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SSC CPO Quantitative Aptitude Quiz 23

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SSC CPO Quantitative Aptitude Quiz 23

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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 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.


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1. Which of the following has fractions in ascending order?


    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

1. 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


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) = \(\frac{ n(E)}{ n(S)}\) = \(\frac{18}{50}\)
= \(\frac{ 9}{25}\)

1. 4 men and 5 boys can do a piece of work in 20 days while 5 men and 4 boys can do the same work in 16 days. In how many days can 4 men and 3 boys do the same work?


    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.


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