Multiple choice qestions

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

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.

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
Hence our answer is C.

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
Hence our answer is A.
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

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%

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.

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
Hence our answer is B.

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

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
Hence our answer is D.
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

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

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.

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.

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

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

C. 4
D. 6
E. 11
Solution;
1/211
517
=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.

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 80 feet of frontage each
8/9 of lots=40lots 100 feet of frontage each
x=5*80+40*100=400+4000=4,400

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 T
Q 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

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

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

D. 50
E. 52
Solution; 5 Square S
5 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.

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.

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

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.

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.

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

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 8 choices
can be from 0 to 9 except 5 9 choices
can be 1,3,5 or 7 4 choices

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

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 22
33
55
66
will end in how many zeroes?
A. 0
B. 2
C. 3
D. 5
E. 6
Solution;
22
33
55
66
=22
33
55
(2*3)6
=22
33
55
26
36
=28
39
55
=23
25
39
55
=23
105
39
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

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

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

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

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

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

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

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

Solution;
34% 34%
2%14% 14% 2%
400 500 600
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

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

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

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 2 3 4 5 6
1 (1,1) (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)
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

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;

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) (4,y)
(-2,-1) (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

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

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;

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.

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.

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.

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%

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.

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.

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

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

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.

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

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)

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 meadian and the
mode of the wages during the six days?
A. $5.5
B. $6.5
C. $7.5
D. $8.5

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

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)

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;

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 (i)
for Paul, 3nx+ny=12.60 (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

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

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?

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?

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;

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;

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

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