Quantitative Aptitude - SPLessons

SSC CPO Quantitative Aptitude Quiz 22

SPLessons 5 Steps, 3 Clicks
5 Steps - 3 Clicks

SSC CPO Quantitative Aptitude Quiz 22

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 22 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 22 will assist the students to know the expected questions from Quantitative Aptitude.


shape Quiz

1. At an election, where there are two candidates only, a candidate who gets 43 per cent of the votes is rejected by a majority of 420 votes. Then total number of votes recorded assuming that there was no void vote are


    A. 2700
    B. 2800
    C. 3200
    D. None of these


Answer: Option D


Explanation:

Difference in % of votes = 57-43 = 14%
14% is represented by 420
Total number of votes = 420*(\(\frac{100 }{14}\))= 3,000


2. The population of rats in a locality increases by 20% in one year. Observing this, the pest control committee decided to use a special kind of pesticide `xyz’ which effectively kills 160 rats in 3 months. Just after 2 years, what is the net increase or decrease in the population of rats if, initially the population of rats is 3200 and pesticide is used effectively?


    A. Increase of 128 rats
    B. Decrease of 128 rats
    C. Neither an increase nor a decrease in the population
    D. None of these


Answer: Option C


Explanation:

Growth rate of rat population in 3 months = 20 *(\(\frac{3 }{12}\)) = 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. A man sells sugar to a tradesman at a profit of 20% but the tradesman becoming bankrupt pays only 80 paise in the rupee. How much percentage does the man gain or lose by his sale?


    A. 2.5%
    B. 3%
    C. 4%
    D. 5.2%


Answer: Option C


Explanation:
Let CP be Rs.x
SP = 1.2 x But he gets only 1.2x * 0.8 = 0.96x
Loss = 0.04x
Hence Loss % = 4.


4. A camera costing 2550 is marked to be sold at a price, which gives a profit of 30%. What will be its selling price in a sale when 20% is taken off the marked price?


    A. Rs. 600

    B. Rs. 572
    C. Rs. 35
    D. Rs. 605


Answer: Option B


Explanation:

CP = Rs.550
∴ Marked price = 1.3×550 = Rs.715
∴ sales price = 0.8×715 = Rs.572.


5. A manufacturer marks his goods in advance at 80 percent more than the cost price, but he allows 15 articles to the dozen and also 10 percent discount for cash. What rate of profit on his outlay does he obtain from a customer who pays cash?


    A. 18.2%
    B. 25%

    C. 29.6%

    D. Cannot be determined


Answer: Option C


Explanation:
Let CP be x
∴ MP = 1.8 x
∴ SP of 15 articles = 12 * 1.8x = 21.6x
∴ Discount = 0.9 x 21.6x = 19.44x
∴Profit on 15 articles = (19.44 – 15)x
∴ Profit % = ( \(\frac{4.44x }{15x}\)) = 29.6%

1. Three bachelors, A ,B and C rented a house for a year.. But, A left after 4 months, B stayed for 8 months and only C stayed for the entire year. If the annual rent was Rs. 6000, then share of A is


    A. Rs.4000
    B. Rs.2000

    C. Rs.300
    D. Rs.2500


Answer: Option B


Explanation:

Rate in which the rent is to be divided = 4 : 8 : 12
A’s share of rent= (8/24) * 6000 = Rs.2000


2. A trader buys a certain amount of goods worth Rs.22520. He decides to make a profit of 5.36% on the sale of goods worth Rs.5000 and increase the profit percent by 3.14% for sales up to 215000 and then increase the profit percent for the sale of the remaining lot such that he is able to make a profit of 25% on the sale of the full lot. Then profit that he makes on a third lot of goods is


    A. Rs.5620
    B. Rs.4512
    C. Rs.3212

    D. None of these


Answer: Option B

Explanation:
C.P. of goods for the trader = Rs.22520
S.P. = 125 x 22520 = Rs.28150
∴ Profit = Rs.5630
Now, 5630 = 0.0536 x 5000 + 0.085 x 10000 + (x/100)* 7520
=>5630 =268 – 850 + 75.2x
=>75.2x = Rs.4512, which is nothing but profit from sale of third lot.


3. A number is as much greater than 36 as is less than 80. Find the number.

    A. 32
    B. 36

    C. 49
    D. 58


Answer: Option D

Explanation:
Let the number be x. Then, x – 36 = 80 – x
2x = 80 + 36 =116
x = 58
Hence, the required number is 58


4. The present ages of three persons are in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 76. Find their present ages (in years).

    A. 8, 20, 28
    B. 12, 21, 27
    C. 20, 35, 45
    D. None of these


Answer: Option C

Explanation:
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x – 8) + (7x – 8) + (9x – 8) = 76
20x = 100
x = 5
Their present ages are 20 years, 35 years and 45 years respectively.


5. 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 the 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

1. The ratio between the present ages of A and B is 3:5 respectively. If the difference between B’s present age and A’s age after 4 years is 2 , what is the total of A’s and B’s present ages?


    A. 24 years

    B. 32 year
    C. 48 years

    D. cannot be determined


Answer: Option A

Explanation:
Let the present ages of A and B be 3x years and 5x years respectively.
5x – (3x + 4) = 2
2x = 6
x = 3.
Therefore,
Required sum = 3x + 5x = 8x = 24 years


2. My father was 30 years of age when my sister was born. My mother was 38 years of age when I was born. My sister was 6 years of age when my brother was born who is 3 years elder to me. What was the age of my father and mother during the birth of my brother?

    A. 41, 36
    B. 24, 28
    C. 28, 24

    D. 36, 35


Answer: Option D

Explanation:
My brother was born 6 years after my sister was born and 3 years before I was born. Age of father during my brother’s birth = (30 + 6) = 36 years.
Age of mother during my brother’s birth = (38 – 3) = 35 years.
Hence, during the birth of my brother, my father was 36 years old and my mother was 35 years old.


3. The difference between the ages of two brothers is 8 years. If 12 years ago, the elder one was twice as old as his younger brother, what is the present age of the younger brother?


    A. 20
    B. 28
    C. 24
    D. 26


Answer: Option A

Explanation:

Let the age of younger brother be x years. Hence, the age of the elder brother will be (x + 8) years.
12 years ago, the younger brother was (x – 12) years and the elder brother was (x-4) years of age.
(x – 4) = 2*(x – 12)
x – 4 = 2x – 24
x = 20 years


4. The ratio of the current ages of Ajay and Vijay is 7:4. The ratio between Ajay’s age 6 years ago and Vijay’s age 6 years from now is 1:1. Find the ratio between Ajay’s age 12 years hence and Vijay’s age 12 years ago.

    A. 12:1
    B. 10:1
    C. 1:1
    D. None of these


Answer: Option B

Explanation:

Let the current age of Ajay and Vijay be 7x and 4x respectively.
(7x – 6) / 4x + 6 = 1/1
7x – 6 = 4x + 6
3x = 12
x = 4
Therefore, \(\frac{(7x + 12) }{(4x – 12) }\) = \(\frac{P*4*7 }{2 }\) X \(\frac{(7*4 + 12) }{ (4 * 4 -12) }\) = \(\frac{40 }{4 }\) X \(\frac{1 }{100 }\) = \(\frac{10 }{1 }\)


5. The sum of the ages of 5 children born at a three-year interval each is 50. How old is the youngest child?


    6
    B. 5
    C. 4
    D. 3


Answer: Option D

Explanation:

Let the age of youngest child be X

According to the question,

X + (X+3) + (X+6) + (X+9) + (X+12) = 50

5X + 30 = 50

5X = 50-30

5X = 20

X = \(\frac{20 }{5 }\) = 4

Age of youngest child is 4 years.


IBPS RRB PO – Related Information
Quantitative Aptitude Practice Sets
Quantitative Aptitude Questions and Answers
Quantitative Aptitude Quiz