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**.

- 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.

- 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

- 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