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 **SSC CPO Quantitative Aptitude Quiz 5** provides Quantitative Aptitude questions with answers useful to the candidates preparing for** Competitive exams, Entrance exams, Interviews** etc. The article **SSC CPO Quantitative Aptitude Quiz 5** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. Rs. 4462.50

B. Rs. 8032.50

C. Rs. 8900

D. Rs. 8925

E. None of these

**Answer**: Option D

**Explanation**:

Principal

= Rs. \(\frac{100 \times 4016.25}{9 \times 5}\)

= Rs. \(\frac{401625}{45}\)

= Rs. 8925.

**2. A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:**

- A. 49 m\(^{2}\)

B. 50 m\(^{2}\)

C. 53.5 m\(^{2}\)

D. 55 m\(^{2}\)

**Answer**: Option A

**Explanation**:

Area of the wet surface = [2(lb + bh + lh) – lb]

= 2(bh + lh) + lb

= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m\(^{2}\)

= 49 m\(^{2}\).

**3. If log\(_{10}\) 2 = 0.3010, then log\(_{2}\) 10 is equal to:**

- A. \(\frac {699}{301}\)

B. \(\frac {1000}{301}\)

C. 0.3010

D. 0.6990

**Answer**: Option B

**Explanation**:

log\(_{2}\) 10 = \(\frac {1}{log_{10} 2}\) = \(\frac {1}{0.3010}\) = \(\frac {10000}{3010}\) = \(\frac {1000}{301}\).

**4. In a 200 meters race A beats B by 35 m or 7 seconds. A’s time over the course is:**

- A. 40 sec

B. 47 sec

C. 33 sec

D. None of these

**Answer**: Option C

**Explanation**:

B runs 35 m in 7 sec.

B covers 200 m in (\(\frac {7}{35}\) x 200) = 40 sec.

B’s time over the course = 40 sec.

A’s time over the course (40 – 7) sec = 33 sec.

**5. The price of 2 sarees and 4 shirts is Rs. 1600. With the same money, one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay?**

- A. Rs. 1200

B. Rs. 2400

C. Rs. 4800

D. Cannot be determined

E. None of these

**Answer**: Option B

**Explanation**:

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 …. (i)

and x + 6y = 1600 …. (ii)

Divide equation (i) by 2, we get the below equation.

=> x + 2y = 800. — (iii)

Now subtract (iii) from (ii)

x + 6y = 1600 (-)

x + 2y = 800

—————-

4y = 800

—————-

Therefore, y = 200.

Now apply the value of y in (iii)

=> x + 2 x 200 = 800

=> x + 400 = 800

Therefore x = 400

Solving (i) and (ii) we get x = 400, y = 200.

Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

- A. 35

B. 36\(\frac {1}{3}\)

C. 37\(\frac {1}{2}\)

D. 40

**Answer**: Option D

**Explanation**:

Let distance = x km and usual rate = y kmph.

Then, \(\frac {x}{y}\) – \(\frac {x}{y+3}\) = \(\frac {40}{60}\) => 2y(y + 3) = 9x ….(i)

And,\(\frac {x}{y-2}\) – \(\frac {x}{y}\) = \(\frac {40}{60}\) => y(y – 2) = 3x ….(ii)

On dividing (i) by (ii), we get: x = 40.

**2. 18 men bind 900 books in 10 days. Find how many binders will be required to bind 600 books in 12 days? **

- A. 10

B. 11

C. 13

D. 15

**Answer**: Option A

**Explanation**:

We have to find the number of binders. Let the number of binders be x.

Direct Proportion:Less Books (↓),Less binders(↓)

Indirect Proportion:More days (↑),Less binders (↓)

\[ 18 : x ::

\begin{cases}

900 : 600 – – – (Books) \\

12 : 10 – – – (Days)

\end{cases}

\]

x × 900 × 12 = 18 × 600 × 10

x = \(\frac {18 \times 600 \times 10}{900 \times 12}\) = 10

**3. Find the value of \(^{20}c_{17}\)**

- A. 1260

B. 1140

C. 2580

D. 3200

**Answer**: Option B

**Explanation**:

\(^nc_{r}\) =\(\frac {^np_{r!}}{r!}\)

\(^nc_{r}\) = \(\frac {n!}{(r!) (n – r)!}\)

\(^{20}c_{17}\) =\(\frac {20!}{(17!) (20 – 17)!}\)

\(^{20}c_{17}\) =\(\frac {20 \times 19 \times 18 \times 17!}{(17!) (3)!}\)

\(^{20}c_{17}\) =\(\frac {20 \times 19 \times 18}{3 \times 2 \times 1}\)

\(^{20}c_{17}\) = 1140.

**4. (1331)\(^{- (2/3)}\)**

- A. –\(\frac {1}{11}\)

B. – \(\frac {11}{121}\)

C. \(\frac {1}{122}\)

D. \(\frac {121}{11}\)

**Answer**: Option C

**Explanation**:

Cube root of 1331 is 11. Therefore,

(11)\(^{3 \times -(2/3)}\)

Remember the law of indices (xm)n = xmn

(11)\(^{3 \times -(2/3)}\) = 11\(^{-2}\)

x\(^{-1}\) = \(\frac {1}{x}\)

Hence, 11\(^{-2}\) = \(\frac {1}{112}\) = \(\frac {1}{112}\)

**5. Two pipes can fill a tank in 8 hrs & 6 hrs respectively. If they are opened on alternate hours and if pipe A gets opened first, then in how many hours, the tank will be full?**

- A. 6 hrs

B. 7 hrs

C. 8 hrs

D. 14 hrs

**Answer**: Option B

**Explanation**:

Pipe A’s work in 1 hr = \(\frac {1}{8}\)

Pipe B’s work in 1 hr = \(\frac {1}{6}\)

Pipes (A+B)’s work in first 2 hrs when they are opened alternately = 1/8 + 1/6 = 7 /24

Now,

In 4 hrs they fill : 2 X (\(\frac {7}{24}\)) = \(\frac {7}{12}\)

In 6 hrs they fill : 3 X (\(\frac {7}{24}\)) = \(\frac {7}{8}\)

After 6 hrs, part left empty = \(\frac {1}{8}\)

Now it is A’s turn to open up.

In one hr it fills \(\frac {1}{8}\) of the tank.

So, the tank will be full in = 6 hrs + 1 hr = 7 hrs.

- A. 18

B. 19

C. 20

D. 21

** Answer**: Option C

**Explanation**:

log (2\(^{64}\)) = 64 x log 2

= (64 x 0.30103)

= 19.26592

Its characteristic is 19.

Hence, then number of digits in 2\(^{64}\)6 is 20.

**2. In a race of 800m, if the ratio of the speeds of two participants P and R is 3:4, and P has a start of 120m, then R beats P by:**

- A. 100m

B. 80m

C. 90m

D. 110m

** Answer**: Option B

**Explanation**:

As P has a start of 120m, P has to cover 680m while R has to cover 800m.

Now, when P covers 3m, R covers 4m.

Hence, when R covers 800m, P covers (800 * 3) / 4 = 600m

Therefore, R beats P by 80m.

**3. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?**

- A. 20

B. 30

C. 40

D. 50

** Answer**: Option B

**Explanation**:

There is a meal for 200 children. 150 children have taken the meal.

The remaining meal is to be catered to 50 children.

Now, 200 children = 120 men.

50 children = \(\frac {120 \times 50}{2500}\) = 30 men.

**4. Using all the letters of the word “NOKIA”, how many words can be formed, which begin with N and end with A?**

- A. 3

B. 6

C. 24

D. 120

E. 12

** Answer**: Option B

**Explanation**:

There are five letters in the given word.

Consider 5 blanks

The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.

The number of words = 3! = 6.

**4. (\(\frac {x^{b}}{x^{c}})^{b + c – a}\) . (\(\frac {x^{c}}{x^{a}})^{c + a – b}\) . (\(\frac {x^{a}}{x^{b}})^{a + b – c}\) = ?**

- A. x\(^{abc}\)

B. 1

C. x\(^{ab + bc + ca}\)

D. x\(^{a + b + c}\)

** Answer**: Option B

**Explanation**:

Given Exp.

= x\(^{(b – c)(b + c – a)}\) . x\(^{(c – a)(c + a – b)}\) . x\(^{(a – b)(a + b – c)}\)

= x\(^{(b – c)(b + c) – a(b – c)}\) . x\(^{(c – a)(c + a) – b(c – a)}\)

. x\(^{(a – b)(a + b) – c(a – b)}\)

= x\(^{(b^{2} – c^{2} + c^{2} – a^{2} + a^{2} – b^{2})}\) . x\(^{-a(b – c) – b(c – a) – c(a – b)}\)

= (\(x^{0} \times x^{0}\))

= (1 x 1) = 1.