Revised gre quantitative complete!!!

2014
Worldlink
Communications
Rejan Chitrakar

Rejan Chitrakar

[REVISED GRE QUANTITATIVE]
Complete revised GRE quantitative problems and solutions. The only book you will need to score at or
above 90th percentile .

Quantitative Comparison Questions - Revised GRE
1. Lionel is younger than Maria.
QtyA

QtyB

Twice Lionel's age

Maria's age

Solution;
here, L's age<M's age,
let L's age=5 and M's age=10.
then QtyA=2*5=10=Maria's age=QtyB which implies our answer is C. (since QtyA=QtyB)
But we must check with other numbers also to be sure.
So now let L's age=5 and M's age=6.
then QtyA=2*5=10 and QtyB=6 which implies our answer is A. (since QtyA>QtyB)
Since different assumptions are leading to different answers, our answer is D. That means the
relationship cannot be determined from the given information.

2. y=2x2+7x-3
QtyA

QtyB

x

y

Solution;
Put x=0, then y=2*0+7*0-3=-3. That means x=0 and y=-3 in which case our answer would be
A(since 0>-3)
Next, put x=1, then y=2*1+7*1-3=6. That means x=1 and y=6 in which case our answer would be
B(since 6>1).
Hence our answer is D.

3. y>4
Qty A

QtyB

(3y+2)/5

y
2

Solution;
(3y+2)/5 ? y
or, 3y+2 ? 5y
or, 2 ? 5y-3y
or, 2 ? 2y
or, 1 ? y
Since we are given, y>4, we also know, y>1.
So, 1<y
If we reverse our process, we get
1<y
or, 2<2y (multiplying both sides by 2)
or, 2<5y-3y
or, 3y+2<5y (shifting 3y to the left)
or, (3y+2)/5<y (Dividing both sides by 5)
That means QtyB is greater and that's our answer.

4. QtyA

QtyB

(230-229)/2

228

Solution;
QtyA=(230-229)/2=229(2-1)/2=229/2=229-1=228=QtyB
So our answer is C.

5. QtyA

QtyB

x2+1

2x-1

Solution;
x2+1 ? 2x-1
3

or, x2-2x+1 ? -1
or, (x-1)2 ? -1
Since square of a number cannot be negative, (x-1)2 cannot be negative. So it's greater than -1.
Now reversing the process, we get x2+1>2x-1. That means our answer is A.

6. w>1
QtyA

QtyB

7w-4

2w+5

Solution;
7w-4 ? 2w+5
or, 7w-2w ? 5+4
or, 5w ? 9
or, w ? 9/5
Now we are given w>1. So let w=2. Since 2>9/5, we get 7w-4>2w+5. That means our answer
would be A.
But we can also assume w=3/2 since w has to be greater than 1. Now for this case, 3/2<9/5. That
means our answer would be B.
Hence our answer is D since different assumptions are leading to different answers.

7.
O is the center of the Circle above.
3

x O

QtyA

QtyB

3

x

5

4
Solution; radius of the circle, r=/32+42=5 (since in a right angled triangle, p2+b2 =h2)
Now x=/52-32=4
That means QtyB is greater. Hence our answer is B.
4

8. x<y<z
QtyA

QtyB

(x+y+z)/3

y

Solution;
Let x=1,y=2 and z=3. i.e., QtyB=2
then QtyA=(1+2+3)/3=6/3=2 which implies our answer is C.(since QtyA=QtyB=2)
To be sure, let x=0,y=1 and z=7 i.e., Qty B=1
then QtyA=(0+1+7)/3=8/3 which implies our answer is A.(since QtyA>QtyB)

9.

QtyA
x/y

x

QtyB
1

50
y
Solution;
remaining angle=180-90-50=40
Therefore, x>y (since side opposite to greater angle is always greater in a triangle)
That means, x/y>1
Hence our answer is A.

10. 0<x<y<1
QtyA

QtyB

1-y

y-x

Solution;
Let x=0.2 and y=0.3
then, QtyA=1-0.3=0.7 and QtyB=0.3-0.2=0.1 in which case our answer would be A.
5

Now let x=0.2 and y=0.9
then, QtyA=1-0.9=0.1 and QtyB=0.9-0.2=0.7 in which case our answer would be B.
Hence the relationship cannot be determined from the given information. That is our answer is
D.

11. 'p' is the probability that event E will occur and 's' is the probability that event E will not
occur.
QtyA

QtyB

p+s

ps

Solution;
Here p+s=1
or, s=1-p
or, ps=p-p2
Now p+s ? p-p2
or, s ? -p2
Here right side is always negative and left side is a number between 0 and 1 since it is a
probability of non occurrence of an event.
So s>-p2. Hence, p+s is greater i.e., A is the answer.

12. X is the set of all integers n that satisfy the inequality 2<=|n|<=5
QtyA

QtyB

The absolute value of the greatest

The absolute value of the least integer

integer in X

in X

Solution;
X={2,3,4,5,-2,-3,-4,-5}
A=|5|=5
B=|-5|=5
6

Hence our answer is C.

13. x and m are positive numbers and m is a multiple of 3.
QtyA

QtyB

xm/x3

xm/3

Solution;
xm-3 ? xm/3
or, m-3 ? m/3
or, m-m/3 ? 3
or, 2m/3 ? 3
or, 2m ? 9
or, m ? 9/2
Since m is a multiple of 3, let m=3.
then 3<9/2 in which case our answer would be B.
Again if we assume m=6,
then 6>9/2 in which case our answer would be A.

14. k is a digit in the decimal 1.3k5 and 1.3k5 is less than 1.33.
QtyA

QtyB

k

1

Solution;
1.3k5<1.33
that means, 0<=k<3.
If k=0, B would be greater.
If k=2, A would be greater.
7


15. st=/10
QtyA

QtyB

s2

10/t2

Solution;
s2 ? 10/t2
or, s2t2 ? 10
or, st ? /10
st=/10

16. Three consecutive integers have a sum of -84.
QtyA

QtyB

The least of the three integers

-28

Solution;
x+(x+1)+(x+2)=-84
or, 3x+3=-84
or, 3x=-87
or, x=-87/3=-29
Therefore the three consecutive integers are -29,-29+1 and -29+2
i.e.,-29, -28 and -27
The least of three integers is -29 which is less than -28.
Hence our answer is B.

17. In the xy-plane, the equation of line k is 3x-2y=0.
8

QtyA

QtyB

x-intercept of line k

y-intercept of line k

Solution;
for x-intercept, put y=0. So 3x-2*0=0 i.e., x=0.
for y-intercept, put x=0. So 3*0-2y=0 i.e., y=0.
hence A=B and our answer is C.
OR we can directly know the answer if we see the equation clearly.
3x-4y=0
or, y=3/4*x which is in the form y=mx and which passes through origin. That means it does not
have x and y intercepts or their values are zero.

18. n is a positive integer that is divisible by 6.
QtyA

QtyB

The remainder when n is

The remainder when n is divided by 18

divided by 12
Solution;
Since n is divisible by 6, let n=6.
then when 6 is divided by 12, the remainder will be 6 and when it is divided by 18, the
remainder will be still 6. That indicates answer C.
But if we suppose n=12, since 12 is also divisible by 6.
then, when 12 is divided by 12, the remainder is 0 and when it is divided by 18, the remainder is
12. That indicates answer B.
Therefore our answer is D.

19. 1-x/1+x=1/x
Qty A

QtyB

x

-1/2
9

Solution;
1-x/x-1=1/x
or, x-x2=x-1 (cross multiplication)
or, -x2=-1
or, x2=1
or, x=+-1
Here you might think the answer is D since, if x=+1, then A would be greater and if x=-1, then B
would be greater. But that is not the case. Try to plug in the obtained values of x back into
the given equation, then you will see that x=+1 will give 1-x/x-1=1-1/1-1=0/0, which is an
indeterminate form. So x=+1 is eliminated. and the only value of x remained is x=-1.
Since -1<-1/2, our answer is B. That means QtyB is greater.

20. In a set of 24 positive integers, 12 of the integers are less than 50. The rest are greater than
50.
QtyA

QtyB

The median of the 24 integers

50

Solution;
Here, median=value of (N+1)th/2 item=(24+1)th/2=12.5th item=average of 12th and 13th items.
We just know that the 12th item is less than 50 and the 13th item is greater than 50 since half
of the numbers are less than 50 and half of the numbers are greater than 50. But we don't
know their exact values.
So let 12th number=40 and 13th number=60
then median=(40+60)/2=50
That means QtyA=QtyB i.e., our answer could be C.
But if we assume 12th number=45 and 13th number=65,
then median=(45+65)/2= 55 in which case our answer would be A.
Hence D is the answer.

10

21. QtyA

QtyB

(40% of 50)+60

(60% of 50)+70

Solution;
Don't try to waste your time by computing on this type of questions.
Just by seeing, you can decide which side is greater.
It's B since 60% of 50>40% of 50. In addition, 70>60.
So our answer is B.

22. 0<r<t
QtyA

QtyB

r/t

t/r

Solution;
Since r<t,
r/t<1 and t/r>1. Hence QtyB is greater.

23. m,p and x are positive integers and mp=x.
QtyA

QtyB

m

x

Solution;
If p=1, m=x. That means our answer would be C.
But if p=2, then 2m=x. That means QtyB would be greater or our answer would be B.
Hence the answer is D.

24. Triangular regions T1 and T2 have equal areas and have heights h1 and h2 respectively.
QtyA

QtyB

(The area of T1)/h1

(The area of T2)/h2
11

Solution;
area of T1=b1h1/2
area of T2=b2h2/2
So QtyA=b1h1/2h1=b1/2
and QtyB=b2h2/2h2=b2/2
Now we don't know what relation exists between b1 and b2. So we can't determine which is
greater. Hence our answer is D.

25. x2y>0, xy2<0
QtyA

QtyB

x

y

Solution;
x2y>0. This means y must be positive since x2 cannot be negative.
xy2<0. This means x must be negative since y2 cannot be negative.
This means, x is negative and y is positive. Hence y is greater. Our answer is B.

26.

The diameter of the circle is 10.

QtyA

QtyB

The area of the region

40

enclosed by the quadrilateral
Solution;
Area of the circle=3.14*52=78.5
So area enclosed by quadrilateral<78.5
12

Now we could draw a small quadrilateral whose area is less than 40 as follows.

Or we could draw a big quadrilateral whose area is greater than 40 with the same diameter
of 10.

We can even draw a quadrilateral with area exactly 40 inside a circle of diameter 10.
That means we cannot determine which is greater.

27.

B
The circle shown has a radius of 5. AB=8.
A

C

QtyA

QtyB

The perimeter of triangle ABC

24

Solution;
BC=/AC2-AB2=/102-82=/100-64=/36=6
Therefore perimeter of triangle ABC=AB+BC+CA=8+6+10=24
Hence QtyA=QtyB=24. So our answer is C.
13

28. x, y and z are negative integers.
QtyA

QtyB

The product of x, y and z

The sum of x, y and z

Solution;
Let x=y=z=-1
then QtyA=x*y*z=(-1)*(-1)*(-1)=-1
and QtyB=x+y+z=(-1)+(-1)+(-1)=-3
That means QtyA is greater.
To be sure, let's check with x=y=z=-2
then we have, QtyA=(-2)*(-2)*(-2)=-8
and QtyB=(-2)+(-2)+(-2)=-6
That means QtyB is greater.

29. Team X scored 10 points in the first half of a certain game. In the second half of the game,
team Y scored 15 points more than team X.
QtyA

QtyB

The number of points scored

The number of points scored

by team X in the first half of

by team Y in the first half of

the game

the game

Solution;
We don't know what team Y scored in the first half of the game. That means Qty B is
unknown. Hence our answer is D.

30. x>y>w>0
QtyA

QtyB

xy/w

yw/x
14

Solution;
xy/w ? yw/x
or, x/w ? w/x
Since x>y>w, x>w. That means, x/w>1 and w/x<1. So QtyA is greater.

31. n>1
QtyA

QtyB

(n/n+1)+1

1-(1/n+1)

Solution;
(n/n+1)+1 ? 1-(1/n+1)
or, n/n+1 ? -1/n+1
or, n ? -1
Since n>1, n>-1. Therefore, QtyA is greater.

32.

y

x

p

q

QtyA

QtyB

x+y

p+q

Solution;
Let the remaining two angles be z and r.
Then x+y+z=180 and p+q+r=180. (sum of angles in a triangle is 180)
Again, z=r (vertically opposite angles made by two straight lines are equal)
So, x+y=p+q
That is our answer is C.
15

33. N

Q

40
M

R

T MN//PQ and PR//ST

55 x

y

P

S

QtyA

QtyB

y-x

15

Solution;
angle QPM=40 (corresponding angles made by two parallel lines are equal)
y=angle RPM=55+40=95
x+y=180 (sum of co-interior angles made by parallel lines is 180)
or, x=180-95=85
So y-x=95-85=10. That means QtyB is greater.

34. x is not zero.
Qty A

QtyB

|x|+|-2|

|x-2|

Solution;
if x=2,
then, QtyA=|2|+|-2|=2+2=4
QtyB=|2-2|=0. In this case, QtyA is greater.
But if x=-2,
then, QtyA=|-2|+|-2|=2+2=4
QtyB=|-2-2|=|-4|=4. In this case, QtyA=Qty B.

16

35. QtyA

QtyB

The standard deviation of a

The standard deviation of 5

set of 5 different integers, each of

integers, each of which is between 10 and 20

which is between 0 and 10.
Solution;
Standard deviation is a measure of dispersion and is not affected by the size of data members.
So we cannot determine the answer from the given information.

36. A power station is located on the boundary of a square region that measures 10 miles on
each side. Three substations are located inside the square region.
QtyA

QtyB

The sum of the distances from

30 miles

the power station to each of
the substations.
Solution;
We could make the sum less than 30 miles as shown in the diagram below.
10 miles

In the figure alongside,
A

B

PS(X)

AX+BX+CX<30, in this case our answer would be B.

C

We could make the sum greater than 30 miles by putting the substations far away from the
power house as below.
A

PS(X)

AX+BX+CX>30, in this case our answer would be A.

B C

Likewise, we can make the sum equal to 30 miles. That means our answer is D.
17

37. 6<x<7 , y=8
QtyA

QtyB

x/y

0.85

Solution;
6<x<7
or, 6/y<x/y<7/y
or, 6/8<x/y<7/8
or, 0.75<x/y<0.875
therefore, x/y=0.84 in which case our answer would be B.
Also x/y=0.87 in which case our answer would be A.

38. 'n' is the total number of numbers under 1000 that can be formed by using the digits
3,7,0,4.
QtyA

QtyB

n

The lowest prime number greater than 48

Solution;
n=(number of choices 1000s digit has)*(number of choices 100s digit has)*(number of choices
10s digit has)*(number of choices unit digit has)
or, n=1*4*4*4=64
The lowest prime number greater than 48=53

39. The average of 21 consecutive integers is M. A new set of numbers is created by adding 23
to the smallest number, 22 to the next, 21 to the third etc. The average of the new set is N.
QtyA

QtyB

N-M

13
18

Solution;
N-M=average of 21 new numbers
=(23+22+21+……21 terms)/21
=21/2{2*23+(21-1)(-1)}/21=(46-20)/2=13
(sum of n terms in an arithmetic progression Sn=n/2{2a+(n-1)d}
that means our answer is C.

40. 4v4-1024=0
QtyA

QtyB

v

0

Solution;
4v4=1024
or, v4=1024/4=256
or, v=+-4
If v=+4, our answer would be A.
If v=-4, our answer would be B.

41. ABC is an isosceles triangle. AB=3 inches and BC=4 inches.
QtyA

QtyB

the length of CA

3 inches

Solution;
A

CA could be either equal to AB or BC.

3 inches
B

That means, if CA=3 inches, our answer would be C.
C

4 inches

and if CA=4 inches, our answer would be A
19

42. p>0
QtyA

QtyB

p2+1

1-p

Solution;
p2+1 ? 1-p
or, p2 ? -p
or, p ? -1
Since p>0, p>-1. Hence reversing the process, we get, p2+1>1-p. Hence our answer is A.
OR we can see, in the expression p2 ? -p, p2 is always positive while -p is always negative.
That means p2>-p.

43. A ball is dropped from a certain height, bounces one half the original height. If this ball is
dropped from 80 ft, it will travel a distance of k ft when it hits the ground for the third time.
QtyA

QtyB

k

160 ft

Solution;
80 ft

40 ft
20ft

k=80+(40+40)+(20+20)=200 ft

44. x is a positive integer and y is a negative integer.
Qty A

QtyB

x-y

y-x

Solution;
20

Since x=+ve and y=-ve, x-y becomes addition of x and y.
and y-x becomes addition of -y and -x. So QtyA is greater since it's +ve while QtyB will be -ve.
So our answer is A.

45. The probability that both events E and F will occur is 0.42.
QtyA

QtyB

The probability that event E

0.58

will occur.
Solution;
P(E and F)=P(E)*P(F) , which is the probability of occurrence of both events E and F
or, P(E)*P(F)=0.42
or, P(E)=0.42/P(F)
When P(F)=max i.e. 1, P(E)=min i.e. 0.42 and when P(F)=min i.e. 0.42 (since if P(F)<0.42,
P(E)>1, which is not possible), P(F)=max i.e. 1 So 0.42<=P(E)<=1. Hence our answer is D.

46. a and b are positive integers.
QtyA

QtyB

a/b

a+3/b+3

Solution;
Put a=1 and b=1, then QtyA=1/1=1 and QtyB=1+3/1+3=4/4=1
That implies answer C.
Let's check with other numbers also to be sure about the answer.
Put a=1 and b=2, then QtyA=1/2 and QtyB=1+3/2+3=4/5
That implies answer B.

21

47. The arithmetic mean of 100 measurements is 23 and the average of 50 additional
measurements is 27.
QtyA

QtyB

Average of 150 measurements

25

Solution;
Average of 150 measurements=(100*23+50*27)/150=24.33
That means our answer is B.

48. QtyA

QtyB

Slope of line k

1

k

.(3,4)

Solution;
The line passing from the origin and the point (3,4) has slope=4/3, which is greater than 1. Also,
we can see from the figure that line k has greater slope than the line passing through (3,4).
This implies, slope of line k is greater than 1.

49. The original price of a suit was 30% less than the suit's $250 suggested retail price. The price
at which the suit was sold was 20% less than the original price.
QtyA

QtyB

Price at which the suit was sold

50% of the suit's suggested price

Solution;
Original price=70% of $250=$175
Price at which suit was sold=80% of $175=140

22

And 50% of suit's suggested price=0.5*$250=$125

50. B

8

F

3
A

C

4

D

5

E

G

7

QtyA

QtyB

Area of rectangular region

Area of trapezoidal region

ABCD

H

EFGH

Solution;
QtyA=length*breadth=8*3=24
QtyB=1/2*height(sum of the bases)=0.5*4*(5+7)=24

51. QtyA

QtyB

The sum of all the integers

The sum of all the integers

from 19 to 59 inclusive

from 22 to 60 inclusive

Solution;
QtyA=19+20+……………………………59
QtyB=22+23+……………………………60
Subtracting 22+23+…………………59 on both quantities,
QtyA=19+20+21=60
QtyB=60
(Note: while comparing, we can add or subtract equal quantities on both)

23

52.

R

QRS is an isosceles triangle.
QtyA

7
Q

x
4

QtyB

Perimeter of triangle QRS

17

S

Solution;
Since QRS is an isosceles triangle, the value of x could be either 7 or 4.
If x=7, then perimeter of the triangle=7+7+4=18 in which case answer would be A
and if x=4, perimeter=7+4+4=15 in which case answer would be B.
Hence the answer is D.

53. The average of a set of mean daily temperatures for x days is 70 degrees. When a mean
daily temperature of 75 degrees is added to the set, the average increases to 71 degrees.
QtyA

QtyB

x

5

Solution;
New average= New sum of temperatures/new sum of days
or, 71=70x+75/x+1
or, 71x+71=70x+75
or, x=4
Hence QtyB is greater.

54. The scores for the 500 students who took Mr Johnson's final exam had a normal
distribution. There were 80 students who scored at least 92 points out of a possible 100 total
points and 10 students who scored at or below 56.
QtyA

QtyB

The mean score on the final

87

exam
24

Solution;

34%

34%

14%

14%

2%
56

2%
mean

92

Fig: Standard Bell Curve
80/500*100=16% scored at least 92 points.
10/500*100=2% scored at or below 56.
That means 92 is within 1 standard deviation from the mean and 56 is within 2 standard
deviations from the mean. Hence we can conclude that 56 and 92 are within 3 standard
deviations.
i.e.,92-56=3*sd
or, sd=12
therefore mean=92-12=80
So QtyB is greater.

55. In 2009, the property tax on each home in Town X was p percent of the assessed value of
the home, where p is a constant. The property tax in 2009 on a home in Town X that had an
assessed value of $125,000 was $2,500.
QtyA

QtyB

The property tax in 2009 on

$3,000

a home in Town X that had
an assessed value of $160,000
Solution;
p% of 125,000=2,500
or, p/100*125,000=2,500

25

or, p=2%
Therefore, QtyA=2% of 160,000=$3,200

56. r, s and t are three consecutive odd integers such that r<s<t.
QtyA

QtyB

r+s+1

s+t-1

Solution;
r+s+1 ? s+t-1
or, r+1 ? t-1
or, 2 ? t-r
Since r , s and t are three consecutive odd integers, t-r=4 (for ex 1,3,5 in which 5-1=4)
therefore, 2<t-r
i.e., QtyB is greater.

57. Machine R, working alone at a constant rate, produces x units of a product in 30 minutes
and machine S, working alone at a constant rate, produces x units of the product in 48
minutes, where x is a positive integer.
QtyA

QtyB

The number of units of

The number of units of

the product that machine

the product that machine

R, working alone at its

S, working alone at its

constant rate, produces in

constant rate, produces in

3 hours.

4 hours.

Solution;
R

x units

30 minutes(0.5 hours)

R

6x units

0.5*6=3 hours
26

S
S

x units

48 minutes(48/60 hours)

x*60/48*4=5x units

4 hours

Since number of units i.e. x can't be negative, 6x>5x.
That means QtyA is greater.

58. Frequency Distribution for List X
Number

1

2

3

5

Frequency

10

20

18

12

Frequency Distribution for List Y
Number

6

7

8

9

Frequency

24

17

10

9

List X and list Y each contain 60 numbers. Frequency distributions for each list are given
above. The average of the numbers in list X is 2.7 and the average of the numbers in list Y is
7.1. List Z contains 120 numbers: the 60 numbers in list X and the 60 numbers in list Y.
Qty A

QtyB

The average of the 120

The median of the 120

numbers in list Z

numbers in list Z

Solution;
average=total sum of list X + total sum of list Y/total number of items in X and Y
total sum of list X=1*10+2*20+3*18+5*12=10+40+54+60=164
total sum of list Y=6*24+7*17+8*10+9*9=144+119+80+81=424
total number of items in X and Y=60+60=120
So, avg=164+424/120=4.9
Now, median=value of (N+1)th/2 item=(120+1)/2=60.5th item=average of 60th and 61st
items=(5+6)/2=5.5

27

We can see from the given table that 60th item belongs to list X and its value is 5 and 61st
item belongs to list Y and its value is 6.

59. Among the 9,000 people attending a football game at college C, there were x students
from college C and y students who were NOT from college C.
QtyA

QtyB

The number of people

9000-x-y

attending the game who
were not students
Solution;
There are three categories in which 9000 people have been divided.
9000=number of students from college C + number of students not from college C + number of
people who were not students
That means, number of people who were not students=9000-x-y.

60. x is an integer greater than 1.
QtyA

QtyB

3x+1

4x

Solution;
Put x=2, then QtyA=32+1=33=27 and QtyB=42=16
this implies answer is A.
Again put x=4, then QtyA=34+1=35=243 and QtyB=44=256
this implies answer is B.

28

61. A, B and C are three rectangles. The length and width of rectangle A are 10 percent greater
and 10 percent less, respectively, than the length and width of rectangle C. The length and
width of rectangle B are 20 percent greater and 20 percent less respectively, than the length
and width of rectangle C.
QtyA

QtyB

The area of rectangle A

The area of rectangle B

Solution;
length of C=100, breadth of C=50
length of A=110% of 100=110 and breadth of A=90% of 50=45
area of A=110*45=4950
length of B=120% of 100=120 and breadth of B=80% of 50=40
area of B=120*40=4800
area of A>area of B

62. Set S consists of all positive integers less than 81 that are NOT equal to the square of an
integer.
QtyA

QtyB

The number of integers

72

in set S
Solution;
Set of all positive integers less than 81 that are square of an integer={1,4,9,16,25,36,49,64}
Therefore, S={1,2,3……..80} - {1,4,9,16,25,36,49,64}
Hence, number of integers in set S=80-8=72
That means our answer is C.

29

63. A certain punch is created by mixing two parts soda and three parts ice-cream. The soda is
4 parts sugar, 5 parts citric acid and 11 parts other ingredients. The ice-cream is 3 parts sugar, 2
parts citric acid and 15 parts other ingredients.
Solution;
QtyA

QtyB

Parts sugar in the punch

Parts citric acid in the punch

Solution;
Punch = 2Soda+3Ice-cream

eqn(i)

Soda = 4Sugar+5Citric acid+11Other

eqn(ii)

Ice-cream = 3Sugar+2Citric acid+15Other

eqn(iii)

From eqns(i and ii), we can see
Parts sugar in the punch=parts sugar in the soda + parts sugar in the Ice-cream
=2*4+3*3=8+9=17
Parts citric acid in the punch=parts citric acid in the soda + parts citric acid in the Ice-cream
=2*5+3*2=10+6=16

64. A random variable Y is normally distributed with a mean of 200 and a standard deviation
of 10.
QtyA

QtyB

The probability of the event

1/6

that the value of Y is greater
than 220.
Solution;
We draw a bell curve for this problem.

30

2%

14%

34%

34%

200

14%
210

2%

220

Here the curve has been divided into six regions and the value above 220 covers one region.
So you might think that the probability of the event that Y>220 i.e., P(Y>220)=1/6 and the
answer might be C.
But that's not so.
The percentages you are seeing on the bell curve refer to the percentages of areas covered by
different regions of the curve. Therefore 2% area covered just after the value 220 means there
are 2% values in the distribution that are greater than 220.
Therefore, P(Y>220)=2%=2/100=1/50 which is less than 1/6.

65. In a decimal number, a bar over one or more consecutive digits means that the pattern of
digits under the bar repeats without end. For example, 0.387=0.387387387….
QtyA

QtyB

0.717

0.71

Solution;
Although both look alike when we expand them, QtyA has repetition of 717 after the decimal
whereas QtyB has repetition of 71 after the decimal and since 717/1000>71/100, QtyA is greater.

66. Of 30 theater tickets sold, 20 tickets were sold at prices between $10 and $30 each and 10
tickets were sold at prices between $40 and $60 each.
QtyA

QtyB

The average of the prices

$50

of the 30 tickets
Solution;
The highest possible average=(20*30+10*60)/30=$40
31

That means even if all 30 tickets were sold at maximum prices, the average would be less than
$50. Hence our answer is B.
Note: you might be tempted to choose answer D since we don't know how much each ticket
was sold for. So you must consider the maximum and minimum possibilities for this type of
questions.

67. Car X can come with any of these 5 additional features: sunroof, stereo, tinted windows,
leather seats and cruise control.
QtyA

QtyB

Number of different

25

combinations possible
Solution;
The car has 2 choices for each 5 additional features. That means it will either have a sunroof or
not. It will either have a stereo or not. It will either have a tinted windows or not. It will either
have leather seats or not and finally it will either have a cruise control or not.
Hence total number of possible combinations=2*2*2*2*2=32
That means our answer is A.
We can solve this problem by another way also.
The car can have no additional features at all or 1 or 2 or 3 or 4 or all 5 additional features.
So, total number of possible combinations=5C0+5C1+5C2+5C3+5C4+5C5=32
if the car has no additional feature at all
if it has 1 additional feature
if it has 2 additional features
if it has all 5 additional features
Note:
Combination= Number of ways of selection of 'r' objects out of 'n' objects at a time and is given
by the formula:
32

nCr=n!/{r!(n-r)!}
For example, There are three objects A, B and C and we have to select 2 objects at a time.
Then we could select the objects as follows:
AB
BC
CA
That means, we have 3 choices for selecting 2 objects out of 3 objects at a time.
If we use our formula, then we get
nCr=3C2=3!/{2!(3-2)!}=3!/2!1!=3*2*1/2*1*1=3

68. QtyA

QtyB

Average of integers from

Average of integers from

-50 to -1 inclusive

-50 to 0 inclusive

Solution;
QtyA={(-50)+(-49)+……………………………………….+(-1)}/50=-(1+2+3+……….+50)/50
-1, there will be 50 terms.
QtyB={(-50)+(-49)+………………………………………….+(-1)+0}/51=-(1+2+3+…………+50)/51
to 0, there will be 51 terms.

From -50 to
From -50

Comparing the two quantities, we see that QtyB is less negative than QtyA as its denominator
is higher. That means QtyB must be greater.

69. QtyA

QtyB

The average of

34

32 ,34 and 36
Solution;

33

Average=(32+34+36)/3=32(1+32+34)/3=3+33+35
We can clearly see QtyA is greater.

70. a, b, c and d are different positive numbers. The average of a and b is 30. The average of
a, b, c and d is 40.
QtyA

QtyB

The greatest possible

99

value of d
Solution;
(a+b)/2=30

a+b=60

(a+b+c+d)/4=40

a+b+c+d=160

Solving the two equations,
c+d=100
Since c and d are different positive numbers, the greatest possible value of d is 99.99………….
which is greater than 99. So answer is A.
Note: If a, b, c and d were different positive INTEGERS, then QtyA=99 and our answer would
be C.

71.

D
3030

A

B

C

QtyA

QtyB

AB

BC

Solution;
We can see, if DB were perpendicular to AC, AB=BC in which case our answer would be C.
34

Since only given quantities are two equal angles, we could draw the figure without affecting
the given angles as follows:
D
30 30 C
B

In the figure alongside, we can see the two angles are still 30

A

degrees but now AB>BC.


72.

O

QtyA

QtyB

Area of semi-circular region
A

B

Area of triangular region ABC

C

O is the center of the circle and AO=OB
Solution;
Area of the semi-circle=1/2*3.14*r2=1.57r2, where r is the radius of the semicircle
Area of the triangular region ABC=1/2*2r*r=r2. That means Qty A is greater.

73. 0<a<b<c<1
QtyA

QtyB

ab/c

1

Solution;
ab/c ? 1
or, ab ? c
or, ab<c
Illustration for this:
Let a=0.5, b=0.6 and c=0.7
then ab=0.5*0.6=0.3 and since 0.3<0.7, ab<c. That means our answer is B.
35

74. 0<abc<1
QtyA

QtyB

ab/c

1

Solution;
Let a=1, b=1 and c=1/2 so that abc=1*1*1/2=1/2<1
then, QtyA=1*1/0.5=2>1

The answer is A

Again let a=1/2, b=1/2and c=1 so that abc=0.5*0.5*1=0.25 < 1
then QtyA=0.5*0.5/1=0.25 < 1

The answer is B


75.QtyA

QtyB

The number of primes

The number of primes

that are divisible by 9

that are divisible by 19

Solution;
QtyA=0 (prime numbers are divisible by 1 and by themselves only)
QtyB=1 (19 is a prime number and it is divisible by 19)

76.The first term of a sequence is 2 and each succeeding term is one more than three times the
preceding terms.
Qty A

QtyB

The term just greater than

200

150 in the sequence
Solution;
t1=2
t2=3*2+1=7
36

t3=3*7+1=22
t4=3*22+1=67
t5=3*67+1=202
That means 5th term is just greater than 150 and its value is 202 which is greater than 200.

77.

x<90
y

QtyB

y
20

QtyA

70

x

Solution;
By the property of triangles, we have
20+y=x
or, y=x-20
If x were exactly 90, y=90-20=70
But since x<90, y<70

78. x2=16 and y3=64
QtyA

QtyB

x

y

Solution;
x2=16

x=+-4

y3=64

y=4

If x=+4 then C would be the answer.
If x=-4 then B would be the answer.
37


79. Rectangular region R has an area 30.
QtyA

QtyB

The perimeter of R

25

Solution;
A=30

l*b=30

Let l=5 and b=6 (l*b=30), then P=2(l+b)=2(5+6)=22 which would lead us to answer B
Now, Let l=3 and b=10 (l*b=30), then P=2(l+b)=2(3+10)=26 which would lead us to answer A.

80. n is an odd positive integer.
700<n<800
QtyA

QtyB

The number of prime

The number of prime

factors of n

factors of 2n

Solution;
Here, since 2 is already a prime number, the number of prime factors of 2n is always 1 greater
than the number of prime factors of n
Ex let n=750. then
750=5*5*5*2*3
and 2n=2*750=2*5*5*5*2*3
5.

number of prime factors of 750=5
number of prime factors of 2*750=6, which is 1 more than

81. ab<0, bc>0
QtyA

QtyB

ac

0
38

Solution;
ab<0

When a=+ve, b=-ve and when a=-ve, b=+ve

bc>0

When b=+ve, c=+ve and when b=-ve, c=-ve

Therefore when a=+ve, b=-ve and when b=-ve, c=-ve
and when a=-ve, b=+ve and when b=+ve, c=+ve

ac=+ve*-ve=-ve
ac=-ve*+ve=-ve

In any case, the product ac will be negative which is less than 0.
So QtyB is greater.
m
82. O is the center of the above circle, with radius r (not shown).
0

The circle is tangent to both x and y axes.
QtyA

QtyB

r

m/2/2

Solution;
from the figure,
r2+r2=m2
or, 2r2=m2
or, r=m//2=QtyB

83. xy>0 ; x>y
QtyA

QtyB

x/y

y/x

Solution;
xy>0

when x=+ve, y=+ve and when x=-ve, y=-ve
39

x>y
Let x=2, y=1, then x/y=2/1=2 and y/x=1/2

A

let x=-1 and y=-2, then x/y=-1/-2=1/2 and y/x=-2/-1=2

B


84. Jim is 3 years older than Jonathan. Myra is 5 years older than Melisa. Jonathan is 2 years
older than Melisa.
QtyA

QtyB

Jim's age

Myra's age

Solution;
Jim=Jon+3
Myra=Melisa+5

Melisa=Myra-5

Jon=Melisa+2
therefore, Jim=Jon+3=Melisa+2+3=Melisa+5=Myra-5+5=Myra
Hence our answer is C

85. 750<n<1500
QtyA

QtyB

1500-n

n-750

Solution;
1500-n ? n-750
1500+750 ? n+n
2250 ? 2n
1125 ? n
Let n=850, then 1125>850

A

let n=1450, then 1125<1450

B
40


86. t is an integer
QtyA

QtyB

1/(1+2t)

1/(1+3t)

Solution;
1/(1+2t ) ? 1/(1+3t )
1+3t ? 1+2t
3t ? 2t
Let t=0, then 30=20

C

Let t=1, then 31 > 21

A


87. x>1
QtyA

QtyB

x/x+1

-x/1-x

Solution;
x/x+1 ? -x/1-x
1/x+1 ? -1/1-x
1/x+1 ? 1/x-1
x-1 ? x+1
-1 ? +1
-1<+1
Thinking logically,
QtyB=-x/1-x=x/x-1
41

Now when x/x+1 is compared with x/x-1, we can see numerator is same in both and
denominator is less in QtyB. Hence QtyB must be greater.

88. For each positive integer n, the nth term of the sequence S is 1+(-1)n.
QtyA

QtyB

The sum of the first 39 terms

39

of S
Solution;
S1=1+(-1)1=1-1=0
S2=1+(-1)2=1+1=2
S3=1+(-1)3=1-1=0
S4=1+(-1)4=1+1=2
and so on.
That means for all odd powers of -1, the term is 0
and for all even powers of -1, the term is 2.
From 1 to 39, there are 20 odd terms and 19 even terms.
The sum of 20 odd terms will be zero and the sum of remaining 19 terms will be 19*2=38
Therefore, QtyB is greater.

89. n is a positive integer.
QtyA

QtyB

The remainder when n is divided by 5

The remainder when n+10 is divided by 5

Solution;
Since 10 is exactly divisible by 5, the remainder when n is divided by 5 will be equal to the
remainder when n+10 is divided by 5.
For ex, let n=2.
Then remainder when 2 is divided by 5 is 2.
42

And 10+2=12. The remainder when 12 is divided by 5 is also 2.

90. k is an integer for which 1/21-k<1/8.
QtyA

QtyB

k

-2

Solution;
1/21-k<1/8
8<21-k
23<21-k
3<1-k
k<1-3
k<-2

91. In city X, the range of the daily low temperatures during June 2012 was 20 Fahrenheit and
the range of the daily low temperatures during July 2012 was 25 Fahrenheit.
QtyA

QtyB

The range of the daily

30 Fahrenheit

low temperatures in city X
for the two month period from
June 1,2012 through July 31,2012.
Solution;
for JUNE, highest daily low temperature(A)-lowest daily low temperature(B)=20 F
Let A=60 F and B=40 F so that A-B=20 F
for JULY, highest daily low temperature(C)-lowest daily low temperature(D)=25 F
43

Let C=50 F and D=25 F so that C-D=25 F
Here, in this case, Range of daily low temperatures for the two months=40-25=15
That means our answer would be B.
But, if A=100F, B=80 F and C=60 F , D=35F
then Range of daily low temperatures for the two months=80-35=45
That means our answer would be A. So our answer is D.

92. x>0 and x4=625
QtyA

QtyB

The greatest prime factor of 36x

x

Solution;
x4=625=(+-5)4

x=+-5

Since x>0, x=+5
Now QtyA= The greatest prime factor of 36x=36*5 is 5.
(since 36*5=2*2*3*3*5)
QtyB=x=5

93. n is an integer and n2<39
QtyA

QtyB

The greatest possible value of

12

n minus the least possible
value of n.
Solution;
n=0,+-1,+-2+-3,+-4,+-5,+-6
The greatest possible value of n-the least possible value of n=+6-(-6)=6+6=12
44


94. A rectangle is inscribed in a circle of radius r.
QtyA

QtyB

Half of the perimeter of

2r

the rectangle
Solution;
l+b > 2r (In any triangle, the sum of two sides is greater than the
third side)
b

r .

Perimeter of rectangle=2(l+b)
l

And 1/2*2(l+b)=l+b=half perimeter of rectangle


95. The figure below shows a regular octagon. A diagonal of an octagon is any line segment
connecting two non-adjacent vertices.

QtyA

QtyB

The number of diagonals of

The number of diagonals of

the octagon that are parallel

the octagon that are not parallel

to at least one side of the octagon

to any side of the octagon

Solution;

45

We can see,
For each of 4 sides of the octagon, there will be two lines that are parallel to sides of the
octagon.
So, QtyA=2*4=8
For each of 4 corners of the octagon, there will be three lines that are not parallel to any side
of the octagon.
So, QtyB=3*4=12

96. QtyA

QtyB

678987*12345

678986*12346

Solution;
Don't try to calculate the product on the calculator because the result will be too large for the
calculator to display.
So, 678987*12345 ? 678986*12346
or, 12345/12346 ? 678986/678987
or, 12345/12346 < 678986/678987
Illustration:
If we compare 7/8 with 8/9 then 8/9 will be greater. Although the differences between
numerator and denominator of the fractions are same, the fraction with larger values will be
larger. Likewise, if we compare 100/101 and 10000/10001, the later fraction will be greater.
Hence, QtyB is greater.

97. 13613-9216=x
QtyA

QtyB

The unit digit of x

0

Solution;
1361 will have unit digit 6
46

1362 will also have unit digit 6
1363 will also have unit digit 6
That means no matter what power of 136 be, the result will have the unit digit 6.
Now,
921=92 has unit digit 2
924=………….6 has unit digit 6
That means the pattern of unit digit of powers of 92 is
2,4,8,6,2,4,8,6………..
If we observe this pattern carefully, we'll find that 9216 will have unit digit 6 since it's in 4th
place in the pattern and the 16th digit in the pattern will also be 6.
Hence, QtyA=unit digit of x=unit digit of 13613-unit digit of 9216=6-6=0.
Thus, QtyA=QtyB and our answer is C.

98. x is a positive integer. When x is divided by 2,3,5 and 6, the remainder is 1.
QtyA

QtyB

x

30

Solution;
Here, clearly x=31,61,91 and so on because when these are divided by 2,3,5 and 6, the
remainder will be 1.
Does that mean we select choice A immediately?
Now there lies a TRAP within this question. You might not have thought of any other numbers
which when divided by 2,3,5 and 6 could yield a remainder of 1. But there is indeed a number.
and that's 1. When you divide this number by 2,3,5 and 6, the quotient will be 0 and the
remainder will be 1.
47

That means, we cannot determine the answer.
Hence, our answer is D.

99. A coin is flipped 5 times.
QtyA

QtyB

Probability of getting 2 heads

Probability of getting 3 heads

Solution;
P(HH)=P(TTT)=P(HHH)
That means probability of getting 2 heads is equal to probability of getting 3 heads.
OR we can solve this problem by using combination rule also.
number of ways of occurrence of 2 heads out of 5 flips=5C2
=5!/{2!(5-2)!}=5*4*3*2*1/2!3!
=10
total number of ways of outcomes=25=32
Therefore probability of getting 2 heads, P(HH)=10/32
number of ways of occurrence of 3 heads out of 5 flips=5C3=5!/2!3!=10
Therefore probability of getting 3 heads, P(HHH)=10/32
That means QtyA=QtyB
REMEMBER: Try to minimize calculations as much as possible during GRE exam since you'll
have lots of questions to answer and only little time. So try to use logic as much as possible
rather than wasting time on calculations. So the first method I showed is better as it consumes
no or little time.

100. In a certain college class, each female student has 3 pencils and each male student has 1
pencil and the average number of pencils per student in the class is 1.8. In the same class, each
female student has 1 pen and each male student has 2 pens.

48

QtyA

QtyB

The average number of pens

1.5

per student in the class
Solution;
average number of pencils per student=(1*M+3*F)/(M+F)=(M+3F)/(M+F)=1.8
or,M+3F=1.8M+1.8F
or,1.2F=0.8M
or,F/M=8/12=2/3
That means, there are 3 male students for every 2 female students.
Now, the average number of pens per student=(2*M+1*F)/(M+F)
=(2M+F)/(M+F)=(2*3+2)/(3+2)=8/5=1.6

101. QtyA

QtyB

220

185

Solution;
185=(2*3*3)5=(2*2*1.5*2*1.5)5=215*(2.25)5
Now, 220 ? 215(2.25)5
or, 220/215 ? (2.25)5
or, 25 ? (2.25)5
We know, 25 < (2.25)5

102. Triangle ABC is scalene.
QtyA

QtyB

Length of altitude to side AC

Length of side AB
49

Solution;
Let's draw figure for the problem.
B

A

M

C

Here, BM has to be compared with AB.
Clearly, AB>BM (In right angled triangle AMB, hypotenuse i.e. side AB is the largest side)

103. Three circles touch each other externally.

QtyA

QtyB

The perimeter of the

The circumference of the

triangle connecting the centers

largest circle

Solution;
The largest possible value of the triangle connecting the centers of circles=6R(when all triangles
have same radius R, where, R is the radius of the largest circle)
And at that time, the circumference of the largest circle would be 2*3.14*R>6R.
Hence QtyB is always greatest.

50

104. Alex has a six-sided die with faces numbered 1 through 6. He rolls the die twice.
QtyA

QtyB

The probability that both

The probability that neither

rolls are even

roll is a multiple of 3

Solution;
P(EE)=P(E)*P(E)=3/6*3/6=1/4
P(neither roll multiple of 3)=4/6*4/6=4/9

105. A polygon known as a "polyhedron" has 10,11 or 12 equal sides and angles.
QtyA

QtyB

The degree measure of any

155

interior angle of a polyhedron
Solution;
Each angle of a regular polygon=180(n-2)/n
for n=10,

=180(10-2)/10=144<155

for n=11,

=180(11-2)/11=147.2<155

for n=12,

=180(12-2)/12=150<155

Hence for all figures, QtyB is greater. Hence our answer is B.

106. rt<0<-r
QtyA

QtyB

t

0

Solution;
rt<0<-r
Since 0<-r, r should be -ve.
51

Since rt<0, r=-ve, t=+ve which is greater than 0.

107. One person is to be selected at random from a group of 25 people. The probability that
the selected person will be a male is 0.44, and the probability that the selected person will be
a male who was born before 1960 is 0.28.
QtyA

QtyB

The number of males in the

4

group who were born in 1960
or later
Solution;
P(Male)=0.44
Number of males=0.44*25=11
P(Male and born before 1960)=0.28
Number of males born before 1960=0.28*25=7
Hence, Number of males born in 1960 or later=11-7=4.
That means our answer is C.

108. T is a list of 100 different numbers that are greater than 0 and less than 50. The number x
is greater than 60 percent of the numbers in T and the number y is greater than 40 percent of
the numbers in T.
QtyA

QtyB

x-y

20

If we arrange the numbers in ascending order, then x will be in 61st position while y will be in
41st position.
therefore, x-y=61st number-41st number
Since we don’t know the values of 61st and 41st number, we don't know the value of x-y.
For ex, if x=40 and y=30, then x-y=10, in which case our answer would be B
52

and if x=45 and y=25, then x-y=20, in which case our answer would be C
That means our answer is D.

109. n5-n3<0
QtyA

QtyB

n

n2

Solution;
n5-n3<0
n5< n3
n2<1
Let n=0.5(since 0.52<1)
then A=0.5 and B=0.52=0.25

answer is A

To be sure, let's check with other number also. Let n=-0.5((since (-0.5)2<1)
then A=-0.5 and B=(-0.5)2=0.25

answer is B


110. Joe is twice as old as Ann. Eight years ago, Joe was four times older than Ann.
QtyA

QtyB

Joe's age 6 years ago

Ann's age 6 years from now

Solution;
J=2A (now)
8 years ago,
Joe=J-8
Ann=A-8
By given, J-8=4(A-8)
or, 2A-8=4A-32
53

or, 24=2A
or, A=12
Hence J=2*12=24
QtyA=J-6=24-6=18
QtyB=A+6=12+6=18

111. Julius has a piece of string that is 24 inches long. He is stretching the string around pegs on a
pegboard to create different shapes.
QtyA

QtyB

The area of a regular hexagon

The area of a square

created by the string

created by the string

Solution;
2

/(42-22)=/12

4

Area of the regular hexagon=2*(area of one of trapezoid)=2*6/12=12/12=41.56
(area of trapezoid=0.5*h(b1+b2)=0.5*/12(4+8)=6/12)
Area of the square=l2=62=36
Hence QtyA is greater.

112. Shaundra drove the same route to work each morning, Monday through Friday, in a
particular week. On Monday and Tuesday, she averaged 20 miles per hour and on her three
remaining work days, she averaged 30 miles per hour.

54

QtyA

QtyB

Shaundra's average speed for

26 miles per hour

all five morning commutes
Solution;
Let Shaundra's distance of the route be x.
Then Shaundra drove a total distance of 2x for two days at 20 miles/hr and a total distance of
3x for three days at 30 miles/hr.
Now,
time to drive 2x distance=2x/20 hrs (speed=distance/time)=x/10hrs
time to drive 3x distance=3x/30=x/10 hrs
Average speed for 5 days=(2x+3x)/(x/10+x/10)=5x/(2x/10)=5x*10/2x=25 miles/hr

113. In the figure below, the area of the smaller square region is half the area of the larger
square region.

1 inch

QtyA

QtyB

The positive difference in

/2/2

length between the diagonal of
the larger square and the
diagonal of the smaller square
Solution;
Area of small square=1/2*Area of large square=1/2*12=1/2
or, (length of small square)2=1/2 i.e., length of smaller square=1//2
Diagonal(larger)=/12+12=/2
55

Diagonal(smaller)=/(1//2)2+(1//2)2=/1/2+1/2=/1=1
/2-1 ? /2/2
/2-/2/2 ? 1
/2/2 ? 1
1//2 < 1

114. H is the mid-point of IG. J is the midpoint of IK. IG=GK=10.
I
H

J

G

K

QtyA

QtyB

IJ

5

Solution;
We could draw the figure as below.
I


5

IJ > 5 (In a right angled triangle, hypotenuse is the largest side

H

J

A

5
G

10

K

OR we could draw the figure as below.
5
H 60

I

J

IJ=5 (Triangle IJH being an equilateral triangle)

5
G 60

C
K

That means our answer is D.
56

115. In the figure below, ABCD is a parallelogram.
B 125

C
6

A

4

D

QtyA

QtyB

The area of ABCD

24

Solution;
Area of the parallelogram ABCD=base*height=4*height
B 125
6

C

height

A

D
base=4

From the figure,
height<6 (since in any right angled triangle, hypotenuse is the largest side)
That means area of ABCD=4*height<24 (since height<6)

116. x>y
QtyA

QtyB

|x+y|

|x-y|

solution;
Her, Let x=2 and y=1.
Then, A=|2+1|=|3|=3
B=|2-1|=|1|=1 which points to answer A.
57

And now let x=1 and y=0.
Then A=|1+0|=|1|=1
and B=|1-0|=|1|=1 which points to answer C.

117.

Three circles with their centers on line segment PQ are tangent at points P,R and Q, where
point R lies on line segment PQ.
QtyA

QtyB

The circumference of the largest

The sum of the circumferences of the two

circle.

smaller circles.

Solution;
From the figure, we can see
PR+RQ=PQ
or, Diameter of smaller circle1+Diameter of smaller circle2=Diameter of largest circle
or, d1+d2=D
Now Circumference of the largest circle=3.14D
and Circumference of circle C1=3.14d1
and Circumference of circle C2=3.14d2
So, C1+C2=3.14(d1+d2)=3.14D, which means our answer is C.
58

118.

Line k is parallel to line m.
QtyA

QtyB

x+y

w+z

Solution;
Since alternate angles made by parallel lines are equal, x=z and y=w
Hence adding the two, we get x+y=w+z

119. The length of each side of equilateral triangle T is 6 times the length of each side of
equilateral triangle X.
QtyA

QtyB

The ratio of the length of one

The ratio of the length of one

side of T to the length of another

side of X to the length of another

side of T.

side of X

Solution;
A=1 and B=1 (since the lengths of an equilateral triangle are equal)
So our answer is C.

120. A right circular cylinder with radius 2 inches has volume 15 cubic inches.
QtyA

QtyB

the height of the cylinder

2 inches

59

Solution;
Volume of a cylinder=3.14*r2*h
15=3.14*22*h
h=15/(4*3.14)=1.19 inches < 2 inches

121.

The frequency distributions shown above represent two groups of data. Each of the data
values is a multiple of 10.
QtyA

QtyB

The standard deviation of

The standard deviation of

distribution A

distribution B

Solution;
Average of A=(10*3+20*3+30*6+40*3+50*3)/18=(30+60+180+120+150)/18=30
Average of B=(10*5+20*3+30*2+40*3+50*5)/18=(50+60+60+120+250)/18=30
That means average of both are same.
now square of deviation from mean
for A={3(10-30)2+3(20-30)2+6(30-30)2+3(40-30)2+3(50-30)2}/18=1200+300+300+1200=3000
for B={5(10-30)2+3(20-30)2+2(30-30)2+3(40-30)2+5(50-30)2}/18=2000+300+300+2000=5000
That means standard deviation of B is greater.
60

122.

D

A

C
O

B
O is the center of the circle.
QtyA

QtyB

The average of BC and BD

AB

Solution;
AB is greater than BC or BD
(diameter is the longest chord of a circle)
i.e. AB>BC and AB>BD
So, AB+AB>BC+BD
or, AB>(BC+BD)/2
Hence Qty B is greater.

123.

Q
x

P y

x R
x>y

QtyA

QtyB

Length of arc PRQ

Length of arc QPR

Solution;
Since angle PQR=angle PRQ, arc QP=arc PR

(i)
61

(arcs opposite to equal circumscribed angles of a circle are equal)
Now arc PRQ= arc PR + arc RQ
arc QPR=arc QP + arc PR

(ii)
(iii)

From (i),(ii) and (iii),
arc PRQ ? arc QPR
arc PR + arc RQ ? arc QP + arc PR
arc RQ ? arc QP
arc RQ < arc QP
(arc opposite to greater angle is greater than the arc opposite to smaller angle in the case of
circumscribed angles of a circle and we are given x>y)
That means, our answer is B.

124. xy<0 and yz>0
QtyA

QtyB

xz

0

Solution;
xy<0
When x=-ve,y=+ve and when x=+ve,y=-ve
yz>0
When y=+ve,z=+ve and when y=-ve,z=-ve
That means,
When x=-ve, z=+ve

the product xz is -ve

when x=+ve, z=-ve

the product xz is -ve


62

125. One of the roots of the equation x2+kx-6=0 is 3, and k is a constant.
QtyA

QtyB

The value of k

-1

Solution;
Since one of the roots is 3, it must satisfy the given equation.
So, 32+k*3-6=0
or, 9+3k=6
or, 3k=-3
or, k=-1

126. x<y
QtyA
The average of x and y

QtyB
The average of x, y and y

Solution;
A=(x+y)/2
B=(x+y+y)/3=(x+2y)/3
Now, (x+y)/2 ? (x+2y)/3
or, 3x+3y ? 2x+4y
or, x ? y
x<y

63

127.

ABCD and EFGH are both squares whose sides are 12. All the circles are tangent to one
another and the sides of the square.
QtyA

QtyB

The area of the shaded region

The area of the shaded region

in figure 1

in figure 2

Solution;
radius of each circle in figure 1=12/4=3
So area of each circle in figure 1=3.14*32=3.14*9
area of 4 circles in figure 1,A1=4*3.14*9=36*3.14
area of square ABCD,A2=l2=122=144
Therefore, area of the shaded region in figure 1=A2-A1=144-(36*3.14)

radius of each circle in figure 2=12/6=2
So area of each circle in figure 2=3.14*22=3.14*4
area of 9 circles in figure 2,A3=9*3.14*4=36*3.14
area of square ABCD,A4=l2=122=144
Therefore, area of the shaded region in figure 1=A4-A3=144-(36*3.14)
64

128. The price of a large pizza is 30% more than the price of a small pizza.
QtyA

QtyB

The price of a large pizza

The price of a small pizza

when it is on sale for 30% off
Solution;
Let price of a small pizza be 100.
Then, price of a large pizza=130
A=(100-30)% of 130=0.7*130=91
and B=100
Hence, QtyB is greater.

129. QtyA

QtyB

The area of an equilateral

The area of an isosceles right

triangle whose sides are 6

triangle whose legs are 6

Solution;
Area=/3/4*(side)2

Area=1/2(base)*(height)=1/2*(base)2
6

6

6

A=(/3/4)*62=/3*9
B=1/2*62=18=3*9
We can see, B>A. So our answer is B.

65

130. A bag contains four slips of paper, two of which have the number 1 written on them and
two of which have the number -1 on them. Two of the slips are chosen at random.
QtyA

QtyB

The probability that the product

The probability that the product

of the two numbers chosen is -1

of the two numbers chosen is 1

Solution;
The product will be -1 if the two numbers chosen are (-1,1) or (1,-1).
So P(product=-1)=P(-1,1)+P(1,-1)=(2/4*2/3)+(2/4*2/3)=1/3+1/3=2/3
the product will be 1 if the two numbers chosen are (1,1) or (-1,-1).
So P(product=1)=P(1,1)+P(-1,-1)=(2/4*1/3+2/4*1/3)=1/6+1/6=2/6=1/3
Hence QtyA is greater.

131. The test scores for a class have a normal distribution, a mean of 50, and a standard
deviation of 4.
QtyA

QtyB

Percentage of scores at or above 58

Percentage of scores at or below 42

Solution;
34%

34%

14%

14%

42 46 50

54 58

mean=50, sd=4
So, mean + 1sd=50+4=54
and mean - 2sd=50-(2*4)=42
Hence from the normal distribution curve, we can see
Both 58 and 42 are at two standard deviation from mean.
So percentage of scores at or above 58=2%
66

and percentage of scores at or below 42=2%

132. In the xy-plane, the point (1,2) is on line j, and the point (2,1) is on line k. Each of the lines
has a positive slope.
QtyA

QtyB

The slope of line j

The slope of line k

Solution;
y

line j
(1,2)

line k

(2,1)

The lines could be parallel. That means their slopes
could be equal.
x

line j line k
(1,2)

(2,1)
Here angle made by line j with x-axis is greater than the angle
made by line k with x-axis. That means slope of line j could be
made greater than slope of line k.
Similarly, slope of line k could be made greater by making its angle with x-axis greater than
that made by line j.

133. For each positive integer n, the nth term of the sequence S is 1+(-1)n.
QtyA

QtyB

The sum of the first 39 terms of S

39

Solution;
67

Here, t1=1+(-1)1=0
t2=1+(-1)2=1+1=2
t3=1+(-1)3=0
That means terms with even powers of -1 have value 2 and those with odd powers of -1 have
value 0.
We know, from 1 to 40, there are 20 even numbers and 20 odd numbers.
So from 1 to 39, there are 19 even numbers and 20 odd numbers.
So our sum=2*19+0*20=38

134. For all positive numbers p, the operation ♥ is defined by p♥=p+1/p.
QtyA

QtyB

((2/7)♥)♥

3.5

Solution;
((2/7)♥)♥=(2/7+7/2) ♥={(2*2+7*7)/14}♥=(53/14)♥=53/14+14/15=3.78+0.93

135. x>0 and x4=625
QtyA

QtyB

The greatest prime factor of 36x

x

Solution;
x4=625

x=+-5

Since x>0, x=+5
i.e.,B=5
Now,
A=the greatest prime factor of 36*5(=3*3*2*2*5) is also 5. Hence our answer is C.
68

136. n is an integer, and n2<39.
QtyA

QtyB

The greatest possible value

12

of n minus the least possible
value of n
Solution;
Since n2<39, possible values of n are 0,+-1,+-2,+-3,+-4,+-5 and +-6.
Now A=+6-(-6)=6+6=12
B=12

137. The function f is defined by f(x)=5x+1 for all numbers x.
QtyA

QtyB

f(t+54)-f(t+50)

20

Solution;
A={5(t+54)+1}-{5(t+50)+1}=5t+271-(5t+251)=271-251=20

138. The radius of the circle is 0.1.

69

QtyA

QtyB

AB+BC+CD+DE+EA

1

Solution;
Diameter=0.2
Since diameter is the longest chord, each side of pentagon is less than 0.2.
that means, sum of lengths of 5 sides of pentagon<5*0.2 or 1.

139. QtyA

QtyB

The perimeter of the pentagon

The circumference of the circle

Solution;
Since each chord is shorter than the arc it forms, so the circumference of the circle will be
greater than the perimeter of the pentagon. Hence our answer is B.

140. The median income of a group of college C graduates six months after graduation was
$3,000 higher than the median income of a group of College D graduates six months after
graduation.

70

QtyA

QtyB

The 75th percentile of the

The 75th percentile of the

incomes of the group of

incomes of the group of

college C graduates six

college D graduates six

months after graduation.

months after graduation.

Solution;
Let there be 4 graduates each from college C and college D.
Let incomes of graduates from college C={1000,2000,8000,9000}
Here, median=(4+1)th/2=2.5th item=(2000+8000)/2=5000
75th percentile(C)=8000
Incomes of graduates from college D={1000,2000,2000,9000}
Here, median=(4+1)th/2=2.5th item=(2000+2000)/2=2000
Difference between the incomes=5000-2000=3000 which is as per given.
75th percentile(D)=2000
In this case, our answer would be choice A since 75th percentile of incomes of college C
graduates is greater.
Next,
Let there be 5 graduates from college D.
Let incomes of graduates from D={1000,1500,2000,9000,10000} so that median
income=2000
and incomes of graduates from C is same as above so that median income=5000
75th percentile(D)=3(5+1)th/4 item=4.5th item=(9000+10000)/2=9500
In this case, our answer would be choice B since 75th percentile of incomes of college D
graduates is greater.

71

141. In a quality-control test, 50 boxes-each containing 30 machine parts-were examined for
defective parts. The number of defective parts was recorded for each box, and the average
(arithmetic mean) of the 50 recorded numbers of defective parts per box was 1.12. Only one
error was made in recording the 50 numbers: "1" defective part in a certain box was incorrectly
recorded as "10".
QtyA

QtyB

The actual average number

0.94

of defective parts per box
Solution;
Defective/box(incorrect)=1.12
Total defective(incorrect)=1.12*50=56
That means 56 parts were defective while examining 50 boxes
Since "1" part was incorrectly recorded as "10",
the actual total of defective parts=56-10+1=47
and actual average number of defective parts per box=47/50=0.94

142. QtyA

QtyB

The sum of the odd integers

The sum of the even integers

from 1 to 199

from 2 to 198

Solution;
1+3+5+……….197+199 (100 terms) ? 2+4+6+……..+198 (99 terms)
199 ? (2-1)+(4-3)+(6-5)+…..+(198-197) (99 terms)
199 ? 1+1+1+….99 terms
199 ? 99
199 > 99
Hence choice A is greater.

72

143. Before Maria changed jobs, her salary was 24% more than Julio's salary. After Maria
changed jobs, her new salary was 24% less than her old salary.
QtyA

QtyB

Julio's salary

Maria's new salary

Solution;
Before
Julio's salary=100
Maria's salary=124
After
Maria's salary=0.86*124=106.64

144. p and q are different prime numbers. r is the least prime number greater than p, and s is
the least prime number greater than q.
QtyA

QtyB

r-p

s-q

Solution;
p=2, q=7
r=3, s=11
r-p=3-2=1
s-q=11-7=4 in which case our answer would be B.
p=7, q=2
r=11, s=3
r-p=11-7=4
s-q=3-2=1 in which case our answer would be A.

73

145. Team X scored p points more than team Y, and the two teams together scored a total of
10 points.
QtyA

QtyB

Twice the number of

10-p

points team Y scored
Solution;
Let team X scored X points and team Y scored Y points.
Then, X=Y+p
and X+Y=10
So, (Y+p)+Y=10
2Y=10-p

146. QtyA

QtyB

(x-1)(x)(x+1)

(x)(x)(x)

Solution;
If x=1, then QtyA=0*1*2=0 and QtyB=1*1*1=1
In this case, our answer would be B.
If x=0, then QtyA=-1*0*1=0 and QtyB=0*0*0=0
In this case, our answer would be C.

147. QtyA

QtyB

The number of distinct prime


factors of x

factors of 4x

Solution;
74

Here, the number of prime factors of 4x cannot be less than the number of prime factors of

x since 4x contains all the factors of x. So, the answer must be that either they are equal or
there is not enough information.
let x=2
then prime factors of 2=2 in which number of distinct prime factors is 1.
prime factors of 4*2=2*2*2 in which the number of distinct prime factors is 1.
Here number of distinct prime numbers is same in both cases. i.e., our answer would be C.
Next let x=21
then prime factors of 21=3*7 in which the number of distinct prime factors is 2.
prime factors of 4*21=2*2*3*7 in which the number of distinct prime factors is 3.
Here number of distinct prime numbers is greater in B i.e., our answer would be B.
Hence D is our answer.

148. x is an even integer.
QtyA

QtyB



factors of 4x

factors of x

Solution;
Let x=2,
then factors of 4*2=2*2*2
and factors of 2=2
That means our answer would be C.
Next let x=6
then factors of 4*6=2*2*2*3
factors of 6=2*3
Since 4 is an even number which has factors of 2 and 2 and which are not distinct, so the
number of distinct factors of 4x will be equal to the number of distinct prime factors of x.
75


149.

x>y
QtyA

QtyB

z

60

Solution;
A
x

k

x

z
B

y

C

z>k
2z+k=180
let k=30, then 2z=180-30=150
z=75 in which case our answer would be A.
let k =70, then 2z=180-70=110
z=55 which cannot be, since z>k and here, z<k
That means z cannot be less than 60. If z=60, then z=k which cannot be.
So our answer is A.

76

150. Three tennis balls of identical size are stacked one on the top of the other so that they fit
exactly inside a closed right cylindrical can, as shown.
QtyA

QtyB

The height of the stack of 3 balls

The circumference of one of the balls

Solution;
A=3d
B=3.14*d

151. S is the set of all fractions of the form n/n+1, where n is a positive integer less than 20.
QtyA

QtyB

The product of all the fractions

1/20

that are in S
S={1/2, 2/3, 3/4, 4/5,………….19/20}
A=1/2*2/3*3/4*4/5*……………*19/20=1/20=B

77

152. QtyA

QtyB

The length of minor arc of

The length of minor arc

WX of the circle.

YZ of the circle.

Solution;
Angles circumscribed on equal arcs (or same arcs) of a circle are equal.
So arc WX=arc YZ

153. A retail business has determined that its net income, in terms of x, the number of items
sold, is given by the expression x2+x-380.
QtyA

QtyB

The number of items that must

10

be sold for the net income to
be zero.
Solution;
Net Income= x2+x-380=0
x2+20x-19x-380=0
x(x+20)-19(x+20)=0
(x+20)(x-19)=0
x=-20 or 19
78

Since number of items sold cannot be negative, x=19. Hence our answer is A.

154. PQRS is a parallelogram
QtyA

QtyB

x

y

x-5=y+10 (opposite angles of a parallelogram are equal)
x-y=10+5=15
Hence A is greater.

155. S is the midpoint of segment PR.
QtyA

QtyB

The length of segment QT

The length of segment QR

Solution;
QP=/QS2+SP2=/QS2+SR2=QR
QP>QT (common sense)
So QR>QT
Hence B is greater.
79

156. Segments PA, PB and PC are the angle bisectors of triangle ABC.
QtyA

QtyB

x+y

57

Solution;
A+B+C=180
A/2+B/2+C/2=90
x+33+y=90
x+y=90-33=57

157. Q and T are the midpoints of opposite sides of square PRSU.
QtyA

QtyB

The area of region PQST

3/2

Solution;
Since PRSU is a square,

80

angle R=90
So in right angled triangle QRS,
QR=1 and RS=2 since 12+22=(/5)2
Area of triangle QRS=area of triangle TUP=1/2*1*2=1
Area of square=22=4
Area of region PQST=4-(1+1)=2
Hence A is greater.

158. In circles C1 and C2, the length of segment PR equals the length of segment QR.
QtyA

QtyB

The circumference of circle C1

The circumference of circle C2

Solution;
A=2*3.14*PR
B=2*3.14*QR
PR=QR
So A=B
hence our answer is C.

159. QtyA
x(x+1)(x+2)

QtyB
x*x*x

Solution;
81

For positive numbers, A will be greater.
For x=0, A=0 and B=0
So our answer is D.
160. The probability that both events E and F will occur is 0.25.
QtyA

QtyB

The probability that event E

0.24

will occur
P(E)*P(F)=0.25
P(E)=0.25/P(F)
When P(F) is maximum i.e. 1, P(E) will be minimum.
If P(F)=1, then, P(E)=0.25
When P(F) is minimum i.e. 0.25, P(E) will be maximum
If P(F)=0.25, then P(E)=1
P(E) lies from 0.25 to 1.
So A is greater.
161. The mean of five distinct positive integers is 10.
QtyA

QtyB

The largest possible value of

46

one of the integers
Solution;
a+b+c+d+e+f=10*5=50
1+2+3+4+f=50
f=50-10=40
A=40
So B is greater.
Note: If the positive integers given were not distinct, then
1+1+1+1+f=50
f=50-4=46 i.e. A and B would be equal.

82

162. M={4,6,3,7,x,x}
N={1,5,8,y}
x>y>0
QtyA

QtyB

The mean of set M

The mean of set N

Solution;
A=(4+6+3+7+x+x)/6=(20+2x)/6
B=(1+5+8+y)/4=(14+y)/4
(20+2x)/6 ? (14+y)/4
80+8x ? 84+6y
8x-6y ? 4
4x-3y ? 2
If x=2 and y=1, then 4*2-3*1=8-3=5>2 which points to answer A.
if x=0.5 and y=0.4, then 4*0.5-3*0.4=2-1.2=0.8<2 which points to answer B.

83

MULTIPLE CHOICE QESTIONS - Select one answer choice

1. If 15!/3m is an integer, what is the greatest possible value of m?
A. 4
B. 5
C. 6
D. 7
E. 8
Solution;
15!/3m=15*14*13*…..*1/3m=36*5*4*2*……../3m
Since,
15=3*5, 12=3*4, 9=3*3, 6=3*2, 3=3*1
We can see, for 15!/3m to be an integer, the maximum possible value of m could be 6.
Hence answer is C.

2. Which of the following numbers is farthest from the number 1 on the number line?
A. -10
B. -5
C. 0
D. 5
E. 10
Solution;
Clearly, the answer is either A or E.
distance between numbers 1 and -10=1-(-10)=1+10=11
distance between numbers 1 and 10=10-1=9

84

Hence answer is A.

3. A certain jar contains 60 jelly beans - 22 white, 18 green, 11 yellow, 5 red and 4 purple. If a
jelly bean is to be chosen at random, what is the probability that the jelly bean will neither be
red nor purple?
A. 0.09
B. 0.15
C. 0.54
D. 0.85
E. 0.91
Solution;
P(neither red nor purple)=P(White or Green or Yellow)
=22/60+18/60+11/60=51/60=0.85
Hence answer is D.

4. A certain store sells two types of pens: one type for $2 per pen and the other type for $3 per
pen. If a customer can spend up to $25 to buy pens at the store and there is no sales tax, what
is the greatest number of pens the customer can buy?
A. 9
B. 10
C. 11
D. 12
E. 20
Solution;
For buying greatest number of pens, the customer must buy maximum number of low cost
pens.
Hence she can buy 12 low cost pens for $24 and take $1 in return or she can buy 11 $2 pens for
$22 and a $3 pen for $3. In both cases she can buy a maximum of 12 pens.
85

Hence D is answer.

5. If y=3x and z=2y, what is x+y+z in terms of x?
A. 10x
B. 9x
C. 8x
D. 6x
E. 5x
Solution;
x+y+z=x+3x+2(3x)=10x
Hence A is the answer.

6. A certain shipping service charges an insurance fee of $0.75 when shipping any package
with contents worth $25.00 or less, and an insurance fee of $1.00 when shipping any package
with contents worth over $25.00. If Dan uses the shipping company to ship three packages
with contents worth $18.25, $25.00 and $127.50, what is the total insurance fee that the
company charges Dan to ship three packages?
A. $1.75
B. $2.25
C. $2.50
D. $2.75
E. $3.00
Solution;
0-$25

$0.75

$25.00+

$1.00

Total charge=($0.75*2)+$1.00=$2.50

86

7. If 55 percent of the people who purchase a certain product are female, what is the ratio of
females who purchase the product to the number of males who purchase the product?
A. 11 to 9
B. 10 to 9
C. 9 to 10
D. 9 to 11
E. 5 to 9
Solution;
Total=100
Number of females purchasing the product=55
Number of males purchasing the product=100-55=45
So, Females/Males=55/45=11/9

8.

The figure above shows the graph of the function f in the x-y plane. What is the value of
f(f(-1))?
A. -2
B. -1
C. 0
D. 1
E. 2

87

Solution;
From the graph,
f(-1)=2
therefore, f(f(-1))=f(2)=1
Hence D is our answer.

9. By weight, liquid A makes up 8 percent of solution R and 18 percent of solution S. If 3 grams
of solution R are mixed with 7 grams of solution S, the liquid A accounts for what percent of
the weight of the resulting solution?
A. 10%
B. 13%
C. 15%
D. 19%
E. 26%
Solution;
In 100 grams solution of R, there are 8 grams of liquid A.
In 1 grams solution of R, there are 8/100 grams of liquid A.
In 3 grams solution of R, there are 8*3/100 grams of liquid A.
Similarly,
In 100 grams solution of S, there are 18 grams of liquid A.
In 1 gram solution of S, there are 18/100 grams of liquid A.
In 7 grams solution of S, there are 18*7/100 grams of liquid A.
That means, when the two solutions are mixed, the resulting 3+7=10 grams of solution will have
(8*3/100)+(18*7/100)=0.24+1.26=1.50 grams of liquid A.
Hence percentage of liquid A=1.50/10*100=15%

88

10. Of the 700 members of a certain organization, 120 are lawyers. Two members of the
organization will be selected at random. Which of the following is closest to the probability
that NEITHER of the members selected will be a lawyer?
A. 0.5
B. 0.6
C. 0.7
D. 0.8
E. 0.9
Solution;
P(neither lawyer)=P(first not lawyer)*P(second not lawyer)
=580/700*579/699=0.686(=0.7)
Another method!!!
P(neither lawyer)=580C2/700C2=0.68 (=ways of selection of both non-lawyers/total selection)

11. A manager is forming a 6-person team to work on a certain project. From the 11 candidates
available for the team, the manager has already chosen 3 to be on the team. In selecting the
other 3 team members, how many different combinations of 3 of the remaining candidates
does the manager have to choose from?
A. 6
B. 24
C. 56
D. 120
E. 462
Solution;
From the 11 candidates, 3 have already been chosen. So for forming a 6 person team, the
manager has to select (6-3)=3 candidates out of (11-3)=8 candidates.
So, 8C3=8!/3!5!=56. That means our answer is C.
89

A.

0

B.

0

C.

0

D.

0

E.

0

12. Which of the following could be the graph of all values of x that satisfy the inequality
2-5x<=-(6x-5)/3?
Solution;
2-5x<=-(6x-5)/3
6-15x<=-6x+5
6-5<=15x-6x
1<=9x
1/9<=x

13. If 1+x+x2+x3=60, then the average of x,x2,x3 and x4 is equal to which of the following?
A. 12x
B. 15x
C. 20x
D. 30x
E. 60x
Solution;
(x+x2+x3 +x4)/4=x(1+x+x2+x3)/4=x*60/4=15x

90

14. The sequence of numbers a1, a2, a3,…..an,…. is defined by an=(1/n)-(1/n+2) for each integer
n>=1. What is the sum of the first 20 terms of this sequence?
A. (1+1/2)-1/20
B. (1+1/2)-(1/21+1/22)
C. 1-(1/20+1/23)
D. 1-1/22
E. 1/20-1/22
Solution;
a1=1/1-1/3
a2=1/2-1/4
a3=1/3-1/5
………………..
a1+a2+a3+……….a20=(1+1/2+1/3+…….+1/20)-(1/3+1/4+…….1/22)=(1+1/2)-(1/21+1/22)

15. What is the least positive integer that is NOT a factor of 25! and is NOT a prime number?
A. 26
B. 28
C. 36
D. 56
E. 58
Solution;
25!=25*24*23*22*…………………1
26=13*2
28=14*2
36=12*3
91

56=7*8
58=29*2
That means 58 is not a factor of 25! since none of the factors of 25! is divisible by 29.
Hence our answer is E.

16. If 0<a<1<b, which of the following is true about the reciprocals of a and b?
A. 1<1/a<1/b
B. 1/a<1<1/b
C. 1/a<1/b<1
D. 1/b<1<1/a
E. 1/b<1/a<1
Solution;
a<1<b
Here a and b are both positive. Taking reciprocals,
1/a>1>1/b
Ex,0.1<1<2
taking reciprocals, 1/0.1>1/1>1/2 i.e., 10>1>0.5

17. Of the 750 participants in a professional meeting, 450 are female and 1/2 of the female and
1/4 of the male participants are less than 30 years old. If one of the participants will be
randomly selected to receive a prize, what is the probability that the person selected will be
less than 30 years old?
A. 1/8
B. 1/3
C. 3/8
D. 2/5
92

E. 3/4
Solution;
Female=450, Male=300
1/2 of female=1/2*450=225 (<30 years old)
1/4 of male=1/4*300=75 (<30 years old)
P(less than 30)=(225+75)/750=300/750=2/5

18. From the even numbers between 1 and 9, two different even numbers are to be chosen at
random. What is the probability that their sum will be 8?
A. 1/6
B. 3/16
C. 1/4
D. 1/3
E. 1/2
Solution;
Even numbers between 1 and 9 are 2,4,6 and 8.
There is just one combination for the sum to be 8 i.e. combination of 2 and 6.
Total number of combinations=4C2=6
Hence P(sum=8)=1/6
Hence A is our answer.

19. If x and y are the tens and the unit digit respectively of a product 725,278*67,066, what is
the value of x+y?
A. 12
B. 10
C. 8
93

D. 6
E. 4
Solution;
725278
*67066
.……….68
...……68
+
48
Now 4+8=12

20. What is the least possible value of x+y/xy if 2<=x<y<=11 and x and y are integers?
A. 22/121
B. 5/6
C. 21/110
D.13/22
E. 1
Solution;
x+y/xy=1/x+1/y
value of 1/x+1/y is least when both x and y are greatest.
From the given inequality,
the greatest possible value of x is 10 and that of y is 11.
So least possible value of x+y/xy is (10+11)/10*11=21/110.

94

21. If 998*1002>106-x, x could be
A. 1
B. 2
C. 3
D. 4
E. 5
Solution;
998*1002=(1000-2)(1000+2)=10002-4=(103)2-4=106-4
So, 106-4>106-x
i.e., -4>-x
i.e., x>4

22. To reproduce an old photograph, a photographer charges x dollars to make a negative,
3x/5 dollars for each of the first 10 prints and x/5 dollars for each print in access of 10 prints. If
$45 is the total charge to make a negative and 20 prints from an old photograph, what is the
value of x?
Solution;
A. 3
B. 3.5
C. 4
D. 4.5
E. 5
Solution;
x+3x/5*10+x/5*10=45
x+6x+2x=45
x=5. So our answer is E.
95

23. A certain cake recipe states that the cake should be baked in a pan 8 inches in diameter. If
Jules wants to use the recipe to make a cake of the same depth but 12 inches in diameter, by
what factor should he multiply the recipe ingredients?
A. 5/2
B. 9/4
C. 3/2
D. 13/9
E. 4/3
Solution;
Volume of cake1=3.14*d12*h1/4
Volume of cake2=3.14*d22*h2/4
Now required factor of multiplication=V2/V1=(d2/d1)2=(12/8)2=(3/2)2=9/4

24. A reading list for humanities course consists of 10 books, of which 4 are biographies and the
rest are novels. Each student is required to read a selection of 4 books from the list, including 2
or more biographies. How many selections of the 4 books satisfy the requirements?
A. 90
B. 115
C. 130
D. 144
E. 195
Solution;
Biographies

Novels

Combinations

2

2

4C2*6C2=90

3

1

4C3*6C1=24

4

0

4C4=1

96

Total ways=90+24+1=115

25. If the probability of choosing 2 red marbles without replacement from a bag of only red
and blue marbles is 3/55 and there are 3 red marbles in the bag, what is the total number of
marbles in the bag?
A. 8
B. 11
C. 55
D. 110
E. 165
Solution;
Let n be the number of blue marbles.
Then P(RR)=P(R)*P(R)=(3/3+n)*(2/2+n)=6/(3+n)(2+n)
or, 3/55=6/(3+n)(2+n)
or, 1/55=2/(3+n)(2+n)
or, 6+3n+2n+n2=110
or, n2+5n-104=0
or, n(n+13)-8(n+13)=0
or, (n+13)(n-8)=0
Since n cannot be negative, n=8.
Hence total number of marbles in the bag=8+3=11. So answer is B.

26. If 1/(211)(517) is expressed as a terminating decimal, how many non-zero digits will the
decimal have?
A. 1
B. 2
97

C. 4
D. 6
E. 11
Solution;
1/211517=1/211*511*56=1/1011*56=(1/5)6*10-11=0.26*10-11=(2*10-1)6*10-11=64*10-17
Hence there will be 2 non-zero digits in the decimal value.
Our answer is B.

27.

Distance from Centerville(miles)

Freight train

-10t+115

Passenger train

-20t+150

The expressions in the table above give the distance from Centerville to each of two trains t
hours after 12:00 noon. At what time after 12:00 noon will the trains be equivalent from
Centerville?
A. 1:30
B. 3:30
C: 5:10
D. 8:50
E. 11:30
Solution;
-10t+115=-20t+150
10t=35
t=3.5 hours.
Hence, they will be equivalent from Centerville at 3:30.

98

28. In state X, all vehicle license plates have 2 letters from the 26 letters of the alphabet
followed by 3 one-digit numbers. How many different license plates can State X have if
repetition of letters and numbers is allowed?
A. 23,400
B. 60,840
C. 67,600
D. 608,400
E. 676,000
Solution;
Total combinations=26*26*10*10*10=676000

29. A developer has land that has x feet of lake frontage. The land is to be subdivided into lots,
each of which is to have either 80 feet or 100 feet of lake frontage. If 1/9 of the lots are to have
80 feet of frontage each and the remaining 40 lots are to have 100 feet of frontage each,
what is the value of x?
A. 400
B. 3,200
C. 3,700
D. 4,400
E. 4,760
Solution;
1/9 of lots=40/8=5lots
8/9 of lots=40lots

80 feet of frontage each
100 feet of frontage each

x=5*80+40*100=400+4000=4,400

99

30. Which of the following is not a factor of 1030?
A. 250
B. 125
C. 32
D. 16
E. 6
Solution;
1030=(2*5)30=230*530
250=53*2
125=53
32=25
16=24
6=2*3
Our answer is E since 230*530 does not have a factor of 3.

31.

R

y
Q

T
x

S

In the figure, QRS is an equilateral triangle and QTS is an isosceles triangle. If x=47, what is the
value of y?
A. 13
B. 23
C. 30
D. 47
E. 53
100

Solution;

=60-47=13
y
47 60

32. If k is an integer and 0.0010101*10k>1000, what is the least possible value of k?
A. 2
B. 3
C. 4
D. 5
E. 6
Solution;
0.0010101*10k>1000
If k=4, 0.0010101*10k=0.0010101*104=10.101<1000
Now k=5 too won't be working because it would become 101.01
So if k=6, 0.0010101*10k=0.0010101*106=1010.1
Any value of k greater than 6 would also satisfy the given equation. But 6 is the least integer.
hence our answer is E.

33.If n is a positive integer and k+2=3n, which of the following could NOT be a value of k?
A. 1
B. 4
C. 7
D. 25
101

E. 79
Solution;
k=3n-2
for n=1, k=31-2=1
for n=2, k=32-2=7
That means, 4 could not be the value of k since there are no integers in between 1 and 2.

34.

-10 A

B

-2 C -1

0

D

2

E

10

Five points A,B,C,D and E are shown on a number line. What is the probability of all numbers
being negative if three numbers are selected at random?
A. 1/10
B. 3/5
C. 2/5
D. 4/5
E. 3/10
Solution;
A,B and C are 3 negative numbers and D,E are 2 positive numbers.
P(all negative)=3C3/5C3=number of ways of selection of 3 negative numbers/total ways
=1/10

35. Square T is formed by joining the mid-points of sides of square S. The perimeter of square S
is 40. What is the area of square T?
A. 45
B. 48
C. 49
102

D. 50
E. 52
Solution;

5
5

Square S
Square T

/52+52=/50

10

Perimeter of S=40
4l=40
l=10
Area=(/50)2=50

36. How many different positive integers are there in which the tens digit is greater than 6 and
the units digit is less than 4?
A. 7
B. 9
C. 10
D. 12
E. 24
Solution;
tens digit needs to be greater than 6 i.e. it can be 7,8 or 9 so it has 3 choices.
units digit needs to be less than 4 i.e. it can be 0,1,2 or 3 so it has 4 choices.
Hence total number of ways=3*4=12. Our answer is D.

103

37. If in 1998 there were 10,000 bias-motivated offenses based on ethnicity, how many more
offenses were based on religion than on sexual orientation?
A. 4
B. 40
C. 400
D.4000
E. 40,000
Solution;
10.0%

10000

1%

1000

16%

16000(religion)

15.6%

15600(sexual orientation)

16000-15600=400

Answer C.

104

38. A rectangular game board is composed of identical squares arranged in a rectangular
array of r rows and r+1 columns. The r rows are numbered from 1 through r, and the r+1
columns are numbered from 1 through r+1. If r>10, which of the following represents the number
of squares on the board that are neither in the 4th row nor in the 7th column?
A. r2-r
B. r2-1
C. r2
D. r2+1
E. r2+r
Solution;

r+1 columns

r
rows

Solution;
Here, r(r+1)=r2+r is the area of the board.
Our answer must be less than this.
So we can eliminate choice E.
Sum of squares of 4th row and 7th column=(r+1)+r-1=2r
So required number of squares=r2+r-2r=r2-r

105

39. S is a set containing 9 different numbers. T is a set containing 8 different numbers, all of
which are members of S. Which of the following statements CANNOT be true?
A. The mean of S is equal to the mean of T.
B. The median of S is equal to the median of T.
C. The range of S is equal to the range of T.
D. The mean of S is greater than the mean of T.
E. The range of S is less than the range of T.
Solution;
S={-4,-3,-2,-1,-0.5,0,1,2,3} and T={-4,-3,-2,-1,0,1,2,3}
mean of S=-4.5/9=-1/2
mean of T=-4/8=-1/2
Choice A can be true.
Next, let S={-4,-3,-2,-1,0,1,2,3,4} and T={-4,-3,-2,-1,0,1,2,3}
mean of S=0/9=0
mean of T=-4/9
Choice D can also be true.
If S={-4,-3,-2,-1,0,1,2,3,4} and T={-4,-3,-2,-1,1,2,3,4}
then range of S=4-(-4)=8
and range of T=4-(-4)=8
Choice C can be true as well.
Next, let S={-4,-3,-2,-1,1,2,3,4} and T={-4,-3,-2,-1,1,2,3,4}
then range of S=5-(-4)=8
and range of T=5-(-4)=9
Choice E can also be true.

106

40. What is the greatest positive integer n such that 2n is a factor of 1210?
A. 10
B. 12
C. 16
D. 20
E. 60
Solution;
1210=(2*2*3)10=210*210*310=220*310
If n=20, then 1210/220=220*310/220=310
But if n=21, then 1210/221=220*310/221=310/2
Hence the greatest possible value of n is 20. Hence our answer is D.

41. If x is an integer and y=9x+13, what is the greatest value of x for which y is less than 100?
A. 12
B. 11
C. 10
D. 9
E. 8
Solution;
Try with x=10
then y=9*10+13=103
That means we can eliminate choices A,B and C because, if x=10 gives y>100, then x=11 or 12 will
also give y>100.
Now try x=9, then y=9*9+13=94(<100)
x=8 will also give y<100. But since we need the greatest value of x for which y<100, our answer
is D.

107

42. Each month, a certain manufacturing company's total expenses are equal to a fixed
monthly expense plus a variable expense that is directly proportional to the number of units
produced by the company during that month. If the company's total expenses for a month in
which it produces 20,000 units are $570,000 and the total expenses for a month in which it
produces 25,000 units are $705,000, what is the company's fixed monthly expense?
A. $27,000
B. $30,000
C. $67,500
D. $109,800
E. $135,000
Solution;
total expenses=fixed expense+variable expense
$570,000=fixed expense+k*20,000
$705,000=fixed expense+k*25,000
(variable expense=k*number of units produced)
Solving above equations, we get
135,000=5000k
k=27
So 570,000=fixed expense+(27*20,000)
fixed expense=30,000

43. A team has a record of 12 wins and 13 losses for the season. Three games remain. If the
probability of winning each remaining game is 1/2 and there are no draws, what is the
probability that the team will finish the season with a winning record?
A. 1/5
B. 1/4
C. 3/8
108

D. 1/2
E. 5/8
Solution;
The possible results of remaining 3 games are
WWW (12+3=15 wins and 13 losses

win)

WWL (12+2=14 wins and 13+1=14 losses, no draws)
WLW
LWW
LWL (12+1=13 wins and 13+2=15 losses

loss)

LLW (loss)
WLL (loss)
LLL (12 wins and 13+3=16 losses

loss)

1 winning possibility out of 5 possibilities
so probability of winning the season=1/5, which is answer A.

44. How many three digit integers are odd and do not contain the digit 5?
A. 360
B. 320
C. 288
D. 256
E. 252
Solution;
8*9*4=288
can be from 1 to 9 except 5
can be from 0 to 9 except 5
can be 1,3,5 or 7

8 choices
9 choices

4 choices
109


45. When the fraction 1/37 is converted to a decimal, what is the 24th digit to the right of the
decimal place?
A. 0
B. 2
C. 3
D. 5
E. 7
Solution;
1/37=0.027027027…..
The pattern is repeating after the 3rd digit of the decimal i.e. 7. So the 24th digit after the
decimal is also 7 since 24 is divisible by 3.

46. What is the ratio of surface area of a cube to the surface area of a rectangular solid
identical to the cube in all ways except that its length has been doubled?
A. 1:4
B. 3:8
C. 1:2
D. 3:5
E. 2:1
Solution;
Let length of each edge of the cube be 1. Then, Surface area of cube, S1=6l2=6
for the rectangular solid,
length=2, breadth=1 and height=1
Surface area of rectangular solid, S2=2(lb+bh+lh)=2(2*1+1*1+2*1)=10
110

S1/S2=6/10=3/5 which is choice D.

47. 12,2732+12,2742=
A. 299,235,509
B. 300,568,327
C. 301,277,605
D. 302,435,782
E. 303,053,291
Solution;
12,2732+12,2742=…………..9+…………6=………………5
The result must have unit digit 5. Hence our answer is C.

48. The quantity 22335566 will end in how many zeroes?
A. 0
B. 2
C. 3
D. 5
E. 6
Solution;
22335566=223355(2*3)6=2233552636=283955=23253955=2310539
Hence there will be 5 zeroes. That means our answer is D.

49. Employee at a company is paid fixed salary of $90,000 annually plus 100 shares of that
company. After 9 months, he leaves the job and receives $65,000 and 80 shares. What is the
price of the share?
A. $450
B. $500
111

C. $700
D. $750
E. Cannot be determined from the given information.
Solution;
In 12 months, he gets $90,000 plus 100 shares.
In 1 month, he gets $90,000/12=$7,500 plus 100/12 shares.
In 9 months, he gets $7,500*9=$67,500 plus 100/12*9=75 shares.
Let each share worth $x.
Then 67,500+75x=65,000+80x
x=$500

50. Paint needs to be thinned to a ratio of 2 parts paint to 1.5 parts water. The painter has by
mistake added water so that he has 6 liters of paint which is half water and half paint. What
must he add to make the proportions of the mixture correct?
A. 1 liter paint
B. 1 liter water
C. 1/2 liter water and 1 liter paint
D. 1/2 liter paint and 1 liter water
E. 1/2 liter paint
Solution;
Paint:water=2/1.5=4/3(required)
In 6 liters of paint mixture, there are 1/2*6=3 liters of paint and 1/2*6=3 liters of water.
Now to make the ratio 4:3, he must add 1 liter paint to the mixture.
i.e., paint:water=(3+1)/3=4/3

112

51. Kelly took three days to travel from City A to City B by automobile. On the first day, Kelly
traveled 2/5 of the distance from City A to City B and on the second day, she traveled 2/3 of
the remaining distance. Which of the following is equivalent to the fraction of the distance
from City A to City B that Kelly traveled on the third day?
A. 1-2/5-2/3
B. 1-2/5-2/3(2/5)
C. 1-2/5-2/5(1-2/3)
D. 1-2/5-2/3(1-2/5)
E. 1-2/5-2/3(1-2/5-2/3)
Solution;
1st day=2/5
remain=1-2/5
2nd day=2/3*(1-2/5)
3rd day=1-2/5-2/3(1-2/5)
Hence our answer is choice D.

52. If x and y are integers and x=50y+69, which of the following must be odd?
A. xy
B. x+y
C. x+2y
D. 3x-1
E. 3x+1
Solution;
50y is always even.
69 is odd.
So x must be odd. (even+odd=odd) and y can be either odd or even.
xy=odd*(odd or even)=odd or even
113

x+y=odd+(odd or even)=even or odd
x+2y=odd+even=odd
3x-1=3*odd-1=odd-1=even
3x+1=3*odd+1=odd+1=even
Hence our answer is choice C.

53. In the first half of the last year, a team won 60 percent of the games it played. In the
second half of the last year, the team played 20 games, winning 3 of them. If the team won 50
percent of the games it played last year, what was the total number of games the team
played last year?
A. 60
B. 70
C. 80
D. 90
E. 100
Solution;
Let number of games played in 1st half be x.
1st half=60% of x won=0.6x
2nd half=3 won out of 20 games
Number of games won in 1st half and 2nd half=50% of total number of games played
0.6x+3=0.5(x+20)
0.1x=7
x=70
Total number of games played=70+20=90

114

54. In the sequence a1,a2,a3,……a100, the kth term is defined by ak=1/k-1/k+1 for all integers k
from 1 through 100. What is the sum of the 100 terms of this sequence?
A. 1/10,100
B. 1/101
C. 1/100
D. 100/101
E. 1
Solution;
a1=1-1/2
a2=1/2-1/3
a3=1/3-1/4
a4=1/4-1/5
…………………
…………………
a100=1/100-1/101
a1+a2+a3+a4+……..+a100=1-1/101=100/101

55. Eight hundred insects were weighed, and the resulting measurements in milligrams, are
summarized in the box plot below.

100 105 110 114 120

126 130

140

146

If the 80th percentile of the measurements is 130 mgs, about how many measurements are
between 126 mgs and 130 mgs?
A. 30
B. 32
115

C. 35
D. 40
E. 42
Solution;
75th percentile=126
80th percentile=130
So between 126 mgs and 130 mgs, there are 5% of total data.
5% of 800=5/100*800=40

56. There is a leak in the bottom of tank. This leak can empty a full tank in 8 hours. When the
tank is full, a tap is opened into the tank which intakes water at rate of 6 gallons per hour
and the tank is now emptied in 12 hours. What is the capacity of tank?
A. 28 gallons
B. 36 gallons
C. 144 gallons
D. 150 gallons
E. cannot be determined from the information given.
Solution;
Let capacity of tank be x.
In 1 hour, the leak empties x/8 gallons water.
In 1 hour, the tank intakes 6 gallons water.
In 1 hour, x/8-6 gallons water is emptied.
In 12 hours, 12(x/8-6) gallons water is emptied.
We are given, x gallon tank(full tank) is emptied in 12 hours.
So, 12(x/8-6)=x
1.5x-72=x
116

0.5x=72
x=144
Hence C is our answer.

57. Set S includes elements {8,2,11,x,3,y} and has an average of 7 and a median of 5.5. If x<y,
then which of the following is the maximum possible value of x?
A. 0
B. 1
C. 2
D. 3
E. 4
Solution;
(8+2+11+x+3+y)/6=7
24+x+y=42
x+y=18
Median=5.5
if x=1, then y=17
1,2,3,8,11,17
median=(3+8)/2=5.5
if x=2, then y=16
2,2,3,8,11,16
median=(3+8)/2=5.5
if x=3, then y=15
2,3,3,8,11,15
median=(3+8)/2=5.5
if x=4, then y=14
117

2,3,4,8,11,14
median=(4+8)/2=6
Hence maximum possible value of x is 3. So our answer is D.

58. If x+y=10, and xy=20, what is the value of 1/x+1/y?
A. 1/20
B. 1/15
C. 1/10
D. 1/2
E. 2
Solution;
1/x+1/y=(y+x)/xy=10/20=1/2

59. In a normal distribution, 68 percent of scores lie within one standard deviation of the
mean. If the SAT scores of all the high school juniors in the Center City followed a normal
distribution with a mean of 500 and a standard deviation of 100, and if 10,200 students
scored between 400 and 500, approximately how many students scored above 600?
A. 2,400
B. 4,800
C. 5,100
D. 7,200
E. 9,600

118

Solution;

34% 34%
2%14%
400 500

14% 2%
600

400-500

10,200

34%

10,200

1%

300

16%

4,800


60. John bought a $100 DVD player on sale at 8% off. How much did he pay including 8% sales
tax?
A. $84.64
B. $92.00
C. $96.48
D. $99.36
E. $100.00
Solution;
SP=92% of $100=$92 (excluding tax)
SP=108% of $92=$99.36

119

61. For how many positive integers m<=100 is (m-5)(m-45) positive?
A. 45
B. 50
C. 58
D. 59
E. 60
Solution;
(m-5)(m-45)>0
Two possibilities
m-5>0 and m-45>0
So m>5 and m>45

m>45 and m<=100

So number of possible values of m in this case=100-45=55
m-5<0 and m-45<0
So m<5 and m<45

m<5

So number of possible values of m in this case=4
Hence our answer is 55+4=59, that means D.

62. If the average (arithmetic mean) of 3a and 4b is less than 50, and a is twice b, what is the
largest possible integer value of a?
A. 9
B. 10
C. 11
D. 19
E. 20
Solution;
(3a+4b)/2<50
120

3a+4b<100
a=2b
2a=4b
So 3a+2a<100
5a<100
a<20
Since a needs to be an integer, a=19
So our answer is D.

63. x,y,a and b are positive integers. When x is divided by y, the remainder is 6. When a is
divided by b, the remainder is 9. Which of the following is NOT a possible value of y+b?
A. 24
B. 21
C. 20
D. 17
E. 15
Solution;
x divided by y gives remainder 6 means y>6
Ex, 12)18(1
-12
6
a divided by b gives remainder 9 means b>9
Ex, 10)19(1
-10
9
If y>6 and b>9, then y+b>15
121

Thus, 15 cannot be the sum of y and b.
So our answer is E.

64. A pair of dice is tossed twice. What is the probability that the first toss gives a total of either
7 or 11 and the second toss gives a total of 7 ?
A. 1/27
B. 1/18
C. 1/9
D. 1/6
E. 7/18
Solution;
1

3

4

5

6

(1,2)

(1,3)

(1,4)

(1,5)

(1,6)

2 (2,1)

(2,2)

(2,3)

(2,4)

(2,5)

(2,6)

3

(3,1)

(3,2)

(3,3)

(3,4)

(3,5)

(3,6)

4

(4,1)

(4,2)

(4,3)

(4,4)

(4,5)

(4,6)

5

(5,1)

(5,2)

(5,3)

(5,4)

(5,5)

(5,6)

6 (6,1)

(6,2)

(6,3)

(6,4)

(6,5)

(6,6)

1

2

(1,1)

There are a total of 62=36 possible outcomes in the first toss of a pair of coins as shown above.
P(total=7 or 11 in the first toss)=8/36
Again, there are same number of 36 possible outcomes in the second toss.
P(total=7 in the second toss)=6/36
P(total=7 or 11 in the first toss) and P(total=7 in the second toss)=8/36*6/36=2/9*1/6=1/27

122

65. Three dice are rolled simultaneously. What is the probability that exactly two of the dice
will come up as the same number?
A. 5/12
B. 11/24
C. 25/54
D. 13/27
E. 1/2
Solution;
If we roll three dice simultaneously, then there will be a total of 63=216 possible outcomes.
Let A,B and C be 3 dice.
If A and B come up same number, then there are 5 choices left with C.
Ex If A and B are both 1 or treating A and B as one unit, then C will have 5 choices left for
exactly two of the dice to come up as same number. (that means C cannot be 1)
So there are 6*5=30 ways that A and B come up as same number.
Likewise there are 30 ways that (B and C) and (C and A) come up as same number.
So there are 30*3=90 favorable outcomes.
So probability=90/216=45/108=15/36=5/12

66. Which of the following could be the median for a set of integers {97, 98, 56, x, 86}, given
that 20 < x < 80?
A. 71
B. 86
C. 91.5
D. 97
E. 397.5
Solution;
123

x,56,86,97,98
56,x,86,97,98

67. In the coordinate plane, rectangle WXYZ has vertices at (–2, –1), (–2, y), (4, y), and (4, –1).
If the area of WXYZ is 18, what is the length of its diagonal?
A. 3/2
B. 3/3
C. 3/5
D. 3/6
E. 3/7
Solution;
(-2,y)

(-2,-1)

(4,y)

(4,-1)

Area=18=6*(y+1)
y+1=3
y=2
Diagonal length=/(62+32)=/(36+9)=/45=/(5*9)=3/5
Hence answer is C.

68. In the repeating decimal 0.0653906539..., the 34th digit to the right of the decimal point is
A. 9
B. 6
124

C. 5
D. 3
E. 0
Solution;
The pattern is repeating after 5th digit. The 5th digit after the decimal is 9. So multiples of 5
will have the digit 9. Hence 35th digit is also 9 and a digit before it will be 3.
69. The numbers in data set S have a standard deviation of 5. If a new data set is formed by
adding 3 to each number in S, what is the standard deviation of the numbers in the new data
set?
A. 2
B. 3
C. 5
D. 8
E. 15
Solution;
Let S={0,5,10}
mean=15/3=5
sd= (0-5)2+(5-5)2+(10-5)2 = 50/3
3
S1={0+5,5+5,10+5}={5,10,15}
mean=30/3=10
sd= (5-10)2+(15-10)2+(15-15)2 = 50/3
3
That means, standard deviation of a set of numbers is not affected by adding same number to
each number.
70. Aisha's income in 2004 was 20 percent greater than her income in 2003. What is the ratio
of Aisha's income in 2004 to her income in 2003?
A. 1 to 5
B. 5 to 6
125

C. 6 to 5
D. 5 to 1
E. 20 to 1
Solution;
In 2003, income=100
In 2004, income=120
income(2004)/income(2003)=120/100=6/5
71. Jacob's weekly take-home pay is n dollars. Each week he uses 4n/5 dollars for expenses and
saves the rest. At those rates, how many weeks will it take Jacob to save $500, in terms of n?
A. 500/n
B. 2,500/n
C. n/625
D. n/2,500
E. 625n
Solution;
Expenses=4n/5 $
Save=(n-4n/5) $=n/5 $
In 1 week, save=n/5 $
in 5/n weeks, save=1 $
in 5/n*500=2,500/n weeks, save=500$
72. The operation ♥ is defined for all integers x and y as x♥y=xy-y. If x and y are positive
integers, which of the following CANNOT be zero?
A. x♥y
B. y♥x
C. (x-1)♥y
D. (x+1)♥y
E. x♥(y-1)
Solution;
126

x♥y=y(x-1) If x=1, then its value can be zero.
y♥x=yx-x=x(y-1) If y=1, then its value can be zero.
(x-1)♥y=(x-1)y-y=y(x-1-1)=y(x-2) If x=2, its value also can be zero.
(x+1)♥y=(x+1)y-y=xy+y-y=xy Its value cannot be zero since x and y are positive integers and the
product of two positive integers cannot be zero.
If we see choice E, x♥(y-1)=x(y-1)-(y-1)=(y-1)(x-1) If x=y=1, its value can be zero.

73. P, Q and R are three points in a plane, and R does not lie on line PQ. Which of the
following is true about the set of all points in the plane that are the same distance from all
three points?
A. It contains no points.
B. It contains one point.
C. It contains two points.
D. It is a line.
E. It is a circle.
Solution;
P

Q
S
R

If any three non co-linear points are given, then we can find a point such that it is at same
distance from all given points. A circle can be drawn through points P,Q and R so that S is the
centre and SP=SQ=SR=radii of the circle. But note that the circle is not at same distance from
points P,Q and R. It's point S or the center of the circle which is at same distance from the
given points.
That means, the set of points that are the same distance from all three points contain just a
point.

127

74. X and Y are two points on a plane. Which of the following is NOT true about the set of all
points in the plane that are the same distance from the given two points?
A. It is a straight line.
B. It contains infinite points.
C. It bisects the line joining X and Y.
D. It is perpendicular to the line joining X and Y.
E. It contains two points.
Solution;
X

Y

The set of points that are same distance from X and Y will be perpendicular bisector of the line
joining X and Y.
Hence choice E is not true.
75. If x<y<0, which of the following inequalities must be true?
A. y+1<x
B. y-1<x
C. xy2<x
D. xy<y2
E. xy<x2
Solution;
x<y<0
Let x=-2 and y=-1
then, -1+1>-2 or y+1>x So eliminate A.
and -1-1=-2 or y-1=x So eliminate B.
and -2*(-1)2=-2 or xy2=x So eliminate C.
and -2*(-1)>(-1)2 or xy>y2 So eliminate D.
and -2*(-1)<(-2)2 or xy>x2 So our answer is E.

128

76. For all integers x, the function f is defined as follows.
f(x)=x-1 if x is even
=x+1 if x is odd
If a and b are integers and f(a)+f(b)=a+b, which of the following statements must be true?
A. a=b
B. a=-b
C. a+b is odd.
D. Both a and b are even.
E. Both a and b are odd.
Solution;
a=b
Let a=b=1
then f(a)+f(b)=f(1)+f(1)=(1+1)+(1+1)=4
and a+b=1+1=2 So choice A cannot be true.
Next, a=-b
Let a=1 and b=-1
then f(a)+f(b)=f(1)+f(-1)=(1+1)+(-1+1)=2
and a+b=1-1=0 So choice B cannot be true.
Next, a+b is odd
Let a=1 and b=2 so that a+b=3 which is odd.
then f(a)+f(b)=f(1)+f(2)=(1+1)+(2-1)=3
and a+b=1+2=3 So choice A can be true.
Next, both a and b are even
Let a=2 and b=4
then f(2)+f(4)=2-1+4-1=4
and a+b=2+4=6 So choice D is not true.
Next, both a and b are odd
Let x=1 and y=3
then f(1)+f(3)=1+1+3+1=6
and a+b=1+3=4 So choice E is not true.
Hence, our answer is C.

129

77. If y-2+2y-1-15=0, which of the following could be the value of y?
A. 3
B. 1/5
C. -1/5
D. -1/3
E. -5
Solution;
1/y2+2/y-15=0
1+2y-15y2=0
15y2-2y-1=0
15y2-5y+3y-1=0
5y(3y-1)+1(3y-1)=0
(3y-1)(5y+1)=0
y=1/3 or -1/5

78. The figure shows the standard normal distribution, with mean 0 and standard deviation 1,
including approximate percents of the distribution corresponding to the six regions shown.
Ian rode the bus to work last year. His travel times to work were approximately normally
distributed, with a mean of 35 minutes and a standard deviation of 5 minutes. According to
the figure shown, approximately what percent of Ian's travel to work last year were less than
40 minutes?
A. 14%
B. 34%
C. 60%
D. 68%
E. 84%
130

Solution;
mean=35
sd=5
the number 1 standard deviation above mean=35+5=40
If we see the above bell curve, then percentage coverage above value 40 is 14+2=16%
So value below 40 will be 100-16=84%
79. 4 meters

10
4 meters
10 meters
The figure above shows the floor dimensions of an L-shaped room. All angles shown are right
angles. If carpeting costs $20 per square meter, what will carpeting for the entire floor of the
room cost?
A. $800
B. $1,280
C. $1,600
D. $1,680
E. $2,320
Solution;
4
10
4
10
Area of L shape=4*10+4*10-4*4=40+40-16=80-16=64 m2
So cost of carpeting=Total area of floor*cost per m2=64*20=$1,280. So our answer is B.
131

80. Mario purchased $600 worth of traveler's checks. If each check was either $20 or $50,
which of the following CANNOT be the number of $20 checks purchased?
A. 10
B. 15
C. 18
D. 20
E. 25
Solution;
let number of $20 checks purchased be x and number of $50 checks purchased be y.
then 20x+50y=600
2x+5y=60
Looking at this equation,
Since 5y=multiple of 5 which could be 5,10,15,20…. and the sum needs to be 60, 2x must be
multiples of 5 or, x needs to be multiples of 5.
Among the options given, choice C is not multiple of 5. So our it's our answer.
If we put x=18 in the above equation, then
2*18+5y=60
5y=60-36=24
y=4.8 which is not possible since number of tickets purchased cannot be in decimal.
81. a+b/c
d/e
If the value of the expression above is to be halved by doubling exactly one of the five
numbers a, b, c, d, or e, which should be doubled?
A. a
B. b
C. c
D. d
E. e
Solution;
a+b/c*e/d
For this expression to be halved, d should be doubled.
So our answer is D.
132

82. In the figure above, arcs PR and QS are semi-circles with centers at Q and R respectively. If
PQ=5, what is the perimeter of the shaded region?
A. 5*3.14+5
B. 5*3.14+15
C. 10*3.14+10
D. 10*3.14+15
E. 100*3.14
Solution;
PQ=QR=RS=5
Circumference of each semicircle=3.14*5
Perimeter of shaded region=QP+circumference of semicircle with center Q+RS+ circumference
of semicircle with center Q=5+(3.14*5)+5+(3.14*5)=10*3.14+10
83. How many of the positive integers less than 25 are 2 less than an integer multiple of 4?
A. 2
B. 3
C. 4
D. 5
E. 6
Solution;
4k-2 where k=1,2,3…
for k=2 4k-2=6
for k=3 4k-2=10
for k=4 4k-2=14
for k=5 4k-2=18
133

for k=6 4k-2=22
OR, 4k-2<25
4k<27
k<27/4
that means k=6
84. The charge for a telephone call made at 10:00 a.m. from City Y to City X is $0.50 for the
first minute and $0.34 for each additional minute. At these rates, what is the difference
between the total cost of three 5-minute calls and the cost of one 15-minute call?
A. $0.00
B. $0.16
C. $0.32
D. $0.48
E. $1.00
Solution;
for three 5-minute calls,
total charge=3*(0.5+(0.34*4))=$5.58
for one 15-minute call,
charge=0.5+(0.34*14)=$5.26
So difference=5.58-5.26=$0.32
85. Which of the following is equivalent to the pair of inequalities x+6>10 and x-3<=5?
A. 2<=x<16
B. 2<=x<4
C. 2<x<=8
D. 4<x<=8
E. 4<=x<16
Solution;
x+6>10
x>10-6
x>4
134

Eliminate choices A, B and C since x>=2.
Now
x-3<=5
x<=5+3
x<=8
Hence, 4<x<=8 which is choice D.
86. In Town X, 64 percent of the population are employed, and 48 percent of the population
are employed males. What percent of the employed people in Town X are females?
A. 16%
B. 25%
C. 32%
D. 40%
E. 52%
Solution;
Let total population be 100
then 64 are employed
and 48% of 64=30.72 are employed males.
That means 64-30.72=33.28 are employed females.
employed female %=33.28/64*100=52
87. If p/q<1, and p and q are positive integers, which of the following must be greater than 1?
A. /p/q
B. p/q2
C. p/2q
D. q/p2
E. q/p
Solution;
Clearly, if p/q<1
then q/p>1. We need not try other choices. Our answer is E.

135

88. A factory has 500 workers, 15 percent of whom are women. If 50 additional workers are to
be hired and all of the present workers remain, how many of the additional workers must be
women in order to raise the percent of women employees to 20 percent?
A. 3
B. 10
C. 25
D. 30
E. 35
Solution;
15% of 500=75 women
500-75=425 men
Let required number of women to be hired=x.
By question,
75+x=20% of (500+50)
75+x=110
x=110-75=35
89. A bar over a sequence of digits in a decimal indicates that the sequence repeats
indefinitely.
What is the value of (104-102)(0.0012)?
A. 0
B. 0.12
C. 1.2
D. 10
E. 12
Solution;
(104-102)(0.0012)=102(102-1)(0.0012)=99*0.12
Eliminate choices A and B.
If 10*0.12=1.21>1.2 So 99*0.12>1.2 so Eliminate choice C.
100*0.12=12.12 which is a little less than 12
So 99*0.12=12
136

90. An empty pool being filled with water at a constant rate takes 8 hours to fill to 3/5 of its
capacity. How much more time will it take to finish filling the pool?
A. 5 hr 30 min
B. 5 hr 20 min
C. 4 hr 48 min
D. 3 hr 12 min
E. 2 hr 40 min
Solution;
3/5 parts=8 hours
1 part=8*5/3=40/3 hours
2/5 parts=40/3*2/5=16/3 hrs>5 hrs.
Eliminate C, D and E.
16/3=5 hours 20 min. Hence our answer is B.
91. John has 10 pairs of matched socks. If he loses 7 individual socks, what is the greatest
number of pairs of matched socks he can have left?
A. 7
B. 6
C. 5
D. 4
E. 3
Solution;
For the maximum number of pairs of matched shocks left, 3 pairs of matched socks must be
lost and remaining one individual sock can be from any pair.
Hence in all, he can have maximum 6 matched pairs of socks remaining.
Note: for minimum number of matched pair of socks left, all 7 lost individual socks must be
from different pair so that minimum number of matched pairs of socks remaining=10-7=3
92.
p

q
(4,3)
137

In the x-y coordinate system above, the lines q and p are perpendicular. The point (3,a) is on
line p. What is the value of a?
A. 3
B. 4
C. 13/3
D. 14/3
E. 16/3
Solution;
slope of line q=3/4 ,it passes through origin.
So slope of line p=-4/3 since they are perpendicular to each other, the product of their slopes=-1)
Also slope of line p=(3-a)/(4-3)=(3-a)/1=3-a=-4/3
a=(4/3)+3=13/3.

93. John works for 5 days. His daily earnings are displayed on the above graph, If John earned
$35 on the sixth day, what would be the closest difference between the median and the mode
of the wages during the six days?
A. $5.5
B. $6.5
C. $7.5
D. $8.5
138

E.$9.5
Solution;
60,75,45,35,40,35
For median, arrange given data in ascending order.
35,35,40,45,60,75
median=value of (6+1)th/2=3.5th item=(40+45)/2=85/2=42.5
mode=35
median-mode=42.5-35=$7.5
94. The degree measures of the four angles of a quadrilateral are w, x, y and z respectively. If
w is the average of x, y and z, then x+y+z=
A. 45
B. 90
C. 120
D. 180
E. 270
Solution;
w+x+y+z=360
w=(x+y+z)/3
So (x+y+z)/3+(x+y+z)=360
4/3*(x+y+z)=360
x+y+z=360*3/4=90*3=270
95. In a certain school, special programs in French and Spanish are available. If there are N
students enrolled in the French program, and M students in the Spanish program, including P
students who enrolled in both programs, how many students are taking only one (but not
both) of the language programs?
A. N+M
B. N+M-P
C. N+M+P
D. N+M-2P
139

E. N+M+2P
Solution;
n(French only)+n(Spanish only)=(N-P)+(M-P)=N+M-2P
96. In the sequence of numbers x1,x2,x3,x4,x5, each number after the first is twice the preceding
number. If x5-x1 is 20, what is the value of x1?
A. 4/3
B. 5/4
C. 2
D. 5/2
E. 4
Solution;
x2=2x1
x3=4x1
x4=8x1
x5=16x1
x5-x1=20
16x1-x1=20
15x1=20
x1=20/15=4/3
97. If a, b, and c are consecutive positive integers and a<b<c, which of the following must be an
odd integer?
A. abc
B. a+b+c
C. a+bc
D. a(b+c)
E. (a+b)(b+c)
Solution;
a=1, b=2 and c=3
then abc=1*2*3=6 (even)
140

a+b+c=1+2+3=6 (even)
a+bc=1+(2*3)=7 (odd)
a(b+c)=1(2+3)=5 (odd)
(a+b)(b+c)=(1+2)(2+3)=15 (odd)
Let a=2, b=3 and c=4
then a+bc=2+(3*4)=14 (even)
a(b+c)=2(3+4)=14 (even)
and (a+b)(b+c)=(2+3)(3+4)=35 (odd)
98. If x can have values -3, 0, and 2, and y can have only the values -4, 2, and 3, what is the
greatest possible value for 2x+y2?
A. 13
B. 15
C. 16
D. 20
E. 22
Solution;
For 2x+y2 to be maximum, 2x and y2 both must be maximum.
The maximum value of x is 2. So maximum value of 2x=2*2=4
The maximum value of y2 is at y=-4. So maximum value of y2 is (-4)2=16
So maximum value of given expression is 4+16=20
99. If B is the midpoint of line segment AD and C is the midpoint of line segment BD, what is
the value of AB/AC?
A. 3/4
B. 2/3
C. 1/2
D. 1/3
E. 1/4
Solution;

141

A

B

C

D

AB=BD=BC+CD=BC+BC=2BC
AC=AB+BC=2BC+BC=3BC
AB/AC=2/3
100. For each of n people, Margie bought a hamburger and a soda at a restaurant. For each
of n people, Paul bought 3 hamburgers and a soda at the same restaurant. If Margie spent a
total of $5.40 and Paul spent a total of $12.60, how much did Paul spend just for hamburgers?
(Assume that all hamburgers cost the same and all sodas cost the same.)
A. $10.80
B. $9.60
C. $7.20
D. $3.60
E. $2.40
Solution;
Let a hamburger cost $x and a soda cost $y.
then, for Margie, nx+ny=5.40
for Paul, 3nx+ny=12.60

(i)

(ii)

(ii)-(i) gives
2nx=7.2
nx=7.2/2=3.6
3nx=3*3.6=$10.8=cost of total hamburgers for Paul
101. The average (arithmetic mean) of five numbers is 25. After one of the numbers is removed,
the average (arithmetic mean) of the remaining numbers is 31. What number has been
removed?
A. 1
B. 6
C. 11
D. 24
142

E. It cannot be determined from the information given.
Solution;
sum of 5 numbers=5*25=125
After one number is removed,
sum of remaining 4 numbers=4*31=124
Hence 1 must have been removed.
So our answer is A.
102. C is a circle, L is a line, and P is a point on line L. If C, L and P are in the same plane and P
is inside C, how many points do C and L have in common?
A. 0
B. 1
C. 2
D. 3
E. 4
Solution;
L
P
C

Hence our answer is 2.
103. A board of length L feet is cut into two pieces such that the length of one piece is 1 foot
more than twice the length of the other piece. Which of the following is the length, in feet, of
the longer piece?
A. (L+2)/2
B. (2L+1)/2
C. (L-1)/3
D. (2L+3)/3
E. (2L+1)/3
143

Solution;
L1+L2=L
L2=L-L1
L1=2L2+1 (assuming L1 is longer)
L1=2(L-L1)+1=2L-2L1+1
3L1=2L+1
L1=(2L+1)/3
104. In the sunshine, an upright pole 12 feet tall is casting a shadow 8 feet long. At the same
time, a nearby upright pole is casting a shadow 10 feet long. If the lengths of the shadows are
proportional to the heights of the poles, what is the height, in feet, of the taller pole?
A. 10
B. 12
C. 14
D. 15
E. 18
Solution;

12 ft

x
8 ft

10 ft

x/12=10/8
x=10/8*12=10/2*3=15 ft
105. List R:28, 23, 30, 25, 27
List S:22, 19, 15, 17, 20
Which of the following is true about List R and List S?

144

A. The first quartile of List S is greater than that of List R.
B. The median of List R is less than that of list S.
C. The mean of List S is greater than that of list R.
D. The standard deviation of List R is equal to that of list S.
E. The range of List R is greater than that of List S.
Solution;
List R:23,25,27,28,30
List S:15,17,19,20,22
1st quartile of List R=(5+1)th/4=6/4=1.5th item=mean of 1st and 2nd items=(23+25)/2=24
2nd quartile of List S=1.5th item=(15+17)/2=16
That means 1st quartile of List R is greater. Eliminate A.
Median of list R=27
Median of list S=19
That means median of list R is greater than that of list S. Eliminate B.
Mean of list R=(23+25+27+28+30)/5=26.6
Mean of list S=(15+17+19+20+22)/5=18.6
If you see the two lists carefully, you can see each term of list R is greater than that of list S and
the number of elements are equal in both lists. So mean of list R must be greater.
That means mean of list R is greater than that of list S. Eliminate C.
square of deviation for list R={(23-26.6)2+(25-26.6)2+(27-26.6)2+(28-26.6)2+(30-26.6)2}/5
={3.62+1.62+0.42+1.42+3.42}/5
square of deviation for list S={(15-18.6)2+(17-18.6)2+(19-18.6)2+(20-18.6)2+(22-18.6)2}/5
={3.62+1.62+0.42+1.42+3.42}/5
That means their standard deviations are equal. So our answer is D.
If you look at the two lists carefully, you can see the differences between any two consecutive
terms are same for both lists.
So without calculating, we can say standard deviations of the two lists are equal.
Range of list R=30-23=7
Range of list S=22-15=7
That means their range are equal.
106. If an ant runs randomly through an enclosed circular field of radius 2 feet with an inner
circle of 1 foot, what is the probability that the ant will be in the inner circle at any one time?
145

A. 1/8
B. 1/6
C. 1/4
D. 1/2
E. 1
Solution;

P(ant being in the inner circle)=area of inner circle/total area=3.14*12/3.14*22=1/4

107. In the triangle above, x is equal to
A. 12
B. 16
C. 15
D. 10
E. none of these
Solution;
146

x/20=12/15
x=12*20/15=16

108. The number of men in a certain class exceeds the number of women by 7. If the number of
men is 5/4 of the number of women, how many men are there in the class?
A. 21
B. 28
C. 35
D. 42
E. 63
Solution;
M-W=7
M=5W/4
So 5W/4-W=7
W/4=7
W=28
So our answer is B.

109. The volume of a cube is less than 25, and the length of one of its edges is a positive integer.
What is the largest possible value for the total area of the six faces?
A. 1
B. 6
C. 24
D. 54
E. 150
Solution;
147

Volume<25
length3<25
length<3/25
length<3
length=2
largest possible value of surface area=6(length)2=6*22=24

148

MULTIPLE CHOICE QESTIONS - Select all that apply

1. y
Q(1,3)

P(1,1)
x
Figure shows line segment PQ and a circle with radius 1 and center (5,2) in the x-y plane.
Which of the following values could be the distance between a point on line segment PQ and
a point on the circle?
A. 2.5
B. 3.0
C. 3.5
D. 4.0
E. 4.5
F. 5.0
G. 5.5
H. 6.0
Solution;

(4,2)
Q(1,3)
(1,2)M

O(5,2)
radius=1

N

R(6,2)

P(1,1)

From the figure,
the maximum distance=QR=/(6-1)2+(2-3)2=/(25+1)=/26<5.5

149

That means we eliminate choices G and H which are greater than /26.
And the minimum distance=MN=4-1=3
That means we eliminate choice A which is less than minimum possible distance 3.
Hence our answers are B,C,D,E and F.

2. Which of the following could be the unit digit of 57n, where n is a positive integer?
A. 0
B. 1
C. 2
D. 3
E. 4
F. 5
G. 6
H. 7
I. 8
J. 9
Solution;
571=57
572=…….9
573=…….3
574=…….1
575=…….7
The sequence of the unit digits of 57n is 7,9,3,1,7,9,3,1…..
Hence, our answers are B,D,H and J.

150

3. If the lengths of two sides of a triangle are 5 and 9 respectively, which of the following could
be the length of the third side of the triangle?
A. 3
B. 5
C. 8
D. 15
Solution;
A
5

9

B

C

BC<5+9 or 14 (sum of two sides of a triangle is always greater than the third side)
Eliminate D.
Likewise, BC+5>9 or, BC>4
Eliminate A.
Hence our answers are B and C.

4. Sandy has a husband and two children. She brings at least 1 member of her family to a
dinner for lunch every day. The dinner offers 10 lunch specials. If no one orders the same thing,
how many orders can Sandy's family make for lunch. (Indicate all possible answers)
A. 45
B. 90
C. 120
D. 210
E. 720
F. 5040
Solution;
151

If she brings 1 family member,
10C2=10!/8!2!=45
If she brings 2 family members,
10C3=10!/7!3!=120
If she brings all 3 family members,
10C4=10!/6!4!=210
Hence our answers are A,C and D.

5. In triangle ABC, angle A=25, angle B>90. Which of the following could be the measure of
angle C? Indicate all such measures.
A. 12
B. 15
C. 45
D. 50
E. 70
Solution;
A+B+C=180
25+B+C=180
C=155-B
If B=90,
C=155-90=65
But, since B>90, C<65
Hence our answers are A,B,C and D.

6. The total amount that Mary paid for a book was equal to the price of the book plus a sales
tax that was 4 percent of the price of the book. Mary paid for the book with a $10 bill and

152

received the correct change, which was less than $3.00. Which of the following statements
must be true?
Indicate all such statements.
A. The price of the book was less than $9.50
B. The price of the book was greater than $6.90
C. The sales tax was less than $0.45
Solution;
Let price of the book be $x.
Total price=104% of x=1.04x
By given, 7<1.04x<10
or, 6.73<x<9.61
Price of the book was less than $9.61 (let $9.55>$9.50
Price of the book was greater than $6.73 (let $6.75<$6.90

Choice A is incorrect)
Choice B is incorrect)

Now, for sales tax, 6.73*0.04<x*0.04<9.61*0.04 (since sales tax=0.04x)
or, 0.2692<sales tax<0.3844
Sales tax was less than 0.3844 which means it was less than $0.45 also.
That means the only correct choice is C.

7. The length of AB is 10/3.
B

A

D

C

Which of the following statements individually provides sufficient additional information to
determine the area of triangle ABC above?

153

A. DBC is an equilateral triangle.
B. ABD is an isosceles triangle
C. The length of BC is equal to the length of AD.
D. The length of BC is 10.
E. The length of AD is 10.
Solution;
Area of triangle ABC=1/2* BC* AB=1/2*10/3*BC=5/3*BC
Now if we know the length of BC, then we know the area of the triangle.
So choice D is one of the correct choices.
Now in case of choice A, if triangle DBC is an equilateral triangle, then angle BCD=60.
That means triangle ABC would be a 30-60-90 triangle in which side AB=10/3.
That means we could determine the other sides BC and CA also. That means we could
determine the area of the triangle. So choice A is also one of the correct choices.
Now in case of choice B, knowing that triangle ABD is an isosceles triangle would not provide
any value of side BC. Hence it does not provide enough information to determine the area of
triangle ABC.
Similar is the case with choice C. BC=AD. Since the length of AD is not known, the length of BC
will also be unknown. Hence it is not the correct choice either.
Now our last choice E says AD=10. Just knowing the length of side AD would not provide any
information about length of side BC. Hence it is not the correct choice.
So our answers are A and D.
Note something: here, choices C and E combined would provide enough information to
determine the area of triangle ABC. Since, AD=10 and BC=AD=10. That means we could
determine the area of triangle ABC if choices C and E were combined. So you can see the
question has emphasized the word 'individually'.

8. In triangle ABC, the measure of angle B is 90. The length of side AB is 4 and the length of
side BC is x. If the length of hypotenuse AC is between 4 and 8, which of the following could be
the value of x?
Indicate all such values.
154

A. 1
B. 2
C. 3
D. 4
E. 5
F. 6
Solution;
A

4

B

x

C

AC=/(x2+42=/(x2+16)
We are given, 4<AC<8
or, 4</(x2+16)<8
or, 16< x2+16<64
or, 0<x2<48
or, 0<x</48
or, 0<x<6.92
Hence all given choices are correct.

9.
3.7, 4.1, a, 8.5, 9.2, 2a
The six numbers shown are listed in increasing order. Which of the following values could be
the range of the six numbers?
Indicate all such values.
155

A. 4.0
B. 5.2
C. 7.3
D. 11.6
E. 12.9
F. 14.1
Solution;
Range=2a-3.7
We are given 4.1<a<8.5
We are also given, 9.2<2a or, 4.6<a
Hence we get 4.6<a<8.5
or, 2*4.6<2a<2*8.5
or, 9.2<2a<17
or, 9.2-3.7<2a-3.7<17-3.7
or, 5.5<Range<13.3
Hence our answers are C, D and E.

10. Every year between 70% and 85% of the students at Central High School attend the
homecoming rally. If one year, 1435 students attended the rally, which of the following could
have been the number of students at Central High School that year?
Indicate all possible numbers of students.
A. 1675
B. 1775
C. 1875
D. 1975
E. 2075
Solution;
156

70% of total=1435
or, 0.7*total=1435
or, total=1435/0.7=2050
And 85% of total=1435
or, 0.85*total=1435
or, total=1435/0.85=1688
That means the range of students=1688-2050
Hence our answers are B,C and D.

11. If a-b>a+b, where a and b are integers, which of the following must be true? Indicate all that
apply.
A. a<0
B. b<0
C. ab<0
Solution;
a-b>a+b
-b>b
b is negative.
choice B is true.
a could be positive or negative.
Ex 3-(-4)>3+(-4) or -3-(-4)>-3+(-4)
That means choice A is not necessarily true.
Similar is the case with choice C.
If a=-ve and b=-ve, then ab=+ve
if a=+ve and b=-ve then ab=-ve
That means choice C is not necessarily true either.
157

Hence our correct answer is choice B.

12. x, y and z are positive integers. The average (arithmetic mean) of x, y and z is 11. If z is two
greater than x, which of the following must be true?
indicate ALL that apply.
A. x is even
B. y is odd
C. z is odd
Solution;
x+y+z=3*11=33
x+y+(x+2)=33
2x+y=31
In this equation, the quantity 2x is always even no matter what x be.
Since 2x(even)+y=31(odd), so y is odd. (even+odd=odd….ex 4+3=7)
x can be either even or odd and z=x+2=(even or odd)+even=even or odd
Hence only choice B is true.

13. A quality control analyst collected 200 measurements of length. The range of 200
measurements is 17 centimeters, and one of the measurements is 49.5 centimeters. Which of the
following could be another one of the 200 measurements, in centimeters?
Indicate ALL such measurements.
A. 31.0
B. 32.0
C. 33.0
D. 34.0
Solution;
Range=L-S=17 centimeters
158

{……………………………………….49.5 cms…………………………………………………………}

200 data

Here the possibility is that 49.5 cms could be either the largest among 200 data or the smallest
among 200 data.
If the given one data 49.5 cms was the largest,
Range=49.5-S
17=49.5-S
S=49.5-17=32.5
If it was the smallest,
Range=L-49.5
17=L-49.5
L=17+49.5=66.5
That means the largest possible data is 66.5 cms and the smallest possible data is 32.5 cms.
Hence our answers are C and D.

14. USED CARS SOLD LAST MONTH AT CAR DEALERSHIP X
Price Range

number of cars sold

under $5,000

7

$5,000-$7,409

10

$7500-$10,000

8

over $10,000

6

For the 31 used cars sold last month at Car Dealership X, which of the following could be the
median price?
Indicate ALL such prices.
A. $5,500
B. $6,500
C. $7,000
D. $8,500
159

E. $9,000
F. $10,500
Solution;
Median=value of (N+1)th/2 item=value of (31+1)th/2=value of 16th item
16th item lies in the range $5,000-$7,409
Hence our answers are A,B and C.

15. A survey asked 1,150 people to choose their favorite laundry detergent from brands A,B
and C. Of the people surveyed, x percent chose A as their favorite brand. If x is rounded to the
nearest integer, the result is 3. Which of the following could be the number of people who
chose A as their favorite brand?
Indicate ALL such numbers.
A. 20
B. 25
C. 30
D. 35
E. 40
F. 45
G. 50
Solution;
2.5<=x<3.5
Any number greater than or equal to 2.5 or less than 3.5 will be rounded to 3.
2.5% of 1150=28.75
3.5% of 1150=40.25
28.75<=x<40.25
Hence our answers are C,D and E.

160

16. If x,y and z are positive integers such that 3x<2y<4z, which of the following statements could
be true? Indicate All such statements.
A. x=y
B. y=z
C. y>z
D. x>z
Solution;
3x<2y
If x=y, 3x>2y Ex x=1 and y=1, then 3*1>2*1
That means choice A is incorrect.
2y<4z
if y=z, 2y<4z Ex y=z=2, then 2*2<4*2
if y>z, 2y<4z Ex y=5, z=3, then 2*5<4*3
That means choice B and C could be correct.
3x<4z
if x>z, 3x<4z Ex x=6, z=5, then 3*6<4*5
That means choice D could also be correct.
Hence correct answers are B,C and D.

17. Roney's policy is such that his spending does not exceed 35% of his income. If one month, he
spent $1000, what could be his income per year?
Indicate ALL such values.
A. $30,000
B. $34,000
C. $35,000
D. $36,000
E. $36,500
161

Solution;
spending<=0.35*income
1000<=0.35*income
income>=1000*12/0.35 per year
income>=$34285.7
Hence our answers are C,D and E.

18. If a positive integer n, divided by 5 has a remainder 2, which of the following must be true?
Select ALL such statements.
A. n is odd
B. n+1 cannot be a prime number
C. n+2 divided by 7 has remainder 2
D. n+3 is divisible by 5
Solution;
n=5k+2 (k=0,1,2…..)
Let k=0. then n=5*0+2=2 which is even.
Choice A is incorrect.
Let k=0. then n+1=(5*0+2)+1=3 which is a prime number.
So choice B is incorrect.
Let k=0. then n+2=(5*0+2)+2=4 and 4 divided by 7 gives a remainder 4 not 2.
So choice C is incorrect.
Let k=0. then n+3=(5*0+2)+3=5 which is divisible by 5.
So choice D is correct.

19. One side of a triangle has length 8 and a second side has length 5. Which of the following
could be the area of the triangle? Select ALL that apply.
162

A. 24
B. 20
C. 5
Solution;
C
5
N
A

M

8

B

Area of triangle ABC=1/2*AB*CM=1/2*8*CM=4*CM<20
CM<5 (in right angled triangle AMC, AC=5 which is hypotenuse is the longest side)
or, area of triangle ABC=1/2*AC*BN=1/2*5*BN=2.5*BN<20
BN<8 (in right angled triangle ANB, AB=5 which is hypotenuse is the longest side)
So eliminate choice A.
We could draw a triangle with sides 5 and 8 and height 5 so that area will be 20.
C

A

B

and we could draw a triangle with sides 5 and 8 and height 1.25 so that area will be 5.
In the above triangle, AB=8,AC=5 and AB is perpendicular to CA so that area=1/2*8*5=20

C
A

M

B

In the above triangle, AB=8,AC=5 and CM=1.25 so that area=1/2*8*1.25=5
Hence our answers are B and C.
Another simple way to solve this problem is to have a little knowledge about trigonometry.
163

C

A

M

B

The area of the triangle ABC in trigonometric form=1/2*AB*CM=1/2*AB*CA*sinA
So area=1/2*8*5*sinA=20*sinA
The value of sine of a function ranges from from 0 to 1.
So minimum possible area=20*sin0=0(when the angle between two given lines is 0)
and maximum possible area=20*sin90=20*1=20(when the angle between two given lines is 90)
So choices B and C are true.

20. A triangle has a perimeter 13. The two shorter sides have integer lengths equal to x and x+1.
Which of the following could be the length of the other side?
Select as many as are correct.
A. 4

C

B. 6
C. 8

x
A

x+1
y

B

Solution;
Let y be the longest side.
x+x+1+y=13
2x+1+y=13
2x+y=12
x=(12-y)/2
If y=4, then x=4 and x+1=5
sides will be y=4,x=4 and x+1=5 which don't satisfy the question because we need y to be the
longest.
Choice A is not true.
164

Here, y is the longest and the sum of any two sides is also greater than the remaining side. So
choice B is true.
here, 2+3<8 i.e. the sum of two sides of is not greater than the third side. So the sides don't form
a triangle.
Hence our answer is B only.

21. In the xy-plane, line k is a line that does not pass through the origin. Which of the following
statements individually provide(s) sufficient additional information to determine whether the
slope of line k is negative?
Indicate ALL such statements.
A. The x-intercept of line k is twice the y-intercept of line k.
B. The product of the x-intercept and the y-intercept of line k is positive.
C. Line k passes through the points (a,b) and (r,s), where (a-r)(b-s)<0.
Solution;
x/a+y/b=1
x/2b+y/b=1(x-intercept=2*y-intercept)
x+2y=2b
y=-(1/2)x+b in which slope=-1/2 i.e. negative.
So choice A is true.
a*b=positive
When a=+ve,b=+ve and when a=-ve,b=-ve
x/a+y/b=1
bx+ay=ab
y=(-bx+ab)/a=-(b/a)x+b in which the slope is -(b/a) i.e. -ve.
x/(-a)+y/(-b)=1

165

-bx-ay=ab
y=(-bx-ab)/a=-(b/a)x-b in which the slope is -(b/a) i.e. -ve.
So choice B is true.
Slope of the line k=b-s/a-r=-ve
(a-r)(b-s)<0
When a-r=+ve, then b-s=-ve
when a-r=-ve, then b-s=+ve
So choice C is also true.
Hence all are correct choices.

22. Each employee of a certain company is in either Department X or Department Y, and
there are more than twice as many employees in Department X as in Department Y. The
average salary is $25,000 for the employees in Department X and is $35,000 for the
employees in Department Y. Which of the following amounts could be the average salary for
all of the employees in the company?
Indicate ALL such amounts.
A. $26,000
B. $28,000
C. $29,000
D. $30,000
E. $31,000
F. $32,000
G. $34,000
Solution;
The combined average salary would be greater than $25,000 since greater average salary has
been combined with less average salary and the combined average will be greater than the
less average salary.

166

If there were exactly twice as many employees in Department X as in Department Y, then
combined average salary=(2*25,000+1*35,000)/(2+1)=$28,333
But, since there are more employees with less average salary, the combined salary of
employees of two Departments must be less than $28,333.
Eliminate choices C,D,E,F and G.
Hence our answers are A and B.

23. The integer v is greater than 1. If v is the square of an integer, which of the following
numbers must also be the square of an integer?
Indicate ALL such statements.
A. 81v
B. 25v+10/v+1
C. 4v2+4/v+1
Solution;
v=22,32,42,52,…….. so let v=k2 (where k=2,3,4…..)
Now,
81v=92*k2=(9*k)2
Choice A is square of an integer.
25v+10/v+1=52*k2+10*k+1=(5k)2+2*5k*1+12=(5k+1)2
Choice B is also square of an integer.
Let v=22 then 4v2+4/v+1=4*(22)2+4*2+1=64+8+1=73
Choice C is not the square of an integer.
Hence A and B are the correct answer choices.

24. A and B are two independent events and the probability that both events occur is 1/2.
Which of the following could be the probability of occurrence of event A?
Select ALL that apply.

167

A. 0
B. 1/4
C. 1/2
D. 3/4
Solution;
P(A and B)=P(A)*P(B)=1/2
We know, 0<P(A)<=1 and 0<P(B)<=1
If P(A)=0, then P(A)*P(B)=0 so it cannot be the correct answer.
If P(A)=1/4,
then, 1/4*P(B)=1/2
P(B)=1/2*4=2, which is not possible. So it is not correct choice either.
If P(A)=1/2,
then, 1/2*P(B)=1/2
P(B)=1, which is possible. So it can be the probability of occurrence of event A.
If P(A)=3/4,
then, 3/4*P(B)=1/2
P(B)=1/2*4/3=2/3, which is also possible. So it also can be the probability of occurrence of event
A.
Hence our answers are C and D.

25. If the area of a rectangle is 40, which of the following could be the perimeter of the
rectangle? Indicate all such perimeters.
A. 20
B. 40
C. 200
D. 400
E. 2,000
168

F. 4,000
Solution;
Perimeter will be the smallest when the rectangle is a square.
i.e., when l=b
So A=l2=40
or, l=/40=6.324
So, smallest perimeter possible=4*6.324=25.29
Eliminate A.
Now let's see the largest perimeter possible.
If l=0.001 and b=40,000 so that l*b=40 as given.
Then Perimeter=2(0.001+40,000)=80,000.002
That means, we can make the perimeter as large as we desire, by making the difference
between length and breadth of the rectangle large.
So all choices except A are correct.

26. Which of the following is an equation of a line that is perpendicular to the line whose
equation is 2x+3y=4?
Indicate ALL such equations.
A. 3x+2y=4
B. 3x-2y=4
C. 2x-3y=4
D. 4-3x=-2y
E. 4+2x=3y
Solution;
Slope of given line=-2/3
So the line perpendicular to given line must have slope=3/2 (m1*m2=-1 for two perpendicular
lines)
169

Slope of line A=-3/2
Slope of line B=-3/-2=3/2
Slope of line C=-2/-3=2/3
Slope of line D=-(-3)/2=3/2
Slope of line E=-2/-3=2/3
Note: Slope of a line in general form Ax+By+C=0 is -A/B.
Hence our answers are B and D.

27. Which of the following inequalities have at least one positive solution and at least one
negative solution?
Indicate all such inequalities.
A. 5x/3<x
B. x3<x
C. x-6<x-7
Solution;
5x/3<x
5x<3x
For this inequality, x cannot be positive otherwise 5x>3x
For ex, let x=1, then 5*1>3*1 which means it has negative solution only.
A is not the answer.
x3<x
Here, x can be positive as well as negative.
Ex, let x=0.5, then 0.53=0.125<0.5 which means x can be positive.
now, let x=-2, then (-2)3=-8<-2 which means c can be negative also.
So B is the answer.
x-6<x-7
170

-6<-7 which is not true. This inequality doesn't have any solution.
Hence our answer is B only.

28. The range of the heights of the female students in a certain class is 13.2 inches, and the
range of the heights of the male students in the class is 15.4 inches.
Which of the following statements individually provide(s) sufficient information to determine
the range of heights of all the students in the class?
A. The tallest male student in the class is 5.8 inches taller than the tallest female student in the
class.
B. The median height of the male students in the class is 1.1 inches greater than the median
height of the female students in the class.
C. The average height of the male students in the class is 4.6 inches greater than the average
height of the female students in the class.
Solution;
A. Tallest female-shortest female=13.2
tallest male-shortest male=15.4
To determine the range of all the students, we must know the heights of tallest student and
the shortest student.
from above, (tallest male-tallest female)-shortest male+shortest female=2.2
5.8-shortest male+shortest female=2.2
shortest male-shortest female=5.8-2.2=3.6
That means the shortest student in the class is female.
So range=tallest male-shortest female

171

29.

Y
X

Z
W

W,X,Y and Z are four distinct digits from 0 to 9, inclusive, and W+X+Y=5Z(where 5Z is the
product of 5 and Z).
Under the given conditions, which of the following could be values of Z? Indicate all such
values.
A. 1
B. 2
C. 3
D. 4
E. 5
F. 6
G. 7
Solution;
Since W,X,Y and Z are distinct digits from 0 to 9, the largest possible sum of W+X+Y=7+8+9=24.
By given, W+X+Y=5Z
So, for choice E, Z=5, W+X+Y=5*5=25 which is greater than the largest possible sum. So eliminate
choice E.
Similarly choices F and G too won't be working since they are greater than 5.
So eliminate them also.
And the smallest possible sum of W+X+Y=0+1+2=3
Hence our answers are choices A,B,C and D.
Illustration;
0+2+3=5*1 (Z=1 can be true since W+X+Y=5Z and 0,2,3 and 1 are distinct digits)
0+3+7=5*2 (Z=2 can also be true since W+X+Y=5Z and 0,3,7 and 2 are distinct digits)
172

0+7+8=5*3 (Z=3 can also be true since W+X+Y=5Z and 0,7,8 and 3 are distinct digits)
Finally, 5+7+8+0=5*4 (Z=4 can also be true since W+X+Y=5Z and 5,7,8 and 4 are distinct digits)

30. Given three integers a, b and 4. If their average (arithmetic mean) is 6, which of the
following could NOT be the value of the product ab?
Indicate all that apply.
A. 13
B. 14
C. 40
D. 48
E. 49
F. 51
Solution;
a+b+4=6*3=18
a+b=18-4=14
a

b

a+b

ab

1

13

14

13

2

12

14

24

3

11

14

33

4

10

14

40

5

9

14

45

6

8

14

48

7

7

14

49

Hence our answers are B and F.

173

REVISED GRE - Numeric Entry (No Choice Given)

1. The item was listed at $65 and sold at $56.16. Two successive discounts were given on the
item and the first discount was 10%, Find the second discount percent.

Solution;
let second discount be x%.
then, 65*0.9*(1-x%)=56.16
1-x%=0.96
x%=0.04
x=4%
Hence the second discount was 4%.

2. Two numbers are in the ratio 6:13. The least common multiple of the two numbers is 312.
Find the sum of the two numbers.

Solution;
Let the numbers be 6x and 13x.
Then, LCM of 6x and 13x=78x=312
So, x=312/78=4
Sum=6*4+13*4=19*4=76

3. Find the value of x3/y3+y3/x3 if x/y+y/x=6.

Solution;
x/y+y/x=6

174

Taking cube on both sides,
(x/y+y/x)3=63
(x/y)3+3*(x/y)2 (y/x)+3*(x/y)(y/x)2+(y/x)3=216 since (a+b)3=a3+3a2b+3ab2+b3
x3/y3+y3/x3+3x/y+3y/x=216
x3/y3+y3/x3+3*6=216
x3/y3+y3/x3=216-18=198

4. How many sides does a polygon with 44 diagonals have?

Solution;
Number of diagonals in a polygon of sides n is given by
n(n-3)/2 where n>3
For ex, for a rectangle, there are 4(4-3)/2=2 diagonals.

For a hexagon, there are 6(6-3)/2=9 diagonals and so on.
So by given, 44=n(n-3)/2
88=n2-3n
n2-3n-88=0
n2-11n+8n-88=0
n(n-11)+8(n-11)=0
(n-11)(n+8)=0
n=11 or -8
Sides cannot be negative so n=11.

175

5. What is the greatest value of a positive integer n such that 3n is a factor of 1812?

Solution;
1812=(2*3*3)12=324*212
Here, the power of 3 is 24. So for 3n to be an integer of 324*212, the greatest possible value of n is
also 24.

6. A merchant made a profit of $5 on the sale of a sweater that cost the merchant $15. What is
the profit expressed as a percent of the merchant's cost? Give your answer to the nearest whole
percent.

Solution;
Profit%=Profit/CP*100%=5/15*100=33.333….=$33 (to the nearest whole percent)

7. RESULTS OF A USED-CAR AUCTION
Small Cars

Large Cars

Number of cars offered

32

23

Number of cars sold

16

20

Projected sales total for cars offered(in '000)

$70

$150

Actual sales total(in '000)

$41

$120

For the large cars sold at an auction that is summarized in the table above, what was the
average sale price per car?
$
Solution;
$120*1000/20=$6,000

8. A list of numbers has a mean of 8 and a standard deviation of 2.5. If x is a number in the list
that is two standard deviations above the mean, what is the value of x?
176

Solution;
x=mean+(2*standard deviation)=8+(2*2.5)=13

9. Line k lies in the x-y plane. The x-intercept of line k is -4 and line k passes through the midpoint of the line segment whose end-points are (2,9) and (2,0). What is the slope of line k?
Give your answer as a fraction.

Solution;

(2,9)

(2,4.5)

(-4,0)

(2,0)

Slope of line k=4.5-0/2-(-4)=4.5/6=9/12=3/4
(slope of a line passing through two points (x1,y1) and (x2,y2) is given by y2-y1/x2-x1)

10.
Y

Frequency

1/2

2

3/4

7

5/4

8

3/2

8

7/4

9

The table shows the frequency distribution of the values of a variable Y. What is the mean of
the distribution? Give your answer to the nearest 0.01.
177

Solution;
mean=(1/2*2+3/4*7+5/4*8+3/2*8+7/4*9)/(2+7+8+8+9)=1.2941…=1.29(to the nearest 0.01)

11. Of the 20 light-bulbs in a box, two are defective. An inspector will select two lightbulbs
simultaneously and at random from the box. What is the probability that neither of the lightbulbs will be defective?

Solution;
P(neither defective)=P(1st not defective)*P(2nd not defective)=(18/20)*(17/19)=153/190
Another method
P(neither defective)=18C2/20C2=(18!/16!2!)/(20!/18!2!)=(9*17)/(10*19)=153/190

12. Twenty percent of the sweaters in a store are white. Of the remaining sweaters, forty
percent are brown and the rest are blue. If there are 200 sweaters in the store, then how
many more blue sweaters than white sweaters are in the store?

Solution;
white=20% of 200=40
remaining=200-40=160
brown=40% of 160=64
blue=160-64=96
blue-white=96-40=56

13. In a graduating class of236 students, 142 took algebra and 121 took chemistry. What is the
greatest possible number of students that could have taken both algebra and chemistry?

178

Solution; 21
A

142-121

121
0

C

236-21-121=94

For the greatest number of students that could have taken both algebra and chemistry, the
number of students taking chemistry only must be zero. Then all 121 students who took
chemistry took both classes.
Hence maximum number of students who took both classes is 121.

14. In the given sequence, each term after the first term is equal to the preceding term plus the
constant c. If a1+a2+a3=27, what is the value of a2+a4?
a2+a4=
Solution;
a2=a1+c
a3=a2+c=a1+2c
a4=a3+c=a1+3c
a5=a1+4c
a1+a3+a5=27
a1+a1+2c+a1+4c=3a1+6c=3(a1+2c)=27
a1+2c=9
a2+a4=2a1+4c=2(a1+2c)=2*9=18

15. In the xy-plane, the point with coordinates (-6,-7) is the center of circle C. The point with
coordinates (-6,5) lies inside C and the point with coordinates (8,-7) lies outside C. If m is the
radius of C and m is an integer, what is the value of m?
m=
Solution;
179

(-6,5)

O

(-6,-7)C

m

From the figure, 5+7<m<8+6
12<m<14
m=13

180

(8,-7)

16. Jordan has taken 5 math tests so far this semester. If he gets a 70 on his next test, it will
lower the average (arithmetic mean) of his test scores by 4 points. What is his average now?

Solution;
Let current average be x.
then new average=new total sum/new number of tests
or, (5x+70)/6=x-4 or, 5x+70=6x-24 or, x=94

17. A

B

D

C

In the figure above, if ABCD is a rectangle what is the sum of marked angles?
degrees
180+180+90-180=270

18. Soltech employs 20 programmers for every 3 managers, and 5 managers for every director.
If the total number of employees at company is between 300 and 400, what must the
number of managers who work at the company equal to?

Solution;
P:M=20:3=100:15
M:D=5:1=15:3
P:M:D=100:15:3=200:30:6=300:45:9=400:60:12
So number of managers must be 45.

181

19. How many 2 digit integers can be chosen such that none of the digits appear more than
once, and none of the digits equals zero?
integers
Solution;
9 choices*8 choices=72 choices
The tens digit can be any integer from 1 to 9 i.e. 9 choices and the unit digit has 8 choices left
since repetition is not allowed.

20. 90 students represent x percent of the boys at the Ardmore Elementary School. If the boys
at Ardmore Elementary make up 40 percent of the total population of x students, what is x?

Solution;
total=x
Boys=40% of x=4x/10
x% of boys=90
x/100*4x/10=90
4x2=90*103=9*104
2x=3*100
x=150

21. Given that the sum of the odd integers from 1 to 99 inclusive is 2500, what is the sum of the
even integers from 2 to 100 inclusive?

Solution;
1+3+5+……..+99=2500
(2-1)+(4-1)+(6-1)+……+(100-1)=2500
(2+4+6+…..+100)-(1+1+1+….50 terms)=2500

182

2+4+6+…..+100=2500+50=2550

22. A confectioner has 500 mint, 500 orange and 500 strawberry flavored sweets. He wishes
to make packets containing 10 mint, 5 orange and 5 strawberry sweets. What is the maximum
number of packets of this type can he make?

Solution;
Here each packet must have 10 mint and there are equal number of each flavored sweets.
That means mint will be used up first and it is the limiting factor.
So maximum number of packets that he can make=500/10=50

23. What is the sum of all positive factors of 12?

Solution;
Positive factors of 12 are 1,2,3,4,6 and 12, the sum of which is 1+2+3+4+6+12=28

24. Of 60 students in a class 2/3 are girls and 2/5 of the class are taking music lessons. What is
the maximum number of girls that are not taking music lessons?

Solution;
2/3*60=40 girls and 60-40=20 boys
2/5*60=24 take music lessons.
60-24=36 don't take music lessons.
Since there are 40 girls, it's possible that all 36 students that don't take music lessons are girls.
So the maximum number of girls not taking music lessons is 36.

25. If |x+1|<=5 and |y-1|<=5, what is the least possible value of the product xy?

183

|x+1|<=5
-5<=x+1<=5
-6<=x<=4
|y-1|<=5
-5<=y-1<=5
-4<=y<=6
for the product xy to be the least, x=-6 and y=6 so that xy=-36

26. What is the length of the largest stick that can be filled in a rectangular box of length 12
cm, breadth 5 cm and height 84 cm.

Solution;

84

longest stick length=/122+52+842=85 cm
5
12

27. 35 percent of senior students in a certain college are absent. Two-fifth of those absentees
went to watch a movie. What fraction of total senior students are neither present nor went for
watching movie? Give your answer as fraction.

Solution;
Let senior students=100
then absentees=35
2/5*35=14 went to watch a movie.
35-14=21 seniors were neither present nor went for watching movie.
184

So required fraction=21/100

28. The total amount of Judy's water bill for the last quarter of the year was $40.50. The bill
consisted of a fixed charge of $13.50 plus a charge of $0.0075 per gallon for the water used in
the quarter. For how many gallons of water was Judy charged for the quarter
gallons
Solution;
Let he was charged for n gallons.
total charge=fixed charge+variable charge=13.50+(n*0.0075)=40.50
27=n*0.0075
n=27/0.0075=3600

29. The average (arithmetic mean) of 11 numbers in a list is 14. If the average of 9 of the
numbers in the list is 9, what is the average of the other 2 numbers?

Solution;
Sum of 11 numbers=11*14=154
Sum of 9 numbers=9*9=81
Sum of remaining 2 numbers=154-81=73 and average=73/2=36.5

30.

185

The circles shown are tangent at point B. Point A is the center of the larger circle, and line
segment AB (not shown) is a diameter of the smaller circle. The area of the smaller circle is
what fraction of the area of the larger circle?

Solution;
Area of smaller circle As=3.14*d2/4
Area of larger circle Al=3.14*(2d)2/4
As/Al=1/4

31. 10,10,10,10,8,8,8,8,12,12,11,y
The twelve numbers shown represent the ages, in years, of the twelve houses on a certain city
block. What is the median age, in years, of the twelve houses on the block?
years
Solution;
The possible positions of y are as follows
y,8,8,8,8,10,10,10,10,11,12,12

median=(12+1)th/2=6.5th item=average of 10 and 10=10

8,8,8,8,y,10,10,10,10,11,12,12

median=10

8,8,8,8,10,10,10,10,y,11,12,12

median=10

8,8,8,8,10,10,10,10,11,y,12,12

median=10

8,8,8,8,10,10,10,10,11,12,12,y

median=10

Hence median age of 12 houses is 10 years (no matter what y be)

32. Working alone at its constant rate, machine A produces k car parts in 10 minutes. Working
alone at its constant rate, machine B produces k car parts in 15 minutes. How many minutes
does it take machines A and B, working simultaneously at their respective constant rates, to
produce k car parts?

186

minutes
Solution;
In 1 min, A produces k/10 car parts.
In 1 min, B produces k/15 car parts.
In 1 min, A+B produce k/10+k/15=k/6 parts.
In 6 min, A+B produce k/6*6=k car parts.

ANNUAL PERCENTAGE CHANGE IN DOLLAR
AMOUNT OF SALES AT FIVE RETAIL STORES
FROM 2006 TO 2008
Store

Percent change from 2006 to
2007

Percent change from 2007 to
2008

P

10

-10

Q

-20

9

R

5

12

S

-7

-15

T

17

-8

33. At store T, the dollar amount of sales for 2007 was what percent of the dollar amount of
sales for 2008?
Give your answer to the nearest 0.1 percent.
%
Solution;
Let the dollar amount of sales at store T for 2006 be 100.
Then, dollar amount for 2007=100+17=117

187

dollar amount for 2008=92% of 117=107.64
117/107.64*100=108.69565…=108.7(to the nearest 0.1%)

34. List L: 2,x,y
List M: 1,2,3,x,y
If the average (arithmetic mean) of the 3 numbers in list L is 10/3, what is the average of the 5
numbers in list M?


Solution;
2+x+y=10/3*3=10
x+y+2=10
1+2+3+x+y=10+4=14
Therefore, average=sum/total number=14/5

35. Thirty five percent of senior students in a certain college were absent. Two-fifths of those
absentees went to watch a movie. What fraction of total senior students were absent and did
not go for movie? Give your answer as fraction.

Solution;
total seniors=100
absent=35
movie=2/5*35=14

present=100-35=65
no movie=35-14=21

Hence our answer is 21/100.
188

36. In the figure shown, what is the value of x?
30

40
x

Solution;

A

B
D

E

C
Solution;
In triangle CAE, angle AEC=180-90-30=60
In triangle DBC, angle BDC=180-90-40=50
x=360-60-50-90=160

37. What is the length of a diagonal of a rectangle that has width 5 and perimeter 34?

Solution;
P=2(l+b)
34=2(l+5)
17=l+5
l=12
Diagonal= /122+52=13

189

38. Among the people attending a convention in the Europe, 32 percent traveled from Asia
and 45 percent of those who traveled from Asia are women. What percent of the people at
the convention are women who traveled from Asia?

Solution;
Total=100
Asia=32
Asian Women=45/100*32=14.4
Hence women from Asia=14.4%

39. A candidate is required to answer 6 out of 10 questions which are divided into 2 groups
each containing 5 questions and he is not permitted to attempt more than 4 from any group.
In how many different ways can he make up his choice?

Solution;
Group A

GroupB

1

6

2

7

3

8

4

9

5

10

He can make up the choices as follows
4 from group A and 2 from group B=5C4*5C2=5*10=50 or
3 from group A and 3 from group B=5C3*5C3=10*10=100 or
2 from group A and 4 from group B=5C2*5C4=10*5=50
So total number of ways of choices=50+100+50=200

190

40.

A

120
O

a
a

P

B
In the figure above, PA is tangent to circle O at point A, PB is tangent to circle O at point B.
Angle AOB measures 120 and OP=24/pi. What is the length of minor arc AB?

Solution;
angle PAO=angle PBO=90
( the line joining the centre and point of contact of the tangent and the circle are at right
angles)
Triangle AOP is 30-60-90 degrees, so OA=radius=0.5*OP=0.5*24/pi=12/pi
circumference of circle=2*pi*OA=2*pi*12/pi=24
So, minor arc AB=120/360*circumference of circle=24/3=8, which is our answer.

41. Average of five positive integers is 10. What is the greatest possible integer among them?

Solution;
a+b+c+d+e=10*5=50
1+1+1+1+46=50
So the greatest possible integer among them is 46.

42. 10,5,x,1,18
Median of the above 5 numbers is 5. What is the greatest possible value of x?

191

Solution;
x,1,5,10,18

median=5 and x<1

1,x,5,10,18

median=5 and 1<x<=5

1,5,x,10,18

For median to be 5, the value of x must be 5.

1,5,10,x,18

median=10

1,5,10,18,x

median=10

From above results, the maximum possible value of x is 5.

43. If mean and median of 5 different positive integers are 10 and 12 respectively, what is the
greatest possible integer among them?

Solution;
sum of 5 integers=10*5=50
1+2+12+13+x=50
x=50-28=22

44. Susan received scores of 75,80 and 85 out of 100 on three math exams. The pass grade is 50
and the score is given only in integer values. If she appeared for 2 more math exams and
passed all five exams with an average of 80, what was the highest possible score she received
on her five math exams?
Solution;
For scoring an average of 80 on 5 exams,
total=5*80=400
total on her initial 3 tests=75+80+85=240
total on her next 2 tests=400-240=160
If the scores on the next two tests be x and y,
x+y=160
192

Since pass grade is 50,
let x=50(minimum) for y to be maximum.
then y=160-50=110 which is not possible since the maximum scores in an exam is 100.
Next if x=59, then y=160-59=101 which is still not possible.
If x=60, then y=160-60=100 which is the highest possible score obtained.

45. During a party attended by 3 females and 3 males, 3 people at random enter a previously
empty room. What is the probability that there are exactly 2 males in the room?

Solution;
3 males=A,B,C
3 females=D,E,F
Possibilities:
ABD+ABE+ABF+BCD+BCE+BCF+CAD+CAE+CAF

2 males+1 female

ADE+AEF+AFD+BDE+BEF+BFD+CDE+CEF+CFD

1 male+3 females

ABC

all males

DEF

all females

P(exactly 2 males)=9/20
P(exactly 2 males)=P(2 males and 1 female)=3C2*3C1/6C3=3*3/20=9/20

46. Set H contains five positive integers such that the mean, median, mode, and range are all
equal. The sum of the data is 25. What is the least possible number in set H?

Solution;
H={a,b,5,5,e} in which median=5 and mode=5
We are given, a+b+5+5+e=25
So a+b+e=15

(i)
193

Also, range=5 i.e. e-a=5

(ii)

(i)-(ii) gives
2a+b=10
a=(10-b)/2=5-b/2
Since all five must be positive integer, b must be even.
if b=2, then a =5-1=4 in which a>b but we must have b>a
if b=4, then a=5-2=3 in which a<b
So H={3,4,5,5,8}
Hence our answer is 3.

194

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