# Common Quantitative Comparison Practice Test 4

Question – 1

1.

Column-A Column-B
The maximum possible integral value of x if 2x-3<5 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

2x-3<5
2x<5+3
2x<8
x<4
The maximum possible integral value of x is 3.Option C is correct.

Question – 2

2. The lines represented by x+2y+7=0 and 2x+ky+14=0 are coincident.?

Column-A Column-B
k -k

• 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+2y+7=0 and 2x+ky+14=0
Since the lines are coincident, the equations have infinite solutions.
1/2=2/k=7/14
k = 2*2=4
k=4
-k=-4
Option A is correct.

Question – 3

3. 3x^2-2, 3x^2 +3, 5x^2-10 are in A.P. and x^2+15x-54=0.?

Column-A Column-B
x. -18

(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

Since the terms 3x^2-2, 3x^2 +3 and 5x^2-10 are in AP, we have
2(3x^2+3) = (3x^2-2)+(5x^2-10)
6x^2+6=3x^2-2+5x^2-10
6x^2-3x^2-5x^2=-2-10-6
-2x^2=-18
x^2=9
x=-3,3x^2+15x-54=0
x^2+18x-3x-54=0
x(x+18)-3(x+18)=0
(x-3)(x+18)=0
x=3, -18

The value of x is 3.
Option A is correct.

Question – 4

4. In a 500 m race, the ratio of the speeds of A to B is 3:4. A has a start of 140 m.?

Column-A Column-B
Time taken by A to finish the race Time taken by B to finish the race

• 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 speeds of A and B be 3x m/s and 4x m/s respectively.
A has a start of 140 m. Hence, A has to cover (500-140) = 360 meters
Time taken by A to cover 360 meters = distance/speed = 360/3x
= 120/x
Time taken by B to cover 500 meters = 500/4x
= 125/xHence, B takes longer to finish the race.
Option B is correct.

Question – 5

5. A two-digit number can be obtained by multiplying the sum of its digits by 10. The number can also be obtained by multiplying the digit in the tens place by 9 and adding 5 to it.?

Column-A Column-B
The digit in the hundreds place The digit in the units 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

Let the digit at the tens place be x and the digit at the units place be y.
The number will thus be (10x + y).
According to the given condition,
10x + y = 10(x + y)
10x – 10x + y -10y = 0
y = 0     …(1)
Also, we have10x + y = 9x+5
10x – 9x + y = 5
x + y = 5     …(2)y = 0
x + y = 5
Hence x = 5
The number is 50.

The digit in the units place is 0, the digit in the tens place is 5 and the digit in the hundreds place is 0.
Option C is correct.

Question – 6

6. The total surface area of a cylinder is 231 sq.cm and its curved surface area is 2/3 of the total surface area.?

Column-A Column-B
Diameter of its base Height of the cylinder

(Note: [pi = 22/7])

• 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 radius of the cylinder be r and the height be h.
Total surface area = 2pi*rh+2pi*r^2
According to the conditions, we have
2pi*rh+2pi*r^2=231 and 2pi*rh=(2/3)231=154     …(i)
154+2pi*r^2=231
2pi*r^2=231-154 = 77
r^2=77/2*7/22
= 49/4
r=sqrt(49/4)
r=7/2
Diameter = 2r = 2*7/2
Diameter = 7 cmPut r=7/2 in (i), we get
2*pi*7/2*h=154
h=154*7/22*2/7*1/2
h=7 cm
Option C is correct.
[r^2=r*r]

Question – 7

7. The distance between the points (x,-1) and (3,2) is 5 and the point (x,5) lies in the first quadrant.?

Column-A Column-B
Distance of (-x,0) from the origin Distance of (x,x) from the origin

• 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

Distance between (x,-1) and (3,2) = sqrt[(x-3)^2+(-1-2)^2]
5 = sqrt[x^2-6x+9+9]
Squaring both sides, we get
25=x^2-6x+18
x^2-6x-7=0
x^2-7x+x-7=0
x(x-7)+1(x-7)=0
(x+1)(x-7)=0
x=-1, 7
Since (x,5) lies in the first quadrant, x is positive.
Hence, x = 7Distance of (-7,0) from the origin = Sqrt[(-7-0)^2+(0-0)^2]
= Sqrt(49)=7

Distance of (7,7) from the origin = Sqrt[(7-0)^2+(7-0)^2]
= Sqrt(49+49)
=7*sqrt(2)

Option B is correct.
[x^2=x*x]

Question – 8

8. In a game, A can give B 12 points in 60 and A can give C 10 in 90.?

Column-A Column-B
The points that C can give B in a game of 70 The points A can give B in a game of 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

When A scores 90, C scores (90-10) = 80
When C scores 70, A scores 90/80*70 = 630/8 = 315/4
When A scores 315/4, B scores 48/60*315/4= 6*105/10=63
When C scores 70, B scores 63
C can give B (70-63) = 7 in a game of 70When A scores 60, B scores (60-12) = 48
When A scores 50, B scores 48/60*50 = 40
A can give B 50-40=10 points in a game of 50
Option B is correct.

Question – 9

9. The lengths of two parallel chords are 6 cm and 8 cm and the diameter of the circle is 10 cm.?

Column-A Column-B
The maximum possible distance between the two chords The radius of the 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

The chords are of lengths 6cm and 8cm.
A line segment perpendicular to a chord, passing through the centre bisects the chord.
Let these two line segments be x and y cm long for the chords of 6cm and 8cm respectively.
On joining the centre of the circle with one end of each chord, we get two right angled triangles.
The radius is the hypotenuse (h) and half the length of each chord (6/2=3 and 8/2=4) is the base of a triangle.
Applying Pythagoras theorem to each triangle, we get
h=sqrt(x^2+3^2) and h = sqrt(y^2+4^2)
Putting h = 10/2 = 5 and squaring each side, we have
5^2= x^2+3^2 and 5^2=y^2+4^2
25=x^2+9 and 25=y^2+16
x^2= 25-9=16 and y^2=25-16=9
x=4 and y=3
The maximum possible distance between the chords is 4+3 = 7cmThe radius of the circle is 5 cm.
Option A is correct.
[x^2=x*x]

Question – 10

10. Three times the larger of the two numbers, when divided by the smaller one gives quotient 4 and remainder 3. Seven times the smaller number when divided by the larger one gives quotient 5 and remainder 1.?

Column-A Column-B
The difference between the two numbers. Remainder when the sum of the numbers is divided by their difference

• 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 two numbers be x and y, x>y
According to the conditions, we have
3x=4y+3
3x-4y=3    ….(1)
7y=5x+1
5x-7y=-1    ….(2)Multiplying (1) by 5 and (2) by 3 and subtracting one from the other, we get
15x-20y-15x+21y=15+3
y = 18
Putting this value of y in (1), we get
3x-4*18=3
3x=3+72
x = 75/3=25
The numbers are 25 and 18. Their difference is 25-18 = 7

(25+18)/(25-18)=43/7 which gives us remainder 1
Option A is correct.

Question – 11

11. In a km race, A runs at 10 km/hr and B at 13 km/hr. A has a start of 100 m.?

Column-A Column-B
Time taken by A to finish the race Time taken by B to finish the race

• 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

Total distance = 1 km = 1000 m
Time taken by A to finish the race = Distance/speed
=(1-0.1)/10 =0.9/10
=0.09 hours
Time taken by B to finish the race = 1/13=0.076 hours
Hence, A takes more time than B does to finish the race.
Option A is correct.

Question – 12

12. 22<3x+1<34?

Column-A Column-B
The least possible integral value of x The number of sides of a heptagon

• 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

22<3x+1<34
22-1<3x<34-1
21<3x<33
21/37

Question – 13

13.

Column-A Column-B
Number of ways in which 10 people can stand in a queue Number of ways in which 10 chairs can be occupied by 10 people

• 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

10 people can stand in a queue in 10! ways10 people can occupy 10 chairs in 10! ways.
Option C is correct.

Question – 14

14.

Column-A Column-B
Slope of the line through the points (9,-3) and (7,-2) Slope of the line perpendicular to the line through (9,-3) and (7,-2)

• 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

Slope of the line through the points (9,-3) and (7,-2) = (-2+3)/(7-9)
= -1/2
Slope of the line perpendicular to the given line = -1/(-1/2)
= 2
Option B is correct.

Question – 15

15. 10 persons finish 2/5 of the work in 8 days.?

Column-A Column-B
Number of persons required to finish the remaining work in 12 days 14

• 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

10 persons finish 2/5 of the work in 8 days
Remaining work = 1-2/5 = 3/5
Work done by 10 persons in 8 days = 2/5
Work done by 1 person in 8 days = 2/(5*10)=1/25
Work done by 1 person in one day = 1/(25*8)
=1/200
Work done by 1 person in 12 days = 12/200 = 3/50
Total work = 3/5
Number of people required = total work/work done by 1 person in 12 days
= (3/5)/(3/50)=10
10 people are required to complete the remaining work in 12 days
Option B is correct.