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IBPS RRB PO Quantitative Aptitude Quiz 13

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IBPS RRB PO Quantitative Aptitude Quiz 13

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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 13 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 13 will assist the students to know the expected questions from Quantitative Aptitude.

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1. What percent of 120 are 90?

    A. 25%
    B. 50%
    C. 75%
    D. 33%


Answer – Option C

Explanation –


\(\frac {?%} {100}\) = * 120 = 90
? = 75%


2. After decreasing 24% in the price of an article costs Rs.912. Find the actual cost of an article?

    A. 1400
    B. 1300
    C. 1200
    D. 1100


Answer – Option A

Explanation –

CP* \(\frac {76} {100}\)
= 912
CP= 12 * 100 => CP = 1200


3. How much 60% of 50 is greater than 40% of 30?

    A. 18
    B. 13
    C. 15
    D. 20


Answer – Option A

Explanation –


\(\frac {60} {100}\) * 50 – \(\frac {40} {100} –\)* 30
30 – 12 = 18


4. 40% of a number is more than 20% of 650 by 190. Find the number?

    A. 600
    B. 700
    C. 800
    D. 900


Answer – Option C

Explanation –

\(\frac {40} {100}\)* X – – \(\frac {20} {100}\) * 650 = 190

\(\frac {2} {5}\)X = 320
X = 800


5. 60% of a number is added to 120, the result is the same number. Find the number?

    A. 300
    B. 200
    C. 400
    D. 500


Answer – Option A

Explanation –

\(\frac {60} {100}\) * X + 120 = X
2X = 600
X = 300

1. A can do a piece of work in 12 days. When he had worked for 2 days B joins him. If the complete work was finished in 8 days. In how many days B alone can finish the work?

    A. 18 days
    B. 12 days
    C. 24 days
    D. 10 days


Answer – Option A

Explanation –

\(\frac {8} {12}\) + \(\frac {6} {x}\)
x = 18 days


2. A can do a piece of work in 15 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in?

    A. 8 days
    B. 10 days
    C. 15 days
    D. 11 \(\frac {3} {7}\) days


Answer – Option B

Explanation –

\(\frac {x – 5} {15}\) + \(\frac {x} {20}\) = 1
11 \(\frac {3} {7}\) days


3. A, B and C can do a piece of work in 24, 30 and 40 days respectively. They start the work together but C leaves 4 days before the completion of the work. In how many days is the work done?

    A. 15 days
    B. 14 days
    C. 13 days
    D. 11 days


Answer – Option D

Explanation –

Let the investment be Rs. x. Then

\(\frac {x} {24} \) + \(\frac {x} {30} \) + \(\frac {x} {40} \) = 1
x = 11 days


4. A, B and C can do a piece of work in 7 days, 14 days and 28 days respectively. How long will they taken, if all the three work together?

    A. 3 days
    B. 4 days
    C. 5 days
    D. 6 days


Answer – Option B

Explanation –

\( \frac 1} {7}\) + \( \frac {1} {14}\) + \( \frac {1} {28}\) = \( \frac {7} {28}\)
= \( \frac {1} {4}\) => 4 days


5. A is twice as good a work man as B and together they finish the work in 14 days. In how many days A alone can finish the work?

    A. 20
    B. 21
    C. 22
    D. 23


Answer – Option B

Explanation –

WC = 2:1

2x + x = = \( \frac {1} {21}\)
<2x = \(\frac {100 * 2x} {x * 10}\)years = 20 years

1. The sum of three consecutive integers is 102. Find the lowest of the three?

    A. 40
    B. 53
    C. 29
    D. 33


Answer – Option D

Explanation –

Three consecutive numbers can be taken as (P – 1), P, (P + 1).
So, (P – 1) + P + (P + 1) = 102
3P = 102 => P = 34.
The lowest of the three = (P – 1) = 34 – 1 = 33.


2. The sum of the two digits of a number is 10. If the number is subtracted from the number obtained by reversing its digits, the result is 54. Find the number?

    A. 34
    B. 28
    C. 12
    D. 17


Answer – Option B

Explanation –

Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 10 —– (1)
(10Q + P) – (10P + Q) = 54
9(Q – P) = 54
(Q – P) = 6 —– (2)
Solve (1) and (2) P = 2 and Q = 8
The required number is = 28


3. The sum of three consecutive even numbers is 42. Find the middle number of the three?

    A. 14
    B. 16
    C. 18
    D. 24


Answer – Option A

Explanation –

Three consecutive even numbers (2P – 2), 2P, (2P + 2).
(2P – 2) + 2P + (2P + 2) = 42
6P = 42 => P = 7.
The middle number is: 2P = 14.


4. The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence?

    A. 50
    B. 60
    C. 70
    D. 80


Answer – Option C

Explanation –

A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
5 years, their ages will be 45 and 25.
Sum of their ages = 45 + 25 = 70.


5. The ratio of the present ages of P and Q is 3:4. 5 years ago, the ratio of their ages was 5:7. Find the their present ages?

    A. 30, 40
    B. 25, 30
    C. 50, 60
    D. 20, 40


Answer – Option A

Explanation –

Their present ages be 3X and 4X.
5 years age, the ratio of their ages was 5:7, then (3X – 5):(4X – 5) = 5:7
X = 35 – 25 => X = 10.
Their present ages are: 30, 40.


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