# TS CAB Recruitment Quantitative Aptitude Quiz 6

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# TS CAB Recruitment Quantitative Aptitude Quiz 6

### Introduction

What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.

A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.

The article TS CAB Recruitment Quantitative Aptitude Quiz 6 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. The article TS CAB Recruitment Quantitative Aptitude Quiz 6 will assist the students to know the expected questions from Quantitative Aptitude.

### Quiz

1. The banker’s discount on Rs. 1600 at 15% per annum is the same as the true discount on Rs. 1680 for the same time and at the same rate. The time is:

A. 3 months
B. 4 months

C. 6 months

D. 8 months

Explanation:
S.I. on Rs. 1600 = T.D. on Rs. 1680.

Rs. 1600 is the P.W. of Rs. 1680, i.e., Rs. 80 is on Rs. 1600 at 15%.
Time = $$\frac {(100 Ã— 80}{1600 Ã— 15}$$year = $$\frac {1}{3}$$ year = 4 months.

2. Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:

A. Rs. 1 : 3
B. 3 : 2

C. 3 : 4

D. None of these

Explanation:
Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, the length of the first train = 27x meters,

and length of the second train = 17y meters.

z = $$\frac {27x + 17y}{x+ y}$$ = 23
27x + 17y = 23x + 23y

4x = 6y
$$\frac {x}{y}$$ = $$\frac {3}{2}$$

3. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?

A. 160
B. 175
C. 180
D. 195

Explanation:
Let the number of hens be x and the number of cows be y.

Then, x + y = 48 …. (i)

and 2x + 4y = 140 x + 2y = 70 …. (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

4. A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.

A. 2 hours
B. 3 hours
C. 4 hours
D. 5 hours

Explanation:
Speed downstream = (13 + 4) km/hr = 17 km/hr.
Time taken to travel 68 km downstream = $$\frac {68}{17}$$ hrs = 4 hrs

5. A man’s speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man’s speed against the current is:

A. 8.5 km/hr
B. 9 km/hr
C. 10 km/hr
D. 12.5 km/hr

Explanation:

Man’s rate in still water = (15 – 2.5) km/hr = 12.5 km/hr.

Man’s rate against the current = (12.5 – 2.5) km/hr = 10 km/hr.

1. A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B’s share?

A. Rs. 500
B. Rs. 1500
C. Rs. 2000

D. None of these

Explanation:
Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x – 3x = 1000

x = 1000.

B’s share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

2. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:

A. 17 kg
B. 20 kg
C. 26 kg

D. 31 kg

Explanation:
Let A, B, C represent their respective weights. Then, we have:

A + B + C = (45 x 3) = 135 …. (i)

A + B = (40 x 2) = 80 …. (ii)

B + C = (43 x 2) = 86 ….(iii)

Adding (ii) and (iii), we get: A + 2B + C = 166 …. (iv)

Subtracting (i) from (iv), we get : B = 31.

B’s weight = 31 kg.

3. 1, 4, 9, 16, 20, 36, 49

A. 1

B. 9

C. 20

D. 49

Explanation:

The pattern is $${1}^{2}$$, $${2}^{2}$$, $${3}^{2}$$, $${4}^{2}$$, $${5}^{2}$$, $${6}^{2}$$, $${7}^{2}$$. But, instead of $${5}^{2}$$, it is 20 which to be turned out.

4. At a game of billiards, A can give B 15 points in 60 and A can give C to 20 points in 60. How many points can B give C in a game of 90?

A. 30 points
B. 20 points
C. 10 points
D. 12 points

Explanation:

A : B = 60 : 45.

A : C = 60 : 40.

$$\frac {B}{C}$$ = $$\frac {B}{A}$$ x $$\frac {A}{C}$$ = $$\frac {45}{60}$$ x $$\frac {60}{40}$$ = $$\frac {45}{40}$$ = $$\frac {90}{80} =$$ = 90:80
B can give C 10 points in a game of 90.

5. How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?

A. 5600

B. 6000
C. 6400

D. 7200

Explanation:
Number of bricks = $$\frac { Volume of the wall}{Volume of 1 brick}$$ = $$\frac {800 Ã— 600 Ã— 22.5}{25 Ã— 11.25 Ã— 6}$$ = 6400

1. If 2994 Ã· 14.5 = 172, then 29.94 Ã· 1.45 = ?

A. 0.172
B. 1.72
C. 17.2
D. 172

Explanation:
$$\frac { 29.94}{1.45}$$ = $$\frac { 299.4}{14.5}$$
$$\frac {2994}{14.5}$$ X $$\frac { 1}{10}$$
$$\frac {172}{10}$$ = 17.2

2. The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:

A. 0.02
B. 0.2
C. 0.04
D. 0.4

Explanation:
Given expression =$${11.98}^{2}$$ + $${0.0}^{2}$$ + 11.98 x x.

For the given expression to be a perfect square, we must have

11.98 x x = 2 x 11.98 x 0.02 or x = 0.04
Then, Sachin’s age = (x – 7) years.

3. A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall is:

A. 720

B. 900
C. 1200
D. 1800

Explanation:
2(15 + 12) x h = 2(15 x 12)
h = $$\frac {180}{27}$$m = $$\frac {20}{3}$$m
Volume = 15 x 12 x $$\frac {20}{3}$$$${m}^{3}$$

4. A man purchased a cow for Rs. 3000 and sold it the same day for Rs. 3600, allowing the buyer a credit of 2 years. If the rate of interest be 10% per annum, then the man has a gain of:

A. 0%
B. 5%
C. 7.5%
D. 10%

Explanation:

C.P. = Rs. 3000.

S.P. = Rs. $$\frac {3600 Ã— 100 }{100 + (10 Ã— 2) }$$ = Rs. 3000.
100 + (10 x 2)
Gain = 0%.

5. In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay’s speed is:

A. 5 kmph
B. 6 kmph
C. 6.25 kmph
D. 7.5 kmph

Explanation:

Let Abhay’s speed be x km/hr.

Then, $$\frac {30 }{X }$$ – $$\frac {30}{2X }$$ = 3
6x = 30
x = 5 km/hr.