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 12** 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 12** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. 170

B. 160

C. 150

D. 165

**Answer**: Option D

**Explanation**:

Given : \(\frac{40}{20}\) – 5 * 10 + 5 = ?

Substituting the coded symbols for mathematical operations, we get,

40 * \(\frac{20}{5 }\) +10 – 5 = ?

40 * 4 + 10 – 5 = ?

160 + 10 – 5 = ?

170 – 5 = 165

**2. If Suresh distributes his pens in the ratio of 1/2 : 1/4 : 1/5 : 1/7 between his four friends A, B, C and D, then find the total number of pens Suresh should have?**

- A. 153

B. 150

C. 100

D. 125

**Answer**: Option A

**Explanation**:

Here, A : B : C : D = \(\frac{1}{2}\) : \(\frac{1}{4 }\) : \(\frac{1}{5 }\) : \(\frac{1}{7 }\)

1) L.C.M of 2, 4, 5, 7 is 140

2) Find the number of pens each friend received ——— (To find no. of pens each friend has, multiply the ratio with the L.C.M. calculated)

A = (\(\frac{1}{2}\)) x 140 = 70

B = (\(\frac{1}{4 }\)) x 140 = 35

C = (\(\frac{1}{5 }\)) x 140 = 28

D = (\(\frac{1}{7 }\)) x 140 = 20

3) Total number of pens = (70 x + 35 x + 28 x + 20 x) = 153 x

Minimum number of pens (x) = 1

Therefore, total number of pens = 153 pens

**3. When folded into two equal halves a rectangular sheet had a perimeter of 48cm for each part folded along one set of sides and the same is 66cm when folded along the other set of sides. Find the area of the sheet.**

- A. 1584

B. 1120

C. 792

D. 1320

**Answer**: Option B

**Explanation**:

Let the sheet be folded along its breadth and its perimeter = 48cm

Therefore, (\(\frac{L}{2 }\) + b) = 48 …… (i)

Now, let the sheet be folded along its length, and the perimeter = 66cm

(l + \(\frac{b}{2 }\))= 66 …… (ii)

Solving (i) and (ii), we get,

l = 56cm, b = 20cm

Area = l*b

Area = 1120 c\({m}^{2 }\)

**4. A clock was set right at 2 p.m. It gains 5 seconds in 3 minutes, and it indicates 8.30 a.m. the next morning, then the true time is:**

- A. 8.00 a.m.

B. 7.45 a.m.

C. 8.15 a.m.

D. 7.30 a.m.

**Answer**: Option A

**Explanation**:

Time elapsed from 2 p.m. to 8.30 a.m. = 18 hours 30 minutes = 1110 minutes.

Now, 3 minutes and 5 seconds of the given clock is 3 minutes of the correct clock.

Therefore, 1110 minutes of this clock is \(\frac{(1110 * 3)}{(37/12) }\) minutes of the correct clock.

= 1080 minutes = 18 hours of the correct clock

Hence, the correct time is 8 a.m.

**5. Find the number of times the hour hand and the minute hand of a clock are at right angle in a day.**

- A. 48

B. 24

C. 22

D. 44

**Answer**: Option D

**Explanation**:

In 12 hours, the hands of a clock are at right angle 22 times.

Hence in a day they are at right angle 44 times.

- A. 270°

B. 77.5°

C. 282.5°

D. 300°

**Answer**: Option C

**Explanation**:

Angle traced by the hour hand in 12 hours = 360°

Angle traced by the hour hand in 9 hours 35 minutes = \(\frac{(360 * 115) }{(12 * 12) }\)= 287.5°

Angle traced by the minute hand in 60 minutes = 360°

Angle traced by the minute hand in 35 minutes = 210°

Therefore, the angle between the hour hand and the minute hand at 9.35

= (287.5 – 210) = 77.5°

Reflex angle = 360 – 77.5 = 282.5°

**2. Find the angle between the hour hand and the minute hand of a clock when the time is 5.45.**

- A. 97.5°

B. 90°

C. 100°

D. 95°

**Answer**: Option A

**Explanation**:

Angle traced by the hour hand in 12 hours = 360°

Angle traced by the hour hand in 5 hours 45 minutes = \(\frac{(360 * 23) }{(12 *4) }\)= 172.5°

Angle traced by the minute hand in 60 minutes = 360°

Angle traced by the minute hand in 45 minutes = 270°

Therefore, the angle between the hour hand and the minute hand at 5.45

= (270 – 172.5) = 97.5°

**3. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then some of the number is:**

- A. 28

B. 32

C. 40

D. 64

**Answer**: Option C

**Explanation**:

Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

**4. Two trains moving in same direction run at a speed of 60 km/hr and 40 km/hr respectively. If a man sitting in slow train is passed by the fast train in 10 seconds, then what is the length of the faster train?**

- A. 53.2 m

B. 55.6 m

C. 150 m

D. 200 m

**Answer**: Option B

**Explanation**:

Given: Speed of slow train = 60 km/hr, speed of fast train = 40 km/hr

Here both the trains move in same direction. Hence their relative speed is obtained by subtracting the individual speeds of trains.

Relative speed = 60 – 40 = 20 km/hr

1) Convert km/hr into m/s

20 x 5 = 100 = 5.56 m/s

18 18

2) Distance (Length of faster train) = Speed x Time

Length of faster train = 5.56 x 10 m = 55.6 m

**5. Find the largest 4 digit number which is divisible by 88.**

- A. 8844

B. 9999

C. 9944

D. 9930

**Answer**: Option C

**Explanation**:

We know that the largest 4 digit number is 9999.

Simply divide 9999 by 88.

After dividing 9999 by 88 we get, 55 as remainder.

The number is said to be completely divisible, only if the remainder is zero.

Hence, we can find the required answer by subtracting the remainder obtained from the 4 digit number.

Therefore, required number = 9999 – 55 = 9944

- A. 3

B. 4

C. 7

D. 5

** Answer**: Option D

**Explanation**:

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 remainder as 5, which is the required answer.

OR

X = 56 × Y + 29

= (8 × 7Y) + (8 × 3) + 5

5 is the required remainder

**2. If 3 spiders make 3 webs in 3 days, then 1 spider will make 1 web in how many days?**

- A. 1 day

B. 1.5 days

C. 3 days

D. 6 days

** Answer**: Option C

**Explanation**:

Indirect proportion: Less spiders (↓) More days (↑)

Direct proportion: Less webs (↓) Less days (↓)

We are given, that 3 spiders make 3 webs in 3 days. This means that 3 spiders make a web in each day. 3 days for 3 webs. Hence, work done by each spider is \(\frac{1 }{3 }\).

Now a complete web by a single spider can be done by working for 1/3rd each day.

1st day spider will work for 1/3rd

2nd day = 1/3rd

3rd day = 1/3rd

Therefore one complete web can be made in = \(\frac{1 }{3 }\) + \(\frac{1 }{3 }\) +\(\frac{1 }{3 }\) = \(\frac{3 }{3 }\) = 1 complete web

This means a spider took 3 days to complete 1 complete web.

**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 C

**Explanation**:

To earn Rs.15, investment = Rs.50.

Hence, to earn Rs.1500, investment = \(\frac{(1500*50) }{15 }\)

= Rs.5000

**4. A company pays 12.5% dividend to its investors. If an investor buys Rs.50 shares and gets 25% on investment, at what price did the investor buy the shares?**

- A. 6.25

B. 25

C. 50

D. 12.5

** Answer**: Option B

**Explanation**:

Dividend on 1 share = \(\frac{(12.5 * 50) }{100 }\) = Rs.6.25

Rs.25 is income on an investment of Rs.100

Rs.6.25 is income on an investment of Rs. \(\frac{(6.25 * 100) }{25 }\) = Rs.25

**5. Find the number of shares that can be bought for Rs.8200 if the market value is Rs.20 each with brokerage being 2.5%**

- 450

B. 500

C. 400

D. 410

** Answer**: Option C

**Explanation**:

Let the initial investments of A and B be 3x and 5x.

A : B : C = (3x x 12) : (5x x 12) : (5x x 6) = 36 : 60 : 30 = 6 : 10 : 5.