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 1** 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 1** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. Rs. 420

B. Rs. 540

C. Rs. 260

D. Rs. 640

**Answer**: Option B

**Explanation**:

Let the average expenditure per personnel = Rs. x

⇒ Original total expenses = Rs. 45x

Now total expenses = Rs. (45x + 54)

& new average expenditure per personnel = Rs. (x – 1) \(\frac {45x + 54}{45 + 9}\)hrs = \(\frac {45x + 54 }{54}\) = (x – 1)

⇒ 45x + 54 = 54x – 54

or, x = 12

∴ The original expenditure = 45 × 12 = Rs. 540

**2. The ratio of the quantity of water in fresh fruits to that of dry fruits is 7: 2. If 400 kg of dry fruits contain 50 kg of water then find the weight of the water in the same fruits when they were fresh?**

- A. 175 Kg

B. 100 Kg

C. 150 Kg

D. 125 Kg

**Answer**: Option A

**Explanation**:

Here, the weight of water in 400 kg of dry fruits is 50 kg.

So, the weight of fruits alone = (400 – 50) kg = 350 Kg

Here, Ratio of water in fresh fruits to the dry fruit = 7: 2

⇒ 2 = 50 Kg

∴ 1 = 25 kg

So, the water in fresh fruits = 7 × 25 = 175 kg

**3. If the ratio of simple interest and principal is 8 ∶ 25/2 and rate of interest is equal to the time invested then find the time of investment?
**

- A. 12 years

B. 16 years

C. 10 years

D. 8 years

**Answer**: Option D

**Explanation**:

Let the rate of interest be r% per annum.

We have a ratio of simple interest to the principal is = 16: 25

then SI = 16x and Principal 25 ×

And, rate of interest = time = r

16x =

\(\frac {25 × r × r}{100}\)

\({r}^{2}\) = 16 × 4

∴ r = 8

Here, the time of investment is 8 years.

**4. 4 men can develop a mobile app in 3 days. 3 women can develop the same app in 6 days, whereas 6 boys can develop it in 4 days. 3 men and 6 boys worked together for 1 day. If only women were to finish the remaining work in 1 day, how many women would be required?**

- A. 9

B. 12

C. 8

D. 10

**Answer**: Option A

**Explanation**:

∵ 4 men can develop a mobile app in 3 days.

3 men can develop it in \(\frac {3 × 4}{3}\) 4 days

6 boys can develop the mobile app in 4 days

App work is done by 3 men and 6 boys in one day = \(\frac {1}{4}\) + \(\frac {1}{4}\) = \(\frac {1}{2}\)

Remaining work = 1 – \(\frac {1}{2}\) = \(\frac {1}{2}\) of the work

∵ 3 women can develop the app in 6 days

∴ 18 women can develop the work in 1 day

Hence, to finish \(\frac {1}{2}\) app development in 1 day by women

We require only = 18 × \(\frac {1}{2}\)

**5. A student rides on a bicycle at 5 km/hr and reaches his school 3 minutes late. The next day he increased his speed to 7 km/hr and reached school 3 min early. Find the distance between his house and the school.
**

- A. 3 km

B. 2.75 km

C. 5 km

D. None of these

**Answer**: Option D

**Explanation**:

We know when distance remains constant then product of time and speed will also be same.

Let the time at which a student should reach his school bet hrs.

Therefore we have, 5 (t + \(\frac {3}{60}\)) = 7(t – \(\frac {3}{60}\))

On solving above equation ,we get

300t + 15 = 420 t – 21

t = \(\frac {36}{120}\)hrs.

t = \(\frac {3}{10}\)hrs.

Distance = Speed × time

Distance = 5 (\(\frac {3}{10}\) + \(\frac {3}{60}\)) = 1.75 km

- A. Rs. 123000

B. Rs. 112500

C. Rs. 138500

D. Rs. 132500

**Answer**: Option B

**Explanation**:

Following the information given in the question, we get

5 sons get 15000 × 5 = Rs. 75000

∴ 1 daughter gets 15000 ÷ 2 = Rs. 7500

∴ Wife gets = 15000 × 2 = Rs. 30000

Total amount = 75000 + 7500 + 30000 = Rs. 112500/-

**2. In covering a distance of 30 km Amit takes 2 hours more than Suresh, If Amit doubles his speed, he would take 1 hour less than Suresh. Amit’s speed is :
**

- A. 5 km/hr

B. 7.5 km/hr

C. 6 Km/hr

D. 6.25 km/hr

**Answer**: Option A

**Explanation**:

Let Amit’s speed = x km/hr and Suresh’s speed = y km/hr.

\(\frac {30}{x}\) – \(\frac {30}{y}\) = 2 & \(\frac {30}{y}\) – \(\frac {30}{2x}\) = 1

putting \(\frac {1}{x}\) = u and \(\frac {1}{y}\) = v , these questions become:

15u – 15v = 1 and –15u + 30v = 1

On solving these equations, we get u = \(\frac {1}{5}\) & v = \(\frac {2}{15}\)

\(\frac {1}{x}\) = \(\frac {1}{5}\) and \(\frac {1}{y}\) = \(\frac {2}{15}\) So, x = 5 & y = \(\frac {15}{2}\)

Hence, Amit’s speed = 5 km/hr.

**3. A driver of a auto rickshaw sees a lorry 60 meters ahead of him. After 30 seconds the lorry is 90 meters behind. If the speed of the auto-rickshaw is 38 kmph, then what is the speed of the lorry?**

- A. 23 kmph

B. 25 kmph

C. 20 kmph

D. 18 kmph

**Answer**: Option C

**Explanation**:

Relative speed = \(\frac {Total distance }{Total time }\) = \(\frac {60 + 90 }{30 }\) 5 m/s

Total time 30

∴ 5 m/s = 5 × \(\frac {18 }{5 }\) = 18 kmph

Now relative speed = speed of auto rickshaw – speed of lorry

or, 18 = 38 – the speed of the lorry

∴ Speed of lorry = 38 – 18 = 20 km/h

**4. Salman was traveling on one side of the Yamuna expressway with a constant speed of 120 mph in his car. Govinda was traveling with a constant speed of 80 km in the opposite direction. When they crossed each other, Salman decided to take a U-turn and meet him. But before taking a U-turn, Salman had to travel for another 3 minutes. How long will it take for Salman to meet Govinda? [Assume time taken by Salman to take U-turn is negligible]
**

- A. 29 minutes

B. 28 minutes

C. 30 minutes

D. 18 minutes

**Answer**: Option D

**Explanation**:

Distance Travelled by Salman in 3 minutes

= \(\frac {3 }{60}\) X 120 = 6 km

Distance travelled by Govinda in 3 minutes

\(\frac {3}{60}\) × 80 = 4 km

Hence Salman will have to travel a distance of 10 km before catching up with Govinda.

The relative speed of Salman to that of Govinda after taking U-turn = 120 – 80 = 40 kmph

∴ Time is taken by Salman after taking U-turn to Meet Govinda = \(\frac {10}{40}\) = 15 minutes

Total time taken by Salman to meet Govinda = (3 + 15)minutes = 18 minutes

**5. The taxi charges in a city consist of fixed charges and additional charges per kilometer. The fixed charges are for a distance of up to 5 km and additional charges are applicable per kilometer thereafter. The charge for a distance of 10 km is Rs. 350 and 25 km is Rs. 800. The charge for a distance of 30 km is-
**

- A. Rs. 800

B. Rs. 750

C. Rs. 900

D. Rs. 950

**Answer**: Option D

**Explanation**:

Let the fixed charges = Rs. x (for the first 5 km)

and the additional charges = Rs. y /km

As per the question,

x + 5y = 350 ….(i)

x + 20y = 800 …(ii)

On solving Eqn. (i) and (ii), we get

x = 200, y = 30

∴ Charge for a distance of 30 km.

= x + 25y = 200 + 30 × 25 = Rs. 950

- A. 20

B. 19

C. 25

D. 17

**Answer**: Option B

**Explanation**:

If R is the reduction in maximum speed and N is the number of passengers, then R is proportional to N

⇒ R = NK where K is constant of proportionality,

Given data: R = 20, when N = 5, hence K = 4

Maximum number of passenger that bus can’t move = (80 – N × K) = 80 – N × 4 ⇒ N = 20

So number of passenger for bust just move = 20 – 1 = 19

**2. In an exam of 100 marks, the average marks of a class of 40 students are 76. If the top 3 scorers of the class leave, the average score falls “down by 1. If the other two toppers except “the highest topper scored not more than 85. “then what is the minimum score the topper can score?**

- A. 86

B. 98

C. 95

D. 92

**Answer**: Option C

**Explanation**:

Total score of 40 students = (40 × 76) = 3040

Total score of top 3 scorers = 3040 – (37 × 75) = 265

To minimize the score of the top scorer, we assume the other two top scorers score the maximum they can = 85 marks each.

So, the top scorer scored = 265 – 170 = 95 marks.

**3. The average weight of three friends Amar, Visera, and Daman is 70 kg. Another person Vishal joins the group and now the average is 66 kg. If another person Tahir whose weight is 6 kg more than Vishal, joins the group replacing Amar, then the average weight of Visera, Daman, Vishal, and Tahir becomes 75 kg. What is the weight of Amar (in kg)?
**

- A. 18

B. 20

C. 22

D. 24

**Answer**: Option D

**Explanation**:

Total weight of Amar, Visera and Daman = 70 × 3 = 210 kg

Again, Amar + Visera + Daman + Vishal = 66 × 4 = 264 kg ……… (i)

∴ Weight of Vishal = 264 – 210 = 54 kg

∴ Weight of Tahir = 54 + 6 = 60 kg

Now, as per the question

Visera + Daman + Vishal + Tahir = 75 × 4 = 300 kg. ………………. (ii)

Subtracting (i) from (ii), we get

Tahir – Amar = 60 – Amar = 300 – 264 = 36

Therefore, weight of Amar = 60 – 36 = 24 kg

**4. The average salary of a company increases by 100 when the salary of the manager, which is Rs. 9500, is included. If the number of employees excluding the manager is the smallest cube divisible by 16, what is the final average of the company?
**

- A. Rs. 4000

B. Rs. 3400

C. Rs. 3700

D. Rs. 3100

**Answer**: Option D

**Explanation**:

The smallest cube divisible by 16 is 64.

Let’s assume the average salary before the manager’s salary is included is x

After addition of the Manager’s salary the average increases by 100

We can write down the above information in form of an equation as:

64x + 9500 = 65 × (x + 100)

Solving for x, we get x = 3000

The final average is 3000 + 100 = Rs. 3100

**5. An exam was conducted in a state over 222 centers. The average number of applicants percenter was found to be 1560. However, it was later realized that in one center, the number of applicants was counted as 1857 instead of 1747. What was the correct average number of applicants per center (up to two decimals)?**

- A. 1557.87

B. 1558.20

C. 1558.92

D. 1559.51

**Answer**: Option D

**Explanation**:

Number of applicants that have been counted extra = 1857 − 1747 = 110

Hence, decrease in average = \(\frac {110}{22}\) = 0.495

∴ Correct average = 1560 − 0.495 = 1559.505 = 1559.51