Common Quantitative Comparison Practice Test 5

Question – 1

1. On day one, Smith ran to a shop and came back walking. He took 8 minutes to cover this trip. On the next day, he takes 6 minutes as he runs both ways.?

Column-A Column-B
Time taken to complete his trip if he walks both ways Time taken to complete his trip two times around if he runs both ways.

• 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 his distance one way be k meters.
Speed = distance/time
Let his speed while walking be x m/min and while running be y m/min.
On day two, he runs both ways
y = 2k/6=k/3
k = 3y
On day one, he walks one way and runs the other way
k/x+k/y=8
k(x+y)=8xy
Putting k = 3y in this equation, we get
3y(x+y)=8xy
3x+3y=8x
3y=8x-3x=5x
y/x=5/3When he walks both ways
Time taken = k/x+k/x
= 3y/x+3y/x = 6y/x
= 6(5/3) = 2*5=10 mins

When he runs both ways
Time taken for going two times around = 2*6mins =12 mins

Option B is true.

Question – 2

2. A semi-circle is rolled and a cone is formed. The radius of the circle is 21cm.?

Column-A Column-B
Lateral surface area of the cone Total surface area of the semi-circle

• 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

Since the semi-circle is turned into a cone, the surface area of the semi-circle becomes the lateral surface area of the cone. The two quantities are equal.
Option C is correct.

Question – 3

3. A, B and C invest \$80000 in a business. A invests the most and B and C invest equal amounts. The profit earned by each is proportional to his share of investment. A gets \$4000 from the total profit.?

Column-A Column-B
Investment of A Total investment of B and C

• 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

We know the total investment, but we do not know the ratio in which A, B and C invested. Hence, we cannot calculate the total investment of A, B and C.
Option D is correct.

Question – 4

4. From a rectangular cardboard of length 25 cm and breadth 10 cm, Adam cuts a square whose diagonal is of length 4*sqrt(2) cm.?

Column-A Column-B
Number of such squares 30

• 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

Area of the cardboard = length*breath
= 25*10 = 250 sq.cm.
Diagonal of a square of side x cm is equal to x*sqrt(2) cm.
Hence, side of square = 4 cm
Area of square = side*side
= 4*4 = 16 sq.cm.Number of squares that he can possibly cut = 250/16
= 15.625
Option B is correct.

Question – 5

5. A dishonest shopkeeper uses an 800 gm weight instead of a 1 kg weight. He claims to sell the items at their cost price.?

Column-A Column-B
Profit \$50

• 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

We can only calculate the profit percent and we cannot calculate the profit earned by the shopkeeper since we do not have sufficient data.
Option D is correct.

Question – 6

6. 585 has to be converted to a perfect square.?

Column-A Column-B
The smallest number by which it should be multiplied The smallest number by which it should be divided

• 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

585=3*3*5*13
The smallest number by which it should be multiplied is 5*13=65
The smallest number by which it should be divided is 5*13=65
Option C is correct.

Question – 7

7. Consider three numbers a<=b<=c.?

Column-A Column-B
LCM of the three numbers HCF of the three numbers

• 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 LCM of three numbers is always greater than the HCF
When a=b=c
LCM=a and HCF=a
Hence, the relationship cannot be established.
Option D is correct.

Question – 8

8. log[(x+y)/3]=1/2(logx+logy).?

Column-A Column-B
(x/y+y/x) [(x+y)^2-2xy]/xy

• 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)/3]=1/2(logx+logy)
log[(x+y)/3] = log[sqrt (xy)]
(x+y)/3 = sqrt(xy)
Squaring both sides, we get
(x^2+y^2+2xy)/9=xy
x^2+y^2= 9xy-2xy
x^2+y^2=7xy
Dividing by xy, we get
x/y+y/x=7Also, (x^2+y^2)=7xy
(x+y)^2-2xy=7xy
[(x+y)^2-2xy]/xy=7

Option C is true.
[x^2=x*x]

Question – 9

9. x+1/x=3,?

Column-A Column-B
x^2+1/x^2 x^2+6/x-1/x^2

(Note: [x^2=x*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

x+1/x=3
Squaring both sides, we get
x^2+1/x^2+2*x*1/x=3^2
x^2+1/x^2=9-2=7
x^2+1/x^2=7x+1/x=3
x=3-1/x
Squaring both sides, we get
x^2=9+1/x^2-6/x
x^2-1/x^2+6/x=9
Option B is correct.

Question – 10

10. 2x+y=35 and 3x+4y=65.?

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

2x+y=35     ….(1)
3x+4y=65..(2)
Multiply (1) by 4 and subtracting (2) from it
8x+4y-3x-4y = 140-65
5x=75
x = 75/5=15Put x = 15 in (1)
2*15+y=35
30+y=35
y= 35-30=5
x/y=15/5=3
y/x=5/15=1/3
Option A is correct.

Question – 11

11.

Column-A Column-B
The sum of the first 100 even natural numbers divisible by 5. The sum of numbers between 35 and 95 divisible by 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 even natural numbers divisible by 5 are 10, 20, 30,….
These form an AP with the first term a = 10 and the common difference d = 20-10=10
Sum of n terms of an AP is given by
Sn=n/2[2a+(n-1)d]
=100/2[2*10+(100-1)*10]
=50(20+990)
=50*1010
= 50500The numbers divisible by 3 and lying between 35 and 95 are 36, 39, 42,….
These form an AP with the first term a = 36 and the common difference d = 3
The last term is 93 and it is given by 93 = 36+(n-1)*3
(93-36)/3=n-1
19=n-1
n=19+1=20
Sum of n terms of an AP is given by
Sn=n/2[2a+(n-1)d]
=20/2[2*36+(20-1)*3]
=10(72+57)
=10*129
= 1290
Option A is correct.

Question – 12

12. Seven-letter words are formed by the letters of the word EQUATION.?

Column-A Column-B
Number of words that end in a vowel Number of words having a vowel in the third place

• 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

There are 5 vowels and 3 consonants in the word EQUATION.
The last letter of the 7 letter words can be filled in 5 ways
The remaining 6 places can be filled by remaining 7 letters in P(7,6) ways
P(7,6) = 7!/(7-6)!
= 7!/1!=7*6*5*4*3*2= 5040 ways
Total words = 5*5040 = 25200Similarly, the third place can be filled in 5 ways and the remaining places can be filled in P(7,6) ways.
Option C is correct.

Question – 13

13. The parallel sides of a trapezium are 28m and 12m. The area of the trapezium is 180 sq.m.?

Column-A Column-B
Distance between the parallel sides 18 cm

• 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

Area of trapezium= 1/2(sum of parallel sides)*distance between parallel sides
180= 1/2*(28+12)*Distance
180*2/40= Distance
Distance = 9 m
9m>18cm
Option A is correct.

Question – 14

14. The volume of a solid sphere A is 87.5% less than the volume of solid sphere B.?

Column-A Column-B
Percent of surface area of A that the surface area of B is Percent of radius of A that the radius of B is

• 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 radii of sphere A and sphere B respectively.
Volume of sphere A = 4/3*pi*r^3
Volume of sphere B = 4/3*pi*R^3
According to the conditions, we have
Volume of A = volume of B – 87.5% volume of B
4/3*pi*r^3 = 4/3*pi*R^3 – 87.5/100*4/3*pi*R^3
r^3=R^3-875/1000*R^3
r^3=(1000-875)/1000*R^3
r^3= 125/1000*R^3
Taking cube root on both sides, we get
r = 5/10*R
= 2r/r*100
= 200%

Percentage of surface area = Surface area of B/surface area of A *100
= 4pi*R^2/(4pi*r^2)*100
= R^2/r^2*100
= (2r)^2/r^2*100
= 4r^2/r^2*100
= 400%

Option A is correct.
[r^3=r*r*r]

Question – 15

15.

Column-A Column-B
The maximum possible value of x when 10^x divides 10*11*…*20 The maximum possible value of x when 2^x divides 20*21*…*30

(Note: [10^x=10*10*10…x times])

• 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

Consider the product 10*11*…*20
The number of zeros can be calculated by multiplying the digits in the unit’s place in the product
10*12*15*20 = 10*180*20
= 36000
No other combination of numbers shall give a product ending in 0.
The product is divisible by 10^3
The maximum value of x is 3.20*21*22*…*30 = (2*2*5)*21*(2*11)*23*(2*2*2*3)*25*(2*13)*27*(2*2*7)*29*(2*15)
2^10 can divide the product.

Option B is correct.