Rejan's gre ebook

Quantitative Comparision 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

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

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.

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 QtyB

x x/y 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.

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

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.


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.

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

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.

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

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

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.

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

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 p

x 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.

33. N Q R T MN//PQ and PR//ST

40 55 x y

M 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.


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

B C

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

In the figure alongside,

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.

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

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

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

4 inches Hence our answer is D.

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;

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)

Since P(F) i.e., probability of occurrence of event F is unknown, we cannot determine P(E).


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.


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

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


50. B 8 C F 5 G

3 4

A D E 7 H

QtyA QtyB

Area of rectangular region Area of trapezoidal region

ABCD 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)

52. R QRS is an isosceles triangle.

QtyA QtyB

7 x Perimeter of triangle QRS 17

Q 4 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

Solution;

34% 34%

14% 14%

2% 2%

56 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

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

S x units 48 minutes(48/60 hours)

S 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

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.


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.

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.

2% 14% 34% 34% 14% 2%

200 210 220

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

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:

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 From -50 to -1, there will be 50 terms.

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

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;

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.

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. B QtyA QtyB

Area of semi-circular region Area of triangular region ABC

A O 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.

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

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 QtyA QtyB

y 70

20 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.


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 number of prime factors of 750=5

and 2n=2*750=2*5*5*5*2*3 number of prime factors of 2*750=6, which is 1 more than 5.

81. ab<0, bc>0

QtyA QtyB

ac 0

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 ac=+ve*-ve=-ve

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

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


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

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.

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

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


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;

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

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.

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.

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

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.

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.

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

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

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.

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

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 In the figure alongside,

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 I In the figure alongside,

H 60 J IJ=5 (Triangle IJH being an equilateral triangle)

5 C

G 60 K

That means our answer is D.

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 C

6 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.

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.

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

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.

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 f arc QPR

Solution;

Since angle PQR=angle PRQ, arc QP=arc PR (i)

(arcs opposite to equal circumscribed angles of a circle are equal)

Now arc PRQ= arc PR + arc RQ (ii)

arc QPR=arc QP + arc PR (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


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 QtyB

The average of x and y 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


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)


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.

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%

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;

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.

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.

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.

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.


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.

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.


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 The number of distinct prime

factors of x factors of 4x

Solution;

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

The number of distinct prime The number of distinct prime

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.


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.

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


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

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.

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,

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 QtyB

x(x+1)(x+2) x*x*x

Solution;

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.

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.


Rejan's gre ebook

Education

Transcript of Rejan's gre ebook