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SSC CPO Quantitative Aptitude Quiz 14

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SSC CPO Quantitative Aptitude Quiz 14

shape 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 14 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 14 will assist the students to know the expected questions from Quantitative Aptitude.


shape Quiz

.1. A, B and C enter into partnership. A invests 3 times as much as B invests and B invests two third of what C invests. At the end of the year, the profit earned is Rs. 8500. What is the share of B?


    A. Rs. 1245.50
    B. Rs. 1623.53
    C. Rs. 1545.45
    D. Rs. 1145.15


Answer: Option C


Explanation:

Let C’s capital = Rs. y. Then, B’s capital = Rs. (2/3)y
A’s capital = Rs. (3 × (\(\frac{2 }{3 }\))y) = Rs. 2y
Therefore, ratio of their capitals
= 2y : (\(\frac{2 }{3 }\))y : y
= 6 : 2 : 3
Hence, B’s share = Rs. (8500 x \(\frac{2 }{11 }\)) = Rs. 1545.45


2. If 36 men can do a piece of work in 25 hours, in how many hours will 10 men do it?


    A. 100 hours
    B. 90 hours
    C. 60 hours
    D. 30 hours


Answer: Option B


Explanation:

Let the required numbers of hours be X.
Less men, More hours (Indirect proportion)
Therefore, 10 : 36 :: 25 : X = (10 x X) = (36 x 25) = X = \(\frac{(36 × 25) }{10 }\) = 90.
Hence, 10 men can do it in 90 hours.


3. By purchasing an article at a 20 % discount on the original price and then selling it at a price of 25% above the original price, a trader earns Rs. 200 as the profit. What was the original price of the article?


    A. Rs. 444.44
    B. Rs. 255.50
    C. Rs. 100.10
    D. Rs. 810


Answer: Option A


Explanation:

Let the original price of the article be Rs. 100. Hence the purchase price for the trader would be Rs. 80 and his selling price would be Rs. 125. Thus, he would earn a profit of Rs. 45 (125 – 80). Therefore,
Profit is Rs. 45 if the original price is Rs. 100
Hence, if the profit is Rs. 200, then the original price will be…
= 100x \(\frac{200 }{45 }\)
= Rs. 444.44


4. Shruti purchased several numbers of three articles P, Q and R in the proportion 3: 2 : 3. If the unit costs of the articles P, Q and R are 200, Rs. 90 and Rs. 60 respectively, how many articles of Q must have been purchased in the total purchases of Rs. 4800?


    A. 8
    B. 10
    C. 12
    D. 14


Answer: Option B


Explanation:

Let the number of articles of types P, Q and R be 3a, 2a and 3a respectively.
Thus, we get,
(200 x 3a) + (90 x 2a) + (60 x 3a) = 4800
960a = 4800
a = 5
Hence, the number of articles of type “Q” = 2×5 = 10


5. A train travels a certain distance by taking 3 stops of 20 minutes each. Considering the period of stoppage, the overall speed of the train comes to 40 kmph; while without consideration of the stoppage, it is 60 kmph. How much distance must the train have traveled?


    A. 170 kms
    B. 120 kms

    C. 270 kms

    D. None of these


Answer: Option C


Explanation:
Let the time taken to travel the distance without taking stops be “b” hours.
Thus, we get,
60 x b = 40(b + 1)
b = 2
Thus, we get 60 x 2 = 120 kms

1. Rohan was 4 times as old as his son 6 years ago. After 6 years Rohan will be twice as old as his son. What is son’s present age?


    A. 10 years
    B. 12 years

    C. 14 years
    D. 18 years


Answer: Option B


Explanation:

Let son’s age 6 years ago be X years. Then, Rohan’s age 6 years ago = 4X years.
Son’s age after 6 years = (X + 6) + 6 = (X + 12) years
Rohan’s age after 6 years = (4X + 6) + 6 = (4X + 12) years
Therefore 2(X + 12) = 4X + 12
2X = 12
X = 6
Hence, son’s present age = (X + 6) = 12 years
Rohan’s present age = (4X + 6) = 30 years


2. Find the average of first 30 natural numbers.


    A. 12
    B. 15.5
    C. 14.5

    D. 16


Answer: Option B

Explanation:
Sum of first n natural numbers = \(\frac{n(n+1) }{2}\)
Hence, sum of first 30 natural numbers = \(\frac{n(n+1) }{2}\)
\(\frac{30 × 31}{2}\) = 465
Therefore, required average of = \(\frac{465 }{30}\)= 15.5


3. 796.21 + 498.05 = 215.50 – 425.01

    A. 71.81
    B. 81.71

    C. 84.74
    D. 88.65


Answer: Option D

Explanation:
a – 796.21 + 498.05 = 215.50 – 425.01
a = 88.65


4. A boy pays Rs. 369 for an article marked at Rs. 600, by enjoying two successive discounts. If the first discount is of 25%, how much should be the second discount?

    A. 20 %
    B. 10 %
    C. 25 %
    D. 18 %


Answer: Option D

Explanation:
First discount = 25% of 600 = Rs. 150.
Thus, the reduced price = 600 – 150 = Rs. 450.
Since the person actually paid Rs. 369, the value of the second discount must be equal to Rs. 81 (450 – 369).
Let the second discount be x
Thus, we get, 81 = x of 450
( \(\frac{81 }{450}\))*100 = 18%


5. A housewife wishes to purchase three articles A, B, and C from a sum of Rs. 200. The unit prices of the articles A, B, and C are Rs. 20, Rs. 35 and Rs. 25 respectively. If she spends the entire amount by purchasing 5 numbers of articles of type C, what is the ratio of the number of articles purchased of type A to that of, type B?


    A. 1 : 2
    B. 2 : 1
    C. 1 : 1
    D. None of these


Answer: Option B

Explanation:
After spending Rs. 125 (25 * 5) for articles of type C, the housewife is left with Rs. 75 (200 – 125). Since this amount has to be spent in totality, she must have purchased 2 articles of type A (equivalent to Rs. 40) and 1 article of type B (equivalent to Rs. 35). Thus, the required ratio is 2 : 1.
Time taken to fill the tank is (\(\frac{35 }{6 }\)) = 5 hours 49 min

1. A briefcase has a number-lock system containing a combination of 3 digits (Each digit can be of numbers 0 to 8). If the correct combination is unknown, how much maximum time would be required to open the bag if each “trial” of combination takes 3 seconds?


    A. 45.23 minutes

    B. 36.45 minutes

    C. 60.34 minutes

    D. 90.15 minutes


Answer: Option B

Explanation:

Maximum number of trials required = 9 * 9 * 9 = 729. Since for each combination trial, 3 seconds are required to open the briefcase is given as 3 * 729 = 2187 seconds = 36.45 minutes.


2. Out of the three numbers A, B, and C, A exceeds B by 20 and C exceeds A by 55. If the sum of all the numbers is 230, What is the difference between the largest and the smallest number?

    A. 52
    B. 58
    C. 63

    D. 75


Answer: Option D

Explanation:
We have , A = B + 20 & C = A + 55 = B + 75
Thus, (B + 20) + (B) + (B + 75) = 230
B = 45
Thus, A = 65 & C = 120. Hence, (C) – (B) = 120 – 45 = 75
Short-Cut Method : Since C is larger than A by 55 and A is larger than B by 20, we get, the Required difference = 55 + 20 = 75 (without calculating the actual values of A, B and C).


3. If 148 is multiplied first by 163 and then by 236, Which of the following numbers would come at the unit’s place?


    A. 8
    B. 6
    C. 4
    D. 3


Answer: Option D

Explanation:
148 * 163 * 236 = 8 * 3 * 6 = 4 at the unit’s place.
When the numbers are multiplied, the unit digit of the final product is obtained as the product of the unit digits of individual numbers.


4. A trader purchases several articles at the rate of 13 for Rs. 10 and sells them at the rate of 10 for Rs. 13. What is his gain/loss?

    A. 69 % gain
    B. 56.25 % gain
    C. 56.25 % loss
    D. 25 % loss


Answer: Option A

Explanation:

We have, C. P of 13 articles = Rs. 10
& S. P of 10 articles = Rs. 13
S. P of 13 articles = ?
= \(\frac{(13 * 13) }{10 }\) = 16.9
Gain = [\(\frac{(16.9 – 10) }{10}\) * 100 = 69 %


5. On a shelf, 2 books of Geology, 2 books of Sociology and 5 of Economics are to be arranged in such a way that the books of any subject are to be together. Find in how many ways can this be done?


    3846
    B. 2880
    C. 900
    D. 90


Answer: Option B

Explanation:

There are books of 3 subjects (Geology, Sociology and Economics), hence they can be arranged in 3! (3 * 2 * 1) = 6 ways.
Further, in each category (subject), books are to be arranged in different order, we get,
Required number of ways: 3! * [2! * 2! * 5!] = 2880


IBPS RRB PO – Related Information
IBPS RRB Quantitative Aptitude Practice Set 1
SBI Clerk Mains Quantitative Aptitude Practice Set 1
SBI Clerk Mains Quantitative Aptitude Practice Set 2
Book for Quantitative Aptitude