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TS CAB Recruitment Quantitative Aptitude Quiz 3

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TS CAB Recruitment Quantitative Aptitude Quiz 3

shape 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 3 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 3 will assist the students to know the expected questions from Quantitative Aptitude.


shape Quiz

1. A certain sum of money at simple interest amounted Rs.840 in 10 years at 3% per annum, find the sum?

    A. Rs.575
    B. Rs.675
    C. Rs.585
    D. Rs.685


Answer: Option C

Explanation:
90 = \(\frac {(450*4*R)}{100}\)
R = 5%
I = \(\frac {(450*6*5)}{100}\)= 135
450 + 135 = 585


2. The area of a base of a cone is 30 cm\(^ {2}\). If the height of the cone is 6cm, find its volume?

    A. 120 cm\(^ {3}\)
    B. 40 cm\(^ {3}\)
    C. 50 cm\(^ {3}\)
    D. 60 cm\(^ {3}\)


Answer: Option D

Explanation:
πr\(^ {2}\) = 30 h = 6
1/3 * 30 * 6 = 60.


3. In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?

    A. 57.5 m
    B. 127.5 m
    C. 150.7 m
    D. 98.6 m


Answer: Option B

Explanation:
When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = \(\frac {(900 * 700)}{800}\) = \(\frac {6300}{8}\) = 787.5 m.
In a race of 1000 m, A beats C by (1000 – 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = \(\frac {(600 * 212.5)}{1000}\) = 127.5 m.


4. [(3.572)\(^ {3}\) + (2.428)\(^ {3}\)] / [(3.572)\(^ {2}\) – (3.572) * (2.428) + (2.428)\(^ {2}\)] is equal to -.

    A. 1.144
    B. 4
    C. 5
    D. 6
    E. None of these


Answer: Option D

Explanation:

[(3.572)\(^ {3}\) + (2.428)\(^ {3}\)] / [(3.572)\(^ {2}\) – (3.572) * (2.428) + (2.428)\(^ {2}\)]

= {(3.572)\(^ {2}\) – (3.572) * (2.428) + (2.428)\(^ {2}\)}{3.752 + 2.428} / [(3.572)\(^ {2}\) – (3.572) * (2.428) + (2.428)\(^ {2}\)]
= 3.752 + 2.428 = 6.


5. Two trains of equal length, running with the speeds of 60 and 40 kmph, take 50 seconds to cross each other while they are running in the same direction. What time will they take to cross each other if they are running in opposite directions?

    A. 10 sec
    B. 9 sec
    C. 8 sec
    D. 7 sec


Answer: Option A


Explanation:

RS = 60 – 40 = 20 * \(\frac {5}{18}\) = \(\frac {100}{18}\)

T = 50

D = 50 * \(\frac {100}{18}\) = \(\frac {2500}{9}\)

RS = 60 + 40 = 100 * \(\frac {5}{18}\)

T = \(\frac {2500}{9}\) * \(\frac {18}{500}\) = 10 sec.

1. A ship was stocked with food to last for 40 days for 2500 sailors. However, some sailors could not board the ship and the food could last for 50 days. How many sailors could not board the ship?


    A. 400
    B. 500
    C. 700
    D. 1000


Answer: Option B

Explanation:
Men = M; Days = D; Time/Hours = T; Work = W
M1D1T1W2 = M2D2T2W1
Note that – W2 is on the left side and W1 is on the right side
Take work done = 1
Let the number of sailors who could not board = S.
So sailors who boarded = 2500-S
∴ 2500 sailors x 40 days x 1 = (2500-S) sailors x 50 days x 1
∴ S = 500 = Sailors who could not board the ship.


2. (3\(^ {a+1}\) 9\(^ {a+2}\) 27\(^ {a}\)) / (3\(^ {a-1}\) 9\(^ {a }\) 27\(^ {a+1}\)) = ?

    A. 3
    B. 9
    C. 27
    D. 1
    E. None of these


Answer: Option C

Explanation:
(3\(^ {a+1}\) 9\(^ {a+2}\) 27\(^ {a}\)) / (3\(^ {a-1}\) 9\(^ {a}\) 27\(^ {a+1}\)) = [3\(^ {a+1}\) 3\(^ {2(a+2)}\) \((3^{3})^{a}\)] / [3\(^ {a-1}\) \((3^{2})^{a}\) \((3^{3})^{a + 1}\) ] = 3\(^ {6a+5}\) / 3\(^ {6a+2}\) = 3\(^ {3}\) = 27.


3. Tap B is 5 times slower than Tap A in filling the same tank. Also, tap B takes 32 minutes more than Tap A to fill the same tank completely. How long will the tank take to get full, if both the taps are opened simultaneously?


    A.\(\frac {5}{32}\)minutes

    B.\(\frac {32}{5}\)minutes

    C.\(\frac {20}{3}\)minutes

    D.\(\frac {32}{3}\)minutes


Answer: Option C

Explanation:

Let Tap A take T minutes to fill the tank alone.
Since Tap A is 5 times faster than Tap B, Tap B takes 5 times more time.
So time taken by Tap B = 5T minutes
Also, 5T-T = 32 ——————- Given
∴ T = 8 minutes = Time taken by A
Time taken by B = 5 x 8 = 40 minutes.
In 1 min, A + B fills =\(\frac {1}{8}\)+\(\frac {1}{40}\) =\(\frac {3}{20}\)part of the tank

So the entire tank is filled in \(\frac {20}{3}\)minutes ————-> Inverse or reciprocal of\(\frac {3}{20}\)


4. To swim the upstream English Channel, a swimmer takes twice the time he takes to swim downstream. His speed in still water is 12km/hr. Find the speed of the English Channel.


    A. 3 km/hr
    B. 4 km/hr
    C. 6 km/hr
    D. 8 km/hr


Answer: Option B

Explanation:

Man’s/Boat’s Speed = X
Stream/Current/River Speed = Y

∴ Downstream speed = X + Y
Upstream speed = X – Y
X+Y = (12+Y) km/hr and X-Y = (12-Y) km/hr

Let time taken downstream be T hours.
∴ Upstream time = 2 x T = 2T hours

Distance is same
∴ D=D
∴ (12+Y) x T = (12-Y) x 2T
∴ Y=4 km/hr = Speed of English Channel.


5. A farm rears geese and dogs. The headcount in the farm is 84 and the leg count is 282. How many geese are there?


    A. 27
    B. 30
    C. 54
    D. 57


Answer: Option A

Explanation:
Let geese be denoted by ‘G’ and Dogs by ‘D’
Geese have 2 legs; Dogs have 4 legs.
Total Heads = G + D = 84 ————————- (1)
Total Legs = 2G + 4D = 282 ——————— (2)
Divide equation 2 by 2, we get,
G + 2D = 141 ————————————– (3)
Equation 3 – Equation 2
G + 2D – G – D = 141 – 84
∴ D = 57
So, Geese = 84 – 57 = 27

1. Ronit invests Rs. 40,000/- in a car wash center and starts a business. After 4 months, Ram joins the business with an investment of Rs.50,000. At the end of the year, they make a profit of Rs. 1,87,000/-. What will be Ram’s share in this profit?


    A. Rs. 38800
    B. Rs. 64666.67
    C. Rs. 85000
    D. Rs. 97000


Answer: Option C

Explanation:
Ratio of Investment x Time = Ratio of Profit
∴ (A’s investment x Time) : (B’s investment x Time) = Profit of A : Profit of B
∴ (Ronit’s Investment x Time) : (Ram’s Investment x Time) = Ronit’s Profit : Ram’s Profit
∴ 40000 x 12 : 50000 x 8 = Ronit’s Profit : Ram’s Profit
∴ Ronit’s Profit : Ram’s Profit = 4,80,000 : 4,00,000 = 6:5
∴ Ram’s Profit = \(\frac {5}{11}\)x 187000 = Rs. 85000


2. Ram saves Rs 3395/- from his salary. He needs to pay this money as milk bill, electricity bill and mobile phone bill in the ratio 42: 32: 23. Find the money to be paid for each bill.


    A. Rs 1245/-, Rs 1150/- and Rs 1000/-
    B. Rs 1470/-, Rs 1120/- and Rs 805/-
    C. Rs 1550/-, Rs 1235/- and Rs 610/ –
    D. Rs 1764/-, Rs 1022/- and Rs 529/-


Answer: Option B

Explanation:
Common factor helps in finding actual values easily
So, take ‘A’ as common factor.
∴ 3 numbers will now be 42A, 32A and 23A
∴ 42A + 32A + 23 A = 3395
∴ 97A = 3395
∴ A = 35
3 parts of 3395 are
42A = 42 x 35 = 1470;
32A = 32 x 35 = 1120
23A = 23 x 35 = 805
These are the amounts to be paid.


3. The two given numbers A and B are in the ratio 5:6 such that their LCM is 480. Find their HCF.


    A. 12
    B. 16
    C. 96
    D. 240


Answer: Option B

Explanation:
If A and B are two numbers,
A x B = HCF of A and B x LCM of A and B
Let K be a common factor. So 2 numbers are 5K and 6K
Also, K is the greatest common factor (HCF) as 5 and 6 have no other common factor
∴ 5K x 6K = 480 x K
K = 16 = HCF


4. The banker’s discount of a certain sum of money is Rs. 72 and the true discount on the same sum for the same time are Rs. 60. The sum due is:


    A. Rs. 360
    B. Rs. 432
    C. Rs. 540
    D. Rs. 1080


Answer: Option A


Explanation:

Sum =\(\frac {B.D. \times T.D.}{B.D. – T.D.}\)= Rs.\(\frac {72 \times 60}{72 – 60}\) = Rs. \(\frac {72 \times 60}{12}\) = Rs. 360.


5. A sum of Rs. 725 is lent at the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?


    A. 3.6%
    B. 4.5%
    C. 5%
    D. 6%
    E. None of these


Answer: Option E


Explanation:

Let the original rate be R%. Then, new rate = (2R)%.

Note:
Here, the original rate is for 1 year(s); the new rate is for only 4 months i.e. \(\frac {1}{3}\) year(s).

(\(\frac {725 \times R \times 1}{100}\) ) + (\(\frac {362.50 \times 2R \times 1}{100 \times 3}\)) = 33.50

(2175 + 725) R = 33.50 x 100 x 3

(2175 + 725) R = 10050

(2900)R = 10050

R = \(\frac {10050}{2900}\) = 3.46
Original rate = 3.46%


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