# Common Quantitative Comparison Practice Test 2

Question – 1

1. 56 times a two-digit number is greater than its square by 17 times 15.?

Column-A Column-B
The number The additive inverse of the number

• 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

• Answer:A

• Answer Explanation:
Let the number be x.
According to the given conditions
56x-x^2=17*15
56x-x^2=255
x^2-56x+255=0
x^2-51x-5x+255=0
x(x-51)-5(x-51)=0
(x-51)(x-5)=0
x=51,5
The number is 51
The additive inverse of the number is -51
Option A is correct.
[x^2=x*x]

Question – 2

2. There are 30 students in a class.?

Column-A Column-B
The number of ways in which three leaders can be chosen The number of ways in which the top three academic positions can be

• 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

• Answer:B

• Answer Explanation:
Three leaders out of 30 can be chosen in C(30,3) ways
C(30,3)=30!/[(30-3)!3!]
= 30!/(27!3!)
= 30*29*28/(3*2)
= 4060Top three positions can be secured in P(30,3) ways
P(30,3)=30!/(30-3)!
= 30*29*28
= 24360
Option B is correct.

Question – 3

3.

Column-A Column-B
The average of the cubes of the natural numbers from 1 to 35. The cube of the average of natural numbers from 1 to 35.

• 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

• Answer:A

• Answer Explanation:
The sum of the cubes of n natural numbers is given by {n(n+1)/2}^2
The average of the cubes of n numbers = Sum of cubes/n
= [{n(n+1)/2}^2]/n
= n(n+1)^2/4
Required average = 35*(1+35)^2/4
= 35*36*36/4
= 11340The average of n natural numbers = (n+1)/2
The average of 35 natural numbers = (35+1)/2
= 36/2
= 18
Cube of the average = 18^3= 5832
Option A is correct.
[n^2=n*n]

Question – 4

4.

Column-A Column-B
P(n,r) C(n,r-1)

• 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

• Answer:A

• Answer Explanation:
P(n,r)=n!/(n-r)!
C(n,r-1)=n!/[(n-r+1)!(r-1)!]
= n!/[(n-r+1)(n-r)!(r-1)!]
= P(n,r)/[(n-r+1)(r-1)!]
Hence, P(n,r)>C(n,r-1)
Option A is correct.

Question – 5

5. The radius of a sphere increases by 20%.?

Column-A Column-B
Percentage increase in its surface area Percentage increase in its volume

• 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

• Answer:B

• Answer Explanation:
Let the radius be r units.
New radius= r+20% of r
= r+(20/100)*r
= 12r/10Original surface area = 4*pi*r^2
New surface area =4*pi*(12r/10)^2
= 4*pi*144r^2/100

Percentage increase in surface area =(new surface area-original surface area)/original surface area*100
= [4*pi*144r^2/100 – 4*pi*r^2]/(4*pi*r^2)*100
= [144/100-1]/1*100
= 44%

Original volume = 4/3*pi*r^3
New volume = 4/3*pi*(12r/10)^3
= 4/3*pi*1728/1000*r^3

Percentage increase = [4/3*pi*1728/1000*r^3 – 4/3*pi*r^3]/[4/3*pi*r^3]*100
= 728/1000*100
= 72.8%
Option B is correct.

[r^2=r*r]
[pi=22/7]

Question – 6

6. There are three events A, B and C, one of which must happen and at a time, only one can happen. The odds are 8 to 3 against A, 5 to 2 against B.?

Column-A Column-B
Probability of occurrence of C Probability of occurrence of 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

• Answer:A

• Answer Explanation:
Probability of occurrence of A, P(A)=3/(8+3) = 3/11
Probability of occurrence of B, P(B)=2/(5+2) = 2/7
Since one and only one event can occur at a time,we have
P(A)+P(B)+P(C)=1
3/11+2/7+P(C)=1
P(C)=1-2/7-3/11
=(77-22-21)/77
=34/77
P(A)=3/11 and P(C)=34/77
P(A)

Question – 7

7.

Column-A Column-B
Distance travelled by the tip of a minutes’ needle in 35 minutes Distance travelled by Rob in 35 minutes at the speed of 5 km/hr

• 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

• Answer:D

• Answer Explanation:
We cannot calculate the distance travelled by the minutes’ needle since we do not know the length of the needle.
Hence, the comparison cannot be made.
Option D is correct.

Question – 8

8. A child runs along the four sides of a square playground at speeds 200, 300, 600 and 800 m/hr.?

Column-A Column-B
Average speed of the child around the playground m/hr 384 m/hr

• 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

• Answer:B

• Answer Explanation:
Let each side of the square be a km.
Speed = Distance/time
Time = a/200+a/300+a/600+a/800
= a/100(1/2+1/3+1/6+1/8)
= a/100*
12+8+4+3)/24
= a/100*27/24
= 9a/800
Average speed = Total distance/time
= 4a/9a/800
= 4*800/9
= 355.55 m/hr
Option B is correct.

Question – 9

9. P(x,-1) is the centroid of the triangle ABC with vertices A(3,-5), B(-7,2x) and C(5x,-2).?

Column-A Column-B
PA PC

• 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

• Answer:B

• Answer Explanation:
Since (x,-1) is the centroid of the triangle, we have
x = (3-7+5x)/3 and -1=(-5+2x-2)/3
3x=-4+5x and -3 = -7+2x
5x-3x=4 and -3+7=2x
2x=4 and 4=2x
x=2P(2,-1), A(3,-5) and C(10,-2)
PA=Sqrt[(3-2)^2+(-5+1)^2] and PC=Sqrt[(10-2)^2+(-2+1)^2]
PA=Sqrt(1+16) and PC=Sqrt(64+1)
PA=Sqrt(17) and PC=Sqrt
(65)
Option B is correct.

Question – 10

10.

Column-A Column-B
Revolutions made by a wheel of diameter 35 cm to cover a distance of 11 km. Revolutions made by a wheel of diameter 11cm to cover a distance of 35 km

• 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

• Answer:B

• Answer Explanation:
The wheel with smaller diameter will require more revolutions to cover the same distance as covered by a wheel with greater diameter. The wheel with diameter 35cm will take lesser revolutions than the wheel with diameter 11 cm to cover the same distance.
The distance covered by the smaller wheel is greater and hence it makes more revolutions to cover the distance required.Option B is correct.

Question – 11

11.

Column-A Column-B
Surface area of a cuboid that has diagonal sqrt(3)*x cm long Surface area of a cube that has diagonal sqrt(3)*x cm long

• 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

• Answer:D

• Answer Explanation:
The diagonal of a cube is given by Sqrt(3)*side
Hence, each side of the cube will be x cm long.
Surface area of the cube is 6*x^2 sq.cm.The side of the cuboid cannot be calculated and hence we cannot compare the surface areas.
Option D is correct.

Question – 12

12. A man sold an article for \$725 to earn a profit of 45%. He marks other similar articles 75% above the cost price and gives a discount of 20%.?

Column-A Column-B
Selling price after the discount is given Selling price for 45% profit

• 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

• Answer:B

• Answer Explanation:
Let CP, SP and MP be the cost price, selling price and marked price of the articles.
CP=100SP/(100+profit%)
CP=100*725/(100+45)
CP = 500MP = (100+profit%)CP/100
=(100+75)500/100
=175*500/100
= 875
SP = (100-20)*875/100
= 80*875/100 = 700
The article was sold for \$700

Selling price for 45% profit is \$725
Option B is correct.

Question – 13

13.

Column-A Column-B
Angle at which the diagonals bisect each other in a square Acute angle at which diagonals intersect in a rectangle

• 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

• Answer:A

• Answer Explanation:
The diagonals of a square bisect each other at right angles.
The diagonals of a rectangle intersect to give one pair of acute angles and one pair of obtuse angles. An acute angle is less than 90 degrees. Hence, option A is correct.

Question – 14

14. The equations (3-k)x+3y-2=0 and (k-3)x-(k-2)y-5=0 have no solution.?

Column-A Column-B
k 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

• Answer:D

• Answer Explanation:
The equations will have no solution if
(3-k)/(k-3) = 3/-(k-2) is not equal to -2/-5
(3-k)*(-k+2) = 3(k-3) and 3/-(k-2) is not equal to -2/-5
-3k+k^2+6-2k = 3k-9
k^2-8k+15=0
k^2-5k-3k+15=0
k(k-5)-3(k-5)=0
k=3, 5For k = 3, 3/-(k-2) = 3/-1=-3, which is not equal to -2/-5
For k = 5, 3/-(k-2) = 3/-3=-1, which is not equal to -2/-5
The values of k are 3 and 5
There are two values of k and hence we cannot compare k with 3.
Option D is correct

Question – 15

15. In a triangle ABC, E lies on AC and D lies on AB such that AE = AD and BD = CE.?

Column-A Column-B
BE CD

• 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

• Answer:C

• Answer Explanation:
We know that
AE=AD and CE=BD
Adding the two we get
AE+CE=AD+BD
AC=AB       (1)In triangles CBE and BCD,
CE = BD
CB is the common side
Angle ECB = angle DBC (since angles opposite equal sides in a triangle are equal and AC=AB)
Hence, triangle CBE is congruent to triangle BCD by SAS congruence condition.
Consequently, BE=CD.
Option C is correct.