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 23** 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 23** will assist the students to know the expected questions from **Quantitative Aptitude**.

- A. \(\frac{2}{3}\), \(\frac{3}{5}\), \(\frac{1 }{3}\), \(\frac{4 }{7}\), \(\frac{5}{6}\)

B. \(\frac{1}{3}\), \(\frac{2}{5}\), \(\frac{3 }{5}\), \(\frac{5 }{6}\), \(\frac{4}{7}\)

C. \(\frac{1}{3}\), \(\frac{2}{5}\), \(\frac{5}{6}\), \(\frac{4 }{7}\), \(\frac{3}{5}\)

D. \(\frac{1}{3}\), \(\frac{2}{5}\), \(\frac{4 }{7}\), \(\frac{3 }{5}\), \(\frac{5}{6}\)

**Answer**: Option D

**Explanation**:

Convert fractions into decimal form. These calculations are to be solved quickly in mind otherwise will require a lot of time.

Option 1: 0.4, 0.6, 0.33, 0.5, 0.8 — (wrong)

Option 2: 0.3, 0.4, 0.6, 0.8, 0.5 — (wrong)

Option 3: 0.3, 0.4, 0.8 , 0.5, 0.6 — (wrong)

Option 4: 0.33, 0.4, 0.5, 0.6, 0.8 — (correct)

0.33 < 0.4 < 0.5 < 0.6 < 0.8 —- (Ascending order)

**2. Which of the following has fractions in descending order? **

- A. \(\frac{5}{6}\), \(\frac{4}{7}\), \(\frac{2 }{5}\), \(\frac{3 }{5}\), \(\frac{1}{3}\)

B. \(\frac{5}{6}\), \(\frac{2}{5}\), \(\frac{4 }{7}\), \(\frac{2 }{5}\), \(\frac{1}{3}\)

C. \(\frac{4}{7}\), \(\frac{1}{3}\), \(\frac{2}{5}\), \(\frac{5 }{6}\), \(\frac{3}{5}\)

D. \(\frac{1}{3}\), \(\frac{2}{5}\), \(\frac{4 }{7}\), \(\frac{3 }{5}\), \(\frac{5}{6}\)

**Answer**: Option C

**Explanation**:

Growth rate of rat population in 3 months = 20 *(3/12) = 5%

Increase in first 3 months = 3200 x 1.05 = 3360

Also, net decrease in 3 months = 160

Rat population = 3360 – 160 = 3200

In the same way, after every 3 months, rat population remains the same

Hence, even after 3 x 8 months i.e., 2 years, the population is maintained

**3. Convert 0.737373… into vulgar fraction? **

- A. \(\frac{73}{99}\)

B. \(\frac{73}{100}\)

C. \(\frac{73}{90}\)

D. \(\frac{73}{900}\)

**Answer**: Option C

**Explanation**:

In a decimal fraction, if there are n numbers of repeated numbers after a decimal point, then just write one repeated number in the numerator and in denominator take n number of nines equal to repeated numbers you observe after the decimal point.

0.737373… is written as \(0. \bar{73}\)

Numerator = 73 —- (one repeated number)

Denominator = 99 —- (73 is the number which is repeated)

Vulgar fraction = \(\frac{73}{99}\)

**4. 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%.**

- A. 450

B. 500

C. 400

D. 410

**Answer**: Option B

**Explanation**:

Cost of each share = (20 + 2.5% of 20) = Rs.20.5

Therefore, number of shares = \(\frac{8200}{20.5}\) = 400

**5. Find the market value of the stock if 6% yields 10%.**

- A. 60

B. 70

C. 80

D. 100

**Answer**: Option C

**Explanation**:

Let the investment be Rs.100 for an income of Rs.10

Therefore, for an income of Rs.6, the investment = \(\frac{600}{10}\) = Rs.60

- 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

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

- A. 6.25

B. 25

C. 50

D. 12.5

**Answer**: Option A

**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

**3. Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 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) = 2/5

P(B) = 4/7

E = {A and B both get selected}

P(E) = P(A)*P(B)

= \(\frac{2}{5}\) * \(\frac{4}{7}\)

= \(\frac{8}{35}\)

**4. From a pack of 52 cards, one card is drawn at random. Find the probability that the drawn card is a club or a jack? **

- A. \(\frac{17}{52}\)

B. \(\frac{8}{13}\)

C. \(\frac{4}{13}\)

D. \(\frac{1}{13}\)

**Answer**: Option C

**Explanation**:

n(S) = 52

n(E) = 16

P(E) = \(\frac{ n(E)}{ n(S)}\) = \(\frac{ 16}{ 52}\)

= \(\frac{ 4}{ 13}\)

**5. Tickets numbered 1 to 50 are mixed and one ticket is drawn at random. Find the probability that the ticket was drawn has a number which is a multiple of 4 or 7?**

- A. \(\frac{ 9}{25}\)

B. \(\frac{ 9}{ 50}\)

C. \(\frac{ 18}{ 25}\)

D. None of these

**Answer**: Option A

**Explanation**:

S = {1, 2, 3, … , 49, 50}

E = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 7, 14, 21, 35, 42, 49}

n(S) = 50

n(E) = 18

P(E) = \(\frac{ n(E)}{ n(S)}\) = \(\frac{18}{50}\)

= \(\frac{ 9}{25}\)

- A. 10 days

B. 15 days

C. 20 days

D. 25 days

** Answer**: Option C

**Explanation**:

Assume 1 man’s 1 day work = x & 1 boy’s 1 day work = y

From the given data, we can generate the equations as : 4x + 5y = 1/20 —(1) & 5x + 4y = 1/16 —(2)

By solving the simultaneous equations (1) & (2),

X = \(\frac{1}{80}\) & y = 0

Therefore, (4 men + 3 boys ) 1-day work = 4 X \(\frac{1}{80}\) + 3 X 0 = \(\frac{1}{20}\)

Thus, 4 men and 3 boys can finish the work in 20 days

**2. If in a race of 80m, A covers the distance in 20 seconds and B in 25 seconds, then A beats B by: **

- A. 20m

B. 16m

C. 11m

D. 10m

** Answer**: Option B

**Explanation**:

The difference in the timing of A and B is 5 seconds. Hence, A beats B by 5 seconds.

The distance covered by B in 5 seconds = \(\frac{(80 * 5)}{25}\) = 16m

Hence, A beats B by 16m.

**3. If X can run 48m and Y 42m, then in a race of 1km, X beats Y by: **

- A. 140m

B. 125m

C. 100m

D. 110m

** Answer**: Option B

**Explanation**:

When X runs 48m, Y runs 42m.

Hence, when X runs 1000m, Y runs \(\frac{(1000 * 42)}{48}\) = 875m

X beats Y by 125m.

**4. 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 \(\frac{(800 * 3)}{4}\) = 600m

Therefore, R beats P by 80m.

**5. If in a race of 120m X can beat Y by 20m and Z by 35m, then Y can beat Z by **

- 12m

B. 10m

C. 15m

D. 18m

** Answer**: Option D

**Explanation**:

In a race of 120m, as X beats Y by 20m, when X runs 120m, Y runs 100m.

Similarly, as X beats Z by 35m, when X covers 120m, Z covers 85m.

Hence, when Y runs 100m, Z runs 85m.

When Y runs 120m, Z runs \(\frac{(120 * 85)}{100}\) = 102m

Hence, Y beats Z by 18m.