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

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```5 ab 1 ac 1 b2 1 bc

17. (C) (a 1 b)(a 2 b) 5
(a 1 b)(a 2 b) 5 a2

(a 1 b)(a 2 b) 5 a2 2 ab 1 ba \u2026

(a 1 b)(a 2 b) 5 a2 2 ab 1 ba 2 b2

a b ab baa b a b2 2+ - = - + -^ ^h h

(a 1 b)(a 2 b) 5 a2 2 b2 MEMORIzE
18. (A) (a 1 b)2 5 (a 1 b)(a 1 b)
(a 1 b)(a 1 b) 5 a2 \u2026

(a 1 b)(a 1 b) 5 a2 1 ab 1 ba \u2026

(a 1 b)(a 1 b) 5 a2 1 ab 1 ba 1 b2

(a 1 b)2 5 a2 1 2ab 1 b2 MEMORIzE
19. (D) 2(a 2 b) 5 2a 2 (2b)
2(a 2 b) 5 2a 1 b

2(a 2 b) 5 b 2 a MEMORIzE
20. (A) a(b 1 c) 5

a(b 1 c) 5 ab 1 ac

21. (C) 2a(b 2 c) 5

2a (b 2 c) 5 2ab 2 a(2c)

5 2ab 1 ac 5 ac 2 ab
D. Exponents
22. (C) 105 5 100,000
5 zeros
23. (B) 107076.5 5 1 0 7 0 7 6 . 5

5 4 3 2 1
5 1.070765 3 105
24. (B) Add exponents:

a2 3 a5 5 a7
am 3 an 5 am 1 n
25. (C)(ab)7 5 a7b7
(ab)m 5 ambm
26. (A) c
a
c
a8
8
8
=` j
c
a
c
am
m
m
=` j
27. (A) a4 3 b4 5 (ab)4
am 3 bm 5 (ab)m
28. (A) a23 3 b5 5 a
b
3
5
a2m 3 bn 5 a
b
m
n
29. (C)(a 3)5 5 a15

MULTIPLY
ExPONENTS
(am)n 5 amn
30. (A) 2a23 5 2a3
ax2b 5
x
a
b since a
2n 5 1an

31. (B) 2a a a a3
1
3
2\u2013 \u2013m n m n# =
a3
2 or\u2013m n= 3
2
a
a
n
m
32. (D) 32 1 322 1 41 1 60 5
32 5 3 3 3 5 9
322 5 3
1
9
1
2 =
41 5 4
60 5 1 (any number to 0 power 5 1)
32 1 322 1 41 1 60 5 9 1 9
1 1 4 1 1 5 14 9
1
GRE2015_P03.indd 56 4/18/14 11:27 AM
THE 101 MOST IMPORTANT MATH QUESTIONS YOU NEED TO KNOW HOW TO SOLVE \u2022 57
E. Percentages
Translate: is ä 5
of ä 3 (times)
percent (%) ä

100
what ä x (or y, etc.)
33. (B) 15% of 200 5
å å å å å
15
100
200
15
100
200
15
100
200 30
34. (C) What is 3 % of 5?
å å å å å å
3100 5
100
3 5
100
15
20
3
x
x
#
#
=
=
x = =
35. (C) What percent of 3 is 6?
å å å å å å
x 100 3 3 5 6

100 3 6
100
100
3
200
x
x
x
# =
3 600x =
3 6
6
# =
=
x =
F. Equations
36. (C) y2 5 16

y
16
16
4
y
y
2
2 !
!
=
=
=
37. (A) x 2 y 5 10
x 2 y 1 y 5 10 1 y
x 5 10 1 y
Subtract 10:
x 2 10 5 10 2 10 1 y
x 2 10 5 y
38. (B) Add equations:
x 1 4y 5 7
x 2 4y 5 8
2x 1 4y 2 4y 5 15
2x 5 15
2
15x =
39. (A) x 2 2y 5 2 1
2x 1 y 5 4 2
Multiply 1 by 2:
2(x 2 2y) 5 2(2)
We get:
2x 2 4y 5 4 3
Subtract 2 from 3 :
2x 2 4y 5 4
2 (2x 1 y 5 4)
0 2 5y 5 0
y 5 0 4
Substitute 4 into either 1 or 2 :
In 1 :
x 2 2y 5 2
x 2 2(0) 5 2
x 5 2
40. (A)
5 12
7 ,x x= =
Cross-multiply x:
5
x
1
7
2
12x 35
Divide by 12:
12
12
12
35x
=
12
35 212
11x = =
G. Angles
Questions 41242 refer to the diagram.
a°
b°30°
41. (B) a° and 30° are supplementary angles (they
add up to 180°).
So a 1 30 5 180; a 5 150 .
GRE2015_P03.indd 57 4/18/14 11:27 AM
58 \u2022 GRUBER\u2019S COMPLETE GRE GUIDE 2015
42. (A) b° and 30° are vertical angles (vertical angles
are equal).
So b 5 30 .

25° a° b°
43. (A) a°, b°, and 25° make up a straight angle, which
is 180°.
a 1 b 1 25 5 180
a 1 b 5 180 2 25
a 1 b 5 155
44. (C) The sum of the angles in the diagram is 360° ,
the number of degrees around the circumference of
a circle.
H. Parallel Lines
1
1
2
2
a°
b°c°
d°e°
f °g°
50°
45. (B) a 1 50 5 180
a 5 130
46. (A) b 5 50 (vertical angles)
47. (B) c 5 a (vertical angles)
5 130
48. (B) d 5 c (alternate interior angles are equal)
5 130
49. (A) e 5 b (alternate interior angles)
5 50
50. (A) f 5 e (vertical angles)
5 50
51. (B) g 5 d (vertical angles)
5 130
I. Triangles
52. (A)
y
xx
a°70°
(Note: Figure is not drawn to
scale.)
If two sides are equal, base angles are equal. Thus
a 5 70° .
53. (A)
50° 50°
x3
(Note: Figure is not drawn to
scale.)
If base angles are equal, then sides are equal, so
x 5 3 .
54. (B)
a
4
3
(Note: Figure is not
drawn to scale.)
The sum of two sides must be greater than the third
side. Try choices:
(A) 1 1 3 5 4: (A) is not possible
(B) 3 1 4 . 6: 6 1 3 . 4; 4 1 6 . 3\u2026OK
(C) 3 1 4 Ú 10: (C) is not possible
(D) 3 1 4 5 7: (D) is not possible
(E) 3 1 4 Ú 8: (E) is not possible
55. (B) Using similar triangles, write a proportion with x.
12
15
20
a°
a°
x
(Note: Figure is not drawn to scale.)
x
x
x
x
20
12
15
15 12 20
12 20
15
12
4
20
4
15
5
16
=
= ×
=
×
=
×
=
GRE2015_P03.indd 58 4/18/14 11:27 AM
THE 101 MOST IMPORTANT MATH QUESTIONS YOU NEED TO KNOW HOW TO SOLVE \u2022 59
In general:
n
m
p
q
r s
r
= =
+
n
m
r
q p
s
(Note: Figure is not drawn to scale.)
56. (B) The greater angle lies opposite the greater side
and vice versa.
If B . A, b . a
a
B° A°
b
57. (C) The greater side lies opposite the greater
angle and vice versa.
If b , a, then B , A
a
B° A°
b
58. (B) Sum of angles of triangle 5 180°.
So 40 1 60 1 x 5 180
100 1 x 5 180
x 5 80

60°
40° x°
59. (C)
A
x
B C
4
C°
80°
50°
(Note: Figure is not drawn to scale.)
First calculate +C. Call it y.
80 1 50 1 y 5 180 (Sum of +s 5 180°)
y 5 50
Since +C 5 y 5 50 and +B 5 50, side AB 5 side
AC.
AB 5 x 5 4
60. (C) x° 5 20° 1 40° (sum of remote interior angles
5 exterior angle)
x 5 60
40°
20° x°
In general,
x
y
z
z 5 x 1 y
61. (B)
c b
a
In right \ue06e, a2 1 b2 5 c2
So for
x 12
5
52 1 122 5 x2
25 1 144 5 x2
169 5 x2
169 5 x
13 5 x
GRE2015_P03.indd 59 4/18/14 11:27 AM
60 \u2022 GRUBER\u2019S COMPLETE GRE GUIDE 2015
Note: Specific right triangles you should memorize;
use multiples to generate other triangles.
Example of multiples:
5
3
4

5 3
2 =
10
4 3 2 = 8
3 3 2 = 6

10
8
6
Memorize the following standard triangles (not
drawn to scale):
5
12 13
7
24 25
3
4 5
8
15 17
2
1
30°
60°
3
1
1
45°
45°
2
40 41
9
145°
45°
2
2
\u221a
\u221a
\u221a
2
2
\u221a
62. (B) Perimeter 5 sum of sides
10 1 17 1 21 5 48
21
10 17
CB
A
8
63. (D)
Area of \ue06e 5 2
1 bh
Area of \ue06e 5 2
1 (21)(8) 5 84
64. (A) Area of any triangle 5 2
1 base 3 height

4
3
Here 4 is base and 3 is height. So area 5 2
1 (4 3 3)
5 2
1 (12) 5 6 .
65. (B)

4
3
x
To find perimeter, we need to find the sum of the
sides. The sum of the sides is 3 1 4 1 x.
We need to find x. From the solution in Question
61, we should realize that we have a 3\u20134\u20135 right
triangle, so x 5 5.
The perimeter is then 3 1 4 1 5 5 12 .
Note that you could have found x by using the
Pythagorean Theorem:
32 1 42 5 x2; 9 1 16 5 x2; 25 5 x2; 25 5 x; 5 5 x.
GRE2015_P03.indd 60 4/18/14 11:27 AM
THE 101 MOST IMPORTANT MATH QUESTIONS YOU NEED TO KNOW HOW TO SOLVE \u2022 61
J. Circles

7 7
0
center
\u2022
66. (B) Area 5 p r 2 5 p(7)2
5 49p
67. (A) Circumference 5 2p r 5 2p(7)
5 14p
68. (B) Inscribed angle 5
2
1 arc
x° 5
2
1 (70°)
5 35°

70°x°
69. (A) Central angle 5 arc
30° 5 x°
Note: The total number of degrees around the
circumference is 360°. So a central angle of 30°, like
the one below, cuts 360
30
12
1
= the circumference.
center
30° x°
70. (C) The diameter cuts a 180° arc on the circle, so
an inscribed angle y 5
2
1 arc 5
2
1 (180°) 5 90° .
Here is a good thing to remember:
Any inscribed angle whose triangle base is a
diameter is 90°.

diameter
y°
K. Other Figures
4 55
10```