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# Gruber's Complete GRE Guide 2015 Gruber, Gary R

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```100 100#=
Simplify by multiplying numerator and denominator
by 4:
250
100 100
,
,
40
x 4
4
1 000
40 000
#
# #= =
=
example 4
John is now m years old and Sally is 4 years older than
John. Which represents Sally\u2019s age 6 years ago?
(A) m 1 10
(B) m 2 10
(C) m 2 2
(D) m 2 4
(E) 4m 2 6
Choice C is correct.
Translate:
John is now m years old
\u2193 \u2193 \u2193
J 5 m
Sally is 4 years older than John
\u2193 \u2193 \u2193 \u2193 \u2193
S 5 4 1 J
Sally\u2019s age 6 years ago
\u2193 \u2193
S 2 6
So we get: J 5 m
S 5 4 1 J
and find: S 2 6 5 4 1 J 2 6
S 2 6 5 J 2 2
S 2 6 5 m 2 2 (substituting m for J)
See Math Strategy 7, Example 2 (page 92) for an
alternate approach to solving this problem, using a
different strategy: Use Specific Numerical Examples
to Prove or Disprove Your Guess.
example 5
Phil has three times as many DVDs as Sam has. Even
after Phil gives Sam 6 DVDs, he still has 16 more DVDs
than Sam has. What was the original number of DVDs
(A) 20
(B) 24
(C) 28
(D) 33
(E) 42
Choice E is correct.
Translate:
Phil has three times as many DVDs as Sam has
\u2193 \u2193 \u2193 \u2193 \u2193
P 5 3 3 S
Even after Phil gives Sam 6 DVDs, he still has 16
{ \u2193 \u2193 \u2193 \u2193 \u2193
P 2 6 5 16
more DVDs than Sam has
{ \u2193 \u2193
1 S 1 6
Sam now has (S 1 6) DVDs because Phil gave Sam 6
DVDs. So we end up with the equations:
P 5 3S
P 2 6 5 16 1 S 1 6
Find P; get rid of S:
P 5 3S; P S3 =

6 16 6
62
P P
P P
3
3
48 18
2 = + +
= + +
3P 2 18 5 48 1 P 1 18
2P 5 84
P 5 42
example 6
If q is 10% greater than p and r is 10% greater than y, qr
is what percent greater than py?
(A) 1%
(B) 20%
(C) 21%
(D) 30%
(E) 100%
Choice C is correct.
GRE2015_P04.indd 78 4/18/14 11:28 AM
STRATEGY SECTION \u2022 79
Translate:
If q is 10% greater than p
{ \u2193 \u2193 \u2193
q 5
100
10 p 1 p
and r is 10% greater than y
{
\u2193 \u2193 \u2193
r 5
100
10 y 1 y
qr is what percent greater than py?

\u2193 \u2193 \u2193
qr 5 x
100
py 1 py
So we have three equations:

q p p p
r y y y
qr x py py x py
100
10
100
10 1
100
10
100
10 1
100 100 1
= + = +
= + = +
= + = +
c
c
`
m
m
j
1

2

3
Multiply 1 and 2 :
qr py100
10 1
2
= +c m 4
Now equate 4 with 3 :
qr x py py100 1 100
10 1
2
= + = +` cj m
You can see that 1x100
10 1 100
2
+ = +c m , canceling py.
, ,
,
x
x
x
100
10 1 10 000
100 2 100
10
100 1
10 000
100
100
20
100
21
100
21
So
2
+ = + = +
+ = =
=
c m 1+c m
The answer is x 5 21.
Alternate approach: Choose numbers for p and for y:
Let p 5 10 and y 5 20
Then, since q is 10% greater than p:
q 5 10% greater than 10
10 10 11q 100
10
= + =c m
Next, r is 10% greater than y:
r 5 10% greater than 20
Or, r 5 10 10 11q 100
10
= + =c m20 1 20 5 22
Then:
qr 5 11 3 22
and py 5 20 3 10
So, to find what percent qr is greater than py, you would
need to find:
11 22 20 102
py
qr py 100
20 10 100
or2 #
#
# #
#
This is:
100 21200
42
# =
example 7
Sales of Item X
Jan2Jun, 2004
Month Sales (\$)
Jan 800
Feb 1,000
Mar 1,200
Apr 1,300
May 1,600
Jun 1,800
According to the above table, the percent increase in
sales was greatest for which of the following periods?
(A) Jan2Feb
(B) Feb2Mar
(C) Mar2Apr
(D) Apr2May
(E) May2Jun
Choice A is correct.
The percent increase from Month A to Month B 5
Sales (Month A)
Sales (Month B) Sales (Month A) 100#
-
Month Sales (\$) Periods % Increase in Sales
Jan 800 Jan2
Feb
1,000 800 100 100800 800
2002
# #=
Feb 1,000 Feb2
Mar ,
1,200 1,000 100 , 100
2
1 000 1 000
200
# #=
Mar 1,200 Mar2
Apr ,
1,300 1,200 100 , 100
2
1 200 1 200
100
# #=
Apr 1,300 Apr2
May ,
1,600 1,300 100 , 100
2
1 300 1 300
300
# #=
May 1,600 May2
Jun ,
1,800 1,600 100 , 100
2
1 600 1 600
200
# #=
Jun 1,800
You can see that 800
200 100# (Jan2Feb) is the greatest.
GRE2015_P04.indd 79 4/18/14 11:28 AM
80 \u2022 GRUBER\u2019S COMPLETE GRE GUIDE 2015
Know how to Find Unknown Quantities (areas, Lengths,
arc and angle Measurements) from Known Quantities
(the Whole Equals the Sum of Its Parts)
When asked to find a particular area or length,
instead of trying to calculate it directly, find it
by subtracting two other areas or lengths\u2014a
method based on the fact that the whole minus
a part equals the remaining part.
This strategy is very helpful in many types of geometry
problems. A very important equation to remember is
The whole 5 the sum of its parts 1
Equation 1 is often disguised in many forms, as
seen in the following examples:
example 1
X
Y
Z
In the diagram above, \u394XYZ has been inscribed in a
circle. If the circle encloses an area of 64, and the area of
\u394XYZ is 15, then what is the area of the shaded region?
(A) 25
(B) 36
(C) 49
(D) 79
(E) It cannot be determined from the information given.
Choice C is correct. Use equation 1 . Here, the whole
refers to the area within the circle, and the parts refer to
the areas of the shaded region and the triangle. Thus,
Area within circle 5
Area of \u394XYZ
64 5 Area of shaded region 1 15
or Area of shaded region 5 64 2 15 5 49
Math
StratEg
y 3
example 2
In the diagram below,
\u2014
AE
\u2014
is a straight line, and F is a
point on
\u2014
AE
\u2014
. Find an expression for m+DFE.
A
B
C
D
E
x° y° ?
60°
F
(A) x 1 y 2 60
(B) x 1 y 1 60
(C) 90 2 x 2 y
(D) 120 2 x 2 y
(E) 180 2 x 2 y
Choice D is correct. Use equation 1 . Here, the whole
refers to the straight angle, +AFE, and its parts refer to
+AFB, +BFC, +CFD, and +DFE. Thus,
m+AFE 5 m +AFB 1 m+BFC 1
m+CFD 1 m+DFE
180 5 x 1 60 1 y 1 m+DFE
or m+DFE 5 180 2 x 2 60 2 y
m+DFE 5 120 2 x 2 y (Answer)
example 3
In the diagram below, AB 5 m, BC 5 n, and AD 5 10.
Find an expression for CD.
(Note: Diagram represents a straight line.)
B C DA
(A) 10 2 mn
(B) 10 2 m 2 n
(C) m 2 n 1 10
(D) m 1 n 2 10
(E) m 1 n 1 10
GRE2015_P04.indd 80 4/18/14 11:28 AM
STRATEGY SECTION \u2022 81
Choice B is correct. Use equation 1 . Here, the whole
refers to AD, and its parts refer to AB, BC, and CD. Thus,
AD 5 AB 1 BC 1 CD
10 5 m 1 n 1 CD
or CD 5 10 2 m 2 n (Answer)
example 4
The area of triangle ACE 5 64. The sum of the areas of
the shaded triangles ABF and FDE is 39. What is the
side of square BFDC?
A
B
C D E
F
(A) 5
(B) 4
(C) 5
(D) 44
(E) Cannot be determined.
Choice A is correct.
Since we are dealing with areas, let\u2019s establish the area
of the square BFDC, which will then enable us to get
its side.
Now, the area of square BFDC 5 area of triangle ACE 2
(area of triangles ABF
1 FDE)
Area of square BFDC 5 64 2 39
5 25
Therefore, the side of square BFDC 5 5.
example 5
B
A
3
O
In the figure above, O is the center of the circle. Triangle
AOB has side 3 and angle AOB 5 90°. What is the area
(A) 9 4 2
12rc m
(B) 9 122
r` j
(C) 9(p 2 1)
(D) 9 24 4
1rc m
(E) Cannot be determined.
Choice A is correct.
Subtract knowns from knowns:
Area of shaded region 5 area of quarter circle AOB 2
area of triangle AOB
Area of quarter circle AOB 5 ( )4
3 2r (since OA 5 3 and
area of a quarter of a circle 5
4
Area of triangle AOB 5 2
3 3# (since OB 5 3 and
area of a triangle 5 2
1 base 3 height).
Thus, area of shaded region 5 9 24
9
2
9
4 2
12 =r rc m.
example 6
The sides in the square above are each divided into five
equal segments. What is the value of