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TS CAB Recruitment Quantitative Aptitude Quiz 2

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TS CAB Recruitment Quantitative Aptitude Quiz 2

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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 2 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 2 will assist the students to know the expected questions from Quantitative Aptitude.


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1. The ratio of Rohan’s age 4 years ago and Rahul’s age after 4 years is 1 : 1. If at present, the ratio of their ages is 5 : 3, then find the ratio between Rohan’s age 4 years hence and Rahul’s age 4 years ago.


    A. 1: 3
    B. 3: 1
    C. 4: 3
    D. 3: 4


Answer: Option B

Explanation:

If ages in the numerical are mentioned in ratio A: B, then A: B will be Ax and Bx
1) At present: Ratio of their ages = 5 : 3. Therefore, 5 : 3 will be 5x and 3x.
Rohan’s age 4 years ago = 5x – 4
Rahul’s age after 4 years = 3x + 4

2) A ratio of Rohan’s age 4 years ago and Rahul’s age after 4 years is 1: 1
Therefore,

\(\frac {(5x – 4)}{(3x + 4)}\) = \(\frac {1}{1}\)

Solving, we get x = 4

3) We are asked to find the ratio between Rohan’s age 4 years hence and Rahul’s age 4 years ago.
Rohan’s age : (5x + 4)
Rahul’s age: (3x – 4)
The ratio of Rahul’s age and Rohan’s age
\(\frac {(5x – 4)}{(3x + 4)}\) = \(\frac {24}{8}\) = \(\frac {3}{1}\)
= 3: 1


2. \(\sqrt {3}^{n}\) = 2187, then , find the value of n

    A. 7
    B. 9
    C. 11
    D. 14


Answer: Option D

Explanation:

2187 = 3\(^{7}\)
\(\sqrt {3}^{n}\) = 3\(^{7}\)

Squaring both the sides, we get
\(\sqrt {3}^{n}\) = (3\(^{7})^{2}\)
\(\sqrt {3}^{n}\) = 3\(^{14}\)
Therefore, n = 14


3. What is the investment made if one invests in 15% stock at 50 and earns Rs.2000?

    A. 8000
    B. 7000
    C. 5000
    D. 6000


Answer: Option A

Explanation:
To earn Rs.15, investment = Rs.50.
Hence, to earn Rs.1500, investment = \(\frac {1500 × 50}{15}\)
= Rs.5000


4. Two friends A and B apply for a job in the same company. The chances of A getting selected is \(\frac {2}{5}\) and that of B is \(\frac {4}{7}\). What is the probability that both of them get selected?

    A. \(\frac {8}{35}\)
    B. \(\frac {34}{35}\)
    C. \(\frac {27}{35}\)
    D. None of these


Answer: Option A

Explanation:
P(A) = \(\frac {2}{35}\)
P(B) = \(\frac {4}{7}\)
E = {A and B both get selected}
P(E) = P(A) X P(B)
= \(\frac {2}{5}\) X \(\frac {4}{7}\)
= \(\frac {8}{35}\)


5. Find the value of \(\frac {(.555 \times .555 – .555 \times .020 + .020 \times .020)}{(.555 \times .555 \times .555) + (.020 \times .020 \times .020)}\)


    A. 1.55
    B. 1.74
    C. 2.36
    D. 5.02


Answer: Option D

Explanation:
The given numerical is in the form \(\frac {({a}^{2} – ab +{b}^{2})}{({a}^{3} + {b}^{3})}\) = \(\frac {({a}^{2} – ab + {b}^{2})}{(a + b) ({a}^{2} – ab + {b}^{2})}\) = \(\frac {1}{(a + b)}\)

\(\frac {(0.555)^{2} – (0.555 × 0.020) + (0.20)^{2}}{(0.555)^{3} + 0.020^{3}}\) = \(\frac {1}{(0.555 + 0.020)}\) = 1.74

This type of numerical can be easily solved if all basic formulae are known.

1. The average of first five prime numbers greater than 20 is?

    A. 31.00
    B. 31.01
    C. 32.00
    D. 32.2


Answer: Option D

Explanation:
23 + 29 + 31 + 37 + 41 = \(\frac {161}{5}\) = 32.2


2. Monika is twice as good as Sonika and together they complete a piece of work in 20 days. In how many days will Monika alone will finish the work?


    A. 22 days
    B. 30 days
    C. 37 days
    D. 52 days


Answer: Option B

Explanation:
If ‘A’ is ‘x’ times as good a workman as ‘B’, then the ratio of work done by A & B = x: 1
Monika’s 1-day work: Sonika’s 1-day work = 2:1
(Monika’s + Sonika’s ) 1-day work = \(\frac {1}{20}\)

Divide 1/20 in the ratio 2:1 ———( To divide the number ‘a’ into ratio x & y , we have first part = \(\frac {ax}{x + y}\))

Therefore, Monika’s 1 day work = (\(\frac {1}{20}\)) x (\(\frac {2}{2}\) + 1) = \(\frac {1}{30}\)

Hence, Monika will alone finish the work in 30 days.


3. The ratio of the areas of a square and rhombus whose base is same is:

    A. 1:2
    B. 2:1
    C. 1:1
    D. 3:1


Answer: Option C

Explanation:
The square and the rhombus are equal in the area if they have a common base.


4. A certain sum amounts to Rs. 7000 in 2 years and to Rs. 8000 in 3 years. Find the sum.


    A. Rs. 6959.37
    B. Rs. 6459.37
    C. Rs. 5359.37
    D. Rs. 5759.37


Answer: Option C

Explanation:
We are given,
1. Amount = Rs. 7000, Time = 2 years
2. Amount = Rs. 8000, Time = 3 years
To calculate the sum, we must first calculate the rate of interest.

Simple interest = \(\frac {PRN}{100}\)

S.I. on Rs. 7000 for 1 year = Rs. (8000 – 7000) = Rs. 1000

1000 = \(\frac {7000 × R × 1}{100}\)

R = 11 \(\frac {23}{7}\)%
Let sum be Rs. X
7000 = P \((1 + \frac {100}{7 \times 100})^{2}\)
\(\frac {8}{7}\) × \(\frac {8}{7}\) × P = 7000

P = 5359.37


5. If the banker’s discount on a certain amount of money is Rs.80 and the true discount on the same amount for the same time duration is Rs.64, then the amount due is:

    A. 320
    B. 360
    C. 400
    D. None of these


Answer: Option A

Explanation:
Amount due = (\(\frac {B.D. \times T.D.}{B.D. – T.D.}\))
(\(\frac {80 \times 64}{80 – 64}\))
= Rs.320

1. Find L.C.M. of \(\frac {2}{3}\),\(\frac {8}{9}\),\(\frac {64}{81}\),\(\frac {10}{27}\)


    A. \(\frac {250}{9}\)
    B. \(\frac {160}{3}\)
    C. \(\frac {128}{9}\)
    D. \(\frac {320}{3}\)


Answer: Option D

Explanation:
L.C.M. = \(\frac {L.C.M. of Numerator}{H.C.F. of Denominator}\)
L.C.M. of numerators = 2, 8, 64, 10
2 = 2\(^{1}\)
8 = 2\(^{3}\)
64 = 2\(^{6}\)
10 = 2 × 5
L.C.M of 2, 8, 64, 10 = 2\(^{6}\)6 × 5 = 320

H.C.F. of denominators = 3, 9, 81, 27
3 = 3\(^{1}\)
9 = 3\(^{2}\)
81 = 3\(^{4}\)
27 = 3\(^{3}\)
H.C.F. of 3, 9, 81, 27 = 3
L.C.M. of \(\frac {2}{3}\),\(\frac {8}{9}\), \(\frac {64}{81}\) \(\frac {10}{27}\), = \(\frac {320}{3}\)


2. In a mixture of 13 liters, the ratio of milk and water is 3: 2. If 3 liters of this mixture is replaced by 3 liters of milk, then what will be the ratio of milk and water in the newly formed mixture?


    A. 10 : 3
    B. 8: 5
    C. 9: 4
    D. 1: 1


Answer: Option C

Explanation:
Given: Total quantity of mixture = 13 liters
3 liters of the mixture is removed from the container – So, let’s forget this altogether!
Now, you have left with only 10 liters of the mixture in 3:2 ratio.
Milk in 10 litres mix = 10 x \(\frac {3}{(2 + 3)}\) = 6 litres
Water in 10 litres mix = 10 x \(\frac {2}{(2 + 3)}\) = 4 litres
We add 3 liters milk to this.
So, milk in new mix is = 6 liters + 3 litres = 9 litres
Water= 4 litres
Ratio of milk : water = 9 : 4


3. Smith and Kate started a business investing Rs. 84,000 and Rs. 28,000 respectively. In what ratio the profit earned after 2 years be divided between Smith and Kate respectively?


    A. 2 : 3
    B. 3 : 1
    C. 13 : 3
    D. None of these


Answer: Option B

Explanation:
P’s share of profit = \(\frac {x}{y}\) – – – – – – (x and y are investments)
Q’s share of profit

x : y = P’s share of profit : Q’s share of profit
Therefore,
\(\frac {Smith’s share of profit}{Smita’s share of profit}\) = \(\frac {84000}{28000}\) =\(\frac {3}{1}\)

The profit earned after 2 years will be divided between Smith and Kate in the ratio of 3: 1.


4. The remainder is 29, when a number is divided 56. If the same number is divided by 8, then what is the remainder?

    A. 3
    B. 4
    C. 7
    D. 5


Answer: Option D

Explanation:
We know that,
Dividend = [(Divisor × Quotient)] + Remainder
It is given that, the remainder is 29, when a number (dividend) is divided 56(divisor).
Dividend and quotient are unknown, hence assume dividend as X and quotient as Y.
Therefore,
X = 56 × Y + 29
56 is completely divisible by 8, but 29 is not completely divisible and we get the remainder as 5, which is the required answer.
OR
X = 56 × Y + 29
= (8 × 7Y) + (8 × 3) + 5
5 is the required remainder.


5. There is a road besides a river. Two friends Ram & Shyam started their journey from place P, moved to the garden located at another place Q & then returned to place P. Ram moves by swimming at a speed of 15 km/hr while Shyam sails on a boat at a speed of about 12 km/hr. If the flow of water current is at the speed of 6 km/hr, what will be the average speed of boat sailor?


    A. 6 km/hr
    B. 9 km/hr
    C. 12 km/hr
    D. 18 km/hr


Answer: Option B

Explanation:

Average Speed =\(\frac {Downstream Speed \times Upstream Speed}{Speed in still water}\)
=\(\frac {(x + y) (x – y)}{x}\)km/hr

Speed of boat in still water = y \(\frac {(t2 + t1)}{(t2 – t1)}\)km/hr

As Ram swims both ways at the speed of 15 km/hr, the average speed of swimming is 15 km/hr.
Being a boat sailor, Shyam moves downstream at speed = 12 + 6 = 18 km/hr & upstream at speed = 12 – 6 = 6 km/hr

Therefore, average speed of boat sailor = Downstream speed x Upstream speed / speed in still water
=\(\frac {[Downstream Speed \times Upstream Speed]}{[(1/2) \times ([Downstream Speed + Upstream Speed])]}\)

=\(\frac {(18 \times 6)}{[(1/2) \times (18 + 6)]}\)
=\(\frac {2 \times 18 \times 6}{18 + 6}\)
= 9 km/hr


IBPS RRB PO – Related Information
IBPS RRB Quantitative Aptitude Practice Set 1
SBI Clerk Mains Quantitative Aptitude Practice Set 1
SBI Clerk Mains Quantitative Aptitude Practice Set 2
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