GRE numeric entry

1. 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

2. 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
+3a2
b+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.

3. 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?

4. 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 mid-
point 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.
Solution;

5. 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 light-
bulbs will be defective?
Give your answer as a fraction.
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?
Solution; 21 121
A 142-121 0 C 236-21-121=94

6. 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;

7. (-6,5)
O
(-6,-7)C m (8,-7)
From the figure, 5+7<m<8+6
12<m<14
m=13
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

8. 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.
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 units 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?

9. 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
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

10. 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?
|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.

11. 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.
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

12. 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.
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

13. 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?
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.

14. 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
Q
R
S
T
10
-20
5
-7
17
-10
9
12
-15
-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
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?
Give your answer as a fraction.
Solution;
2+x+y=10/3*3=10

15. 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 present=100-35=65
movie=2/5*35=14 no movie=35-14=21
Hence our answer is 21/100.
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

16. 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
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?
ways
Solution;

17. 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
40. A
120 a
O 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)
In triangle AOP,

18. 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?
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

19. 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
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

20. 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.