# SSC CPO Quantitative Aptitude Quiz 5

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# SSC CPO Quantitative Aptitude Quiz 5

### 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 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.

### Quiz

1. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?

A. Rs. 4462.50
B. Rs. 8032.50
C. Rs. 8900
D. Rs. 8925
E. None of these

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$$^{2}$$

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

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

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

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.

1. A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

A. 35
B. 36$$\frac {1}{3}$$
C. 37$$\frac {1}{2}$$
D. 40

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

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

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. $$\frac {121}{11}$$

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

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.

1. If log 2 = 0.30103, the number of digits in 2$$^{64}$$ is:

A. 18
B. 19
C. 20
D. 21

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

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

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

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$$^{a + b + c}$$

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.