# Common Quantitative Comparison Practice Test 3

Question – 1

1.

Column-A Column-B
(a^2-ab+b^2)/(a^2+ab+b^2) when a = -b (a^2-ab+b^2)/(a^2+ab+b^2) when a = b

(Note: [a^2=a*a])

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

When a =-b
(a^2-ab+b^2)/(a^2+ab+b^2)
=[(-b)^2-(-b)b+b^2]/[(-b)^2+(-b)b+b^2]
=(b^2+b^2+b^2)/(b^2-b^2+b^2)
= 3b^2/b^2
= 3When a=b
(a^2-ab+b^2)/(a^2+ab+b^2)
=[(b)^2-(b)b+b^2]/[(b)^2+(b)b+b^2]
=(b^2-b^2+b^2)/(b^2+b^2+b^2)
= b^2/3b^2
= 1/3
Option A is correct.

Question – 2

2. A task was to be completed in 20 days by a team. 12 members of the team were dropped and the remaining completed the work in 32 days.?

Column-A Column-B
Half of the number of members initially in the team Number of members dropped from the team

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

Let there be x members in the team initially
x members complete the work in 20 days
Work done by 1 person in 1 day = 1/20x
Work done by 1 person in 32 days = 32/20x = 8/5xNumber of persons after 12 members were dropped = (x-12)
Work done by (x-12) persons in 32 days = work done by 1 person in 32
days*number of persons
1=(8/5x)*(x-12)
5x=8x-8*12
5x=8x-96
8x-5x=96
3x=96
x=32

There were 32 persons initially
Half of 32=32/2=16
Number of members dropped from the team = 12
Option A is correct.

Question – 3

3. (a/b)^(x-6) = (b/a)^(2x-3)?

Column-A Column-B
-x^2. x

(Note: [a^2=a*a])

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

(a/b)^(x-6) = (a/b)^[-(2x-3)]
(a/b)^(x-6) = (a/b)^(3-2x)
x-6=3-2x
2x+x=3+6
x=9/3=3
-x^2=-3^2=-9<3

-x^2

Question – 4

4. The average of x and y is 25. x is 200% of the average of 5, 22, 13 and y.?

Column-A Column-B
x (x+y)/3

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

The average of the numbers is (5+22+13+y)/4
= (40+y)/4
200% of this average is x
x = 200/100*(40+y)/4
x = (40+y)/2
2x = 40+y
2x-y = 40     …….(1)
Also the average of x and y is 25.
(x+y)/2=25
x+y = 25*2=50     ….(2)Add (1) and (2)
2x-y+x+y=40+50=90
3x=90
x=30
Putting x=30 in x+y=50, we get
y=50-30=20

x=30
(x+y)/3=50/3=16.67

x>(x+y)/3
Option A is correct.

Question – 5

5. x+1/y=1 and y+1/z=1.?

Column-A Column-B
z1/x y1/z

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

x+1/y=1
x=1-1/y
=(y-1)/y
1/x=y/(y-1)     …..(1)y+1/z=1
1/z=1-y
z=1/(1-y)     …..(2)

Substituting values from (1) and (2), we get
z+1/x = 1/(1-y)+y/(y-1)
= 1/(1-y) – y/(1-y)
= (1-y)/(1-y) = 1

z+1/x=y+1/z
Option C is correct.

Question – 6

6. Consider a regular polygon.?

Column-A Column-B
Number of diagonals Number of interior angles

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

PConsider a triangle:
Number of diagonals = 0, number of interior angles = 3Consider a quadrilateral:
Number of diagonals = 2, number of interior angles = 4

Consider a pentagon:
Number of diagonals = 5, number of interior angles = 5
Hence, a relationship cannot be established.
Option D is correct.

Question – 7

7. A pipe can fill a tank with water in 6 hours. Due to a leak, the tank is filled in 7 hours.?

Column-A Column-B
Time taken to empty a full tank through the leak Two days

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

Let the time taken to empty the tank due to the leak be x hours.
Tank filled in one hour = 1/6
Tank emptied in 1 hour = 1/x
1/7=1/6-1/x
6x=7x-42
7x-6x=42
x=42
Time taken to empty the tank through the leak = 42 hoursTwo days = 24*2=48 hours
Option B is correct

Question – 8

8. log[(x+y)/5] = 1/2 (logx+logy).?

Column-A Column-B
x/y y/x

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

log[(x+y)/5] = 1/2 (logx+logy)
(x+y)/5 = sqrt(xy)
(x+y)=5*sqrt(xy)
Squaring both sides
(x+y)^2 = 5^2*xy
x^2+y^2+2xy = 25xy
x^2+y^2=25xy-2xy = 23xy
Dividing both sides by xy
x/y+y/x=23We cannot determine which of x/y and y/x is greater.
Option D is correct.

Question – 9

9. A person covers half of his journey at 30 km/hr and the other half at 20 km/hr.?

Column-A Column-B
His average speed for the whole journey His average speed for the first half of his journey

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

Let the distance be 2x km
Time = Distance/speed
Time for first half of journey = x/30
Time for second half of journey = x/20
Total time = x/30+x/20 = (20x+30x)/(20*30)
= 50x/(20*30)
= x/12Average speed = total distance/total time
= (x+x)/(x/12)
= 2x*12/x
= 24 km/hr

Average speed for the first half of his journey = 30 km/hr
Option B is correct.

Question – 10

10.

Column-A Column-B
The number of chords of a circle formed by joining the vertices of a cyclic pentagon. The number of diameters of a circle formed by joining

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

Any two points of the pentagon can be joined to form a chord.
The number of points is given by C(5,2)
C(5,2) = 5!/(2!3!)
= 5*4/2 = 10
10 chords can be drawn by joining the vertices of a cyclic pentagon.Of these 10 chords, 5 are the sides of the pentagon and 5 are the diagonals of the pentagon. None of these diagonals passes through the centre of the circle. Hence, the number of diameters is zero.
Option A is true.

Question – 11

11. A stick of length 30cm is at a distance of 8 cm from the centre of a hemi-spherical urn.?

Column-A Column-B
Length of the longest stick that can be placed in the urn. Diameter of the urn

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

The longest stick that can be placed shall be equal to the diameter of the urn.
Trivially option C is correct.

Question – 12

12. \$20000 is lent out in two parts at 8% and 15% simple interest respectively. The net interest in a year is \$2300.?

Column-A Column-B
First part Second part

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

Let the amount lent at 8% be \$x and the amount lent at 15% be \$(20000-x)
Simple interest, SI=PRT/100, where P, R and T are principal, rate and time respectively.
According to the conditions, we have
x*8*1/100+(20000-x)*15*1/100=2300
8x+300000-15x=230000
7x=70000
x=\$10000
First part = \$10000 and Second part = \$20000-\$10000=\$10000
Option C is correct.

Question – 13

13.

Column-A Column-B
The number of consecutive multiples of both 2 and 3 that lie between 200 and 350 The number of consecutive multiples of both 2 and 3 that

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

The numbers will be multiples of 6 since they are multiples of both 2 and 3.
The numbers lie between 200 and 350 and form an A.P. with common difference 6.
The first term of the A.P. will be 204 and the last term will be 348.
The nth term of an AP with first term ‘a’ and common difference ‘d’ is given by
Tn=a+(n-1)d
348 = 204 +(n-1)6
144 = (n-1)6
n = 24+1=25
25 such numbers exist.The numbers will be multiples of 6 since they are multiples of both 2 and 3.
The numbers lie between 350 and 500 and form an A.P. with common difference 6.
The first term of the A.P. will be 354 and the last term will be 498.
The nth term of an AP with first term ‘a’ and common difference ‘d’ is given by
Tn=a+(n-1)d
498 = 354 +(n-1)6
144 = (n-1)6
n = 24+1=25
25 such numbers exist.

Option C is correct.

Question – 14

14. The difference between the areas of two concentric circles is 500*pi. The sum of the radii of the circles is equal to 50 cm.?

Column-A Column-B

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

Let R and r be the two radii such that R>r
Area of a circle = pi*r^2, where r is the radius
pi*R^2-pi*r^2=500*pi
R^2-r^2=500
and R+r=50 cm
(R^2-r^2) = (R+r)(R-r) = 500
(50)(R-r)=500
R-r=500/50 = 10 cm
The difference between the radii is 10 cm.R+r=50 and R-r=10
Subtracting one relation from the other, we get
R+r-R+r=50-10=40
2r=40
r=20
Radius of the smaller circle = 20cm

Option B is correct.
[pi=22/7]
[r^2=r*r]

Question – 15

15. In a school function, the girls have to raise one hand and the boys have to raise both their hands holding colored pieces of cloth in each. The boys hold three pieces, one green and two red and the girls hold one green piece each.The number of green cloth pieces is 120 and the number of red cloth pieces is 40.?

Column-A Column-B
Half the number of girls among the group Twice the number of boys in the group

• A. If the quantity on the left is greater

• B. If the quantity on the right is greater

• C. If both are equal

• D. If the relationship cannot be determined without further information

Let the number of boys be x and the number of girls be y.
According to the conditions, we have
x+y = 120     …(1)
2x=40     …(2)
x = 40/2=20
y = 120 – x
= 120 – 20 = 100
Number of girls = 100
Number of boys = 20
Half the number of girls = 100/2=50
Twice the number of boys = 20*2=40
Option A is correct.