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SSC CPO Quantitative Aptitude Quiz 21

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SSC CPO Quantitative Aptitude Quiz 21

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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 21 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 21 will assist the students to know the expected questions from Quantitative Aptitude.


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1. What is John’s present age, if after 10 years his age will be 5 times his age 5 years back.


    A. 6.2 years
    B. 7.7 years
    C. 8.7 years
    D. 10 years


Answer: Option C


Explanation:

1) Let John’s present age be x
2) John’s age before 5 years = (x – 5)
3) John’s age after 10 years = (x + 10)
We are given that, John’s age after 10 years (x + 10) is 5 times his age 5 years back (x – 5)
Therefore,
(x + 10) = 5 (x – 5)
Solving the equation, we get
x + 10 = 5x – 25
4x = 35
x = 8.75 years


2. Rahul is 15 years elder than Rohan. If 5 years ago, Rahul was 3 times as old as Rohan, then find Rahul’s present age.


    A. 32.5 years
    B. 27.5 years
    C. 25 years
    D. 24.9 years


Answer: Option B


Explanation:

1) Let age of Rohan be y
2) Rahul is 15 years elder than Rohan = (y + 15). So Rahul’s age 5 years ago = (y + 15 – 5)
3) Rohan’s age before 5 years = (y – 5)
5 years ago, Rahul is 3 times as old as Rohan
(y + 15 – 5) = 3 (y – 5)
(y + 10) = (3y – 15)
2y = 25
y = 12.5
Rohan’s age = 12.5 years
Rahul’s age = (y + 15) = (12.5 + 15) = 27.5 years


3. One year ago, ratio of Harry and Peter age’s was 5 : 6 respectively. After 4 years, this ratio becomes 6 : 7. How old is Peter?


    A. 25 years
    B. 26 years
    C. 31 years
    D. 35 years


Answer: Option C


Explanation:
Hint: If ages in the numerical are mentioned in ratio A : B, then A : B will be Ax and Bx.
We are given that age ratio of Harry : Pitter = 5 : 6
1) Harry’s age = 5x and Peter’s age = 6x
2) One year ago, their age was 5x and 6x. Hence at present, Harry’s age = 5x +1 and Peter’s age = 6x +1
3) After 4 years,
Harry’s age = (5x +1) + 4 = (5x + 5)
Peter’s age = (6x +1) + 4 = (6x + 5)
4) After 4 years, this ratio becomes 6 : 7. Therefore,

\(\frac{Harry’s Age}{6}\) = \(\frac{Peter’s Age}{7}\)
\(\frac{(5x + 5) }{(6x + 5)}\) = \(\frac{6 }{7}\)

7 (5x + 5) = 6 (6x + 5)
X = 5
Peter’s present age = (6x + 1) = (6 x 5 + 1) = 31 years
Harry’s present age = (5x + 1) = (5 x 5 + 1) = 26 years


4. The age of mother 10 years ago was 3 times the age of her son. After 10 years, the mother’s age will be twice that of his son. Find the ratio of their present ages.


    A. 11 : 7
    B. 9 : 5
    C. 7 : 4
    D. 7 : 3


Answer: Option D


Explanation:

We are given that, age of mother 10 years ago was 3 times the age of her son
So, let age of son be x and as mother’s age is 3 times the age of her son, let it be 3x, three years ago.
At present: Mother’s age will be (3x + 10) and son’s age will be (x + 10)
After 10 years: Mother’s age will be (3x + 10) +10 and son’s age will be (x + 10) + 10
Mother’s age is twice that of son
(3x + 10) +10 = 2 [(x + 10) + 10]
(3x + 20) = 2[x + 20]
Solving the equation, we get x = 20
We are asked to find the present ratio.
(3x + 10) : (x + 10) = 70 : 30 = 7 : 3


5. Sharad is 60 years old and Santosh is 80 years old. How many years ago was the ratio of their ages 4 : 6?


    A. 10 years
    B. 15 years

    C. 20 years

    D. 25 years


Answer: Option C


Explanation:
Here, we have to calculate: How many years ago the ratio of their ages was 4 : 6
Let us assume x years ago
At present: Sharad is 60 years and Santosh is 80 years
x years ago: Sharad’s age = (60 – x) and Santosh’s age = (80 – x)
Ratio of their ages x years ago was 4 : 6
\(\frac{(60 – x) }{(80 – x)}\) = \(\frac{4 }{6}\)

6(60 – x) = 4(80 – x)
360 – 6x = 320 – 4x
x = 20
Therefore, 20 years ago, the ratio of their ages was 4 : 6

1. Which of the following year is not a leap year?


    A. 1960
    B. 2080

    C. 2024
    D. 2100


Answer: Option D


Explanation:

The two conditions that decide that a year is a leap year or not is:
• For a year to be a leap year, it should be divisible by 4.
• No century is a leap year unless it is divisible by 400.
Hence, the year 2100 is not a leap year as it is not divisible by 400.


2. The last day of the century cannot be:


    A. Sunday
    B. Wednesday
    C. Friday

    D. Saturday


Answer: Option D

Explanation:
100 years have 5 odd days. Hence the last day of the 1st century is a Friday.
200 years have 10 odd days or 1 week + 3 odd days. Hence, the last day of the 2nd century is a Wednesday.
300 years have 15 odd days or 2 weeks + 1 odd day. Hence, the last day of the 3rd century is a Monday.
400 years have 0 odd days. Hence, the last day of the 4th century is a Sunday.


3. 5 times a positive number is less than its square by 24. What is the integer?

    A. 5
    B. 8

    C. 7.5
    D. 9


Answer: Option B

Explanation:
Let the unknown number be x.
5 times a positive number = 5x
5 times a positive number is less than its square by 24
\({x }^{2}\) – 5x = 24
\({x }^{2}\) + 3x – 8x – 24
x (x + 3) – 8(x + 3)
(x – 8) (x + 3)
x = 8
8 is the required integer.


4. H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers.

    A. 230, 140
    B. 215, 130
    C. 195, 143
    D. 155, 115


Answer: Option C

Explanation:
Hint: Product of two numbers = Product of their H.C.F. and L.C.M.
Given:
1) H.C.F. of two numbers = 13
2) The numbers are in the ratio of 15: 11

Let the two numbers be 15y and 11y
H.C.F. is the product of common factors
Therefore, H.C.F. is y. So y = 13

The two numbers are:
15y = 15 × 13 = 195
11y = 11 × 13 = 143
We can cross-check the answer using the trick. (Product of two numbers = Product of their H.C.F. and L.C.M.)
Product of H.C.F. and L.C.M. = 13 × 2145 = 27885
Product of two numbers = 195 × 143 = 27885
Hence, the calculated answer is correct.


5. If the product and H.C.F. of two numbers are 4107 and 37 respectively, then find the greater number.


    A. 111
    B. 222
    C. 332
    D. 452


Answer: Option A

Explanation:
4107 is the square of 37.
So let two numbers be 37x and 37y.
37x × 37y = 4107
xy = 3
3 is the product of (1 and 3)
x = 1 and y = 3
37x = 37 × 1 =37
37y = 37 × 3 = 111
Greater number = 111
OR
Hint:
Product of two numbers = Product of their H.C.F. and L.C.M.
Product of two numbers = 4107
4107 = 37 × L.C.M
L.C.M. = \(\frac{4107 }{37}\) = 111
The greater number = 111

1. In a kilometer race, A beats B by 50 meters or 10 seconds. What time does A take to complete the race?


    A. 200 sec

    B. 190 sec
    C. 210 sec

    D. 150 sec


Answer: Option B

Explanation:
Time taken by B run 1000 meters = \(\frac{(1000 * 10)}{50 }\)= 200 sec.
Time taken by A = 200 – 10 = 190 sec.


2. A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?

    A. 111.12 m
    B. 888.88 m
    C. 777.52 m

    D. 756.34 m


Answer: Option A

Explanation:
A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= \(\frac{(1000 * 800)}{9 }\) = \(\frac{8000}{9 }\) = 888.88 m.
B can give C = 1000 – 888.88 = 111.12 m.


3. In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?


    A. 57.5 m
    B. 127.5 m
    C. 150.7 m
    D. 98.6 m


Answer: Option B

Explanation:

When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = \(\frac{(900 * 700) }{800 }\) = \(\frac{6300 }{8 }\) = 787.5 m.
In a race of 1000 m, A beats C by (1000 – 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = \(\frac{(600 * 212.5) }{1000 }\) = 127.5 m.


4. In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?

    A. 50
    B. 40
    C. 25
    D. 15


Answer: Option C

Explanation:

A scores 60 while B score 40 and C scores 30.
The number of points that C scores when B scores 100 = \(\frac{(100 * 30) }{40 }\) = 25 * 3 = 75.
In a game of 100 points, B gives (100 – 75) = 25 points to C.


5. What sum of money will produce Rs.70 as simple interest in 4 years at 3 \(\frac{1 }{2}\) percent?


    1584
    B. 1120
    C. 792
    D. 1320


Answer: Option B

Explanation:

70 = \(\frac{P*4*7 }{2 }\) X \(\frac{1 }{100 }\)

P = 500


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