbers.
8,9,4,5,7,12
(b) The mean of five numbers is 36. Tltee of the numben are 28, 3I and 39. If each of
the other two numbers is equal to r, find the value ofr.
(c) The mean of six numbem is 21. lf another number is added, the mean of the seven
numbers is 24. What is the number added?
SOLUTION:
45
(b) The sum of the 5 numbers = 5 X 36 = 180
28 +31 +39+-r+.r = 180
2r+98 = 180
\ =82
The sum of the 6 numbers = 6 X 21 = 126
The sum of the 7 numbers = 7 ) WORKED EXAMPLE 2:
The mean of a at\d b is 24. The mean of d' b, x ar,d ) is 3t Find the mean ot
and )-
SOLUTION:
Thesumof .tandb=2 x 24=48
Thesumofa'D,-IandY=I I3l = 124
. . The sum of .x and ] = 124 48 = '/6
16
The mean of r and Y = ;1 = 38
3. Given a set of data, t!, ,t2, t3, .. 
' 
r,, occuring wjth corespondjng ftequencies'i'i'-i
...,/,, its mean, t is given bY
v",", o =4rlfiifrf
_ t"ft- r.f
(@ WORKED EXAMPLE 3:
A survey was carried out to find the number of hours a group of 50 students spent on Ih
jntemet in a particular week.
Find the mean number of hours spent on the internet per student-
SOLUTION:
0x8+ I x 12+2x9 +3x ll +4x6+5 x4
Mean = 
-1A
50
=2.14h
50
@ r"*"'",n. 
^.. 'u
Altemative m€thod:
tft
Mean = -7
107
50
= 2.t4h
WORKED EXAMPLE 4:
A survey was conducied to find the number of siblings each of a group of students has. The
rcsults are shown in the table beloq
(a.l
(b)
(c)
(d)
If the median is 3, find
(i) the largest possible value of .x,
(ii) the smallest possible value ofr.
If the mode is 2, write down the smallest possible value of r.
If the mode is 5, write down the Iargest possible value of-r.
If dre mean number of siblings is 2.9, find the value of r.
soLuTtoN:
gj4' l' ;' I'3'7
WW
2. . 2.S. .@.4. ....4. s. .... s
(i) 3+7+r=3+6+8
l0+x=17
. . The large(l possible \alue of .r = ;.
(4,
Chaeler lr: Statistics 
@
(b)
(ii) 3+7+r+3=6+8
13+'I =14
. - The smallest possible value oft = 1
If the mode is 2,
the smallest possible value of -t = 9. - ..--S Sffiii
(c) lf the mode js 5,
rhe larsesr possibte vatue of x = ?. .-iii.*ifffi,Sl
0x3 + Ix l +2tx+ 3 x4+4 x 6+5x 8
3+7+.r+4+6+8
83+2r
(o) Mcan number of siblings = 2.9 (Given)
28+i
83+2{=2.9(28+r)
83 + 2ir = 8l.2 + 2.9.r
1.8 = 0.9-r
1.8
'' 0.9
WORKED EXAMPLE 5:
The number of goals scorcd during 30 soccer matches in an inter school toumament
season is shown in the table below
(a) Showthatn+)=8
ibj t{ th" In.on nu.ber of goals scored is 2 9, show that t + 5) = 20
(c) Find the values of; and 1.
(d) Hence, state the modal number of goals scored
soLuTloN:
(a) Total number of matches = 30
3+i+7+5+2+l+5=30
22+r+l =30
x+J =8 (Shown)
@ "*n".",*. 
t*- tu
0x3+l
Mean number of goals
xr+ 2x'7 +3 x 5 +4x2+5x]J+6 x 5
30
0+:r+14+15+8+5)+30
67+r+5)
30
6'7 +r+5!
t+51 = 20 (Shown)
(c) r+) =8
r+5)=20
(2) - (1): 4y = 12
Substitute ] = 3 into (1)'
i+3 =8
..r=5 and )=3
_fl)
_ (2)
(d)
Further Worked Examples
WORKED EXAMPLE 1:
The table below shows tle scores obtained by 40 players in a cefiain game.
Find
(a) the modal score,
(b) the median,
(c) the mean.
SOLUTION:
(a) Modal score = 4
4n+1(b) MiddJe position = 2::
= 20.5th position
Modal number of goals scored = 2
t 2 3 1 5 6 7
8 7 5 10 3 2 5
a*r*,, ,*,*u* @
Median = Mean of 20th and 21st values
3+1
2
1 x 8+2 x 7+3x 5 +4x l0 +5 x3+6x 2+ 7 x5
Mean =
139
40
= 3.475
(K> WORKED EXAMPLE 2:
The diagram illustrates the number of occupants per flat for a sample of 100 flats iq
particular HDB block.
(a) State the modal number of occupants per flat-
(b) Calculate the mean number of occupants per flat.
(c) Find the medjan number of occupants per flat.
40
:10
35
l0
flats
20
l5
10
5
0 012345
Number of occiLpanh per flat
SOLUTION:
(a) Modal number of occupants per flat = 4
(b) Mean number of occupants per flat
0 x 5 +1 x 10+2x 15 + 3 x 25 + 1x40+5 x5
300
100
@ ""*..*"'^',tu
100
lsi
(c)
lno+l
Middle position = -j
= 50.5th Position
Median number of occupalts per flat = Mean of 50fh and 5 1 st values
3+3
2
online in January with her sister who is stltdying ovelseas.
I
2
3
4
5
2
5
5
I
5
3
6
8
3
6
9
3
,7
4
8 9
l{:.", ,i51"!1"*o 1511 I
Find (a) the mean,
(b) the mode,
(c) ihe median.
soLuT|0N:
WORKED EXAMPLE 3:
The stem ard leaf drasram below shows the number of minutes each time Caroline chatted
Stem Leaf
(al Mean =
5
1
5
6
3
6
8
3
9
3
,7
a3+162+292+45
2l
5
0
2
5
8
9
@
8
582
.21
- 27.7 min (colrect to 3 sig. fig.)
rb) Vode = 2r min .-'6ffi1"$#-.
d! :Lr".'1l(4. ii.
Chaeterlr:statsdcs @
(c) Middle position =
Median = 24 min
2l+1
2
11th position
@ Comparison between the Mean, Median and Motle
(D 1rygp659 gx66p1g 1'
The daily eamings, in dollars, of Mr Tan in a week are:
80, 88, 86, 89, 87, 470, 80
(a) Find the
(i) mean,
(ii) mode,
(iii) median
of the ditdbution.
(b) Which avenge gives the best picture of Mr Tan's daily earnings?
SOLUTION:
(a) (i) Mean d?ily eamings
$(80 + 88 + 86 + 89 + 87 + 470 + 80)
l. Although the mean is the most cormronly used average, it may not always be
most appropdate choice.
$980
1
= $140
(ii) Mode = $80 *-ffi
(iiD 80, 80, 86, 87,88,89,4?o
t
Middle losiiion
Median = $87
(d) The median gives the best picture because six ofthe daily eamings for the week
between $80 and $89.
The mean is $140, but six of the daily eamings were less than $90.
The mode is $80 but five of the daily eamings werc morc than $80.
(O 
""*"."0- ^.. 
o
WORKED EXAIIIPLE 2:
The sizes ofjeans sold by an apparel shop on a certain day are shown below:
25, 28, 28, 26, 25, 28. 28, 26, 28, 26
(a) Find the
(1) mean,
(ii) median,
(iii) mode
of the sizes ofjeans sold that day.
O) Which average gives the best picture of the sizes ofjeans sold on that day?
soLuT|0N:
25+28 r28+2b+25 | '8 r28'2b- '8 -26
(a) (i) Mean = 10
264
t0
= 26.8
25, 25, 26, 26, 26, 28, 28, 28, 28, 28
Middle oosition= :#
= 5.5th Position
Median = Mean of 5th and 6th values
26+28
2
(jD
(a)
(iii) Mode=28 .---W
(b) The mode gives the best pictule since it shows the most popular size ofjeans sold.
WORKED EXAMPLE 3r
The table below shows the number of people living in each of 50 flats in an HDB block'
Find the
(i) mean,
(ii) mode,
(iii) me.dian
of the disdbution.
Which average gives the best pictue of the number of people livinB in each flat?(b)
chaprer 12: sdiinics 
@
soLurl
(a) (i
2.
(b)
TION:
l\2+2reliable
representatrve
of numerical
data Fovided
there aie no
Median Not affected by
exfteme values.
Its calculation uses
only one or two
middle values.
Does not use the
total quantity
represented by the
oata.
It is a better
representation
of average tha
the mean wh€!
thete are
extreme valu6
Mode Not affected by
exteme values.
There may be more
than one mode or it
may not exlst,
Does not use the
total quantity
represented by the
data.
Useful as a
measurc of
opinion and
popularity.
Tutor 28
Mean for Gmuped Data
1 . To calculate the mean for grouped data (where data arc grouped into inteflals). use
f,A"un, ; = #
Mean. .i = +LJ
5030
60
= 83.8 (correct to 3 sig. fig.)
. . the mean score of the group of sfudents is 83 8.
where r = mid-value of the class interval
and / = frequency of the class inte$al
WORKED EXAI',PLE 1:
The table below shows the scores obtained by a group of students playing the latest
computer game.
10 '74 '75'79 80-84 85-89 90 94 95 99
8 l2 l0 l6 8 6
Estimate the mean of the disdbution.
SOLUTION:
c{
idd
no
ia
of
ld
ty-
Chaeter 12: Statisti6 @
(ffi> WORKED EXAMPLE 2:
A survey was caried out to ftnd the nomber ofhours a student used the internet in a cert:r'-
month- The table below shows the disftibution
(a) Write down the modal class of this distribution
(b) Calculate an estimate ofthe mean number ofhours a student used the internet in ||.
soLuTloN:
(a) Modal class is 10leaf diagram shows the time' in seconds, taken by the group of studenfs to
complete the mce.
ll
ll
t2
12
l3
E:'-l lF;"**:'! llf :"{: i
(a) Find the number of students who took part in the selection ftce.
(b) Find the time taken by the fasrest student
(c) The top 8 students will be selected to compete on spofis day Find the range of tlme a
student needs to clock to qualify for the 100-metre ftce on sports day'
of the woms arc shorter than t mm. Find the value of ,tI
34
56 6 8 8
00 01r
JI
0l
5
9
34
Chu*.. r:, Sturi.ti"" @
16. The stem and leaf diagrarn shows the montlily wages of the employees in company 48C-
7889
(a) Find the number of people employed by Company AiC
O) Find the highest monthly wage.
(c) Find the percentage of employees whose monthly wage are less than $2500
(d) Describe briefly the distribution of the data.
17. The stem and leaf diag&m below shows the masses, in grams' of each mango in a box'
I
2
3
4
5
0t 2 2 4 6 7
00 0 3 5 ? 8
24 5
57
9
89
.0 I
00 2 2 2 6 6 8
03 4 6 6 6 6 8
00 5 7 ? 8 8
J)
36
37
38
39
40
899
I
I rP':J518 represents Js8 g
(a) Fird the number of mangoes in the box
O) Find the mass of the heaviest and lightest mango.
(c) Find the most common mass of the mangoes
(d) The grades of the mangoes ale detemined by theh masses Use the table below to
the percentage of manSoes which are
ti) Crade A.
(ii) Grade B,
(iii., Grade C.
(e) Describ€ briefly the shape of the distribution.
A 400 ] and the mode of the numben is 9, find the values ofr and ).
When another number, ? is added to the above numbers, the medial is ?.75, find the value
of a.
{a)
(b)
tc)
Cfraoter fZ: Statisrics 
@
29. Find (
below
(i) the mode, (ii) the median and (iii) the mean of each of the frequency distributi
(a) ffi|;-f,TtTt ETill;#
L.%ff2 il7 el8l'
(D
(ii)
(iiD
Median =
Mean =
(i) Mode =
(ii) Median =
(iii) Mean =
9
9
7
l
!
96
I
7
mIeTtT,o frr EIl3lElru
(i) Mode =
(ii) Median =
(iii) Mean =
(i) Mode =
(ii) Median =
(iii) Mean =
(1)
(ii)
(iii)
Mode =
Median =
Mean =
(D
(iD
(iii)
Mode =
Median=
l''er ]
l 2 3 1 5
:f' 9 l6 p l3 I5
Given that the mode is 2,
find the largest possible value ofp.
Givcn that the median is 15.
find the smallest possible value
of p.
Given that the mean is
find thc value ofp.
38 39 10 41
52 19 36
Civen that the mode is 38,
find the smallest possible value ofp_
Given that the median is 13,
find the iargest possible vahe ofp.
Gilen that the meaD is 9.5,
fiid the valuc of/.
i
(c) 5 l0 20 25
2.1 )9 32 11
12 l3 l1 l-s
t1 8 6
l0 20 30 40
26 32 l1
9 IO t1 I2
16 l0 ,7
5
Crr,rc,rz,StutlrU* @
Given that the mean is 2,
find the values of a and ,.
, @ ""*".",", ^.. r"
The results obtained in a survey of 100 schools on the number of drinls vending machmes rn
eacb school is shown below.
Calculate
fa) the total number of vending machine.
(b) the mean number of vending machines per school,
(c) the modal number of vending machines per school,
(d) the median number of vending machines per school.
The rable shou \ rhe number of mobile phone\ in some households.
(a) Find
(i) the mean,
(ii) the median,
(iii) the mode
of the distribution.
(b) w1lich average gives the best picture of the distribution?
Cr,"Ot., lZ. St"r,s'"" @
33. The numb€r of hours of wolk of each of the
given below
15 part-time workers in a fast-food restaurant
(a) Find
(i) ihe mean,
(ii) the median,
(iii) the mode
of the distdbution.
(b) Which average gives the best pictue of the data?
(a) Calculate
(i) fte mean,
(ii) the mode,
(iii) the median
of the distribution.
(b) When the similar survey was carried out on 100 students of anothel school' the
number of books boughi per student was 'x. Given that the mean of the 300 students
these two schools was 1.78, calculate the value ofr'
34. A suvey was carried out to final the number of books bought at a book fair by each of a
of 200 itudents from a school. The results are shown in the table below'
@ t*"'"*' ^'""
(a)
(b)
Five coins were tossed 24 dmes. Tbe table below shows the number ofheads obtained in each
of the throws.
Find, for this distribution,
(i) the mean,
(iD the mode,
(iii) the median.
The coins were tossed thrice morc. The mean number of heads obtained in all the 27
throws was exactty 3. Find the number of heads obtaned in each of the extla three
throws.
Two factodes, each with 25 worke$ assemble toys for export. The number of foys
employee can assemble in 20 minutes is given below.
Factory A
ns$t{ihisff l.. 4 5 6 7 >8
^4ffihnl o 8 -1 7 2
Factory I
N
N
8
4 ,7
2 t0
l^)
(c,
Is it possible to calculate the mean number of toys produced by the
factory?
Find, for each factory
(i) the median,
(ji) the modal
number of toys that each employee can assemble in 20 minutes.
Which average is best used to comparc ihe workers of both factories?
workers in each
o)
Chaeter 12:slansdcs @
31. Eachschoolsent5sltlclentstoparllcrpareuranintel-schoo]Mtthematjcsolynpiad'ThelNh.:
shows the scorcs of each student fiom thc top 3 schools
Ifthe schools arc ranked by their mean scorcs' which school \lould wiD thc Mathenetr'
Ol)n'pi. Jl
iii"tla,ft. schools are rankcd by lheir nredian scores' which school would uin L:'
Nlathematics OlYmPi.ld?
iiui" ,t. 
"tt" "i,rtJ ".,npetito$ 
from school B, would you choose the Inean \core or tl -
medirn scorc to rank the schools?
The nunber of absentces of 5 classes frotn Jatuary to May erc given bclow
(a)
(b)
(c)
-{1.
(.l) Find the class with the most absentees
(i) their moctes,
(ii) thef nrediaDs
lJ\e their mftlns to rank tle classcs ln
least numbcr of absentees.
per month bY compa ng
order fl{n the mosl number of absentces nt r"
(b)
2A 3 '1 l 3
2B 2 4 I l I
2C -l 2 6 2
)D 4 -l 5
2E I 5 2 ) 3
@ *,,n"*'"''*-.u
(a)
(b)
(c)
,A group of students was asked how many music CDs they had bought during the prcvlous
weekend. The table below sbows the resulls.
Wdte down the laryest possible value of j!, given that the mode is L
Wdte down the largest possible value ofr, given that the median is l.
Calculate the value of n, given that the mean is l.
The table below shows the number of hours spent doing community work of each of a gloup
of people in a week.
(i) If the mode is 3, write an inequality satisfying x.
(ii) Using the largest possible value of .x in (a)(i), find the mean and the median of the
distribution.
If the median is 4, write down
(i) the largest possible valoe of -t.
(ii) the smallesl possible value of .d.
(a)
(b)
Ch.eter12:slatistrs @
41. A survey was conducted to find the number of books some students read last month.
(a) Wite down the largest possible value ofr, given that the mode is L
(b) Write down ao equality that r must satisfy if the median is 2.
(c) Calculate the value ofr given thaf lhe mean is 1.72.
42. The frequency table shows the number of pets in each household.
(a) If the mean number of pets is 1.7, find the value of -r.
(b) If the distdbution is bimodal, state the possible value(s) of .r.
(c) If the median number of pets is 2, find the largest possible value of .x-
43. The number of pens that each studenl in a class has in his pencil case is shown in the tabl-'
beIow.
Given that there are 20 students in the class, calculate the mean
sludenL
If the mode of the distribution is 3, find the range of values of .n.
If the median is 3, find the largest possible value ofr.
number of pens pe(a)
(b)
(c)
@ t",0"."0*^,-r"
(a)
(b)
(c)
The table shows the distribution of the number of siblings of a group of students.
Find the smallest possible value of .x if the modal number of siblings is 3.
Find the value of ,I if the mean number of siblings is 2.5.
Find the Iargest possible value of .r if the median number of siblings is 2.
The table below shows the number of guests in each of the 100 chalets on an jsland resort
during a particular day.
(a) Find the value ofr + ).
(b) If the mean number of guests per chalet is 2.65, show that r + 4] = 135
(c) Find the values of i and ) by solving the appropriate equations.
(d) State
(i) the median number of guests per chalet,
(ii) lhe modal number of $re.ts per chdlel.
Chaeb 12: Sratistics 
@
46. The table below shows the number ofnovels purchased by each ofa group of 100 studenls i!:
d monlh,
(a) Sbowthrtr+]=48.
(c)
(d)
Given that the mean number of novels purchased per student is 3.5. show tha.
3r+5-r=196.
Solve the equations jn (a) and (b) simultaneously to find thc values of .r and I.
Hence. state the modal number of novels purchased per student.
(a) Given that the mean is 7.2, calculate the values ofr and ).
(b) Hence, state the modal mark and the median mark.
47. The table below shows the marks obtained bv 40 s$dents in a Mathematics quiz.
48. A fair die is thrown 25 times. The results are recorded in the table below.
(a) Show that -r + ) = 10.
(b) lf the mode is 5, find the maximum value of.I-
(c) If the mean is 3.44, find the value ofr.
@ t"""-",,* ^.. r"
The nuftber of goals scored by a hockey team in each of 30 matches played during the pasf
year was as follows:
(a) Complete the table below.
O) Draw a histogram to represent this information.
(c) Find, for this distribution,
(i) the mode,
(ii) the median,
(iii) the mean.
(a) Use the given text to complete the freqDency distribution of the number of letters in each
(b) Find, for ftis disffibution,
(i) the mode,
(ii) the median,
(iii) the mean.
cr'rnr",LZ,Suirti., @
51. A group of students was asked the number of watches they owned. The histogram illu
the rcsults ofthe survey. Fi[d
(a) the number of students in the group,
(b) the modal number of watches per student,
(c) the median number of watches per student,
(d) the mean number of watches per student.
The number of goals scored in a hockey match
shown in the bar graph below.
(a) Find, for this distribution,
(i) the mode,
(ii) the median,
(iii) the mean.
(b) ff tiis information is to be represented in a
pie char| calculate the angle of the sector
represenling 2 goals scored per malch.
in each of 36 matches of a hockey= 65 mm' (Given)
,65
jo)
'{ 3"7
- 2.63 mm (correct to 3 sig. fig.)
.. The radiu. of lhe hemisphere is 2.63 mm.
(b)
t hapFr R Vehiar.on.r P\rmr(L Cob d \rtFs
D WORKED EXAMPLE 3:
A solid is made up of a cone and a hemisphere as shown.
Calculate
(a) the volume,
(b) the total surface area
of the solid in tems of z
SOLUTION:
(^) Radius of cone =12-2=6at
Height of cone =14 6=8cm
^t,volume oI cone = J)tr 
-/?
= ! x xx 6'x
a
= 962 cm'
2
volume oI hemlspnele = J,7tr 
'
2-
=_X7rX6,
= 14'1. cm3
. . Volume of solid = Volume of cone + Volume of hemisphere
= 96n + l44n
= 240r cml
(b) Let the slant heighr of rhe cone be I cm.
I' =6'+82
= 100
I = nrl00
= 10 cm
Curved surface area of cone = ttl
=1tx6x
= 602 cm'z
Curved surface area of hemisphere = 2zr':
=2xEX
= 72x cmz
6'
10
. . lolJl surlace arcd of solid =/Cuned rur'| 'race]+ / (uned.urlace 
l
\ tu-ea ot cone , ldreJ ol hemisphere/
= 60tt + 728
= ljlrrcm
. : Uct?lis Turor 28
(D WORKED EXAMPLE 4.
Forty solid metal balls of diameter 1 .2 cm each were melted and recast to form a solid cfie
of height 6 cm. The density of the metal used is 8.4 g/cm3.
Find
(a) the radius,
(b) the mass
ol rhe cone formed.
t)
lTaiteam i
@ "",0"."u* ^.. 
r"
The following data shows the shoe sizes of a group of 24 students'
(a) Complete the dot diagram below
l0
(b) Find, for this distdbution,
(i) the mode,
(ii) the median,
(iii) the mean.
The ages, in years, of 21 patients in a hospital ward are reprcsented in the stem and leaf
diagram below.
I
4
5
6
'l
9
6
9
Find, for this distdbution,
(a) the mode,
(b) the medlan,
(c) the mean.
]J 88
id33
24
chaprer 12, sratistics @
56.
55. The heights, in cm. of 20 children are given below
33
u5
4
t
l
0
3
1
2
3
4
5
(a)
(b)
389
57',7',l8
Boys cirls
(a) Find the number of
(i) boys,
(ii) girls
in Class 2A.
(b) Discuss the disdbution of each gender
(c) Which gender perfonned better?
(d) Find
(i) the highest and lowest mark,
(ii) tle modal mark,
(iii) the median mark
of Class 24.
Draw a slem and leaf diagram to represent this infonnation.
Find, for this distriburion,
(i) thc mode,
(ii) the median,
(iii) the mean.
The stem and leaf diagram shows the Mathematics marks of a class test of the boys and girl\
of Class 2A.
3
02'7
003
225
02
@ r'r",r,".",i., r".. :r
Mean for Grouped Data
The table below shows the ages. in years. of 30 people at a beach carnival
(a) Complete the following table.
(b) Hence, calculate the mean age of the people :lt ihe camival
(a) Complete Ihe table beiow.
f+
'7 -9
10 t2
13 l5
16 18
19 2l
22,21
The lenqth of 40 lishes. measured correct to the nealesl mm, luc glven l]clow
l-encth (mm,l l-10 1I 20 2.r l0 31 40 4t -50 51 60 6t 70
uency 2 3 12. 9 5
-lld-**-" I FrequencY !f r 
t 
Ir 
-
I l0
1l -20
21 -30
31 40
,{l ,50
5r -60
61 ',70
(b) Calculale the mean lenglh of these 40 fishes.
"n.0.,', 
O.u*.' @
(b) Hence, calculate the nean mass of each pench in the box'
60. The lifespaDs, in days- of 60 ameobas in a colony are shown in the table below
(a) Conpletc the following lable
(b) Hence. calculate the mean lifcspan of the colony of ameobas
(c) Write down the modal lifespan.
The tabLe below shows the mass. in grams' of each of the peaches in a box
I10 lI1
Complete the ibllo$iDg table.
ll0 Ir4
0ftnd the value of d
@ *"*".",., *.. r"
Given that the mean, the median and the mode of the set of ordered numbers below are 84, 85
and 88 respectively, find the values ofp, 4 and r..
p, 81, 81, 82, q, r, 88, 88, 89. 93
The median of seven diflerent positive integers shown below is 8.
(21 + 1), 6, 4, 9, 8, 1s, r
Find the value of .I.
l0 students each fiom two schools, A and B, took part in an inter school Mathemari,js
competition. Their sco.es are displayed in the stem and leaf diagram below.
SchoolA Stem School B
024
5x6
023
531
8654
2vo
'7
8
9
The sum of the scores of all the 20 students is 1669.
(a) Find the value of.I and the value ofj',
(b) The school with the higher mean score won the competition. Which school was the
winner?
"n"Oo,,r 
a"*,,' @
Probability
l
2.
2.
l.
1.
l
5.
Probability is the branch ot' statistics which allows you to work out how likeiy rl
unlikely an outcome or rcsuk might be. lt is used in areas such as predicting sale.
rnaldng investmenls decisions and planning political campaigns
Some everyday probability conveNations are given below:
Whal is thc prcbability that my school will win the Science competition?
There's a 50 percent chance that it will not rain today.
ES oetnitio.,
Probabilities are ratios. It is usually written as a proper fraction. lt can also b!
expressed as a decimal or as a percentage,
A probability experiment is an aclivity where the result depends on chance.
E.g. Tossing a coin and drawing a card from a pack of shuffled playinS cards.
Each o[ the possib]e results of an experiment is called an outcom€ When ea.:
outcome has cxactly the same chance of happening, then we say that the outcom:!
arc equally likely.
A possible outcom€ is a result fhat could possibly occut even though it lnay n:
occur this time round.
E.g. I The possible outcomes of tossing a coin are 'Hftds' and 'Tails'.
E.g. 2 The possible outcoDres of rolling a fair die arc l, 2, 3, 4, 5 and 6
A favourable outcomc is a result thal we arc intercsled in.
E.g. I If we want to get a 'Head' iD a coin toss, then getting a 'Head' would be .
favourable outcomc.
E.g.2 If we want to ge1 an even number in rclling a die, then the favourab':
outcomes would be rolling a 2, 4 or 6.
@ v"rr..."ti., ruto.:t
6.
1.
8.
The sc! of aII possible outcomes for an expedment is called the samlle space lt is
denoted by S and the outcomes are put in braces { ).
E.g. 1 The sample space oftossing a coin, S= {Head, TailJ.
E-g. 2 The sample space of rolling a die, S = { 1, 2, 3, 4, 5, 6}.
The total numb€r of possible outcom€s is the same as the total number of elements
in the sample space. It is denoted by ,?(S).
E.g. 1 If S = {Head, Tail}, then n(S) = 2
E.g.2 If S = { 1, 2, 3, 4, 5, 6}, then 
'?(s) 
= 6.
An event is a set of outcomes in which we arc interested in. Any subset of a sample
space is :m event.
E.g.l when we toss a coin, we may define getting a head as an event.
The outcome of the event is 'head'.
E.g.2 When we roll a die, we may define getting an even number as an event.
The outcomes of the event are 2, 4 and 6.
rE> WORKED EXAMPLE 1:
An experimenf consists of drawing a card fiom a j ar containing I 2 cards Each card has a
dilferenl monfi ol lhe )ear wrilren on il
(a.) List the sample space for the expedment
List the following events.
(b) The event A consists of drawing a card haling a monlh beginning with I
(c) The event B consjsts of alrawing a card having a month beginning with M or N'
(d) The event C consists of dnwing a card having a month that has 4 letters or less
SOLUTION:
S = lJanuary, February, March, April, May, June, Jult August, September,
October, November, December J
(b) A = {January June, July}
(c) B = {March, May, Novemberl
(d) C = {May, June, July}
9. Probability is a measure of how likely it is for an event E to happen. It is denoted by
P(r).
The probability of an event E, P(4, in an experiment with equally liLely outcomes is
Nurnber of favourable outcomes for event ,E
Told' number of Po.tible oulcome'
_ n(E)
n(s)
where n(E) is the number of favoumble outcomes jn event E and
,(t is the total number of possible outcomes in the sample space' J^'
Chapter l3: Probability @
E.g. Find the probability of rclling a number which is more than 3-
Possible outcomes = {1,2,3,4,5,61
Favourable ourcomes = {4, 5, 6}
Number of favoumble outcomes
P(rolling a number more than 3) = Total number of possible outcomes
1
I
2
10. The probability of an event caD be a number from 0 to l(both inclusive).
0 r t6pK59 6xan plg 3'
A box contained 20 coloured balls, where 8 are green, 10 are red and the rest arc yello$.
A ball is selected at ftndom from the box. Calculate the probability of getting
(a) a green ball,
(b) a ball that is not green,
(c) either a green or a yellow ball,
(d) either a green, a rcd or a yellow ball,
(e) a blue ball.
SOLUTION:
Number of yellow balls = 20 8 10 = 2
R(ar P(u sreen ballJ = ::' 2t)
2
5
P(not a green ball) = I - P(green ball)
.2
5
(b)
3
(c) P(a green or yellow ball)
t0
20
I
2
(d) P(a green, red or yellow ball) = 
fi -ffi
(e) P(a blue ball) =
0
20
@ ""*".",,". 
t'- r"
g> WORKED EXAMPLE 4:
A box contains 10 cards numbered 1' 2' 3' 5' 6' 8 9' l0' 13 and 15 A card is drawn at
random liom the box-
{a) Write dowll the sanple space' 'S aDd the total number of possib]e outcomcs' n(t)-
iij plni trt" p.ou'tilityihat the nr.rmber on the card drawn is
(j) r 15. (ii) an odd number' (iji) a prime oumber'
(iv) a composile numbe! (v) a perfect square'
SOLUTION:
(n) .! = {l' 2' 3,5,6 8' 9' 10' 13' lsi
,r(S) = I t)
lbl lir P(a l5J =
I
to
6
(ii) P(m odd numtrer) = l0
3
a
4
(jii) P(a prime runber) = t0
2
5
(i\,) P(n composite number) =
(v) P(a pertect square) =
cl[pLer l3: fiobabill! @
(a) a 4, (b) an even number, (c)
WORKED EXAMPLE 5:
A six-sided die is rclled. Find the probability of getting
(d) a multiple of 3, (e) a number less than 4,
(g) a 7.
soLuTtoN:
Sample space = { 1, 2, 3, 4, 5, 6}
a pnme numoer,
a positive integer,
I(a) P(a 4) = -
(b) P(an even number)
3
I
2
(c) P(a prime number)
(d) P(a multiple of 3)
(e) P(a no.sectorA) (b) P(outside sector A)
= 1-P(inside sectorA)
90"
360"
1
(c) P(inside sector B)
Area of sector B
Areaofcircle
360"
I
.1
3
(d) P(inside sector C or D)
Toial area of sector C and sector D
Area of circle
120" + 105'
360"
225'
Area where E occurs
Totai area
360"
5
8
qwter I r, rrcutiritr @
(D WORKED EXAMPLE 7:
TWo farr coins are tossed together
(a) List the sample space.
(b) find lhe probabilily of gertin8
(i) 2 heads,
(ii) t head and 1 tail,
(iii) at least I head.
soLurloN:
(a) S = {HH, HT, TH, TTI where H = Head and T = Tail'
Q) (i) r(two head$ = ]
(ii) P(1 head and 1 tail) =
2
n
r
2
Ca> WOBKED EXAMPLE 8:
A two digit number is chosen at random. Find the Fobability that the dumber chosen
(a) is smaller than 30,
(b) is odd,
(c) is a multiple of 5.
SOLUTIONI
Sample space,,t = {10, 1l' 12, 13, 14' ,99}
n(s)=99 l0+l
=90
(iii) P(at least t n*ol = ] *ffi
la) Pla no.Find the probability that the number on the card
drawn is
(a) a 9, (b) an even numbet
(d) a composite numbet (e) a perfect square,
(c) a prime number,
(f) a 3-digit number
Chapter 13: Bobabilily 
@
8. A box contains l0 cards as shown below- A card is druwn at random ftom Ihe box. Find thr
probability that the number on the card drawn is
(a) an 8,
(d) less than 3,
(c) cn odd number.(b) greater th:m 5,
(e) a positive integer
a t a--.] --u l. )tsEsE 3t trL!trEitr
(a) a B,
Each of the letters of the word 'PROBABILITY' is w.itten on a card. The cards
shuftled and placed jnto a box. A card is then drawn at random from the box.
Find the probability that the lettcl on the card drawn is
I0.
11.
Each of the 6 cards shown below has a letter of the alphabet wfitten on one side of the caJd
The cards are then shuffled and laid face down on the table. A card is chosen at random lrom
these cards. Find the Drobabilitv that Ihe letter on card drawn is
(b) aP,
(d) a coosonant,
(b) an A,
(d) either an A, a P or a l.
h)aP.
(c) an E,
(a) gold,
(c) either silver or bronzc,
Hffi8ffiffiffi
A bag contains ,l gold medals. 8 silver medals and 6 brcnze mcdals. A medal is selected ar
random from the bag. Find the probabilily thal the medal selected is
(b) nol gold,
(d) either gold, silver or bronze.
@ r",n..",n.,*-,u
12. A can contains 10 red. 8 white and 6 yellow sweets A sweet is dmwn at random from the can
Find rhe probahility tbiLt the sweet Llrawn i\
rar rellor.r. lb) not ]ellos'
ici irot a colour of the rainbo*, (d) red or yellow'
(ej red, white or yellow, (0 green'
13. Ma{c conalocted a probabilily experiment using a box of 3O.ma$les He co-unted. the number
" ;;ttbtl'fe;'"oloo' *i '"iotata 
tn"rn ii the table below He then draws one marble at
random from the box. Find the probability rhat
(a) a yellow marble is alrawn from the box'
(b) the marble drawn is not green
l:1. (a) According to the weather report' there's a 20'l' chance that it will
Find the probabitity that it will not lan tomolrow'
(b) There are 49 narbles in a can Given that the probability of dmwing a green marble from
)le' in lhe can fial i\ nol green
the can 1s ;, calculate rhe numt]er or malt
(c) The probabilitY
on time is 0.52
that a train arrives early at a station is 0 3 The probability that rt arnves
Find the probabiiity that the train is late-
@
615
1-5.
(d) A manufacturcr makes switches for electric circuits. The probability that a switch t
faulty is 0.05. Mr Lim buys 200 of these switches. How many ol lhese switches can be
exDected Io be fhultv?
(e) A manufacturer produces microchips. ln a sample of 100 microchips, it is found thaI8 o:
these microchips arc faulty. Find the probability that a microchip chosen at random frorE
the samDle is not faultv.
(1) A printing company fbund that the prcbability that a parcel will be delivered wirhin
'7
2 working days of being posted is t. Find the probability that a parcel will be delivered
nore than 2 working days after being posted.
Two fair coins are tossed together Find the probability of gettilg
(a) exacdy one tail,
(b) at least one lail,
(c) at most one tail.
Three lair coins are tossed logether Find the probability of getting
(a) 3 heMs,
(b) 2 heads and 1 tail,
(c) at least 2 heads.
@ u"tr,".,ri* r"o. ze
18.
17. There are 50 children at a games camival. 35 of them arc boys 28 of the children wear
spectacles. A child is chosen at random from this group. Find the probability thir lhe child
(a) is a girl,
(b) wears spectacles.
A box contnins 30 chocolates. 9 arc milk chocolates, 15 are mint chocolates.and the rest are
dark chocolates. A chocolate is picked at randon from the box. Find the probability of
(a) picking a milk chocolate,
(b) not picking a milk chocolate,
(c) picking either a milk or dark chocolate,
(d) picking either a milk, mint or dark chocolate.
!9, The diagram shows a spirmer with 8 equal secton. The pointer is spun Find the probability
that it will stop at a sector containing
(a) a 5,
(c) an odd number,
(e) a multiple of 3,
(g) a prime number,
(rl neilhef a prime nof a compo\ile number
(b) a 9,
(d) ejthera2oraT,
(1) a factor of 35,
(h) a composite number,
chqter 13:Prcbabiliiy @
20. The diagam shows a spinner in the shape ofa regular octagon. When the pointer is spun, fitd
the Fobability that it will stop at a triangle conlaining
(a) an even number,
(c) a number less than 5,
(e) a number no less than 9.
(a) a 3,
(d) eithera2ora3,
(b) a multiple of 4,
(d) a number more than 8,
(c) a whole numbet
21. The diagmm shows a spinner having 9 equal ftiangles. When the pointer is spun, find the
probability that it will stop at a tdangle containing
(b) a positive integer,
(e) a number less than 8.
22. There are 500 students in a school. Each of them bought one ticket for the school anniversary
p.rze draw. The 500 tickets, numbered I to 500 will be put into a barel and the wiming ticket
drawn,
(a) Marc has the ticket numbered 288. w}lat is the probability that he will win the prize?
(b) Find the probability that the winning ticket number will be greater than 325.
(c) Esther says that either a girl or a boy will win the prize, so the probability that a 9tu1 will
. t_
win lhe Dfi/e i\ -. T\ \he correcrl
A lette. is selected at nndom from the set of alphabets. Let E be the event that the letter
selected can be found in the word 'SUNDAY'.
(a) Li\l lhe sample space.
(b) Find P(t') and P(t').
23.
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.1.!- Sandra chose a month of the year at mndom.
(a) Write down the sample space.
(b) L.t A b" ,h" .u"n, lhat the name of the montb begins with the letter M Express A usrng
the listing method. Hence find P(A).
(c) Let I be-the event fhat the name of the montb begins with the letter A or '/- Express B
using the listing lnethod- Hence nnd P(B)
(d) Let - be the eveni that the name of the month contains the letter R' Express C using the
listing method Hence find P(C).
:5. A bag contains 30 balls numbered 1 to 30. A baU is dlawn at nndom from the bag and the
drawn is a multiple of 5 List all the
3' A 3 digit number is fomed at random using the numbers 0, 3, I and 5 The digirs cannot be
repeated.
(a) List the sample space.
(Uj I-et C Uc tle event lhat the number formed is an even number Express C using the listing
method.
(c) Find P(C) and P(C).
iaj l"t O tt" ttt" 
"u"nt 
that the number forried is at least 350 Express D using the listing
metnoo.
(e) Find P(D) and P(D').
number is recorded,
(a) List the sample space.
(b) Let I' be the event that the number on the ball
possible outcomes of the event
(c) Find P(F) and P(F').
Chaptei 13: Prcbabilitl 
@
27. Trckets are numbered I 1() 50 They are then shufflcd and placed into a bag A licket is the:
drawn at raodom liom the bag. Find lhe probability that the ticket drawn is
28. A bag cortains some btlls numbered 30 to 43. A ball is drawn at random from thc bag Fin'l
the Drobabilitv Ihat the number on the ball drawn is
(a) an even number,
(c) a nunbef less than 33,
(e) not a perfect cube.
(a) odd.
(d) divisible by 8,
(a) less than 20,
(d) a pdme number,
(b) a multiple of 7.
(d) a pedeci cube,
29. Car{ls with the numben 3 to 102 are placed into a box A card is then drawn aI random fioE
the box. Find the probability of dnwing a number that is
(b) a prime nuDber,
(e) a pefect cube.
(b) grcater than 30,
(e) not a prime number.
(c) morc than 33,
(a) even. (b) less than ll,
(c) a prime number less lhan 30, (d) a mulliple of 11 greater than 100
30. There are 4 cards in a box numbered l' 2' 3 and 4. Two cards are drawn at random from lh:
box to a fo.m a 2 digit number Find the probability that the number forned is
(c) a mulliple of 12,
@ 
Mathematics fttor 28
31. A two digit number is selected at random. Write down the probability that the number selected
rar i. a pedecr \quare.
(b) is greater than 80,
(c) has at least one digit which is 2,
(d) is a muttiple of 11.
32. A card is drawn at random frcm a deck of 52 playing cards
Find fie probabilrlythal lhe card draun is
la) an ace, (b) an ace of clubs, (c) a spade,
(d) a picture card, (e) not a pictue card.
-i3. A card is dmwn at rardom from a deck of 52 playing cards.
Find rhe probabiliry lhat the card dra$ n is
(a) a red card,
(d) a king,
(b) a 3,
(e) not a queen,
(c) eitheraTora9,
(0 either a pictue card or a diamond.
34. All the picture cards are removed frcm a pack of 52 playing cards. A card is then drawn at
random from fhe rcmaining cards in the pack. Find the probability that the card drawn is
(a) a rcd ca.rd, (b) not a heart, (c) not a 5.
ch.pler 13: Prcbrbilily 
@
35. A card is drawn at random from a pack of 54 playing cards. Find the probability that fhe card
drawn is
(a) a red king,
(d) aioker,
(b) a black card,
(e) not a joker
(c) eithef , jrck or a queen.
36. The diagram shows
(a) a heart,
(d) an 8.
(b) a club,
(e) either a 4 or an 8,
12 cards. A card is selected at mndom. Find the Drobabilitv of selectins
(c) a picture card,
(1) a queen of hearts.
The diagram shows a circle divided into sectors with different colours. A point is selected ar
random inside the circle. Find the probability that the poinl
(a) lies inside the blue secfor,
(c) lies inside the orange sector,
(e) lies inside the red or yellow sector
(b) does not lie inside the blue sector,
(d) lies inside t}Ie red sector,
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In the diagam, O is fhe centre of the larger circle of mdius 14 mm. ZAOB = 90" and the radius
of the smaller circle is 4 mm. A point is selected at random inside the larger circle. Fi[d the
probability that it lies inside
(a) the smaller circle,
(b) the sector AOB,
(c) the shaded region.
The pie charl shows the propol'tion of T-shirt sizes
Small, Medium, Large and Extra Large.
(a) Find the value of,I.
(b) Ar employee is chosen at random fiom lhe
employee chosen wears a T-shirt of size
(i) small,
(ii) medium,
(iii) large.
wom by the employees of a company:
company. Find the probability that *re
The diagmm shows two circles of radius x cm and 4r cm. A point is selected at random inside
the larger circle. Find ihe prcbability that it lies inside the shaded region.
Chapter 13: Probability 
@
4l . The diagram shows a circle of radius 20 cm and a scmicircle- centre O and radiu s 8 cm. A ooiT
is selected at randorn inside the circle. Find the Eobabilitv that it will lie inside
(a) the semicircle,
(b) Ihe shaded region.
12. The diagram shows 3 concentdc circles, centre O of radius 2 c1n,4 cm end 8 cm respecdveh:
A poinl is selected at random inside the largcst circle. Find the probability tbat the point rl:l
lic inside
(4.) region A,
(b) region C,
(c) region A or B.
The diagram shows
point is selected at
APQ.
I squafe ABCD. P and 0 are the midpoints ofAD and AB respectivel)'. r-
nndom insidc the square. Find the probability that it lies inside rriangl:
AQB
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The diagnm shows a dalt board. Each quadrilateral in the diagram is a square. A dafi is thrown
at random and hits a point on rhe board. Find ihe Fobability that it will land inside
(a) region A,
A bag coltains some balls labelled X y and Z When a ball is selected at random ftom the bag,
t7
the probability that it is labelled x is i and the probability that it is labeued f is 
Ls 
.
(a) If a ball is selected at random fro-m the bag, find the probability that it is labelled Z
(b) If there are 86 balls labelled Z in the bag, find the total number of balls in t}le bag.
(b) rcgion B,
There are 20 grcen balls and r brown balls in a bag. A ball is drawn at random from the bag.
(a) Write down, in tems ofx, an expression for the prcbability ftat the ball drawn js brown.
rbl Given thdl lhi. probabilily i' t9. find lhe \alue ofr'
(c) region C.
Chaprer13:Prcbabirily @
17. A box contains 120 red, green and blue marbles The probability of drawing a rcd marble and
x green marble Irom lhe bo\ clc 5 irnd : le'nccli\el]
(a) Find the number of blue malblcs in thc hox-
itl wnen " 
blue marbles nle removed fiom the box' the prcb:rbility of drawing a blue
m:uble frcm the rcmainirg marbles in the box becom"s ft Find tht value ot ''
48. There are 27 boys and r girls in Class 24
(a) Find the value ol i if the probability of selecting a boy from Class 2'A is
(b) Fiid the value of ir if the probability of selectiig a girl from Class 14 is
49. There are 32 puQle creyons end 28 brown crayons in a box Alter 'rt puryle clayons ani
(l + 2) brown irayons afe removcd fron the box, the probilbilily of sclecting a brown crayon
'I
fiom the box becomes t. Find the value ofr-
3
l'
2
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Abag contains 50 balls. n ofthem are black and the rest of them are red. When 6 more rcd halls
5
are added to the bag, the prcbability of&awing a fed bcll ftom the bag becomes t Find the
\ tiue o1\.
There are 30 studeDts in the hall. t of them are girls
(a) Write down in terms of .r, the probnbility that a student chosen at random from the
grouP rs
(i) a girl, (ii) a boy.
(b) 8 more girls and 14 more boys entered the hall. The probability of choosing a girl lrcm
3
the hall becomes |3. Find the value ofn.
A bag contains 8 red balls and 4 blue balls.
(a) A ball is drawn at random from the bag. Find the p.obability of getting
(i) a red ball, (ii) a blue ball, (jii) a grcen ball.
(b) One red ball is removed liom the bag. A ball is then drawn at rundom fiom the bag. Find
the probability of getting
(i) a red ball,
(ii) a blue ball,
(iii) eitler a red or a blue ball.
Cbrpd 13:Probabilriy 
@
53. Each of the numben from 000 to 999 is written on a ball and ali the balls are p]aced insrde z
conlainer A ball is then druwn at nndom from the container. Find the probability that the bBl
drau n
(a) bas 3 digits that are the sane,
(b) hes the lasr digir 8.
(c) has the last digir which is no1 8.
54. A packet contains 30 lollipops. 12 lollipops are strawberry flavoured and have red w.appeE
I lollipops are vanilla flavoured and have white wfappels.s lollipops arc strawbeffy flavoural
and have white wrappers. ,+ lollipops are vanilla flavoured and have red wmppers. A lollips!
is picked at random from the packer. Calculale the prcbability that rhe lollipop picked
(a) is vanilla flavoured,
(b) bas white wrappers,
(c) is strawberry flavoured and has red wrappers,
(d) is vanilla flavourcd and hns green wrappers.
55. A roulette wheel hrs 38 slots around the rim. The first 36 slots are numboed 1 to 36. Halfd
these 36 slots are red and the other half are black. The remaining 2 slots are numbered 0 asd
00 and are green. As the roulette wheel is spun in one direction, a small ivory ball is rolled
along the rim in an opposite direction. The ball has an equally likely chance of falling into alrl
one of the 3E slors. Find the probabiliry that the ball
(a) lands in a red slot,
(b) lands on 0 or 00,
(c) does nol land on a number from 1 to 16,
(d) lands on an odd number.
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There are 18 red balls and 22 blue balls in a bag. 10 of the red balls and 4 of the blue balls have
a star painted on it.
(a) A ball is drawn at random from the bag and placed into a box. Find the probability that
rhe ball in lhe box is
(il red.
(ii, has a qlar painted on it.
(b) The first ball dlawn is a red ball with a star painied on if. It is not retumed to the bag. A
ball is then drawn at ra[dom from the remaining balls in the bag. Find the probability that
lhe ball drawn
(i) is a led ball witi a star painted on it,
(ii) is a blue ball with a star painted on it.
A bag cortains 5 bl;ck cubes and 13 ihite cubes.
(a) A cube is dr.iwn at randam from the bag. Find the prcbability that it is
(b) 7 green cubes are then added to the bag. The cubes are thoroughly mixed before a cube
is drawn at random from the bag. Find the probability that the cube dmwn is
(ii) white,
(iv) either black or white.
(ii) greer.
at random frcm the goup. Find the probability
(b) 2 pets, (c)
(i) blnck,
(iii) rcd,
(i) black,
The number of pets each of a Broup of 20 children in a kindergarten has is given below.
A child ischosen
(a) no pets,
that child chosen has
2 pets or more.
Chapter 13: Probability 
@
59. The table Derow shows the number of poted plants each household has in a block r{
apartments.
.Numbei o 0 I 2 l 1 5
er o[ housefioldd -t2 21 8 6
A household is chosen at random from the apartment block. Find the probability tlat rh
household chosen has
(a) 2 potted plants,
(b) more than 3 potted plants,
(c) no potted plants.
(d) I potted plant or less.
(e) more than 5 potted plants.
60. A die is rolied 200 times and the results are rccorded in the tablq, below
(a)
(b)
Find the probability of getting
(i) a 2.
(ii) a 4.
(Give your answcr in decimals.)
Is the die biascd?
61.
A student is chosen at mndom flom the group. Find the probability that the student chos.r
werghs
(a) over 50 kg,
(b) morc than ,15 kg but less than or equal to 55 kg,
(c) not morc than 60 kg.
The tnble shows the masses, in kilograms, ol a group of 60 students.
55@
17. The time taken, in seconds, by a group of students to send an SMS are given below.
Write down the modal class.
Complete the table below
(c) Estimate the mean of the distribution.
(d) Calculate the percentage of students who took I minute or less to send the SMS.
(e) A student is chosen at random iom the group. Find the probability that the
chosen takes
(i., more lhan 55 \econds Lo send lhe SMS.
(ii) 50 seconds or less to send the SMS-
(4,
(b)
40that ), is direcdy proportional 10 the squaxe root of j and if ] = 2.5 when r = 9" fui(a) an equation cormecting _{ and },
(b) the value of-r when v = 131.'1
The diagram shows the Mathematics marks of a class of 30 students in a quiz.
:.::.
0 510 15 20 25 30 35 46
Mathematrcs mdks
(a) Find the modal mark of the students.
(b) Wriie down the mark of the stlrdent who topped the class.
(c) The passing mark for the quiz ras 25 marks. Find the fraction of students who failed the
quiz.
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(a)
(b)
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(b)
(c)
Fidar Exminatioo sp€cinen p"p". r @
5. A map is drawn to a scale of 1 : 200 000.
6.
'7.
(a) The distance between two fire stations on the map is 6.4 cm. Calculato the actual
between the two fire stations, givirg your answer in kilometres.
(b) The actual area of a reren oir is l5 km:. Calculate lhe aria on the map u hich
the reservoir, giving your answer in square centimetres.
Answer: (a)
(b)
Answer: (a)
(b)
tU
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In the diagram, A,{-BC is mapped onto MB'C'by an enlargement with centre A and
factor ,t. If AB = 6 cm and BB' = 9 cm and CC' = 6.3 cm. find
(a) the value of l,
(b) tlrc length ofAC.
A card is drawn at Endom from a pack of 52 playing cards. Find the probability that the card
dmwn is
(a) a red picture card.
(b) an ace or a 10,
(c) not a diamond.
(a.J
(b)
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(a) Factodse 2(r': 4) x(x + 2).
(b) Expand and simplify () + 3Xy - 5) - (y - 4)'.
(c) Solve6z:- 1= llz+9.
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(a)
(b)
The diaglam shows a metal solid which consists of a cylinder and a cone.
(a) Find the volume of the solid.
(b) The solid is melted and recast into a hemisphere. Fhd the mdius of the hemisphere.
lTake 7t = 3.142.\
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(a)
(b)
Finar Exanination specimen p"r.. r 
@
10. Solve the following equations.
(a\ 3(2r + t) - 2(a - 7\ = s(x -
/l \2 l(br l: -2.J =-
+l=0
ll)
,^, 2t-l .r+l,., 4 - l
Answer: (a)
(b)
(c)
11. Cotn bought r plastic ftles and y ring files.
(a) He bought 17 files altogether Wnte an equation in terms oft and y.
(b) Each plastic file costs $2 and each ring file costs $7. The total cost of the files was
Write down a second equation in terms of .I and ).
(c) Solve these simultaneous equations to ftnd the number of ring files he bought.
Answer (a)
(b)
(c,
(.i) "*0"**r*- 
*
(a)
(b)
Three of the interior angles of a pentagon arc 68", 84" and 122' while the other two are
5x and 9x. Find the value ofr.
The size of each interior angle of a rcgular polygon is five times the size of each exterior
angle. How many sides does the polygon have?
12 students from differcnt families were asked how many
consumed by their families per month. The answe$ are:
5, 11,1, a, tt, 6, 4, 10, 3, 5, 11, 15
Find
(a) the mode,
(b) the median,
(c) the mean
of this set of data.
of milk were usually
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(b)
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ta)
(b)
(c,
Final E\amination Specmen Pt" , 
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14. Tho curve ] = l 3r - 10 cuts the r-axis at the points A aod B, and the:).axio at the point C-
(a) Write down the coordinates ofA, B and C.
(b) Write down the,cootdinates of the lowest point on the gaph.
Answer: (a) A =
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(b)
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(a)
(b)
II 1so martsl
ALL the questions.
Time: t hour 15 minutes
working must be clearly shown.
A (22 marks)
Facrorise l5p.r - 44I - 5qx + l2py.
Express as.a single fraction in its simplest folm.
5_2+l
2r-8 12-3r r-4
If I + y'z = 34 md ry = 7.5, find tho value of 5(r ])'z
The diagram shows a cardboard consisting of 3 semicircles and a circle. The radii of the
semicircles are 14 cm, 14 cm and 28 cm respectively. The diameter of the circle is 14 cm. A
ribbon is then glued onto the edges of the cardboard.
(a) Calculate the area of the cardboard.
(b) Find the lenglh of the dbbon used.
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22t
i'
Finar Exmination specimen P"r" t 
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The diaqram shows the net of a right pyramid with a
i+ 
"-. 
fr i, p"tp"nai"ular to '4' and is of length 25 cm'
Calculate
(a) the total surface arc4
(b) the volume
of the pyramid formed.
square base, ABCD of side3
4. A rcctangular tank with a base measuring-1 6 m OY n8 
T lt l]tfl-- tl-water to a height
Ld ,n*s6 .J;l ;J. ualls of diameter I2 cm each are dropped into th-e^water
3.
", 
"i"iii"lt. 
in lhe uater level in rhe tank giving your anqwer in cendmetre-q 
^.
i;j il"";;;;iil;"maae or materia ot aen-sitv i 25 g/cmr' calculate the mass of all
' ' 
metal balls, giving your answer correct to the nearest kllogmm'
22.
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Section B (28 marks)
5. (a) On the Venn diagram in the answer space, shade the set A' n B'.
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(b.) €= [ r :.:r rs an integef and 2on the map.
(c) The area of a lily pond on the map is 0.025 cm2 Find the actual area of the lily pond
Answer: (a)
(b)
(cl
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The table below shows the marks of a spelling test by a group of students.
Find
(a) the smallest possible value of x if the modal mark is 6,
(b) the possible values ofr if the median is 5.
Answer: (a) tll
t2l(b.)
The heights, in centimetres, of 16 men in an office were measured. The informabon 1s
displayed in the stem and leaf diagram below.
l1
15
16
t'7
18
8
0
I
U
3
9
5
4
6
t'
5
7
'7
9
8
(a)
(b)
Find the mode. median and mean of the distribution.
Calculate the Dercentase of men who are shorter than 157 cm.
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(a)
(b)
Final Exanination Specinen P"t"., 
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13.
14.
If € = {11, 12, 13, 14, 15, ,6, 17, 18, 19,201,
P= {r : r is a primenumber},
0= {i:-r is an odd numberl and
R = {r : ,v js a multiple of 61,
find
(a.) P u R,
(b) P' 
^ Q,
(c) n(.Q' ^ R').
Answer (a)
(b)
(c)
Answer: (a) (i)
(ii)
11,
tr,
A box contains 9 red balls, 6 blue balls and 5 green balls.
(a) A ball is &awn at random ftom the box. Find the probabiliry that the ball drawn is
(i) not red,
(ii) either red or green.
(b) 4 red balls arc then rcmoved from the box. The rcmaining balls are thoroughly mixdi
before a ball is dftwn at random from the box. Find the probability that the ball dras:!
is blue.
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(b)
4. The diagram shows the graph ofl = 9 - t.
(a) Using the same axes, draw the gftphs of
I ._)= .rand)y=Jx+l)
(b) Fmm your graph, solve the simultaneous equations
(ii) 1= 9 a
5r=3r'+25
(c) What is the figure formed by the three lines and the line ,I = 0?
I
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I
Section B (28 marks)
5. The diagam shows a solid cylinder inscribed inside a hollow cone with a base mdius of42 cm.
The height of lhe cone ic 70 cm and lhe beight of the cylinder i\ 52.5 cm.
(a) Find the numerical value of the mtio
volune of cylinder
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(b) Calculate the lateral surface area ofthe cylinder,
giving your answer conect to 1 decimal place.
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(c) The cylinder is melted down and recast to
form a spherical solid. Find the radius of the
solid. t3l
ITake n = 
=.1
rilal ennimtiu srcimo raner z @
6. Mr Tan bought r pocket dictionaries that cost $180 from a supplier
(a) If he wants to sell the dictionaries at a pmfit of $3 for each dictionary, write down atr
expression, ir tefins of.{, for the selling price of each dictionary t2l
(b) H; sold 28 dictionaries at a profit of $3 each while the rcst were sold at $7 each write
down an expression, in tems of i, for the total amount of money he would rcceive for
.clling all the Jiclionarie\. If
(c) If he made a profit of $l00lor selling all the dictionaries, form an equatjon in 'I and shoc
that it reducis to xt 56x +'720 = 0- Hence find the number of dictionaries Mr Tan
bougrrt. t3l
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The time, in minutes, taken by a clerk to type each of 30 repofis are given below
(a) Complete the table below.
Find the mean time taken by the clerk to type each report.
Write down the class interval where the median lies.
Iit
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(b)
(c,
t0- 14
15-19
20-24
25 29
30 34
35 39
Fiial Exmiianon specinm Papd 2 g?
8. Answer the whole of this qu€stion on a sheet of graph papet
A ball is thrown from the top ofa building.Its position during its flight is given by the equaticxl
y = 72 + 6x I where y metres is the height of the ball above the ground and { metres is iL(
horizontal distance from the building.
(a) (i) Solve the equ a|Lon 0 = 72 + 6x - | . Ill
(ii) Descdbe briefly what the positive solution of the above equation represents' [I]
The table below shows some coresponding values ofr and ).
(b) Calculate the values of d and D. - Ill
(c) Using a scale of 2 cm to rcpresent 1 metre on the x-ilxis and 2 cm to represent 5 metre''
on th-e y-axis, draw the graph ofl' = 72 + 6t xzfor0length ofa side ofthe sque baw is 16 bn.
tf,t the height of the pyrmid be /, cm.
Volume of cube + 6 x Volume of py.amid
:'-ox []'o
.. The heighr uf edch p)rnid i\ 5 cm.
(h)
C)
27+18h=l),1
18t = 90
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8. (a)
8cm ' rcfr
9. (a) (i) rectangular pymnid
(ii)
(b) (i)
Total surface dea of pyramid
= Area of base + A.ea of ,1 tnansulaf fac€s
=(8x6)+t2x(+ x8x9)
+2x(;r6xe.4)l
(ii) trt the sldt height of the pyranid be I cm
___+r-+
BC= 10+ 2 =5 cm
Using Plthagor ' rheoren on AdaC,
h1 + 51= t31
h1 l31-5'1
- 1.14
h = 
'\^=12cm
Total suface dea of pyranid
= Base a.ea + 4 x Area of AAAC
=(10 x 10)+:l x (t x 10 x 12)
= 34O cm'
10. Let the height of the squatr pFAmid by ,'? cm.
. . Height of rectanguld plnmid = 2h cm.
let the length of a side of the squde base be r cm.
/ Volune ot recaneuldl = / Volume of souare I
1 _Il 
I\plramrd,lpyramro
*.(9 8r.Zt=t lx ,) J
r = {44
.. ftre lengiuld'ideot'he'ouarebr\eot,5ep)ranid
11. Volume of pyramid
= - t l44X 15
Volome oI water in the tdk
= (30 x 30 x 15) 720
= 12 780 cm3
Let the height of the water level afief the pyramid is
removed be, cm.
30x30xn=12780
l2 730
'' :10 r 30
= 1,4.2.m
.. The required height ls 14.2 m.
12. (a) Tota] surface dea of pymnid fomed
= Base &ea + 4 x Arca of AYCD
=(14 x 14)+4 )is 24xr cmr and irs base ,-rea js 8r cm':. Find t])e heighr of the cone.
ea
iht
Lnt
21. Draw a net of the following open cones and label its dimensions.
22. A \olid cone has a ba= Volume of cylider + 2 x Volume of cone
=ri t' ctt*t t1 ' ; '1' 24\
- 8932 cn'
29. (.) Sldt height of cone tbmed
CuNed surface aea of cone forned
= Area of sector oABC
=;xnx18'
= 162n
= 509 m'z (3 sis. fic.)
Lel lhe base ndius of the cone fomed be n cm.
Curved surface dea of cone fomed = l62t cml
tRt = t62l
n(18) = 162
- 162x= rr
=9cn
.. Tlle base radius of the cone fomed is 9 cm.
30. (a) Ler the base mdius of the cone fomed be r cm.
Cn umference of base ofcore = 16jrcm .
. . The base mdius of tbe aone fomed is E
(b)
Answe^ with cobplete work* *t'.* (rD
(b) Sla!! beight of cone fomed = 12 cm
(c)
Irt the heiSht of the cone fomed be t cm
UsingPythagons lh€orem,
l;=r5o
= 8-94 cn (3 sic. figJ
. . The height of the cone fomed is E.94 cm.
Volune of cone fomed
=;x,'x8'x J8o
= 594 cnoj (3 sis. ffg,)
31 (a)
(b)
L€t the bde radius of the cone forned be r cm
UsingPythagoras theorem.
= 8l
,="EI
tuea of circle lsod to fom base of cone
= 81,11
= 254 cnz (3 sig, ficJ
I tu"u oi . ) rco^.a,o.r"ce \ I Base area I
- rccrms-ord tuea ofcone lofcone
sneer oi I lfomed I Lfomed ltpaper )
=(65x41) (rx9x41)-81,'
- 1250 cn'z (3 siC. fiC.)
@ 
Malho-atisrutd2B
(a) (i) tf,t the base radius of the cylinder be I cn
Bde area of cylinder = 8l n cm' (Giver)
/b;= 81,
,_,ti
=9cm
Base radiN of cone
= B6e radius of cylinder
. . The bde radius of the cone is 9 cm.
(ii) Le1 ihe height of tbe cone be li cm.
Using Pytlagoraj tho€m,
tt + 9'= rs'
h'1=15'-9'
= 144
h = ,Eqc
-12cm
.-. The height of the cone is I 2 cm
Volune of sold
= Volume of cylinder + Volume of cone
r)122-
-t-; r-24t,,i.7 q \r:l
= ?124 cn
(in) Tolal surlace @a of solid
aBase ma I fcDNed surrace )
= [ofcyuaer)+ [dea ofcYlinder,
f cwed surfee )+ \rea of mne ,
=r; e')+(2'7 o z+'
+(;x9xls)
4.
= 2o?'6i .m'
(b) Mas = Densiiy x Volume 
-
= 7.5 s/cm' x 7128 crn'
= 53 :160 
C
= 53.46 Lg
c. spbrs
33. (a) (')
(ii)
(ii)
(ii)
(b) (t
(c) (i)
(d) (i)
=Jx3.142r4.5',
= 38r.8 cnr (1 d.p.)
Surface dea ol sphere
= 254.s cln'(r d.pJ
- \*'
-{r:.t+ztto'
- 41893 mr (r d.p.)
Surrace dea of sphere
=4x3I12xlO1
= 1256.8 cn: (1 d.p.)
=;x3.142x8',
= 2144.9 nlm' (r d.pJ
Surface dea of sphere
= 804.4 Inrn (r d.p)
Volume of hemisPhere
?-t
=; x 3.142 x 6'
- 452.4 cln' (r d.p)
(ii) Surlace area of hemisphere
=ix3.142x6-
= 339.3 cm (r d.p.)
(ii) surrace tuea of hemisPhee
=3x3.1:12xL5'
= 2120.9 cln'(r d.p.)
34. (a) volume oisphere= 960 cn3(civen)
=t =960
. 960x3
VolNe of hemisPhere
2]
=at3.142tt2'
- 3619.6 nr (1 d.p.)
Suface dea of hemisPhere
=3 t 3],42 x Lt
= 13s7.1 In' (r d.pJ
Voldne of hemisphere
?-s
=]r:.t.+zrts'
= 7069.5 mr (1 d.p.)
- 6.12 cn (3 siS. fig.)
.'. The radius ofthe sphere is 6.12 m
volune of sphee = 7328 mmi (Given)
- nl = 1328
. 7328x3
.llzzs v. t
\4n
- 12.0 mn (3 si8. tic.)
.. The radnB of $e sphere is 12.0 !m
volume of sphere= :1500u nl (Given)
= 
fr' = 45AOl
. 4500x1t =__4
.l45oo _r
=15m
. . The radius of tbe spherc is l5 m
(e) (i)
(0 (t
(ii)
AdsweB with conprele work",l sor'ttos @
35. (a) Suface area of spherc =
(b)
. . The radius of the sphere is 6.49 cm.
)ufface .rea ol spncLc = i)o4d m tur\en]
4rl = 664a
, 6648
i4618
= 21.0 m (3 sig. flc.)
.. The radius oirhe sphe.e is 23.0 b.
Surhce ma of sphere - l69u cn'(Given)
4il = \691
, t69
ha,/=i,
= 6.5 cm
. . The radius of the sphere ls 6.5 cm.
sun(e ma ot spnerc = rr44 cm (urven)
22^
4 /. = / t'=5541
. 55'14 x 7
=441
/=1[4,
=21 cm
422
= 38 80E o"
Volume ofsphere= 14 130 cm'(Given)
-x3.l4xl=14 130
16. (a)
= 530 cm'? (civeD)
= 530
530
1530
= 6.49 c6 (3 sig. ficJ
t=,1t:ll4
' - lTrai
@ r.,n".",". ,"'* ,u
Surface ma of spherc
=4x3.1,1x151
= 2826 cm1
Lel the height ofthe cone fomed be, cm.
' Volume of cone = Volume of sphere
1-4
i ,( 8'.h=i n 6
, 4x6r
It'
= i3.5 cm
.. Thc height of tne conc ibmed is 13.5 cm.
Let the base radius of fte hemisphere be r cm.
Volume of hemisphere = 24 x Volune of spb.r.
1.4.
tt=u ; r t5'
r 2,1x4!351
2
- 2058
, = $058
= 12.7 cm (3 sig. fig.)
. . The radius of rhe hemispher€ fomrd i$ 12.7 cn
Let the slot heighl of the cone be I mm.
Using Pythagoras theorem.
= 1369
I = ./i369
Let the mdius of the sphe.e be r hm.
Surface ara of sphere = CNed sudace ea ofcG
\t;=y't12x37
. \2\ 31
'4
= lll
, = ./in
= 10.5 Im (3 sic. fi8.)
.. The radius of the sphere is 10.5 mb.
o Volume of henislhere
= t x Volume of cylindef
=;x(rx8'x24)
Let lhe iadius of the hemisphcrc Lre / cnr.
2.
1/t = 1t52/t
1 ll52xl
/ = 1/i7rs
Total sudice area of henisphere
= 1360 cn'? (3 sic. fic.)
Volumc of 200 br1l bearings
= 2OO t 1fl/
=200^:^z^0.8'
= 428.93 cm'
Ma$ = Density )i Volume
= 7.85 g/cn: x 428.93 cnl
= 3370 e (3 sic. fic)
Volme of cyllnder = 3 x Volume ot sphere
Ern =r - ;tu
lh =41
lr'
; = ..,-
.l
(i) surface dea of sphere = 141z cn'
4y't = 44\t
,r = r 10.25
r = 1ri loli
= 10.5 cn
volume of sphere = Jtr'
=axrx10.5'
t)
(k)
Let the hciett ofthe cone be t cm.
volume of cone = Volume of sphere
- .. 1^ r1.5' ^ h = r543.5f
, 15,1:1.5 x 3
= 15.l2 cn
.. The heieht of the cone formcd is 15.12 cm.
Volune of 70 hemlspheres
=70^-^,'r\1.5
Let the rise iD thc ratcr lcvol be /? cm,
/\52xh=151.5/
. t5?.5
.. The nse ir the water level ls 6.3 cm.
Let rhe height ofthe cylinder be ft cm.
Volume of cylinder = volLde of sphere
-4y'\8 xh= a ty'x4'
, 4x4r
3xn'
I
=13 cm
. . The heisht of the cvlinder is I - cm.
31.
(i)
,,:,@1''*
= volune of hemislhere + Volume of cone
=l " i,o'- \ " ] "s''rz
= r54s; ctu'
Total surface dea of solid
fcurved surracc I lcurved surfacel
= larea of hemislhercl + l"."or"o* .l
=1uxfx11*tf;rsxr:t
= r:r I cm'
(b)
AnsqaN wirh conplere worked sorutio* @
3E /---- -116:\,r*W
39.
40.
41. (a)
Volume of rcmaining solid
= Volune of hemlspherc Volume ofcone
222^t22-
=(a x 7 x14') (; x 7 x14'x 14)
= 287s cn" (correct to nearest cb")
Lat the height of cylinder be n cm.
Volume of solid = 3850 cmr (Civen)
fvolume of I /vulune or\ fvolume or) ^^-
l"*i.pt'"*J '[,.r'"a" J'l i"* .1=tssn
))22t22,, ., ,t'ti 1'Yh\-t t / tr1
x 10) = 3850
718;+15ar+513i=38s0
l54h = 2618
2618
'' - 15.1
-17cm
. . The height of the cylinder is 17 cm.
(a) Radins of hemisphere = 2 + 4.5 = 6.5 cn
Volume of caddle holder
= Volntrof hemispheF Volmeof cylindricalhole
(b)
=ti ir ' 65'r-(r. 2'. 1.5r
= ss6 c# (3 slg. fic.)
Total surface @a of candle bolder
frotal surfee area) fcu^ed surface deal
= lofhemrsphe€ .l 
+ |\ofc]lrndricalhole J
=(3 x rx 6.51+(2 x zx 2 x 1-5)
= 4r7 m? (3 slg, fig.)
(i) Volune of container
= Volume of sphere + Volume of cylind€r
=t;' i . 2.1't+t; /2.1'/ t2.6\
= 213.4 mr G d.p.)
@ 
Mathenatics rutd 28
(ii) External surface area of container
=/s!rfac€ a.ea\ + / cdred sr&e \
\ of sphere / \ de. of cylinder/
22.22
=(4 x t x 2.1')+(2 x 7' x 2.\ x lLa
= 221.8 m'(1d.p)
Honronnl ,urtace ma ot $ arer in conLin-
i Area or circre r fA-Es or.rcctaicr')
= lof.adius 2.1 mJ +
\&o wrdrn 4., m /
)).
=( 21)+(126 . 42)
42. (a) Volume of each conical cup
1 :2...'.o
-l
No. of conical cups filled
= 103.896...
.. t03 coni@l cups cd be filled.
Volum€ of waler left in the container
= 12000- 103 x r15.5
= 103,5 cm'
trt the radius of the conlainer be / cm.
Votune of conlainei = 12 000 cnr
l"|"t= 12 000
12000x3xt
2x22
, = ltqztzt
= 17.892 cm
External surfice area of container
= 2010 m? (3 sic. fig.)
43. (a) Yolme ofhemisphere = 354.68 cn (Given)
;rf = 3s4.68
. 354.68 x 3nr- 2
- 532.02 cm'
= = 
t*(6r\
= 2(532.02)
(b) Total suda@ area of hemispherc = 180 @2 (Given)
1n; = $o
nt = ffi.tt
Lel rhe slant height of the cone be I cm-
Using PythagoraY th€orem.
I=[;7
= 15t' cm
Total surface dea of con€
-Et+d"En
= (t + ,E)nf
= (l + .v5tx6o)
= 42s cn' (3 sig. fig)
trt the radius of the sphere be . m.
. . The radius of the cylinder = / m and
the height of the cylinder = 2/ m.
Volume of cylinder = 204 m' (Given)
df)(2t) = 204
2rd' = 204
# = 1,wm1
Volme of sphere - lD2
- loort
(b) Snrfee Na of sphere = 312 m' (Given)
4tt?=3r2
Total extemal suiface trea of open cylinder
= nt + 2B(2r)
= s(78)
= 390 rtr
Using Pythagorat theorem,
i = \]'1 - r5'z
x =!A
=8+17
=25m
=7nh
=axzx15'x25
- sE90 crnr (3 sic. fic.)
UsingPythagom theorem.
t'=25'+15'
= 850
I = 1650 cin
Cufled surface dea of cone
=rx15x1/850
= 1370 cn (3 sic. fic.)
Volune of 50 marbles
/Volume ofhemisphencdl I ...
= | containei
= r: x 1.14 ^ 12'r- 2910.78
= 706.5 cmr
(a)
(b)
46. (a)
Answes with conpiere worke,l sol"tt"* \.l47)
47.
Let the radius of each maJble be r cm
706.5
vohme of each mtrble = -6
= 1413 cmr
'1^rra^l=14.13
. 1413x3r= 4 \ai
. = t6i?5
= 1.5 cm
.'. The radius of each marble is 1 5 cm
(b) Extdior sDrface arca of 200 containds
=mOxQx3.14x12)
= lS0 864 cm'?
180 864 ,
= 18.0864 n'
Cost of painting the exterior of 200 conB €rs
= 18.0864 x $96
= $1?3629 (cored ro nearest Gnrl
Volune of nodel = Volume of cone
l\iorme of Jr (u9lYT:_:fI 
= u.r,." "r-*!\ spnet / \ LtrINq I
ifr . r',"rtst - tfoatl-- a
Jl*rr"=trol W
/'(jf 45)=o
l/ 4s =0
1/ =4i
r=3375mm
Irt the slanl beight of the cone be I cm'
Using Pythago€s' lh€orem'
t = r&o
@ 
Madema,ics rutu ,B
(a) Reqlired surtace area
I i surr*. m!
=;- ldh..,.ph." - *"'r'
- ; x t2'(r8t + 2z(r8X52) + 418)(30)l
(b) Volume of Nater in container
I rVoumeot \olLmof - voludeot'
- '[hetuspr.e .\linder cone
= ,, I;dlsl + ,dr8)'(52) + i'r118)'(24)l
49.
1".- -ind 
_ 
-- 
3,?5 cn 11.25cn
volume of henisPhere
=ix3.142x8'
= \072.469 .d
Volume of frustum
-,ir't.ra:>8 r5,-'; Jri2 2'(b) o = (0,4)
(c) x = (-3r 4)
(d) (i) Arca of APoR
=t'8'se'HeisbL
=!x+x:
(ii) U8ing Pytnagors' theoren,
PQ' = PR" + QF
=4'+3"
Po = '!8
= 5 uDits
Perimeler of APOn
=PQ+QR+PR
=5+3+4
= 12 un s
(a) SDbstitnte ( 3. p) into 2r + 3) = 9.
2(-3) + 3p =9
O) sub;dture (2q - 3, -l) inro 2r + 3t = 9.
2(2q -3) + 3( r) =9
4q 6-3 =9
4s-18
q=4,5
(c) Substitute (4., 5 - 3/) irto 2r + 3) = 9.
2@t)+3(5-3r)=9
8r+15-9r=9
t =6
Answ6 with comptek worked soludos @
22. (a) Substitute (d, 0) into 51 4) = 8
5d 4(0) = 8
5d=8
.I = 1.6
Substitute (0. z) lnto 5r - 4] = 8.
5(0)-4"=8
1e=8
e=-2
SubstitDte (4, /) into 5r - 't) = 8
5(1) 4/= 8
20 4J=8
4f= 12
(b) 5r-4i'=8
Wher r = -2,
5(-2) 4) = 8
-10 4)' = I
It= ta
,l
.. (-2. -41) fies on the lin€ sr - 4J = E-'2
2l Subsdrute 12.l + I ) into -r + 4r = tl
l(2) + 4{i? + 1) = -t I
L5 + 4t +,1= l1
4h= 16.5
h = -4.125
At thea irtercePt, ) = (J,
r+3(0)=10
t= 10
.. A = (r0,0)
3r- 5), = 15
Atrhe)intercept,r=0,
3(0)-5J=15
..8=(0!-3)
=l^Base Heiqhr
=lxror:
2
@ 
Mddrematics rutor 28
B, Solving Sinullan@us Linear EquatioN Graphi@!
25. (a) -!=r-5
(b) The solutionisr= 3 udt ='2
26. (a) r- 1r-3
3 6
5 -2 I
0 3
3 2 1
(b) Thesolutionisr= 9 and, ={r.
n, r-2f -6
The solution isr = 2 andJ = -2
(ii) )=-r+4
3 I
0
5 7
0
8 0
(c) The solution is x = -2 and) = 6.
Answers rjtb Conplete work.d S"1"1i-, 
@
29. (a) !=u+3
lhe \olurron sr = z andl = /.
3
3 9 15
l
0 3
5
-l 0 3
14 tz t0
The sotution is r = -2 and J = -10.
l^
J=-3i+5
Tbe solution isr = 2ddl =-1.
0 2
2 -t
0 2
@ r.,n".",n, "*- 
t"
The sohnon is x = -2 and I = 0
Zr+r+6=0
| ='Lt 6
l
I= 2t
The solution ,\r = -4 dd.'' = 2
(e) -t=r 2
t 2.y+2=0
2J=r+2
r=lx+1
2
2 0
0
2
,2
I
8 0
10 2
8 0
(i) 3x- 101= 0
10] = 3,
3j= ro"
x+2r=12
2r= '+12
0 8
0 t.2 2.4
0 8
2
The solution is : = 7.5 and J - 2.25
G) 31 2J-9
2'-3t-9
}= :J.4'22
11
0 3
0
0 3
{.5 2.5
The solDtion isr = 4 and)' = 1.5.
Answes with conprere worked sorutioN @
(D
3
)=;x+1
t
2r-])+l=0
:
3
3J=2.+l
2l
33
0 2
7
2
2.5
Tha solution is x = 2.4 dd J = 2.8.
1
0 I J
5 2
-3 -l
The solution is r = -2 and J = -1.
0 2 4
12 1
0)
J+;)=-i
l^
;)=-r+r
l-4x+12
Therc n ar inflnite number of solutions
3)-r+6=0
l^
;u= f'*zs
(l) 3r- 8l = 12
8t=31 12
r= I' r':
5r=r+:
ti
-55
8
0.6 2.2 3.8
0
-1.5 1.5
0 3 6
,2 1
3 6
3.5
(,D r"*'.,",'.,,'
The soluiior is r = 12 andJ = 3.
(m)
2j+l+7=0
t=2x 1
I 0
6 2
I
5.5 6.5
: 0
-5 -l
2 0
I -3 7
The $ohlion is r = -2 and J = -3.
(o) 4r+5)=20
5r=4x+20
4,
]= ;r++
4f=Jt+24
"= f'*o
Ans$eF with Conplete Worked Soludons
2
2.4 0.8
0 2
6 2.5 I
30. (a) l=2 r
(i) wh€nr=2.5.1 =-.0.s,
(ii) wh€n'--1.8,.t=3.8.
(c) )=-1I+3 -t 0 2
5 3 -l
,, The sobtion to the simultaneons equations $ r =
--ffi
*-ffiffili
31. (a) 2)r 3r+6=0
2! =3x-6
3"
'2
0 8
-3 .] I
-1 0
3 2
@ r",n.-"u* ,",. r"
The solution to the
x=6odt=6.
32. (a) I+2)=i0
2f =-t+ l0
=:xBasexHeight
=!rzxz
2
1= l*++ 0
5
simultaneous equarions ir
5
l
_3
(r)
The solution isr =6ddJ =2.
=f^sase^nereht
=fxsxo2
33. (a) (i) )'=xt 1
0 3
I 5 1t
0 l
8 7 6
(ii) 3)+r=24
r- lr+s
0 9
0 2 3
(b)
14. (a)
(i) The solution isr = 3 a;d) = s.
(ii) The solutior is x = 4.5 aldl = 6.5.
itr:i=;ffi
ffi
(b)
35. (a)
5 10
l4
The solntion to the simultaneous equation is r - E
. . Pauls is 8 years old dd Maft is 12 years old
The totat cost of tickets for 2 adults dd 3 childrcn
is $48. (Given)
. . 2($r) + 3($r) = $48
2! + 3J = 48 (Shom)
The rotal cost of tickets for 1 adull dd 2 childrcn
is $28. (Given)
l(tu)+2($))=$28
(b)
Answers with comprete wo.ked sor"tt.* (O
(c)
1- 1r+16
x+2!=24
2t=-x+zz
0 9 l5
16 10
0 8
l4 l0 1
The solulion to the sinrulianeous eqlations is
. . The cost of aD adult ticket is $12 and the cost
of a .hild ticket is $8.
'' lli=ji'.' qflffiffiffiffi#qi-"i'
(b) a+l = 150
x t- =20
) =r-20
0 50 100
150 100 50
0 50 100
:o 30 80
@ vrr'"-"t* n o, zt
The solution to the simultaneous eqDalions is r = 85
His narks for Mathematics was E5 and his ma.k fo.
When.=160,1=4.
160=r+41
..r+4J=160.
When.=190.d=5,
190 =r + 5J
''.t+5J=190
(b) -. + 4l = 160
4)= r+ l6C)
I
),_ ;a+40
a+5)r=190
5)=-t+ 190
I __
40 80
40 30 20
0 40 80
]IJ 30 22
2. (a)
...=40+3(V +-
.= 40 + 30(10)
. . The cost of a chest wilh 10 drawes is $340.
tFt x be the nunber of vases sold and )' be the
prcduction cosL for mating x vases.j.l=&r+30-ffi
t€r r be the nunber of vases sold and I bc ihe
.'r=r.r".ffiffiffiffiffi
r=8r+30
srmultareous eqDatons's
7
0 9E 196
7
30 E6 112
(a) she needs to sell5 vases to break even
(c)
(d)
when 3 vses were made.
tnss - Production cos! - Tot l sales
= $54 - $42
= $12
.. M^ Wory's loss when lhree vases were ma.le
when 12 vases $erc made rnd sold,
goduction cosl = $126
ftofit = To€t sales Prcduction cost
= $i68 $126.
= $42
.. Nrfts Wong\ profit when twelve vases were
nade md sold was $42.
(b) she n@ds to sell at least 6 vases to nake a prcfit
Ais$eA $rh Cnmplete WoiLeO *,,** 6i
1. (d) (i)
= Volume of pyranid + Volune of cuboid
= (! x 32 x 32 x 30) + (32 x 32 t 36)
Total surlace dea of solid(ii)
rLrrnlsrrre I /BNm!\ fluDlqud.dl
= \mnotpym al - lor' ruo,tl - \rs ordboid ,
=4 x (l x 32 x 341+(32 x 32)+4 x
(32 x 36)
- 780E cm?
(b) (i) volume of ntid
= Volune of benlsphere + Volume of corc
= ?ot * ! nfn33
=, I.tlr2 J0'-''.rl42 10r'dn
= 94 260 cm"
(c) (i)
Let the slaDr height of the cone be I cm.
Usiry Pythagorat theorem.
11 =30'z + 101
- 2500
I = .{6500
Totai suface dea of solid
/ Lurve{l \drtace ) , rCuded edeel
= lr,o.t ncmspnereJ - [doorrcre ]
= (2 x 3.142 x 30') + (3.142 x 30 x 50)
= r0 368.6 mz
= Volune of cylindcr + Vo[tme of cone
= rl.l42 l0 8)+r; ll42vl0'"241
= 5027.U cni
@ r,,n"."'* ,'- ,u
(ii) Let the slan! height of the cone be I cm.
Using Ptthagoras' th@rcm,
= 676
I = .Gro
Total surface area of solid
/C!nuradnel lroprE! I fcm'dr{d*
= lJrtiorcl!nd{l 
+ lot(r irtrJ r \tuoor.oit
= 12 >l"_2.
33
=ti - 2.1 lt+t: :=.2t1
No. of solids that can be fomcd
_ 25 924
= 500.928
. . The naximum lumtEr of solids that can be formed
is 500.
7. Radlus of cylinder = 12 - 2= 6cn
Volume of 2 metal spheres
=2 - ,r'25'
= 130.8997 cm'
Ler the incrcase of the water lelel bel1 cm.
4/rn = 130.8997 cml
rx6rxr=130.8997
, B0 8997
= i.16 cn (3 siS. fig.l
. . New depth of waler in cylinde.
=8+1.16
= 9.16 cn (3 siC. fi8,
@ r'n".",* ,*. ,u
.. The dl.merer oltbe scmicircle is l0 cn.
Let the base radiN of lhe soLid be / cn.
volumc of solid = 3465 cm' (civen)
voiume of cone + volume of cylirder = 3465
L Elh + nlH = 3465
; ?1 xt xt:t+r? x I x 8)=3465
^l-, ,. 1..-.,-.
3r; t = 3465
I 3465
:t]
1
= 110;
/ = .lt1olll 4
= t0.5 cm
.. The base radiLs of the solid is 10.5 cln.
Dimeter of hemisphero
=26 4 4
.. Radius ofhenisphere
Total surfa.e arca of naterial used
/CuNed $rf{c I /Areofcncb I aAtaolcirle .
= la,erorhenispheEl 
+ 
loi ndrus ]] cm] lofndiusecn
=(.2 t 7t x 91) + (E x 131 (z x 91
10. (a) Ler the radius of the semicncle be r cm.
e.u or,".i"i."r" = tz j,, 
".'1ciu"n)
(;
.n!j-J
t
Lcngth oi dc ,{f
I=tx7nr
Lct the base radius of Lhe cone ibrmed be R cm
(c)
.. The base diameter ofihe cone fomcd is 5 cm
Let the height of lbo .one formed by , cm
Usnrg PylhaSoEs' rheorem.
= 1E.75
l' = .i1sl5 ."
Volumc of cone fomed
= )= -2.35 orr - 3.35.
(iii) tbe smallesl valDe of I' is -5.2 and the
conesponding value ofr is 0.5.
(iv) the equation of the line of symmetry is
(c) Fron the 8raph.
(i) when:t'= 0, t: -2.?s orr - 0.?s
(ii) when,= l s,.)' - -325
(iil) the equation of line of symmetrv: r = -l
(iv) the coordinates of the marjmum poinr 1s
(-r,3)
13. (a) I=2l'+3' 1
(d)
Equation of line of symmety: x = {.75
(i) whenr=ls,]=8
(ij) whetr ) - 5,r = -2.6s or j = l.rs
(iji) the smnllest value of 2f + 3, 1 is -2.1
14. (a) J=(1 r)(r 3)
(i) when r = 1,x=0.6orr=3.4
(ii) whenr=4.5.1 =-s2s.
(iii) lhe greates! value of ) is 1 and the
conesPonding value of r is 2
(iv) the eqnaiion of the line of symmelrv of lbe
cufle: r = 2.
Answ4 with comprere worked s.rudotr (a
15. (a) )= ztr+7r+a
d= 2(r)'+ 7()l + 4 -9
b= 2(2)'+7\2)+4-ro
c= 2(5)':+7(5)+4=-rr
(b)
(c) Equation of line of syftmelry: r = 1.75
From the graph. the maximum value of
2;+tu+4is10.2.
a=2(0.51,5(0.5)-3=J
b=2\5)1-5(5)-3=22
16. (a)
@ *"*".",., t*- ,u
17. (a)
(i) whenr=4.3.J=12.s.
(ii) when ] = 8.r = -1.4 ort = 3.9.
(lii) the smallest valu€ of ) is -6.25 and the
coreslonding value ofr is 1.25.
d= 1f5l l) l-tr':t = -3
b=;1512)-2')=3
18. (a)
19. (a)
(i) whenr=4.7.t-0.7
(ii) wlen)= 2.5,r- r.4 orx = 36.
(iii) the greaESI value of) is 3.1.
(iv) lhe eqDalion of the line of symmetry; r = 2.5
a = t2(0) 3t(5 0) = -ls
D = t2(6) 3(5 - 6) =-9
(i) wben)= 10,r=0.i1
(ii) whenr=2.5.t*s.
Eqnation of line of symmeLry: x = 325.
a= 18 +5( 1) - 3(-l)'= 10
,=18+5(3)-l(3)r=6
(c) Equaton of litre of symmety: r = 0.85.
(d) From the graph.
(i) when i = 2.5,J = -13.25.
(ii) when) = 15.x =-0.45 orx=2.1s.
20. (;) ),=2;+ar e
(i) when I = 10. r = -4.25 or r = 2.25.
(ii) whcnr=-2.5.1--6.s
(ili) the equation of the line of symmerryi r = -l.
(lv) the solutions arer = 3.35 ort = 135.
2L (a) ))=t -\-s
a = \,0.5)' 21 0.5) 5 = -3,75
b=1'-2(l) 5=-6
AnsweN with comprere work"d s.r'ttds @5)
22. {.n)
From the graph. the least vaiue of ) is -6 and it
Equation of llle of symmotry:r = l.
whe!j=1.8.S--5.35.
whe! I = 5.35., = 0.2 orr = 1.6.
: P=o2
i.p=0.2md4=-5.35
@ r.,n".,,". ,'-,u
(b) Equation of line of synmelry of cune: r = 0.
(c) The soludons are r = -1.7s or r = 1.7s.
23. (a) r=3-sr r'
-6 -5 -3 : 0
3 3 7 9 9 3 l
(i) rhe solutioDs 1o 3 5r _l=0arcr=-5.55
orr = 0.55.
liil .he .ot | .n. o , ., i .5 Je x - _4,S5
or r = -0.45.
(b) Frcnr tbe g.aph,
rrt rtrc m lrmum rort .uix.r ore, u. th..one l
(ii) wher,1=6.'=18.
... When nD to&l surtirce area is 6 cn,
r - l.ti cm.
25. (a) A= t x Sum ofpd.llel sides x Heighr
I-txl(r 2)+([ zr,x(r+2)
I
= 2(e-J)('+2)
= t(er + 18 _l xr)
=t(r,+tu+18)
... A = I :r.+ Jir+9)cm, (Sho$n)
r N max'nn'n Nhen I = 3.5.
AD=3.5-2=l.5cm
,c=ll-2(3.5)=4cn
AE=3.5+2=5.5cm
&) A--l;+3fl+q
Answe$ wirh comptere worked sol,it.m (a
26 (a)
Let the length of the enclosure bel m.
Perimeter bf rectangular en.lorut = l0 n (Given)
2(r+l)=10
r=(5 r)m
A=LenglhxBreadth
A = (sr -i nl] (sho\Yn)
(i) whenA =s,r- l.4ora= 3.6.
.. When thc dea of the enclosnre is 5 m',
x=1.4nofx=3.6n.
(ii) the maximum area oi rhe enclosure is
625 m:.
@ r"r"."u* 
^.. 
r"
' (i) when / = 2.4, Y= 23.
.. The \olume ol rhe tanl r.2J m' trheo iE
radius is 2.4 m.
(ii) when Y= 60. / = 3.9.
..lr'"d.').irherark \ l.q m { \en rlF
vohtue of the lank is 60 mr.
28. (a) r=l-14'+80
(b)
29. (a)
(i) rhen ) is nininun, r = 7.
.. Mr Lim needs ro order a batch of7 cbairs
for the cosr per chair to be minimum.
(ii) whenr= lt_ 2,.1, 5.201
An\{( \ $ irh conrf crc u.rl",i s,,l,t.* (O
(d) E
(0
. A={3,3}
Lt+1-1
.. B= {3}
A+B
l t-l
A=11.2,3lu{4.5}
A = 11.2,3.4,5l
a={1.2.3,4,51
..4=B
ltl
A = ld..t. el.\ lb.cl
A =A
5r- 1= 3
5
..B-A
C=A
D=tal
t= {01
C = r, C and a de equal sets.
G = 1. G and 1 ue equal s€ts.
1l = .r. Il and ./ de equal sets.
(aJ n(A) = 4
(b) No, rub! e d.
G = 12,3, s,7)
a= (5,6.7,8,9, ...1
r = 17,2, s.3l
= {2.3,5,7}
"r = 15,6.7.8,9, ...1
(c) No,lstaplerl ed.-ffi
(d)
(e)
1. l^J
€ = [All stationery itebs]
B = {p€ncil, stapler, eraser}
R = lP,e, tl
r(R) = 3
@ r,,n .o", 
^., 
,"
(a)
(b)
Q= tl, e. a, p)
li) P=Q
0t nco
9. (a)
10. (a)
(d)
(e)
(0
G)
1t
(b)
2E {2.5,7}
oE)a
-r@r".,,rt
{r}8ir,3.5}
2E{0.2.4,61
{a}g(a.r..l
o fq-l {appre, pa, orange}
l8.18lE{ls.8}
E
E t", 't ee
({a., is a lroper subset ofA, i.e. {a,rl cA.,
fl, Z is a sutser or every seL.
ct
ld,,@i+a( ) =a)
E]
E
(") E
G) [tr l4€8
({r} is a prcper subset ofB. i.e. {t} c a-)
(c) E, Z is a subset of eveq, set.
(d) E z(8) + 10, z(B) = s
(e) E {a. r. i... il d B
({a, t, t, ., r} is not a prcper subset ofd
since it contains the sam€ number of
€lemen$ 6 A. Her€, {4, r, t, c, rJ is a slbsr
ofr. i.e. |a. r, t. c. r] E 8.)
...P=1_1.tl
c=trl
f = {0, l,2l
s={ l, 1}
(a) QcP
(b) PER
(.) QcR
(d) PEs or
A = {1.3,5}
a= {1,5}
c = 1r,2,3.4.51
D = {1,51
scP
14.
(a) (i) n(4) = 3
(11) n(B) = 2
(b) (l) Bca
ltl) ACC
(ni)B=D
c)E
(b)E
(c) EJ m e e {r, tu. r, r, 
"}
(d) E
(e) f{] 2 € {2,3,5.7, ll, ...]
(t [fl, 1i..,.] e {r, s,,,..., rl
({,,.. el c {t, s, o,.. e.l})
G) El.5 e l-5.61
61 fl. o.zs x 16 € lr : ' is a factor of 121
1€ 11,2,3.4,6.121
(D E. t:t e {-r, :t (:t c {-r, :lt
E.ze {r,I is a !e'fe.t square}
(O c {r:a ls a perrect sque})
rtr I
Answes wirh conpiele workt *t,... @r)
(l) E] o is a subsel or elery set.
(.rE
(n) E {bN}g I!, u, b, n. a. \ i, n. "l
(.) E
'r' [] tt' aa 1. lvond.D. lue d.').wedne\dry.
Thmday, Fnday, Saturday,
15.
t6.
(b)
l7 (a)
18. (a)
Sundayl
E] {7t E {1, 2, 7. ral
a,I2l
a, \al, IbJ, Ia, bJ
z, {ant}, {bee), {ant, bee}
a, {tl, I2J, I3t, 11, 2t, 1r, 3}, {2, 3}, {1, 2, 3l
o, {4}, {6}, {8}, {{, 6}, {4, 8}, {6,8}, {4, 6,8}
= z, {4}. {6}, {8}, 11,6}. 14.8}, {6, 8l
. . There e 7 proper subsets of.1.
AC6 l^
.. Mdimum ,(A) = 49
IfCEDandDqC,thonC=D.
(i) n(O = 20 (civen)
''' n(Dl = 2o
(ti) c-D
(i) E = |n, a,t,h, e, i,., sl
(b)'
(c) (i) ,(€) = E
(ii) r(A) = s
(iii) ,(A') = 3
19. (a) .
21. (a)
ffi/-lI ll t*'u I
v"t/ \-/
20. (a)
(i) A' = {apricol orange, suaYa}
(il) B'= {sfrawberry, plun, ap.icot, orang€J
They de disjoint s€ts.
(i) € = {1, 2, 3, 4, s, 6, 7, 8, 9, 10, ll, 12, 13,
14, 15]
(ii) A = {s,10,1s}
(iit B = 1r,2,3,6,9I
(iv) A'= {1, 2,3' 4, 6,7, 8,9, ll, 12' 13' l4l
(v) B'= {4, s, 7, E, 10' 11, 12, 13! 14, ls}
(i) ,(s) = ls
(ii) ,(A) = 3
(iii) tr(r') = 10
(i) s = 11,2,3,4,5,6.7.8,9l
(b)
,,1=11,2,3,4.s1
I = {2.4.6,8}
(it
(b) (i)
A u a = 1r, 2,3, 4, s,6, 8]
A^B=12,41
€ = {1, 2, 3,4, 5. 6. 7, 8, 9l
A = {1,3.5,71
f = {2.4,6, 8}
i,--\ /---{
/ ' 5 \/ , 'i \\r!y
9
A u A = {r,2,3, 4,5,6, 7' 8l
A^B=A
(it
@ ru'*-,,n, r*- ,u
(c) (i) €= 11. 2. 3,4, 5. 6, 7, 8, 9)
A= 11.3.5,7,91
B = 12,3.5,71
AoB=11,2,3,5,7,91
A^A={3,5,7}
€= {1. 2, 3. a, 5, 6, 7, 8. 9}
A = {1.2,3.6J
B = It,2l
(i0
(d) (i)
(it
(e) (i)
(iD
O (1)
ArB=11,2,3,61
A 
^R =Ir,2l
€= { 1, 2, 3. 4. 5, 6. 7. 8. 9l
B=11,2,5)
I o \/' ' \
\-7r." /
Aur={r!2,4,5!8}
A^B-A
s= { 1,2, 3.4,5,6.7. E,9J
A= \21
B = \2,1,6, El
13 s 7
(ii),1uB={2,4,6,6}
A^B=I2l
22. (.n)
(b)
(c)
(d)
(i)
A U B = {7,8, 9, r0, llJ 13}
A^B=T9}
Au B = la, b, c, d, e, f, sl
AuB={3,6,9,12,15}
A^B=I6,t2l
Ar) B = Ip,q,t,v,w,r,zl
A^B=A
A u a = {pecil, pen, eraser, ruler, stapl€rl
A=la,o)
B = la, el
A = {1.2,3.61
B= (1,5)
AwB=11,2,3,s,61
Ana={1}
(h) AUB=\k,e,!l
A^B=A
(i) A=ll,ar,s,a.l
Aw B = Ia, d, e, s, t,n,o, sl
A^B=Ie,t,nl
(j) A = {-4.1}
B- {1,2,4,5. 10.201
AoB= .4,1,2,4,5,10,m1
A^B=l4l
(a) A ua= {3,4,6, 8,9, 10, rr]
(b) A u C= {4,6,8,10,12, 16}
(c) , u c = {3,4,6, 8, 9, rr, 12, 16}
(d) A.a={6}
(e) A^c=14.8J
hlA^c)=2
(t B^c=A
,(B.C)=0
(a) (i) €= {1,2,3,4, s,6,7'.,.1
(ii) / = {s, r0, 15,20,25,...}
(iit , = {2,4,6, E, r0, ...}
(b)
25. (a)
d.d = {r0,20,30,40,50,...}
A. a = F : r is a nultiple of 101
(i) € = {r,2, 3,4, s,6, 7r ...,,{9}
(iD A = {4,8, 12, 16! 20! ...,,r8}
(iii) , = {16,30,44}
(i) A^B={16,44}
(ii),(,,{.6)=2
A = {2,3,5,7, rr, n,...]
n = {4, 6,8,9, r0, 12,...}
26. (a)
2'7. (^)
O) A^B-A
(ii),(4.a)=0
oo
(i) AnB={3,s,71
(ii) ,{ u B = {r,2,3,4, s,6,7, 8,9, rr]
(ili) ,(A ^ 
B) - 3
(iv) 
"(A 
u a) = r0
(i) [r] (ii) E
(iil) I-1. , d A, Not an elements in D are in A
(i')lrl
{r, 2. 3,4, 5. 6.7. 8, 9, l0l
{2,4, 6. 8. 10. 12. 14, 16, 18, 20}
l\5. 16, 17. t8. 19, 20. 2t. 22, 23, 24, 25]'
(i) A n a = {2, 4, 6, 8, 10}
(ii) a. c= {16, r8,20}
(iit)A^C=A
(nr,(A^a)=s
(v) ,(O = rr
(vi),(A^O=0
28.
C=
(a)
Answes vith conplete work"d sor.Lio$ (A
29. (^)
l0 (a)
(b)
(c)
(d)
(t 8.7 etr
(il) E,A uB = {1, 2, 3,4, 5, 6, 7, 8, 9, 10, 12,
14, 16. t8,201, 16 e,{ u 
'(iji)[tr
(iv) [tr
€ = {i. 2, 3,4, 5, 6. 7, 8, 9. tol
(i) A = {2,4,6, 8, r0l
( ) B = {3,6,91
(iii)c={4,8}
(iv) A'= {1,3, s,7,9}
(v) a'= {1,2,4, s, 7,8, r0}
(vt c, = {1, 2,3, s, 6,7,9, 10}
(i) .{ u a = {2, 3,4, 6, 8,9, 10}
(ii) auc={3,,t,6,8,9}
(iii)A.a={6}
(i!)A^c-{4,8l
(!) B^c=A
(vi) A'^ a'= U, s,7l
(vii)A'u C= {r,3,4, s,7,8,9}
(i) r(a) = 3
(il) 
"(4=aQ\ Aw B = la,b,c,d,el
(iv)A^B={dl
(i) ,(r) = 6
(ii) n(C) = 2
(in) r(cu D) = s
(iv) c n D= {r,,} ^ lr, r. r}
= {r}
(i) n@^n=2
(11J n(E \) F)'=z
(iii) (E. 4'= {1,2,3, 9, 10, l
(iv) Eu F = {1. 2, 4, 7} u {1, 2, 3, l0}
= tl,2,3,4,7,10)
(v) E 
^,c={3.9, 
10, lll-{1,2.3,r0}
= {3, r0}
(vi) True
(vii)False,3e(Euo
(viii) True
(i) G'= {3, s, 6,9, roL
{li) r(Gu14'=3
(iii) d 
^ 
d = 13. 5, 6, 9, 10j . 12. 6. 8, 101
= {6, 10}
(iv) G u 11= 11. 2,,1, 8l u 11, 3. 4, 5, 9l
= {1,2,3, 4, 5, 8, 9}
(v) dug= {3,5.6.9, l0} u {r,3,a, 5.9}
= u,3, 4, 5,6,9, 101
(vi) lialser(5€11but5 € /t')
(vii) false,2€G^It
(viii) Fals€ ({1, 2, 4, 8J is not a proper subset of
G since iI contains the sme number
of elenents as C.)
@ r"r"."u* 
^., 
r"
32.
33
r- 11.2.l.4. \.6.7.8.q. 10. l.12. 11. 14. 15. l6)
A = {2,3.5,7. 11, 13)
a = {1, 3, 5. l5l
(a) A u d = {1,2, 3, s,7, 11, 13, rs}
(b),a.8={3,s}
,c, 4'^a-11.4.b.8.o l0 l'. 14 la lol . rl.
3,5, l5i
= {r, 15}
td, A B =.2. t.5.7. l l . ll . 
^ 12. 4. o. '. 8. o. lr.
11. 12, 13, 14, 16]
= {2, 7, rr, n}
(a) sinceA =8,2 - ; and, = 1.
(br sin.e,4 _ A - A, rhe r{^ |ne\,. -or I re6a|'
i.e. they @ pdallel.
1
a = t md , cd be any value ercept l.
(r) (i) P = {r€ctansl6, squares}
(ii) 0 = tsquaEs, .hombusdl
(iii) a - {paralldosrams, rhoinbuses,
Ectangls, squares]
(b) € =.{All quad.ilateEls}
(c) (i) P^0={squares}
(ii) 0.x = {squares,rhonbus6}
(d) PcR -
{1,2.3,4, 5.6.7}
12,41
{1,2,4,6}
34. €=
C=
(b) lt) C^D=12,41
(c 
^D)'= {1j 3, s,6,7}
(ii) c u, = {1.2.4,6l
n\Ct)D)=4
Yes, C c D.
.1r € = {t, ,1, 5,6. 7, 8, q. 10, 11, t2. BJ
A = 13,5,6, 101
a = 15,7,9, 10, i1, 13)
(d) (i) A !.r l] = {3, s,6,7, 9, 10, 11, t3}
(ii) A .8 = {s,10}
{iii) a'= {3, 4,6,8, 12}
{iv) (/ u 8)/= {4. 8, 12}
'(,4u8)'=3(v) A'^ A = {4, 7, 8.9. l r, 12, r:l} . {5.7.9,
i0. u, 13]
= {7.9. I l. 131
h(A'^B)=1
(b) (i) False (ii) False
(iv) True (v) Tne
r.= \a. b. c, d, e,l, L h,...1
(lii) False
(vi) Falsc
16.
(l
G)
18. (a)
(r) ,{ - B = {a, I, I, i}
(b) B.\ C = \a,t,i)
(c) A.c={a,r!/,i}
(d) (A . 6) u {A . O = 14. s. l. i} u {4, /, l, l}
= la, t, a s,tl
(e) '(A^B)=4rgr ,rd^tl ='l€) ,rd.tl. i,iia):26
37. (a)
(c)
1'= {r,3,5}
B = 12,4,5l
(')'=A
= 12, 1, 6l
,1u ll'= {2.4, 6} u 12,4,5l
= t2, 4, s, 6l
rl,. a= {1.3,5} . I1, 3,6l
= u,3'
r ! 6,= 11.3, 6l ! i2,.1,5)
= {1,2, 3,4, s, 6}
R . B',= Ir, l, 6l . 12. a. 5l
=O
A u A= {5,7.8,9. l0}
(ArB)'=16l
A ^a= {5.71
(,,1 . B)' = {6, 8,9, r0}
Altemative neihod: Drar a Vem diag.am
(a) (4 !.r d)'= {61
(b) (4 . Bl' = 16, 8,0, rrrl
(c) .1' Lr B' = {6,8,9, r0} '
(d) ,4'- 8'= {6}
39. (a) M. N= {1.ll
n(.M^l!)=2
(b) (i) MLrN={r,..d.c,/}
(M u ,\)' = tel
(1i) ,w. N= {.,.,8} . {..4..J}
=\c,ej
(iii) ,u u Ar = 1., ., s) u {r, s}
= lb,c, e.sl
A' u r' = {6. 9, 101 u {6. 8l
= i6, E, 9, r0l
4' . d'= {6, 9, 101 . {6, El
= {6}
€ = 12, 3, 4. 5. 6, 7, 8. 9. 10. I l, l2l
(a) f, = {2,3,6,9}
(b) 5r l>14
5r> 35
.. r = 8, 9, 10, 11. 12
s = {8, 9, 10, rr, 12}
Ansse, s wirh conprete worrre,j sorutios 
6Ee)
(c) R'. s= {4.5.7,8, 10, I l, 12}. {8,9, 10, I l,l2l
= {8,10,11,12}
(d) R u Y = 12, 3,6.9) U 12.3,4. s.6, 7l
= {2, 3, 4, 5, 6, 7, 91
(e) n'^ y = {4, s.7.8. 10, 11.12). {2.3.4.5.6.71
_ {4.5.71
'(r^s')=3
L/ f s={4.5. .E. l0. ll.l'lL ' . 'r .6 
I
= (2,3.4,5,6,7,8, t0. l l. r2l
a(Ru,5)=r0
E= Vjr.i,g,o.n.n,e.ll
A = ls,e. a. n,t. r, Jl
(a) (i) (A 
^ 
B)' = {s, o, t, n, r,.}'}
(ii),1'.d={t}
(iii) (,a u 8)' = {D}
lirJ A w B' = Ie,s, o, h. t, t,t,rl
"nl
(b) (i) A/. a'= {1. nl .{8,,,n.^tr'l
= lnl
nlA'^B')=r
(ii) A' !, B'= {,. ,l u {8, ,. r. /, )l
= t8, t. o, h, r.r)
n(A' w B') = 6
@ r",*.",". r*- ru
42. E - { 1 , 2. 3. 4, 5, 6. 7, 8, 9, 10, ll. 12. 13, 14. 15}
A - {5, 6.7,8,9. 10, ll, 12. r3]
B = {5,7. ll, 13}
(a) A' = U,2,3, 4, 14, rsl
(b) B' = {r,2,3,4, 6, 8,9, 10, 12, 14, l5}
(c) ,4 
^ 
B = {s, 7, 1r, r3l
(d) A'ua=11.2.3,,+, 14, 15| u 15,7, ll. 1ll
= {1, 2,3, ,{, 5,7, 11, 13, 14, l5}
(e) {1. 1.7} ^8= {1. 1.7} ^ 15.7, ll. lll
- I7l
43. (r) .
44. (a)
(b) €= 17, 8,9, 10, 11. 12, 13. 14, tsl
P = {9. 12. 15}
0 - {7, 9. 11, 13, 15l
R= {10. ll. 12, 131
(i) P \, 0 = {7. 9, 11, 12, 13, 15}
n(PQO=6
(ii) x' = {7,8, 9, 14, 1s}
(iii) O ^ 
R'= {7,9. ll. 11. 15} . {7.8.9, !4, 15}
A ua= {squaresl u {rhombuses}
c u D= {parallelogrdms} u {rectanglesl
= { paralebgran$ }
-c
,,1 
^ 
I = {squlresl - thonbsesl
= {7.9. 15}
(d) A. D = {squarcs} ^ {rectanetesl
= tsquaJesl
X= lb.c,dl r = |d.b,c..1, el
(^) z-Ia,el
X r) Z - lb,c,.tl u Ia, ej
= la. b.c, d, el
lb) z = [b,., d) -
Y ^z = Ia.b, c..1,el ^ 1b,., dl
G)
E
trl
tf t
@.4'na=a
G) E.Acg
(d) ffl(08
(h) E
(b) 5'+ l2'= 169
. . The trimgle wilh sides 5
is a right-angled tridgle.
48. (a) Since A 
^ 
B = O.,4 and a de disjoint sets.
'(A 
uB)= n{A) + "(r)
=E+5
o
(b) Since C c ,, set C is contained inside set D
,c^D)=n(i:)
h(cwD)=n(DJ
=10
@
P^Q=P
(i)
(ii)
(b)
(d)
50. (a)
PL)Q=Q
A = {1, 2. 3,:1. 5, ... }
B = {1.2,3.4,5,...1
G) [fl
A = 12.3,5,7, 11. 13, ...1
a = {2,4, 6. 8, 10, 12, ...1
A^B=12'
..A^B+A
c)E
lalgAbsaeA.
{d E]Pt /- l.zlanda ll.J. efd,o
l e,RbutI+n.
c) [L]
(r [L
G) ElrA =8. theiA E B andB E,1 .
Ansses wirh conplete worked sorurtos @)
(h)
IfACA,thenAud=4.
l-r I
o
IfA n, = A, tlen A !r I - A.
51. (a) A'ffi
(b) AUB
E E.s.IfA = { i } and B = {1,2}. then 4
E
Tit
CB
(i)
l)
(m)
(r)
tf A c B aa.l B 
^ Q = A. $en A 
^ Q - |
flln ta t = a, rrterto a +a.
A
(A Mathemdaruo,28
ffi
(c) Ana
(d) A'.8
(e) (A u B)'
(1) Aud
G) (A . a)'
A'QB'
(i) A' 
^ 
B'
:n;
(A.B)UA,
(l) (Aua)^(A^a)'
1
52. (a) AUB
(b) /t^,
{nswe6 wi!h complere !'vork"d s.r't "", (O
(,,r. o'
(d),4'uB
ffi
(t A^B'
B
@o
x:
@ r"*"."0* 
^., 
,"
(gl 4'. c
(h) (,4 u 8). c
1
(i) (A^a)uc
i
(j) (A^B)u(r^O
(k) (A u B)'^ c
(t) (8.\ c)' .\ A'
v"
t
53. (a) A CB
(b) A^B=A
oo
(c) acA'
(d)
oo
A^B=A
5.1. (a) A^B=A A AQC=A
@o
(c) c 4. a.c=o cGA Jotl,r4. cr+0B
t,lr (4 uC) c A, CeA cndr(4 ^O+0
a c,4, r(d. c) =0and c.A - c
An!sc,\ s rh conprete worked s.ruLroN 6.i)
There ae other Possible answeB
(a) Ant
(b) ,'
(c) AUB
(d) A'^B
(e) (,a na)'
(D IAJ B),
(s) A.r'
(h) A'u B
(t (A.B)u(B^C)
(j) auc
(k) (,4 
^ 
c) n B'
(t) (,{^r)uc
(c) f4uq.(r.O
(d) (B u C).4
(l B'.rC/]A
r. (a) tB. O. A',
Q) (Aua)^c
2. Answers mY vdy
(a) (auB)nc
ft) (a^a)uc
(e) (A 
^a) 
u c
@ r.*..**',- ,u
a
Ibl
Idl
la. bl
la. d
la. dl
lb, dl
\.,.t|
la. b. cj
4. (a) Greatest possible value of r(A n B)
= n(B)
(b) Lqst lossible value of,(d . ,)
=0
(c) Crcatest possjble value of,(A u B)
= r(.{) + ,(8)
=20+8
5. r = { l,2, 3,4.5,6,7. 8, ...}
M*N=(Metr^(M^M'
tf,tM= {1.3,41 andN= {3,4,5.6}
Mu N= 11.3,4,5.61
Mn N= 13.4)
(Mn M'= {1.2,5.6.7. E,...}
(,uuMn(M^^')' = {1,3,4,5,61 ^ {r,2,5.6,7. 8,...}
. M* N= Il,5,6l
.. { 1. 3,41 * {3, 4, 5. 6} = U, s,6}
6. (a) (i) {Pupils who like apples bur not grapes}
= {Pupils who like apples} ^ lPeople who
do not like grEp€s]
=A ^G,
(ii) {Pupils who like gapes btrt not both apples
and pears I
= {Pupils who like grapes} . {Pupils who
do not iike both apples dd peaB)
=c^(A^n'
(iii) { Pupils who only Like two of the thrce ftrits }
= {Pulils who like apples and grapes bul not
peml u {Pupils who like grapes and pears
but not applesl u lPupih aho like apples
and p€ars but not grapesl
6^G)' P .\G' n. A t^ P)^tJ'
L€a-st possible value of ,(A u a)
= n(A)
=24
(b) Dnw a Venn diagram to dswer this section.
LeI l be the number of pupih who like all three
AoqweF rrLh Complete Worked S"lr,l""' 
@
There are 4 pupils inA 
^ 
G-
There re 7 lupils in C . P.
8 pupils llke ody apples.
No. of pupils who like otrly Srapes
=19 (4 I) j (7 r)
=19 4+r-r 7+r
No. ofpupils who like both apples and pears btrt
=2
No. of pupils who like ody pea6
=20 2 r (7 r)
= 1l
..8+r=li j
..,(,4.G.P)=3
.. 3 pupils like all three l'tuits.
(a) 1= (5 - rxl + r)
When jr = 0. .
(5 J)(1 +r)=0
..1 - (-r, 0) and B = (s, 0)
Wheni=0, .
v=(5 0x1+0)
c=(0,s)
@ t"*".",", ,*- ,u
=2
.. Eqmllon of llne of symmery : r = 2
r'=(5-2Xr+2)
=9
(c)
2 (a)
3+1
-5
.. Equation of line of symmelry :r = 5
Substitute (3,0) and (7,0) into ) =r+ d+r.
0=3'+a(3)+,
o=11+a\7)+b
]a+b=49_12)
12) - l1): 4a - 40
S$sttute a = l0 into (1):
3(-10)+D= 9
.-.a=-lOandb=21
) = 5' 10(5) + 21
. . The coordlnates of the midinum pollt is (5, -4).
lar nre rd,,ge.l va ue. ^l ' lo, shich ,te gadienr oi
tlE curve is positive is.r > 5.
,,r 'lffl,rygM.
craaicir; 
1
ff"ii", r'1 
1',;.1 s'10*t 9t',- 
-Nr9)6 @ra.
1=5 l-ine uf\rmmflry
3. (a) l=J: 4r+3
d=( 1t 1( 1)+3=8
,=3' ,l(3)+3=0
(i) when a = {.8.J = 6.65-
(ii) whenl = 2, x =0.25 orr = 3.75.
(iii) the coordinates of the nlnimum point or the
curve is (2, -r).
r.\l J'e equaioo olJ'e Ine ot jme,D i x=2
:1. (a) ]' = 2r' si + I
a=2(t): 5(l)+r--2
, = 2(,1)' 5(4)+r=t3
(c)
5 (a)
(i) when J - 10.: =-1.2 orx - 3.?.
(ii) whenr=2.6,t=1.5.
{iii) the minindm point is (1.25, -2.1).
(iv) the equalion of the line of rhe symmotry is
o=2 3(4.5) (+.5)'= -4.7s
b=2 3(-2:) (-2i = 4
An,sers u,rh compek wurr€d sorm,o", (rE
(i) whe! r = -4.2.) =-3.0s-
(ji) when ] = 2.5,i = -2.8 orr = -0.2
(iii) tbe ertatest talue ofJ = 4.25.
t=3+Zt-21
a=3+zGi)-2(3)7=-2r
@ *"*".",", ^.. ru
7. (a)
8. (a)
(i) when J = 15,r=-2.ss
(ii) the mdimum value of I is 3.5 when r = li
(iii) the equatior of the line of symmery: r = a5
1..irrL:.
Equalion of lire of symmetry:r = 2
Coordimtes of minimum point = (2! 0)
(tr) wnen y =ru,a =-r.r5 orr =5.r5.
(iii) whon,= 3.5,1 =4.s.
) = -t(1 1)(r + 3)
p=-;{ 2 l[ 2+]r=r.s
4=-;d r(1+Jr=0
(c)
e (a)
Fmm the eraph, the greabsr value of ] ls 2 and i!
EquatioD of line of symnetryrr = -1.
whenr=0.7,r=0.55.
wnen I = u.:\),r: r./ or r= u./
.. a - -2,7,
.. a = -2.7 and , = 0.55
i:, ()
(i) when ) = 10. x = -2.75 orr =3.25.
(ii) when r = 1.5,1,=-2s.
(iii) the smallesr value fo 2rr a - ?8 is
app.oximately -28.25.
Equation ofliDe ofsynmeLy: r = 0.25
28+r=t
2,r' a 28=0
. . The solutions a.e x = -3.5 or x = 4.
Areaoftnangle= t x Bare x Hcighr
lA-tx(r+2)x(s r)
t-
= 2(5r r + 10 Zr)
= ,(-j +lj+ l0)
r = (-tr'+ I 2r + 5) cm'bhosn)
(d)
L0 (c)
(b) il= ;;+ r;x+5
Answes wirh conprere work"d s.r,t.N @)
ll. (a)
A is maxinum when r= 15.
base = 1.5+2=3.5cm
heiCht=5 1.5=3.5cm
@ ""t"..,o,... 
r"
12. (a)
(i) wben I is gleatesl, t = 2.
.. The ball is at its g€ates! heighi when
(ii) the Breatest heighi reached bv the ball is 16 m
(iii) when I = 10. t= 0 8 or 1= 3.2
.. The requircd rarye is 0.8. . When the dea of the iemaining sheel of
aluminium is 4l cnf x=0'4cm ori=2'6.n
13. (a) (i) €= {r,2' 3,4! s'...}
(ii) .4 = {4 4, 6,8, 10, -}
(iii) , = {3, 6' 9' u' rs' - }
(iv) A'= {1, 3' 5' 7, 9, . '}
14.
15.
i8.
CA
A = {2,4,6,8, 10, 121
a= {4,8, 12}
c = 1.1,2.3,61
D= {5,7, 1tl
E= {1, 3,5.7,9,...}
F={1,2,3,6}
G = {2. 3, 5, 7. I L, 13, ...)
(a) aEt
@) B [q.4
(c) c El.
(d) DEG
(") E
(b) E
(c) bl {vl 4,1but l}}
(d) ffl
G)E
(o E, {r} dA
G) []l
{b) Ff
G)E
(d) ffl. o is a subset of eve.y set.
(a) (i) A^B={8,20}
(ii) A u a - {L 3,4, s, 8' 12'14,16,2I,241
(b) e = 17,3, 4,5,8,12,74,16,20, Ul
(a) (t P^O={2,3,s,7}
(ii) P v Q = 11,2,3,4,5,6,7,8,91
(ii\) nlP . Q) = 4
(iv) {1, 51.P= {1. 5} ^ {2. 3, 5,?}
= lsl
O) (D E.PeCbutPcO.
(ii) E
t9_
(lii)E,geobure€0.
0v)E]Pn9={2,3,s.71
..3eP^Q
(a) (i) Pun={7,9,rLrsl
(ii) 0^x={11}
(jii) ,(0 
^ 
x) = r
(iv) n(P u R) = 4
(v) (Pu0).R={7.9, 10,1l} ^ {7, ll.
= 17, rrl
(vi) (P . 0) u R = {9. 111 u 17. 11, 15}
= {7, 9, r, 15}
G) (i) EP-{7,e, 1r}
0 u n = {7, 9, 10, 11. 15}
..Pc(OuR)
1i1 fflrap=1e.rt1
.'.9ee^Ql
1ii1 fl 9 un= 12, r, ro, ti, t:1
..10 € (q un)
15)
(c)
21. (a)
(b)
20. (a) 3
(b)
(c)
€ = {1,2. 3, 4, 5, 6, 7. 8, 9, l0}
M = 11.2,',1,9)
N= {2.5,6}
P = {6.7,8. 10}
8
(i) A' = {4, s,6 8}
(ii) Aua={3,4,6,7,9}
(iii)14^B={9}
(iv)(,auE)'={5,8l
nIAUBJ,=2
No,A€t.
B^C={5!t}
A' . a= {4, 6. 7, 9, 101 n {3,4, 5. 8}
= {4}
B' 
^ 
c - It, 2, 6, 7, 9. r0l. \1, 2, 3, 4, 9.
= 11,2,9,1Ol
a u C= {3,4.5,6,7. 8}
(B u C)'= {r,2,9,10}
A u C= {1,2,3.5,6.7,81
101
(d)
Ansves wilh comprete workeu r"*.* (oD
ll (r)
24. (a) .
(r) M^N=l2l
ff'^ N = 13, 4. 5, 6, 8. l0l - {2, 5. 6}
= {5, 6}
M \.r (N^ P) = {1.2,7.91 u 16}
= 1r,2,6,7,91
NUP= {2.5,6,7,8. l0l
(NuP)'=11.3,4,9)
(i) E= {r,3,4, s,7,8,9}
(it a = {3,,r, s, 8}
(iii) B = {3, s, 7}
{iv) B'= {r,4,8,9}
(v) /Ua=13,4,s,7,E}
(vi) A.a={3,s}
(vii) (Aud)'={1.9}
n\AuB)'=Z
(vili) ,1' . B = { l, 7. 9} . ll. 5. 7l
= tll
n(A'.8)=\
Y6, {3.7} c B
c= 13.5.71
(b) (t) B'=Ia,b,e,Il
(11) Aw B = 1.,d, f,s, hj
(ili) (lr u,.1)' = {a, ,, "lli\) {A ^ 
ts)'= |a, b.d, e,l, hl
h( 
^B)'=6
25. (d) (i) € = {pdallelosrans }
A = {rhombuscs}
a = { reclangles l
A.Zr={squaresl
(ii).
(b) P= lisosceles bianslesl
0 = Iequilltern1 rri.ryles l
(i) P 
^ 0 = {equilateral trianslesl
(ii) P u 0 = {isosceles t.ianeles}
@ r",*-,,., '*-,u
26.
21.
28.
29.
(a)
(c)
30. (a) .
{/.,.1, r} .
f- [7. P.']. 10. Ll. 12. ll. 1,1. 15. 16. l-. 18. la.2c.
2t,22]'
A = {8. 16l
d = {9, 16l
c = {7, 11, 13. 17, 19}
(a) A^d={16}
(b) A^c=a
r(A.O=0
(c) (]={8.9. 10. 12, 14, 15, 16, 18.20.21.22}
rlc) = rt
.u 4 c q, t6l . {8.o. t0. t2. t4. 5. 6. P' 20.21.22)
= {8. 16l
Q) ru. = {s. r, u. r,,}
(K r L)' = 1.r. tj
(c) (./ u L) 
^ 
r= {4, r, r. r, !, rI .ls, u, r, v}
= tr, y, yi
4
BL) C = ld, e. t, a, s, tl
n(BrC)=6
A=B
(b) € - {1.2, 3,4,5....}
M = { l. 2. 3.4, 5, 6. 7, rJ, 9l
N= {3. 4. 5. 6, 7, 8, 9,...}
(l) M. N= {3! 4, 5, 6, 7, 8, 9, ...}
(ii) M = {r.21
,-'-:\ Ca-).]
\"/'".-
31. (a)
(b) € = {2. 3,4,5,6,7.8,9, 10. 111
(i) P =12,3,4,61
(ii) zx 1better in the examination s,nce
the ndks of rhe studerts from School ,4 cluster
dound 90 Narks ro 100 ma*s while the marks oi
the students fron Sch@l B clustef alound 60 ndks
to 70 mdks. Tbls meds that the average hdks of
the studenh fion School A is higher than that of
the students in School d.
Bobby made Lhe longest call thal day. (87 nin)
Bobby made rhe sbolresl call that day. (10 niD)
Tbe duration of cals made by Bobby cluster around
30 nin to:10 min. Hc nade ver) l€w calh lasdng
The duEtion of calk hade by Anne clnster dound
60min to 70min. She frade lery few calls hsnng
ThDs, the average duration of call made by Anne
ls longer than the duFtion of cau ftade by Bobby
20. (a)
21,- (a) 12
tl
l5
16
(b)
2t0 6
6
5
302
89
r355
0:1 45
0001
0241
412 t
50r
Mass of heaviesl child = 58 kg
Mdss of lighrest child = 20 kg
No. of chil&en havhg mdsses 36 kg = 13
=;r100%
= 65E
6
5
(c)
889
667 9
35889
Required heiShl = 150 m
No. of studenh whose heiehts de betwen 140 cm
=12
12
2
Ans,es with compreb worked sorutro* 6D
22. (J)
Required lorccntagc
= - ^ rno,;
15+17
= 43.757a
C. M@n. Median and Mode
Middle posiLion
23. (a) 1,2.2,2, 3,.1.5.5, 6, 8
(b)
(c) (i) MosL conmon mtuk = s0
(iil No. of studerts who scorcd > 75 ma*s = 14
057
tt38
34599
238
000157
0112 3 4 88
12345799
023558
2t399
(,)
(ii)
38
=lo
- 3.8
Mlddle posilion
(b) l, 3, 3.4.5.6, 6,7_ tl
(l) Mode=3and6*
(li) Median = s
(iii) Mean =t
= s.r, (r sra. fle.)
Middle posi!ion
(c) 2,,1,6.7, E, Ij,9. 14
(i' ) Medicn =-
(iii ) Mean
l+2+2+2+l+4+5+5+6+8_ 
lo
@ r",n..,n'", ^.,,"
{d)
Middle losilion
63. 75. 75. 80, 80, 82, 88, 90. 92. 92. 92
(i) Mode = 92
(ii) Median = 82
= 82.6 (3 sic. fis)
Middle position
12.0. 12.5. 12.9.13.2, 13.3, 13.8, 1.1.2, 15.8, 15.8-
r6.9, 17.2. 18.7
(i) Mode = r5.E
12+l(ii) Middle pn\lti.n = 
-= 6.5th Position
Mcdian = Mean of6lh and 7th values
l3 8 + 14.2
=,2
- 14.0
= r4.7 (3 sie. fic)
Mlddle position
s23, $29. $35, $38. $3E, $45. $46, $48.1j49
(i) Mode = $lE
(il) Median - $38
$351(iii) Mean = e
= $39
lxl+2x2+4r2+5r4+6xl
Mean = 
-
39
=ro
l0+l
Middle Dosltion = 
-= 5.5th posillon
Mcdian = Meu of 5th and 6th values
24. (^)
(t) Med
_ $(3:1+9:2+l0t I + ll xl+12x2+13t3+ l,l x l)
$1,16
= $11.23 (oNct io nearest ent)
i3+l
Middle positjon= 
2
= 7th Position
Median = $11
Mode = $rr
750(c) Metu = 2,a
= 3125 years
24+1
Middle losition - ---
= l2.5tn Position
Mediu = M€an of l2th dd 13th vaiues
30+32
=2
= 31 Years
539(d) Mean - r9
= 2$.4 m (3 sic. fic)
19+1
Middle FDsition= ---
= 10th position
Mediatr = 29 cm
Mode - 26 o tud 29 cm
43+ll0+218+lll+60
]J
=13
= 41.7 (3 sic, figJ
13+l
Middle posilion = ---
= ?th position
Median = 43
Mode = 45
202 + 188 + 281+ 168 + 91 + lol(i) Mean =
1135
= 70.9375
16+1
Middle position = ---
= 8.5th Posilion
Medid = Mean of 81h and 9th values
10+71
=2
= 70,5
Mode = 50 sDd 71
282+4tA+231+443+271
Mem= 
-
1689
= 112.6 Im
15+l
Middle posinon = ---
= 8th Position
Mode = 121 mb
1.1 + 3.6 + 15 + 10.2 + 16.8
(h) Mean = l9
46.7
=1,
= 2.46 c (3 sig. frgJ
19+1
Middle losition = ---
= 10th Posilion
Medid = 23 g
Mode = 2.3 g
5x3+6:5+7x7+8(i) Mem = 3+5+7+4+l
20+l
Middle position = --
= 10.51h position
135
X
Ansves wirh comprete wdked soruriom (4D
Median= Mean of 10th and llth valoes
=7 ke.
Mode = 7 kC
C) Meai
10 :,1+11r6+ 12t3+13x2+14x6
- ,l+6+3+2+6
252
=t2h
2l+l
Middle posinor = --
= llrh position
Median = 12 h
Mode = 11 h and 1'l h
3x4+4x5+5t3+6;4+7x2
(k) 4+5+3+4+2
0)
85
= 4,72 years (3 sic. fic)
18+l
MidJle @sirion = -
= 9 5th Position
Mediar = M€an of 91h and 10th lalnes
2
20x 3+ 2l x i + 22x 5 + 23 x 3 + U, 2
3+l+5+3+2
= 22 CIrl
14+l
Middle msition= 
-
= 7.5tlr position
Medid = Mear of 7th dd 8th varues
--2
@ uarr-"tics ruto zo
25. (d) Sum of 8 ndnbe6 = 8 x
Sumof9nunbeB=9x35
= 315
Nunber added = 315 - 2?2
(tr) Sum of 5 nunbeN= 5 x 9
l+2+l+4+5l
45
x45
5
Sum ofg Nmbers = 9 x 12
= loE
Sumof 8nuftbeN=8 x t3
= l(t'1
Number renoved = 108 104
291
= 48.5 kC
= 48.5 + 1.5
=50k8
=7x50
=3509
= 350 - 291
=59kC
Sun of 12 numbe6 = 12 x 21 5
= 258
= 12+ i6+ 18+20+23+24+33+ls
= 181
Sum of 4 other numbe6
= 258 - l8l
(0
Mear of 4 other rMbes
Sun of5 number= 5 x 2.3
= I1.5
Sutu of 17 nunbes = 17 x 4.5
Sum of 22 numbeB = 11.5 + 76.5
=88
88
Mean of 22 numbers = ::
Let the scorc for fie fifth exam be x.
Mean > 85
70+92+86+89+r
>85
x+337 > 425
J >E8
Minimun valDe ofr = 88
. . The ninimum scoe she n@ds on the fifih exam
is 88 to rceive a disnncdon.
(h) Let the two numbes be r dd r + 3.
Sun of9 numbeB = 9 x 16
= 144
Su of7 number= 7 x 15
= 105
Sun of oth€r 2 ,urnbers - 144 105
,+(r+3)=39
7r =36
r = 18
a+3 =21
.. The two numbers d€ lE rnd 21.
(i) Suln of5 numbers= 5 x 6
=30
Sum of? numbeB= 7 xx
Sun of 12 nmbes = 12 x 7.75
=93
.-.30+7I=93
(j) Sum of 15 numbes = 15 x:r'
Sun of I 5 nunbes {each in@sed by 30)
=15)+15x30
= 15) + 450
_ l5r +,150
15
,1(}' + 30)
=____
=J+30
(k) Sum of6 numbds= 6 x 28
19+27+35+r+r+r=168
8l +3i= 168
3r= 87
Sumof3numbeG=3x24
a+b+c=72
26.
0)
Sum of5 nunben= 5 )f =7 +16+12+8+2=4s
'15+lMiddle Position = --
= 23rd Position
(iii) Med
25 x ? + 30 x 16+ 35 x 12+ 40x 8+ 45 x 2
= --_ 1s
1,185
=33
(i) (i)
(ii) t/= 52+ 38 +42+ 30+ 18 = 180
130+l
Mid.Ue losilioD = 2
= 90.5th posilion
Median = Mean of 90th dd 91st valu€s
30+35
2
(iii) Med
28 x 52 + 30 x 38+ 35 x 42 + 38 x 30 + 42 r 18
r80
ffc,
w30.
5962
= 33.r (3 si8.
tW
Largest possibie value ofp = 15
(b)
Smallest po$ible value ofp = 53
ffiffiffiffi!ffi!ffis fiffii&&w
13'qasE!;acrr-issi !,?"6ffibs]etssfr*rl
5,..., 5, 10.... 10, 15,..., 15, 20...., 20. 25,...,25
21 29 32 t7p
2l+29=31+1,7+p
50=48+p
Smallst possible value ofp = 2
38 39 40
52 49 36
Anses with Complete work* *n'o-, @
(d) 12 13 t4 15
tl 6
ffiW
JJ
(t
*=30
u+5+r+3=30
d+b=22
Mean = 2
lxa+2x5+3x6+4r3
12, ..., 12, 13, ..., 13, i4...., 14,15,:-., 15
Pll 86
P=10+8+6
Iagest possible vatue of, = 24
Men =24
r0 x 26 + 20 xp+ 30 x 32 +,10 x 17
26+?+32+11
20p + 1900 
^_
20p+19ffi=24(p+75)
20p + 19ff) -Up + laOO
100 = 42
Mean - 9.5
I x 16 +9 x l0 + l0 xp+rl x 7+ 12 x 5
16+10+p+7+5
l0l,+355=9.5(p+38)
ljp +355 =9-5p+361
30
a+3b+22=60
a+3b=34
a+b=22_(t)
a+3r=38_(2)
(2)-\1):2b=t6
b =8
@ r*o*"0* ^,.r"
(h)substirute b = 8 into (1):
a+8 =22
v =10
8+7+a+5+b-40
a+b =20
5x8+6!7+7x4+8x5+9xb
'ld+9b+ 122 ^4r =t
7a+9b+122=zaO
70+9b=158
a+b =20 _(1)
7a+9b=158 _ t2)
11) x 7: 7o + 7b= 144 
-Q)
(2) (3): 2b = la
b =9
sub8rituler=9inlo(l):
d+9=20
a= ll
4 =7O
18+a+16+b+ l=70
a+b=29
2x18+3xa+4x16+5
m = t:t
3a + 5b + 142 =245
3a+5b = 1,03
_ (3)
a+b=29
3a + 5D= 103
(l) x 3: 3a + 3r= 87
(2) - (3): 2b = 16
b =8
Substirute, = 8 into (l)l
a+8-29
a=2\
U]
31. (a)
x/ = 100
15+10+d+/,+ 19 = 100
a+b =56
Med
0r 15 +1 x l0+2xa+3xr+4x 19
=2.3100
2,+3r+86 ^"r00 = -'
2d+3b+86=230
2d+3b=144
4 +, = s6 _(1)
2d+ 3b = 114 
- 
(.2)
(1) x 2: 2d + 2b = ll2 
- 
{.3)
(2) (3): b =32
substitute, = 32 inro (1):
d+32=56
d =24
Tolal number of vending machines
= \ x 29 +2 x 22 +3 t 23+4 x l8+5 x 8
Mean no. of vending machincs per school
254
= 2.51
Modal no. of lelding nachines per school = r
100+l
Middle posinon = ---
= 50.5th posnion
Median do.of verding machines per school
= Mean of 50th and 5lsL values
2+2
2
32. (a) (i) Mean
0x3+l!5+ 2tl+3x9+4x1+s\2
'12
30
30+l
(ii) Middle position = 2
= l5.5th posinon
Median = Med of 15th and l6th vaiues
2+l
(iii) Mode = 3
The mead giles rhe best plcturc of fte disrrlbutior
since i$ calcdallor lnlolves all fte dala dd therc
arc no exlreme valucs in the drla.
Middlc position
5, 36, 37, 39, 40, 40, 42, 45. 45, 46, 46, 48, 48, 48, 65
630
= 42h
15+l(ii) Middle oosinon = -_:-
= 8th Positbn
Median = 45 b
(iii) Mode = 4E h
Themedian givei the best alerage since the data
have two extreme laldes. 5 dd 65.
(i) Mean
0,35+l r64+2 x68+3\ lE+4x15
200
314
200
(ii) Mode=2books
200 + I
(iii) Middle posilio. = 2
= l00.5th posilion
Median = Mean of l00th dd Lolst lalues
l1 (r)
(b)
34. (a)
2+2
= 2 bmks
(b) Total rc. of books bought by 200 studerts = l14
Total no. of books bought by 100 studenls (lion
=l00xr
- 100r
Total no. of books boughr by 300 studenls
= 314 + l0or
Answes wirh comprele worked sor'tr.". 6r)
Nlei .o oi bn,'r.\ houehr b) 100 \ruilenL. = l.7E
31.1+ 100r--{xr = r 78
3l'1+ l0(L = 53+
l00r = 220
, = 2.2.
3s. (!) (i) Nledn
0 x I + | r6+1 !5+1x.1+.1x 2+5 !6
rii) Modc = 1 and 5
2.1+l
liiil Middle Dosltion = 
-
= l2 5ih Positnnr
Mcdia. = Nle.n of 12tb and lllh laloes
2+3
= 2.5
(b) Irt Lhe no. othe.ds oblained nr each ofthe cxL,!
3 rhn)s( be,. , .nd ..
Nlean of27 throqs = 3
66+d+r+. .
t ='
n6+!+r+.=Nl The mNnun
..,r=5.1=rJnd! =5 eJrb tiLox i.:
.. The no. ol heads obtaircd in exch of lhc .l
throws dc 5, 5 aDd 5.
(bl (i) E&lo+l
2i+l
Mlddlc posilion= r
= rr; posilion
Median = 5 (ors
@ r",n..uo, t*., .u
FacLorv a
NLeoLJn = 5 ro]\ Medrd = tjth vatue
(ii) FactoN n
Modal no. ol Loys = :l toys
fict(w A
\'lodal no. of loys = 6 tols
The mode is best uscd otcohpare fte so*es of
both laclodes sincc tnen medirns are equal !trd
Lheir eds cinnot be calculated.
Mean scorc (School,4)
_ h+86+91+85+93
5
.132
= 86..+
Ilean scoie (SdroqL9)
rJ.1+ u2 + ?6 + 91 +33
431
= E6.2
Mcan score (ldlorc)
85+95+90+75+92
= 81..1
lf lhe schools re ranked by ficj. medn scores.
School C {,ould win the Marhematics olynpird
since it hrs lhe highesl mean scorE,
Middie positlon
(b) School A: 75. 85, 86,91,9:l
Mcdian s.o.e = E6
Middlc Po\itnnr
SchooLS: 76- 8.1- 88. 91, !2
Medirn scorc = 8li
Middle posior
S.hooi C:15. 85.90.92,95
Median score = 90.
lr .h- .Ll"'1 
' 
e Jll,c'q) . med.. .nrc
School C would \rl! thc Mathedrtics Oly'npiad
I would choose rhe m€die soE to rank the
schools. B) conpdjng the medid scores, School
t would be ranked secod. If tbe mean score n
used insrsd, School B would tE.mked third.
Middle position
38. (a) Class 24 : 1, 3. 3. 4.7
Clas 28: 1. 1, 1.2.4
Clas 2C: 1.2,2. 1.6
Clas 2D: 1.3,4,4.5
Clas 2t: 1.2,3, 5.5
(ii)
Ra*:Class 2,4, Cl6s 2D, Class 2E Clas 2C,
Chss 2A
Class t has the mos! absenlees per month if
their modes a.e used to conpre the clasles.
Class, har the mosr absentees per month if
tbelr medians ae used io conparc the classes.
39. (a)
(c)
. Lr+est posible value.'fr = 7
0x6+lx8+2x2+3xr
:10. (a)
Requjrcd jnequality:r 5
13x6430
1+3+r=5+4+3+0
.. t{gest possible value ofr = E
(b)
+2x9+3x2r+4x2+5t1
6+2+9+2t+2+7
6x + 63 = 2-5(zx + 26,
6i+63=5r+65
0,....o. r,r, 2, ...,2,3. ...,3, 44 5....,5
6292a27
6+2+a=2x+2+1
.'. Ldgest possible value of x = 35
45. (a) Total no. of glests = 100
j+35+20+)=100
'+f=45
(b) Mm = 2.65 .
t r+2.35+I.20+4\y ^--
r + 4) + 1.10 ..r
r00
x+4]+130=2.65(100)
r+4r+ 130 = 265
J + 4J = 135 (Sbown)
(c) r+t=45 
-(1)
r+4)=135 
-(2)(2) (1): 3] = 90
)= 30
Substitule ) = 30 into (l):
x+30=45
'''m+Fm
100+1
= rn 5ri n^(irlnn
Median = Mem of 50th and 51st values
2+3
Modal no. of guests = 2(it
46. (a)
()
Total no. of studenrs = 100
27+r+25+y=lU)
r +J = 48 (Sho$,n)
(b) Mean
2\21 +3xt+4x25+5xr
100
3r+5y+15,1 --l0o = t'
3r+5.t'+154=350
3i+sJ = 196 (Shorvn)
r+ J =48 
-(l)
3r+5J=196 
-(2)
(l)x3: 3r+3)-144 
-(3)
(2) (3): 2t =52
! =26
Suhritule ) = 26 into (1):
r+26 =48
Answss witb comprete wo.k* *ro-. (.i)
2 5
21 22 25 16
l:6 + 2:r + I x 2 + 4:3 + 5 ! ! + 6: 1
15
l,+tr+ 1ll --,
-15='*
.17 (a)
Modrl no. of novcls rurchased = 2
T{)l.l no. of (udcnLs = 40
ar+r+9+8+)+5=40
r+J=12
2t+51+4lj=86
1r+5) =38
}t+5)=lli
(l) x 2: fr+2t=20
(2) f3): 3r = tti
)=6
- 
1ry-" I pro. pgn
- 
[] 
(i)
Mcan = 7.2
5.ii+o^!+7 ,+\'8+'r')r+ln 5 --
6r+9-r+207 --- 4r = t'
n\.r, Flu 188
D,viLre cd.h 'r-l =ll
rem bY l
r+r =12 
- 
(1)
\+3i=27 
-{.2)
(!) x 2: 2r+2)=2'1 
-(3)
12) (3): |=r
snbsLiLure )'= I into (l ):
r+l=12
-!=9
..'=9andt=3
6 7 9
9 l 5
Modal mark = 6 and 7
Middlc lroenon - 2-
= 20 5th tositron
Median nrrk = Mexo of 20th and 2lst ldues
Subsrirure] = 6 inlo (l):
r+6=10
2
=7
18. (d) Tolal no. of thrcss = 25
6+r+?+3+)+;1=25
r+r + 15 =25
x +J = l0 (sholvn)
nrinimum laluc oit = 7
.. m.xinunr value of, = l0 7 "
30+l
(ii) Middle Postuon- 2-
= l5.5dr posixon
Itedian = Mean.l l5tb rnd 16th lalue\
= 3 sodls
@ r,,n-,,".,*. ,u
(ii) Mean
0x3+lx5+2x6+3i8+4t7+5! |
= 2.47 coals (3 sic. fis)
(b) (i) Mode = 3letters
(iD r/= 31
ll+l
Middle P\iLion = :
= 16th Posilion
Medid = 4 letters
1,18(iii) Mean - x
= 4.77 l€tt€rs (3 siC. IiCJ
51. (a)
=3+4+6+2+4
Modal no. of w ches per student = l
19+l
Middle tosnion = 2
= l0lh posinon
Median no. of watches pd student - 3
Mean no, of waLches per studonl
lxl+2\4+3:6+4!2+5x4
=--1,
t9
52. (a) (i) Mode = l soals
36+l
= 1E 5th Position
Medio = Mean of 18th ard l9th values
=2
30
50.
(ili) Mcan
lx2+2x8+3x5+4rll+i/9
126
= 3.5 Coals
(b)and a c@-
It is exacdy half-filled with water. The radius ofthe cone,is 18 cm and its length is 24 cn The
length of the cylinde! is 52 cm. Find, leaving your answefi in tenns of fi,
(a) the sudace arca of the contarner in contact with water,
fb) the volume oI waler in the conuiner.
49- A solid cone has base diameter 16 cm and height 15 cmr The top part of the cone is cut off. The
rcmaining part of the cone is then attached to a hemisphedcal solid which has the sa.rne base
diameter as the cone as shown in the diasnm below. Find the volume of the solid fomed.
giving your answer corect to 3 significant figures.
[Take n = 3.142.1
52 cm
l l.25 cm
Chapter 8: Mensuration ot htuidr C(s ,nl iScE
L Alph.ifuallcadba ofmass 1.2l.kg is dropped into an empty open conical container dfradius
E co- l5 litres of water is then Doured into the container to fiIl it ro the brim.
Fitrd
(e) the volume of the ball,
O) the radius of the ball,
(c) dle height of the conical container
(rle oensrry or reao rs I I g/cm ,)
5l- Diagram I shows the qoss-section ol the symmetiically shaped hourglass made from hollow
hemispher€s, cylindem and cones joined together. It contains sandjust enough to fill the top
conical aDd cylindrical parts.
(a) Calculate the volume of salld in the hourglass.
Diagram II shows all drc sand which has run through and collected ai the bottom of the
hourglass.
(b) Calculate the height of sand, , cm in
3.3 cm
the hourglass.
| 3cm ra\
ryll
Diagam I
I
O2\
w
Diagrantr
(D *---'"*,"
A rcctangular box is inscribed in a cylinder of height 6 cm
and with a circular base of mdius 2.5 cm. The length of pC
is 4 cm. Find the volume of the box.
"-w
The sudace area of a cube is 384 cmr. The cube just fits exactly inside a sphere. Find the
surface area of the sphere.
A sphere just fits inside a cube and the cube just fits inside a cylinder. Find the fraction of the
cylinder that is occupied by the sphere.
Chapie. 8: Mensumion of rymids_ clIF a!.1 slit: l
t Diagrao I shows a closed conical container with a diameter of 16 cm and a height of 12 cm
etid is filled with water to a depth of 6 cm.
(a) Frnd dte volume of water in the container, leaving your answer in tems of r.
(b) The conical container is then inverted as shown iD Diagmm II. Find the value of r, giving
vour answer conect to 2 decimal Dlaces.
---->
12 cm
5. A cylindrical container of radius 2 m is placed on its side up against a wall. A ball just fits in
the gap beiween the container, the wall and the floor Find the volume of the ball, gjvjng your
answer corect to 2 decimal Dlaces,
Diagram I Diagram Il
Q"'-'-'-'*-,"
a
i"tr$?.:.l1"#:ffi 'J::ff %:U:-1'.:1f L:*Tf#jiJX,i j,'il*";fu.-;*J
j of the volume of water in rhe cylindrical parr
to the brim. The hemispherical part contaim
of the test tube.
(a) Find the values of ft and r.
(b) The water in the test tube is then poured
into a right citcular cone as shown in
Diagram II. If the diameter of the
surface area of the water level is 5 cm,
find
(i) the heighr of the water level in the
(ii) the su.face area of the cone that is
contact with waier,
Diagmm I
ru
Diaelam II
Chapter 8:Medsuralion of qy€nids_ Cor r.rj S:]E-
Graphs of Linear
Equations in Two
Unknowns
EE Graptrs of Linear Equations
Et>
Graphs ol line equations are straight linelt.
Slcps lo dra$ the graph of a linear equation:
WORKED EXAMPLE 'I:
Draw the graph of ) = 2r - 1. From _rour graph, find
(a) the value ofl, when i = 2.1,
(b) the value ofr when r. = -3.8,(c) the value ofp given thnt (1.5, p) is a solrtion of ) = 2r I
i.1) Construct a table of values ofr and )..
You will need the coordinates of three points.
(A sffaight line graph can be drawn using on]l'two points but the third
point is needed to chcck.)
O Choose a suitable scale-
The scale used on thc r-axis does not have 1() be the samc as the )-:rxis.
The scale chosen should allow fo. the laryesl possible gmph to be drawn.
The bigger the graph. the more accurate will be the results obtained from it_
r!] Plot the points on the graph paper and draw a sffaighl line rhrough all rtte
points.
@) Label your graph with rhe equarion of rhe line.
ffi
soLuTtoN:
Constuct a table of values for the equation 1 = 2, - l.
We then plot the points and join them with a straight line to obtain the gaph of y = 2, t.
From dre graph,
(a) whenr=2.1,)=3.2.
(b) when ), = -3.8,, = 1.4.
(c) whenr=1.5,p=2.
ChdoLs o. C aph, o, Lma tCdon\ o leo I blr..d
The graph of a horizontal line passing through the
point (0, c) and paraflel to the .r-axis is of the fom:
E.g. The graphs of ), = 2 and ], = -1 are shown below.
tl
l ::i
t
...1
-L .l
I ii il
l I I
The gradient of the gaph of the folm J = c is zero.
The graph of a vertical line passing through the
point (1', 0) and parallel to the )-axis is of the folm:
E.g. The graphs of jr = 3 and x = -2 are shown below.
The gradient of the graph of the form jr = d is undefined.
i'a '---o-ri,,* 28
4.
5.
6.
of the fom:
tl::y 
I
wheremisaconstant,
If the gradient, m is positive, the line slopes upwards,to the righr.
If the gradient, m is negative, the line slopes upwards to the left.
The bigger the numerical value of m, the steeper the line.
The gaph of a straight line passing through the origin, (0, 0)
. m is positive
. the line slopes upwards to dre right
. m ls negatlve
. the line slopes upwards to the left
and wirh gradientEb
m is of the folm:
lt=.'t;l
The graph of a straight line that cuts the y-a\is ar rhe point (0, c) and has gradient,
When the value of m remains the same with c taking on different values, the graphs
are parallel lines cufting the y-axis at the pohrs (0, c).
chapter 9: craphs of Lined Eq*,.* t *. u"**r^ Gi)
m iS a constant value of 1 while c takes on dillerent values.
2
WORKED EXAMPLE 2:
(a) Draw the graph of each of the following equations on the same axes.
(i) .d+2=0
Ltll, v= .x I
2
(iii) 2y+3"r-6=0
(b) Find the arca of the triangle bounded by these three lines.
soLuT|0N:
(a) (i) r+2=0
(a**--'-**,u
t.
'2
2y+3x-6=Q
2Y= 3r+6
3^J=--X+J
(b) Area bounded b) the three lines
=lxsx+
2
= 16 units'?
-2 0 2
2 -t 0
,2 2
6 3 0
ChapFr'1. Gdol\ or I inear Lqlalrol. In Iso I MqG
iD WORKED EXAMPLE 3:
(a) State whether the points Iie on the given lines.
(i) (4,5),)=2r-3
(ir) (2,7).l,=9-f.{
)
{b) ll ta.a+gtrsa\olurionof}= J.r+5. findrhevalueofd.
soLuTtoN:
(a) (i) )=2r-3
When t = 4,
r = 2(4) 3
(b)
The point (4, 5) lies on the line ] = 2r - 3.
(ii) r=9 lx
2
y=s-lrzt'2
The point (2, ?) does not lie on the line y = 9 l;r.'2
Since (a, d + 9) is a solution of l, = 3r + 5, substitute -I = a and ) = a + 9 into the
equation)=3r+5.
)=3x+5
(a+9)=3(a)+5
In Book 24, we have leamt to solve a pair of simuhaneous linear equations bl using
either the Elimination method or the Substitution method. We now leam to use the
Graphical method to solve a pair of simultaneous linear equations.
Steps to solve a pair of simultaneous linear equations graphically:
@ Draw the graphs of both equations on the same axes.
@ Read off the values df the variables at the point of intersection of the
lines. (The coordinates of the point of inte$ection gives the solution of
the simultaneous equations.)
@ Sot"ing Si-ultaneous Linear Equations Using the Graphical Method
2.
1.
,9 *r**,u
the
od
ing
the
rD woRKED EXAMPLE 1:
Solve the following simultaneous equatioDs gmphically.
x+2Y=8
3x-Y=3
SOLUTION:
Constluct a table of values for each equation.
x+2i=8
2t= x+8
I)=--.rt4-2
3x-y=3
Y=3:( 3
Choose a suitable scale and plot the graphs of r + 2) = 8 and 3r - ) = 3 on the same axes.
The solution is, = 2 andy= 3.
ofLiner Equions in T{. LnIa-E$ i:
I
'I
A
.tl
i
i: 1l
l:
-t
(B> WORKED EXAMPLE 2:
Solve the following simultaneous equations graphically.
x+2i=6
2x+bt=4
soLuTtoN:
)c+2Y=6
2Y=-x+6
l^,=-_r+r-2
2x+4y =4
4Y --2r+4
I1= -t+l'2
.. The simulianeous equations, + 2), = 6 and 2r + 4) = 4 have no solution.
0 2
4 3 2
T I I
I I
+
I
+
lri +4) I
-
,l
l i II
C:i rraa*remricsAngle of sector rcpesenting 2 goals scored pel
=-x360"
= 80'
53.
156
= 6.5
54. (a) Mode = s6 years
(b) Middle position= -:
= lltb posihor
Median = 55 yearc
24+1(iiJ Middle Dosition = 
-= 12.5ft position
Medlan = Mean of 12Ih dd 13th values
1+J
=7
(iii) Mear
4 ! 5 +5 x 2 + 6 x4+ 7x6 + 3!.1+ 9r2 + l{r : I
24
ADsReB with comprete worled sor!1io* @
7, + 235 + 180 + 123 + l,1o .i
2l
= 55.5 yeaN (3 sis. fic.)
'"iT*"*llll-
55. (a)
C)
5
6
1
9
l0
2.
24
l2
z4
5
5
5
1
6
7
(b) (i)
(i)
20+l
Middle position= -l
= l0.5th positton
Medirn = Mern of ioth and l lLh v.lttes
= tr.i*
(ili) Mean =
52 + 126 + 371 + 606 + 378 + 100
2A
l6l5=ro
= 81.75 cm
(i) No. ol boys in Clas z4
=l+9+1+8+l
(ii) No. ofeirls in Ch$ 2A
=l+3+6+8+2
= ztl
The narks of the bols dre clusteied dound 20t
and .10s. The marks of 1he Sirls are wcll spread
14+213+:ll+356+53
=----
669=n
= 13..15
13 + 6r + 201+ 36:l + 102
= -- tIJ
750
'n
@ t,.".".* tn...u
.. Since lbc mean ma r! ofthc gids is hisher than the
mexn ndk of the Lrors, the girls pertbrncd better in
(d) (i)
(ii)
(iii)
Highest mdk = 53
.10+1
Middle posidor = --
= 20.5th positron
Nledian ma.k = Mcar of20th and 2lstrdlrcs
38+39
=2
= 3E.s
D. iUea! ibr Grouoed Data
57. (r)
58. (a)
:n
hr Meao age = t/
= 14.9 yeals
ee.,l*;lMd*.r* * 'l 
r*q"*"v v 'l 7,
8.: 21
l0 t2 11 5 55
13 t5 l4 l0 140
16 18 t1 102
19 21 20 80
22 24 2l 2
,ft=47
110 5.5.
lt 20 15.5 :16.5
21 3t) 255 8 20.1
ll :10 35.5 I2 126
,ll 50 ,109.5
51 60 55.5 211.4
65.5 I 65.5
'f=40
tr4 = 1440
59.
abl Mee leneLh = =
14,14
=36mm
t6
lb, Mean mas! = =;
9610
tm
= 96.1 c
(a)60.
80-84 82 8 656
85-89 87 12 1044
90-94 9Z 2l 1932
95 99 28 2',716
100 104 1,02 t6 t632
105 109 lo7 10 1070
ll0 - ll4 112 5 560
V= 100 lfa = 9610
0 560
=19+5
= - lt)l),i
(d) Middle posnion = 
-= 23rd posnion
.. Thc clars inte,lrlwhlch conl.irs the medianls
60 35 cm
= + x looc,
.2.
=40i /r
Modal class = 40is 85.
. !lt a8
L\>1
Ir>3-
I..3t 6) -
I
P(az) = t
1(b) P(d odd no.)= ;
1
6
I
2
=0
P(win buy a chicken) = ;
P(wiU not buy a chicken)
= I - P(will buy a chicken)
l
5
P(will a@ept the coin) = 0.8
P(will rot acaepl fie coin)
= I - P(will ecept the coin)
= 0,2
{1,2,3,4.5,6}
2
I
=2
0)
(a)
Answm with conprete wortt *to-. @
(d)
3
P(a multille of 2)= ;
I
=1
4
(e) P(a factor of 6) = e
2
(0 P(a naturar rc') = 6'
5. (a) P(a 3) =?
3
(b) P(an evei !o.) = ;
I
=2
2
(c) P(ano.>a)=;
I
6. (a)
(b)
2
(d) P(eitle.2 or 6) = 6
1
2
(e) P(a perfect square)= ;
I
s= {0, 1, 2. 3, 4, 5, 6, 7, 8, 9}
4(i) P(l4t disit 5) = i0
(c) P(an odd no)= lo
3
{d) P(a no. E) - s
I
2
2t. (a) P(a 3) =
(b) P(a losilive inteser) =
(c) P(a whole no.)= t
2
I
2
=t
I
8
9
1t
:
9
(d) P(a 2 or 3)
(e) P(a no. 325)
23. G)
(b)
175
=5oo'
1
20
No, tbe probabilit) that a girl will win the lnze
cannot be detemined sirce we do not know the
number of students in the school thal do girls.
s = la, b,c, d, e,f, s, -, zl
E=ls,bn,d,a,rj
n(E)
3
AaweF with complere work* *t, -, (o*)
u. (a)
(c)
(d)
25. (a)
(c)
(b)
P(t) = I P(E)
3
=1-=
10
l3
s = {Janu{y, February, March, April, Ma&
June, July, August, September, October,
Noeembe., Decemberl
A = {Ma.ch, May}
2=a
I
a - {January, Apdl, Jmq Jury, Angust}
P(A)= -:;-
5
12
C = {Janua.y, F€b.uary, March, April,
September, October, Novmber, De(:mberl
",^ !s)
8
12
t
3
s = {r,2,3,4 5,6,..,,30}
r' = {s, r0, rs,20,25,30}
I
P(r) = I - P(t:)
I
@ r**'",",.., *
(d)
n.
(a) s = {r03, 10s, 130' 150, 135, 153' 301' 30s, 310,
J50, Jl5, 35t, 50t, 501, 510, 5J0, 5tJ, 5Jl;
(t) c = [130, rs0,3r0,350, sr0, s30]
6
3
P(C)=1-P(C)
1
2
(d) D = [3s0,3s1, s0r,503, sro, s30,513, 531]
8
P(D') = 1 P(D)
5
s= {1, 2, 3. 4, 5. 6,.... 50}
,(s) = 50
25(a) P(d even no.)= -
I
2
32
(b)
(c)
50
(e) P(rct a perfect cubo)
= I - P(a perfect cube)
=so
(c) P(Do€ rha. 33) = 1 P( 100) - l6d
=0
s = ll2, 13. 14,2t, 23, 24, 3) . 32. 34, 41. 42, 43j
n(.5) - t2
P(a io. 30) = 12
2
P(a mdlliple of 12)
2=n
I
5
31.
(e) P(not a pnme no.)
= I P(a ?rime no.)
5
=1- =
s= {10, 11, 12, 13,14....,991
n(S) = 99 10+1=90
(n) P(a perfect squde)
6
=ro
I
Ansvs wirb conprete wuk€d s.rtrrtds (e
(b) P(a no
l9
90
{c) P(a no. having a disit which is 2)
> 80)
l8
=ro
(d) P(a nultiple of ll)
90
10
r(s) = 52
(b)
33. ,(s) = 52
ral Praftd ced) = -
i=t
P(d ace ofclubt - s,
l3
P(a spade) = -
(d) P(a picture cdd)= .11
52
3
t3
(e) P(not a picture caJd) = I - P(a licluie .dd)
3
l0
@ uatrrematics nrtor ze
13
I
8
52
2
I
P(not a queen) = I - P(a qu€en)
t2
(f) P(a picrw cdd or a dimond)
22
52
ll
34. n(S)= s2 - 12
=40
?l
l
2
(b)
t0
L
P(not a heart) = 1 P(a heart)
3
=7
(c) P(a 5) = -
1
=ro
P(Dotas)=l-P(a5)
9
t0
35. n(t - 54
(a) P(a red king) =;
=t
{b) Plablack ctud) = -
t3
(c) P(a jack or a aueenJ = -
4
ldl P{r iokerr = :
_ln
(e) P(not ajoket = 1- P(ajoker)
26=E
36. r(D = 12
(a) Pla hean) = -
I
(b) Pla.lub) = -
1
(c) P(a picture card) =
3
n
i
' (e)
P(an 8) = ;
I
P(a 4 or 8) = D
I
.0
=0
(i)
37. (a) P(inside blue sector) =
60'
= 360.
1
(b) P{odside blue sector) = I - P(itrside blue seclor)
5
Area of oMce setur
ronsroe ordse s4tor) =
90"
360'
An8le representing red sector
- 180" 6ilP (adj. zls on a sh. line)
{d)
(d)
(e)
120.
360'
3
P(inside rcd or yellow setor)
_ Afta of red ed yellow sctos
- Area of ciicle
120" + 45'
360"
165"
ll
Arcwds with comprete work"u .t*"^ 6)
38. (a) P(irside smaller cncb)
Area of sfr.ller cn.b
= A,* 
"f 
t,,c...*t"
39. (a)
(b)
r5,f
= A*" 
"it"rg.""l*t"
90'
= 36rr
P(inside shaded region)
= 1 P(unshaded reeion)
- I - [P(inside smal]er circle) + P(inside sector
AOB)l
131
t96
r + l0'+ 190'+r + 90' = 360'(1s ar a poino
1+ 290P = 360"
2.t = 10'
- Aedor;aor rep€sc.rine snrall size
- AEaolcircle
= 360.
35"
7
(li) P(medium size)
Aiea of s4tor represenring medtum size
Area of cncb
9CP
360'
@ r"*"."0....r"
40.
(iii) P(ldse sizc)
AEa of sector repesentine hrge size
Afta ol cncb
= i6rr.
35' + l0'
360"
I
Arca of shaded region
= Area of lfger circle AJea of snallef ci.cle
P(lnside shaded region)
Area of shaded region
15
Area of circle = r(20):
= ,100t cm':
l^
Area of semicircle = - x tl8)"
= 32n cm'
Area of semicnch
la) P(insiJe senilircle)=
$oy'
2
(b) P(shaded region)= | - P(unshaded resio!)
2
42. (a) P(insjde rcsion,4)
_ ArcaofrcgonA
- Area of ldgen cncb
,(2)'
,nl8r
I
16
= Area of ldgest chcle - Area of middle cilcle
= '/.8)' d4)'
P(inside region O
Area olFgion C
- Area of hgesr circlc
3
(.) P(iDside region A or B)
= I - P{not iNide roCion ,4 nor a)
=1 P(inside resion O
3=t_i
I
43. Let the tength of AP = r mits
.. L€ng1h ofAD = 2r uits
P(inside AAPo)
Area ol AAPO "1;ffi"
At Q a
2 ,:.-
I
(a) P(inside resion A)
_ Area ol snall6t squde
Ar€a of ldgest sqnde
I
_ are ofmid,Cle squaE Area or $mlest squm
- AEa of Larsest squde
-:-
8
= 1 P(not inside reSior O
= I tP(inside recion A) + P(inside resion n)l
a5. (a) P(a ball labelled Z
= 1 - P(a ball not labelled Z
= I tP(a ball labelled 11+ P(a ball labelled l)l
(b) Let the total no. of baUs in lhe bag be i.
r=86x3
= 25E
.. There are 258 bals in the bag.
a6. (a) P(a brown ball)=
I
16
. (1 rl
-' 15 rll
2
l
= n*t
I
rbr Pra brown ball) = =
x t.r'
)A+] / 19
19r = 9(20 +a)
lox = 180
Answds with Complete wort* ,t*"* @
47. (a) P(a blue narble)
= 1 P(!ot blue)
= I - tP(a red mdble) + P(a sreen frarble,
ri . r)
30
t9
-;.frx12O
-76
7
P(a blue marble) = j?
No. ofblue nrbles 1
Toral no. of Mbles 18
'16 - ,\,. 7
Do -,n ls
18(76 -r) = 7(120 -')
1368-18,=840-h
528 = 111
''-48
1
48. (a) P(a boy) = i
No.olboys _ 3
Tolal no. ofchilde. ,l
n \y'3
,J *a'i
27(1) = 3(27 +t)
108=81+3a
3x =27
(b) P(a sirl) = ;
No. ofeirls , 2
Toral no. oichildren - ll
,?* u1n
l9
(b)
49.
'ttx =2(21+t)
1,1,x =54 + Zr
9x =54
x =6
No. of purple crayons left = 32 a
No. of trom dayons left - 28 -(r+ 2)
=28 a 2
=26-r
@ r"**o*'",. r"
s2. G) (i)
Total no. of crayols left = (32 !) + (26 x)
=58-2r
3
P(a brown cnyon) = t
No. ofbrcw. dayoN left _ 3
Tolal no. ofcrayos left - ?
26 tr.r3
53 - r.xt
7(26 - ') =3(58 ?jt)
182 7* = I'74 - 6.
-r =-8
..x =E
No. of balls in bas after 6 red balls are added
-50+6
5
P(a red ba!) = t
7(56 -r)= s(56)
392-7r=zaO
112 = ]x
51. (a) (i) P(a sid) = ;
(ii) No. ofboys = 30 -r
(b)
l
P(a grr , = -
30.8. r4 = r3
5' *n
13(r+ 8) = 3(52)
l3r+104=156
13x = 52
P(a red bal) = 12
2
(b) (i) P(a red ball) =
(i i) P(a blue bJU)=
1
ll
!
\2
l
3
(iii) P(a gMn ball) = it
=0
(ii) P(a blue bax) =
(iiir Pr! cd u! blue ball) : -
53. (a) ,(S) = 999 - 0 + I = 1000
P(3 digits are the sue)
l0
t000
I
100
(b) P(tast disit 8) = #
I
(c) P(ldt digit not 8) = 1 P(last digit 8)
=l- lo
t0
(a) Plvrnilla ilavoured) = -:-
l3
30
(b) P(white wralpes) = a-
_tl
30
7
(c) P(stawbenf flavoured md has red sappers)
12
30
2
5
(d) P(vdina fiavoured and has sreen wrappe*)
30
=0
55. (.) P(o red slot)=
(b) P(0 or 00) = L
38
I
t9
(c) Plx no. aom I to 16)= --
8
P(nol a no- ftom I to 16)
= I P(a !o. from 1 to 16)
tl
=1,
(d) P(an odd nubeO = *
t9
.:3
33
9
l9
Adswes with complete worked sorltios @
56. Beibre:
Total no. of red and blne balls
=18+22
(a) (i) P(a red ball) =
q
:
m
(Ljl P(a \td on ball) = 40
7
20
5'1
Total no. of red dd bluo balls left
'9(i) P(a rcd ball with a sta) = ]'
3
4
(ii) P(a blue ball witb a sttr) = :'
Total no. of cubes = 5 + 13 = 18
5
(a) (i) P(a black cube) = is
13rir Pra white cube) = =
0
(iii) P(a rcd cube) = it
=0
l8
(iv) P(either a black or white cube) = rs
@ 
Math"-'tics ruto. 28
Total no. of cubes now = 18 + 7 = 25
5ri) Prr black cubel = -
7riir Pra crcen cube) = -ii
58.
(b)
(c)
I(r) P(io p€ts) = 20
2
5
f
3+2+l
P(2 Des or more) =- -
6
=20
3
Toial no. of households
= 32+?4+ 8 +9+ 6 + I
=80
rar Pl2 ootted Dlets) = =
I
=10
6+l
rbr P(> 3 Do ed olanlsl = :-
7
=80
32(c) P(no potted plmts) = i0
2
59.
32+U
(d) P(l poned plmt or t€ss) =
56
80
7
l0
(e) P(> 5 potted tlanls) =
l0
60. (a) (i) P(a2)= nn
= 0.05
58(ii) P(a a) = ,00
- n.29
61. (a)
(c) P( 50 ks) = --:.
60
I
10
q
0
36+4
rbr Pl> 45 k! but E5 Pearuts) = 11
24 12
4 ldr \u.ol f'oa)a. 4 6Fb o-4'-r
tu *o*
\b/ \4d.o lsnr'r paPdr-=JsoIhlY**'
N4r ' I hed.rr. pdp ). - 40e s ercl :ltm
@ r"*..'. ',* 'u
7. (a)
(c) No. of papdyas $ith nass 65 YcaB = 5
RequLtd p{renra e= - lu+ir
= 18.59. (3 sis, fig)
ABe ol youngest cohpetitof - 2E vea6
Age ofoLdcsL comPetitor = 78 Yea.s
The distribution varies lion 28 veaa to 78 vea$'
The dataclustefaroud the stem eqdal io 50 years
(e) P( 50 kg
=18+12+l
Rcau.eL-l fttucnt.rre = .- 100?
= 66s'
Modalclass=5555t=
'10
2tJ
l5
3
8
(b)
66.t0
=qo
=166h
Modal ch$ = 160 - 164 and l?0 - 174
40+1
Middle posilion = ---
= 20.5th losilion
Median = Mern of 20Ih ed 2lsl values
Clas inlenal where median lies = 165 - 16t
(e) P(litbspan 3)=1
;
I
;
i
25. (a)
(b)
26. (a)
(b)
21. (a)
(b)
J = 1 0.28 0.54 r
= o.lE
No. ol cas $xt tDrned lefi
= 0,l lJ x 450
=81
S = { IIH, HT, TTI, TT }
r, r Prt$o heads) = 1
riir Pro hcad ard a railr = ;
I
2
(iii) P{at leasr one lail) = l
S = { HIIH,IIH! HTH, TIIH, HTT, TH'I', TTH.
TTT }
(i) P(tbJee headsl = l'R
rir Pl2 heads dd I LdrlJ = i
3
liii) P('o heads) = 1
1r\I PrJL ea\t 2 headsl =
2
Answen ivith comprete worked sohti.N (e
2E. n(s)=59 30+l=30
(a) P(a no 30) = ;
'52
P(not a red ace) = I P(a red ace)
I
=l
25
26
_ ^._ J
P(either a 2 of spades or an ace oI hearts)
2
52
=l
26
(f)
34.
(c) P(a composite no.)- I - P(not a composite no.)
= I - P(a prime no.)
=t 30
30
,e' P,nor a perle.r squml = | P(a perfecr .out)
=t-,1
l0
35. r(l') = 9
(a) P(a 5) = l
=!
3
(b) P(a naturar no.) = !
9
(c) P(a Dnme no ) = -'9
=1
3
7(d) P(an inreserl = ;
(e) P(not an inteser ) = 1 P(an integer)
'1
=l 9
=?
3
Answes with comprere worked sorutiom (A
36. (a) P(a sold cojn) = 1 P(notasoldcoin)
= 1 - P(a copper coin)
=l-0.65
=03s
(b) P(lose) = I - P(not lose)
-l-LP(sln)+P(draw)l
. 15 lt
18 5.1
.33
7
17. (a)
90'
360"
125.
160'
12
Argle represeniing yellow sectof
= 360" - 90" - 125' 45' (Zs at a loint)
= 100"
P(inside yellow sectot
_ Afta ol yellow se.lor
100'
360'
5
18
P(inside green oryellow sectot
A€aoleen a.d yeuow sectob
A€a ofctrcle
45. + r00"
360'
360'
29
72
P(insjde secLoa cont ining colous ofthe rairbow)
360"
160"
@ r'*'*.. ",., 
,*
38. Arca of ldgef square - (sr)'
= 25r' cm'
Area of smaller squde = (r)?
Area of shaded reSion
= Area of la€ef squa.e Arcd ot smaller squde
40. (a) Total no. ofballs at fttt = 36 + l9 = 55
P(a sreen ball) = 1
P(inside shaded rcgion) =
*}
39. Area oflarycst circle = r(20):
= '100n cm'
Area of shaded regior
= Area of ldCest circle Area of two snaller circles
=4AOn ft(.6:)1 n(.2)1
P(inside shaded resion) =
A€a ol shaded Esion
NM
9
10
Hxj.-ii1r$iidffiffi
4(19 I) = 1(55 -r)
76 '1J=5s r
(b) Total no. ol paper clips in the bag at lirst
=49+14=123
Total no. of paper clips in the bag afte.
=123+j+(r+6)
= 1,29 + ?-a
No. of blue pape! clips after = 49 + r
Pla blue naper clip) = l
rrq * 2. --i
5(49+r)=2(129+2r)
245+5J=258+4r
41. (a) Tokl no. of balls = 12 + 9 = 21
I2
P(not a gold ball) = I P(a gold ball)
1
(b)
rru I PLneiLher sold nor alver) = 1
=0
Tolal no. ofeold balls now = 12 2 = 10
Total no. of silverbdlh now = 9 1 = 8
ToLl no. of gold and silver brlls low = 10+8= lE
lr) Pta snld ball) = -
5
aiil Pl, silverball) = :' 13
=l
Lrr PLenhcr a eold or ahcr ballr = 11
42. (r) P(4 senior c,tizens)= a
80
3
(br Prdt leasr 2 \cnior citia
!1
3
t0
(c) P(less ihan 3 senior citizens) =
32+24+3
a
t
(ii)
3
12'll(d) P(dot morc tho l senior citizen)=
80
7
l0
43. Total no. of studenLs
=12+36+,1+6+2
=60
12
(b) P(not mofe the $60)
= 1 P(more than $60)
2
29
30
L r+r= _' -
2.A
A.1,Rc = AEDC (Given)
_.. EC=Ar=8cs * -,.
...8t=9+8=17ctu
Anssrb \irh compete wurked sorur.^ (e
Using PlahaSoras' theorem on 4-448,
AB'+a'=n'
{& = 225 iror" oostit. *ooo ,
4a = J225 rcor ol br! sl9::.
ED =AB +
=15m
r (J) l=i!r l
Whenr=9.r=2.5.
z.: = t.i!
2.5 = 3t
3
25
30
=:
, 5-
13; = -ir
t3:I:- 3
_;
r=256{
(c)
Modal ndk = 25 and 32 "
Highest nark = 39
who failed ( 12 years)
l!
150
L
ll+16
67
150
28+9
150
l. (a) l&:50Jr
= 2 o. '... , ir.. _r_ k
=.2lrr,r ,J,lt . i'-l- I :
- 2i.]x + s]^lx - sr, =ta'r ot\4 D'
(b) )2p' 24p t5
= l(4pr 8p 5) -
=3(k + Dl2p 5)
=(l rr(, \r " i;iit o,n:*'".""
Iacqlr: Ir,
,. ': ,,
';,d 
"bt,a. bt.
.r:::':- _ '-._ ':
2. (a) (2rr' + lxlrl'' - xu + r)'::.-'"-,,"
= 6ay 4ir + ar"- + 3i. 2a) + I
= or'y' - r'1' + r
(b) lE5 )rme rdr.en= t h + t h+ i h= r; h
Averaee spe.d fo. elrole joumeY
Total dislan.e ravelled
= Tr"l rl.. t"k-
-T
3
8. (a) A,Atr is similar 10 A, BC.
AB EC
AE EF
BE+3 t64
BE+3=413)
BE+3=12
(b)
Ib) LAEF = LABC =90'
Usnrg Plthagoras Lheoren on 44l!.
AF = ..tE
(c) ACFG is sifrilai to Aal.lD.
63
3
9. (a) volume of pyrmid
=1^Baseaea^ugrtrtl_
=lx5.xr:
Pyramid d
FX=10+2=5cm
Using Pythagoras Lheorem on AY!,I,
vF - 5'+ 12'
= 169
Totals ace area of Pyramid B
= AJea of square,44CD + 4 x Arer of l\v.13
=10'+4x:xl0xl:l
= 360 cn'
0. '", nrcr. " ' 13.
Smallcst possiblc value ofx = 1;l '
:.Mi4q!.4ar*$ ru:iii!.*
r.t!at o4dli mdsf tr.hed.:i:
The sJna bst vaiue The legdt lalqe ot
oI r o(Lus when r occuu whetr rhe
+i
(b)ri*or zo
WORKED EXAMPLE 3:
Solve the following simultaneous equations graphically.
2x-3Y=9
4x=6)+ 18
SOLUTION:
h-3y=9
3Y=2r-g
'3
4r=6y+ l8
6y=hc-18
2^Y=-x-J
.. The simultaneous equations 2r - 3) = 9 and 4r = 6y + 18 have an infinite number of
of solutions.
Chapter 9: Grapns of Lined Eq*,.^ t *" U".-*^ Crr)
A paL of simultaneous linear equations can have one sohtion, no'Sohitio or an
idlnile number of soludons.
a5;) Mad'emdcsruo.28
One solution No solution Inffnite number
of solutions
b-t-q
Lx+Y=8
Solution:r=3and)=2
2r t =+
E.g.
b t=-zb Y=+
*-t=q
4x-2Y=g
Tutorial
Graphs of Linear Equations
Complete the table. Then plot the coordinates and draw the graph of each equation on the axes
provided.
(a) y=x+2 Y=2r(b)
(c) )=_r r ffi:-FTffi
t-lf ilrl, lfirl'jilf.lllEF:Iillll'-ll T lTll-I i
hHrttjl1i.
nflil1tit-,li-ll
F,' lir t-ilLlllii,l;i ll ]l_
f-.-' [. ]li ;t ' llliil ll l| ,lt_
f '*; i1l:I L ri l-f;llltirrlllrIl.l.fl I
Fll:tfjfifftlif
l
l
l
l
-l)=J-ax(d)
cbaptd 9: Cnphs of Unetr Eo*,t^ t 
"*. 
U"*t"t* 
@
(e) -t+1=3
0 2 4
(D Lx-y=4
2 0 2
(g) 2t+5]= l0
0 5
(h) 2y-3.i+6=0
0 2 4
@ u"o..*i* r"m. zo
O) r=-t
tl
t
!-
ll :
,irl
:l
l
(c) r+4=0 (d) l.s-)=0
(e) r+)=0
i
,l'
I
L
I
I
, Zx=O
On the axes provided below, druw and label each of the following graphs.
chapter 9r craphs of Lined EqD"u-. t *" u*.*. Crr)
3.
4.
Write down ihe equations of the following lines
Sketch the graph of l' = mr + c wherc:
(a) m>0,c>0 (b) 'fl0 (d) nof graph paper.
(a) Civen the equatioo 2) - 3r + 6 = 0. complete lhe table below
Draw the graph of 2)' - 3r + 6 = 0, using I cm to reprcsent I unit on each axis, lbr
04y = 12, find the value ofp.
(c) On the same axes, dmw the graph of 3x - 2y = L Hence solve the simultaneous equations:
r+4y=12
3x-2!=l
@ "*.'*",,'-,"
mto
s for
2
21. (^\
(b)
The diagram shows the graph of 21 31 = 24. On the same axes, draw the graph of
Use the gmphs to find the solution of the simultaneous equations:
2x-3Y=24
x+Y=2
(a) r+]=4
2x-Y= 1
(b) l=-2r+0
t,= -!, z'3
(d) 4.i 5) = 10
2x l0Y = 35
Solve the followng simultaneous linear equations graphically.
(c) 3r+1 6=0
5r t,+14=0
RwisionErercisel 
@
23. (a) The equation of the line I is 5) + 26 = 8r. Find the value of k, gjven that the point
(1 - 2*, 6) lies on the line l.
(b) Write down the equation of the vertical line that passes through the point (5,
(c) The line r = 8 meets the r a-\is at A. Wrile down the coordinates of A.
24- The diagram shows the lines wjth equations 3) = 21 9,2! + x = 6,! =3,2j =.r 6 and
3) + 2r = 9. Write down the equation co esponding to each line.
The cost of a mug and a plate is $r and $], respectively.
(a) The cost of two mugs and a plate is til5 while the cost of one mug and two plates is
$16.50. Folrn two equations connecting ir and I.
(b) Draw the graphs of these two equations on the same axes.
(c) Hence, find the price of a mug and the pr;ce of a plate.
25.
@ "**.*..,'-,u
Marc and Esther order pizza fiom Pete's Pizza regularly. Esther lives 5 km further from pete,s
Pizza than Marc does. In a ceftain week, Pete's Pizza alelivered 4 times to Marc and 3 iimes
lo Esther. The total distance travelled by the delivery man to each of their houses for that week
was 36 km.
(a) Marc lives i kln from Pete's Pizza and Esther lives y km from Pete's pizza. Using the
infomation given, write two equations involving .x and ).
(b) On the same axes, dmw the gaphs of the two equations.
(c) From your graphs, find the distance that Marc and Esther each live from pete's pizza.
Mr Smith bought 210 lighl bulbs consisting two brands, A and B. He found that
/ bulbs and : of Brand B bulbs were faulty. The number of faultj, bulbs for each brand was
lhe same.
(a) Letting r and), be the number ofdefective Brand A and Brand B light bulbs respecuvcry,
write down two equations in tems of r and y.
(b) On the same axes, dmw the graphs of these two equations.
(c) Hence, find the number of Bmnd A and Brand B liehr bulbs Mr Smith bousht.
ofBrandI
RwstonErercis4 6i
I
Graphs of
Quadratic Equations
@ craphs of Quadratic Equations
1.
2.
Quadratic graphs are graphs whose equations arc of the form 1 = ar2 + ,I + c where
17, D and c are real numbers but i7 cannot be zero.
E.g y=1,'.\=r +Zr-Jcndv=5 lr'are 
quadratic Eraph\.
The graph of a quadmtic equation is a smooth U-shaped or n _shaped cu e called
a parabola.
The examples below show some quadratic gaphs.
3. The curve of a quadratic gmph is symmetrical about the line of symmetry.
J=Lf -1r+9
@ ""*"'"0* ^., ru
The curve of a quadratic graph has either a maximum or minimum point depending on the
value of d. i.e. the coefficient ofi'.
The Iine of symmetry is a vertical line that passes through the maximum or the minimum
point of the curve.
to:_::l-*u" 
"t::,"*-lc cuehs 
1 
= 
:r' 
+ ,' + ' are siven berow
(r) If a is positive, i-e. .l > 0, the graph has a minimum point.
. The gmph opens upwards i.e. U shape.
. The vertical line through the minimum point is the line of symmetrl
. The smallef the numerical value ofl1, the wider the graph opens.
. The graph may cut the,v axis at0, I of 2points, and the)-atis at only 1 point.
(Hcrca=1)
Minlmun poinL, (2. -9)
If a is negative, i.e. r o)
The smaller the numerical value of d, the wider the gmph opens'
8. Steps to dra\t the graph of a quadratic equation:
i
Construct a table of values oft and y for the quaalratic equation'
Choose a suitable scale.
The scale used on the rr-axis does noi have to be the same as the J-axis The scale
chosen shoulal allow for the largest possible gmPh to be drawn The bigger the
gmph, the morc accumte will be the results obtained from it'
Plot the points on the graph paper and join them up to lblm a smooth curve
Label your gaph with the equation of the cufle.
@ **".** ^.. 
ru
WORKED E)GMPLE 1:
Dra\ lhe graph of) =,r r 2for-3
(c) ftom the graph,
(i) whenx=-1.6,y- 4.8,
whenr=3.2,)=-2.8,
(ii) when)= 9, x - 2.3 ar.d t - 4.3,
(iii) the greatest value of :y is 2 and it occuls when,I = I,
(iv) the equation of the line of symmetry is r = L
WORKED EXAMPLE 4:
The diagram shows the gmph of l, = I - 2r - 8. The
and B, and the )-axis at the point C.
(a) Write down the coodinates ofA, B and c.
(b) D is the minimum point of the gaph
Write down the coordinates of D.
(c) Write down the equation of the line of
symmetry of the curve.
graph cuts the x-axis at the poitts A
SOLUTION:
(b)
Substitute r = I
y=12-2J)-
... D=(t, 9)
(c) Equation of line of symmetry: jr = I
'"' '=1,;f;;lor *-ffi
When ], = 0, --ffi
(x+2)(r 4)=0
x+2=0 or x-4=0
x=2 or x=4
..A=(-2,0)andB=(4,0)
When x = 0, - ffi
rv=0'-2(0)-8= 8
... c= (0, 8)
J+4
rnto]=x--a{-6,
@ ""*.'.o* ^.,,"
@ Sottiog Problems involving Quadratic Graphs
L The following examples show how we solve problems involving quadratic graphs.
(rc WORKED EXAMPLE 1:
The total surface area of a solicl prism, A is given by A = (41 5x + 2) cm'].
(a) Draw the graph of A = 4l - 5x +2forO WORKED EXAMPLE 2:
In the dia$am,,4,BCD is a rectangle.,4'B = (4- i)cm'AD =(Lt +3) cm and CE=rcm'
(a) Show that the are4 A of ABED is Siven bv A = I +l + lr + 6) cm'
rb) Drau rhegraphof A = :./ t -t '6for-4of x that will rcsult itr the maximum area of ABdD'
2r+-l D
C
soLuroN:
(a) BE=BC-EC
=(2r+3)-.t
=(r+3)cm
Area of ABED
-ffi
=lxBasexHeight
=!xnptco
2
=1x(x+:)r(4-r) *#ffi
=Lg, f*tz zr1
=lgf+x+121
= (-ll + lx + 6) cIIf (shown)
(D ""**o*'"-'u
(b) A= lf+ !r+6
22
ti
' 
"l
]
iil .'
fc) from lhe graph.
when A is maximum, r - 0.5 cm.
WORKED EXAMPLE 3:
Wlen an object is thrown upwards fiom the top of a tower, the height, I metres, of the
object fiom the ground after I seconds is given by the formula ft = 30 + I0r - 51.
The table below shows some vallres of l and tbe coresponding values oflr.
.
r.rlrl
ili .i
.)
:::iift
(a)
(b)
Complete the table above.
Using a scale of 4 cm to rcpresent I unit on the t-axis and 2 cm to represent 5 units
on tbe t-axis, draw the graph ofl? = 30 + lOt 5t' for 0 WOP669 sx4t"aa o'
The variables ofr and ), are connected by the equation 1= I - 1- 1. 5ome coresponding
values of -r and ) are given in the table below.
(a) Calculate the values of a and ,.
(b) Using a scale of 2 cm to represent I unit on the i-axis and 1 cm to represent I unit
on the y axis, draw the graph ofy = *'z bt Tfor-2r alues o1'r nncl _r' nre gjven in the following rable.
ib)
Calculate the vnlues ol d, l? and ..
Using a scale of2 cm to repfesent I unit on the r-axis rnd I cm to rcprescnt I unit on thc
) axis, draw the graph of ) = -2r' + 7i +,1 fbr the range 0