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IBPS RRB PO Quantitative Aptitude Quiz 4

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IBPS RRB PO Quantitative Aptitude Quiz 4

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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 IBPS RRB PO Quantitative Aptitude Quiz 4 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. IBPS RRB has released IBPS RRB Officer 2019 Official Notification for Officer Scale(I, II, and II). Quantitative Aptitude plays major role to qualify examination. The article IBPS RRB PO Quantitative Aptitude Quiz 4 will assist the students to know the expected questions from Quantitative Aptitude.

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Q1. P takes 6 days less than Q to finish the work individually. If P and Q working together complete the work in 4 days, then how many days are required by Q to complete the work alone ?

    A. 7 days
    B. 10 Days
    C. 5 Days
    D. 12 Days


Answer: Option D

Explanation:
\(\frac{1}{6}\) + \(\frac{1}{12}\)
= \(\frac{2 + 1}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)


Q2. If X, Y and Z can complete a work in 6 days. If X can work twice faster than Y and thrice faster than Z, then the no of days Z alone can complete the work is

    A. 22 days
    B. 11 days
    C. 33 days
    D. 30 Days


Answer: Option C

Explanation:
\(\frac{1}{x}\) + \(\frac{1}{2x}\) + \(\frac{1}{3x}\)
\(\frac{1}{6}\)
\(\frac{6 + 3+ 2}{6x}\) = 6

\(\frac{11}{6x}\) = 6
X = 11
Z = 3x = 33


Q3. If 20 men can do a piece of work in 42 days working 8 hours per day then how many men are required to complete the work, working 5 hours per day in 33 days ?

    A. 39
    B. 47
    C. 32
    D. 41


Answer: Option D

Explanation:
M = \(\frac{20*42*8}{5*33}\) = 40.72 = 41


Q4. Three men, four women and six children can complete a work in 7 days. A woman does double the work a man does and a child does half the work a man does. How many women alone can complete this work in 7 days ?

    A. 7
    B. 6
    C. 8
    D. 10


Answer: Option A

Explanation:
2 men = 1 woman
1 man = \(\frac{1}{2}\) woman
3 men = \(\frac{3}{2}\) women
2 children = 1 man = \(\frac{1}{2}\) woman
1 child = \(\frac{1}{4}\) woman
6 children = \(\frac{6}{4}\) = \(\frac{3}{2}\) women
Now, three men, four women and six children
= \(\frac{3}{2}\) + 4 + \(\frac{3}{2}\) = 3 + 8 + \(\frac{3}{2}\)= \(\frac{14}{2}\) = 7 women


Q5. If A can complete a work in 30 days, B can do the same work in 36 days, If after doing 5 days, leaves the work.Find in how many days B will do the remaining work ?

    A. 22 days
    B. 28 days
    C. 30 days
    D. 36 Days


Answer: Option C

Explanation:
A = \(\frac{1}{30}\), B = \(\frac{1}{42}\)
6 days work = 5*\(\frac{1}{30}\) = \(\frac{1}{6}\)
Remaining = \(\frac{5}{6}\)
B = 5*\(\frac{36}{6}\) = 30

Q1. The weights of 19 people are in Arithmetic progression. The average weight of them is 19. If the heaviest is 37 Kgs. Then what is the weight of the Lightest?

    A. 1 Kg
    B. 2 Kg
    C. 3 Kg
    D. 4 kg


Answer: Option A

Explanation:

19*19 = \(\frac{19}{2(2a+18d)}\)
38 = 2a + 18d
37 = a + 18d
a = 1


Q2. The average weight of 40 Students is 32. If the Heaviest and Lightest are excluded the average weight reduces by 1. If only the Heaviest is excluded then the average is 31. Then what is the weight of the Lightest?

    A. 30
    B. 31
    C. 32
    D. 33


Answer: Option C

Explanation:

40*32 = 1280
1280 – \(\frac{H}{39}\) = 31
H = 71
1280 – 71 – \(\frac{L}{38}\) = 31
L = 31


Q3. Average of 17 students in a class is X. When their marks are arranged in ascending order it was found to be in Arithmetic Progression. The class teacher found that rank the students who ranked 15th, 11th, 9th and 7th had copied the exam and hence they are suspended. Now the average of the remaining class is Y. Then

    A. X = Y
    B. X > Y
    C. X < Y
    D. X = 2Y


Answer: Option C

Explanation:
17X = \(\frac{17}{2 (2a+16d)}\)
X = a + 8d
13Y = \(\frac{17}{2 (2a+16d)}\) – (4a+26d)
Y = a + 8.46d


Q4. The average monthly expenditure of Mr. Ravi’s family for the first three months is Rs 2,750, for the next three months is Rs 2,940 and for the last three months Rs 3,150. If his family saves Rs 4980 for nine months, find the average monthly income of the family for the 9 months?

    A. Rs. 3800
    B. Rs. 3500
    C. Rs. 3400
    D. Rs. 4200


Answer: Option B

Explanation:
Average monthly expenditure for 3 months = Rs. 2750
Total expenditure for 3 months = Rs 2750 x 3 = Rs. 8250
Average monthly expenditure for 3 months = Rs. 2940
Total expenditure for 3 months = Rs 2940 x 3 = Rs. 8820
Average monthly expenditure for 3 months = Rs. 3150
Total expenditure for 3 months = Rs 3150 x 3 = Rs. 9450
Total savings for 9 months = 4980
Average monthly income for 9 months = \(\frac{(8250 + 8820 + 9450 + 4980)}{9}\)= 3500


Q5. The average age of a family of 8 members is 24 years. If the age of the youngest member be 6 years, the average age of the family at the birth of the youngest member was?

    A. 23.42 years
    B. 21.42 years
    C. 27.42 years
    D. 26.42 years


Answer: Option B

Explanation:
Total present age of the family (8*24) = 192 years
Total age of the family 6 years ago = (192 – 6*7) = 150 years
At that time, Total members in the family = 7
Therefore Average age at that time = \(\frac{150}{7}\) = 21.42 years

1. The sum of the digits of a two-digit number is 6. If the digits are reversed, the number is decreased by 36. Find the number?

    A. 15
    B. 51
    C. 24
    D. 42


Answer: Option B

Explanation: a + b = 6
(10a + b) – (10b +a) = 36, a – b = 4
We get a = 5 and b = 1
So number is 51


2. If the places of last two-digits of a three digit number are interchanged, a new number greater than the original number by 36 is obtained. What is the difference between the last two digits of that number?

    A. 2
    B. 3
    C. 4
    D. 7


Answer: Option D

Explanation:
let the number be 100a + 10b + c
(100a + 10b +c) – (100a + 10c +b) = 36
b – c = 4


3. A number when divided by 837 leaves a remainder of 79. What will be the remainder when the same number is divided by 31?

    A. 11
    B. 13
    C. 15
    D. 17


Answer: Option D
Number = 837*a + 79
when this number is divided by 31, it leaves remainder of 17 (837 is completely divisible)


4. When a number is added to 20 percent of the second number, we get 150 percent of the second number. Find the ratio between the first and second number?

    A. 13:9
    B. 12:10
    C. 13:10
    D. 17:10


Answer: Option C

Explanation:
a + (\(\frac{20}{100}\))*b = (\(\frac{150}{100}\))*b
a:b = 13:10


5. Two different numbers when divided by same divisor leaves remainder 7 and 9 respectively. When their sum is divided by the same divisor remainder was 4. Find the divisor?

    A. 11
    B. 12
    C. 13
    D. 1


Answer: Option B

Explanation:
Let first number N1 = D*a + 7
and second number N2 = D*b + 9
N1 + N2 = (a+b)*D + 16
Remainder is 4, so D will be 12


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