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IBPS RRB PO Quantitative Aptitude Quiz 7

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IBPS RRB PO Quantitative Aptitude Quiz 7

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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 7 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 7 will assist the students to know the expected questions from Quantitative Aptitude.

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Q1. Last year there were 610 boys in a school. The number decreased by 20 percent this year. How many girls are there in the school if the number of girls is 175 percent of the total number of boys in the school this year ?

    A. 854
    B. 848
    C. 798
    D. 782


Answer: Option A

Explanation:
No of boys in a school last year = 610
No of boys in a school for this year
\(\frac{610*80}{100}\) = 122
610 – 122 = 488
No of girls = \(\frac{175}{100}\) * 488 = 854


2. The tank-full petrol in Arun’s motor-cycle last for 10 days. If he starts using 25% more every day, how many days will the tank-full petrol last?

    A. 4
    B. 6
    C. 8
    D. 10


Answer: Option C

Explanation:

Assume – Arun’s motorcycle uses 1L per day and therefore tank’s Capacity = 10L.
25% increased per day= 1+(\(\frac{25}{100}\)) = \(\frac{5}{4}\) ie. 1.25L per day
Days = \(\frac{10}{1.25}\) = 8


3. The price of a car is Rs. 4,50,000. It was insured to 80% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received?

    A. Rs.1,76,375
    B. Rs.3,24,000
    C. Rs.1,82,150
    D. Rs.1,26,000


Answer: Option D

Explanation:

4,50,000*( \(\frac{80}{100}\))*(\(\frac{90}{100}\))= 324000
450000 – 126000 = Rs.1,26,000


4. Deepika went to a fruit shop with a certain amount of money. She retains 15% of her money for auto fare. She can buy either 40 apples or 70 oranges with that remaining amount. If she buys 35 oranges, how many more apples she can buy?

    A. 35
    B. 40
    C. 15
    D. 20


Answer: Option D

Explanation:
Assume Total amount = Rs.100
Auto fare= 15% of Total amount i.e Rs.15
Now the amount is Rs.85
Price of 70 oranges = Rs.85
Price of 35 oranges = (\(\frac{85}{75}\))*35 = Rs. 42.50
Remaining amount to buy apples is =Rs. 42.50
Price of 40 apples = Rs.85
Price of X apples = Rs.42.50
X=(85/42.5)*40 = 20 Apples


5. A does half as much work as B does in one sixth of the time. If together they take 20 days to complete the work, then what is the time taken by B to complete the work independently.

    A. 80 days
    B. 100 days
    C. 120 days
    D. 140 days


Answer: Option A

Explanation:
Let B take X days to complete the work then in one –sixth of the time i.e. x/6 days. Now A do half work as done by B so A will take twice the time i.e. 2*x/6 = x/3 to complete the job alone
So \(\frac{1}{x}\) + \(\frac{3}{x}\) = \(\frac{1}{20}\), x = 80 days

Q1. A contractor undertakes to make a mall in 60 days and he employs 30 men. After 30 days it is found that only one- third of the work is completed. How many extra men should he employ so that the work is completed on time?

    A. 20men
    B. 25men
    C. 30men
    D. 40men


Answer: Option C

Explanation:

Let total work is w and it is given that one-third of the work is completed after 30 days. Means
M*D = 30*30 = \(\frac{w}{3}\), so total work = 30*30*3
2700 = 30*30 + (30 + p)*30, so we get P = 30 (p = additional men)


Q2. 50 men could complete a work in 200 days. They worked together for 150 days, after that due to bad weather the work is stopped for 25 days. How many more workers should be employed so as to complete the work in time?

    A. 25
    B. 35
    C. 50
    D. 60


Answer: Option C

Explanation :

Let additional workers be P,
(50*150)/(50*200) = \(\frac{3}{4}\) of the work is already completed and now only 1/4 of the work is to be done. So,
\(\frac{1}{4}\) = \(\frac{(50 + P) * 25)}{50*200}\), solve for p, we get P = 50


Q3. P and Q were assigned to do a work for an amount of 1200. P alone can do it in 15 days while Q can do it in 12 days. With the help of R they finish the work in 6 days. Find the share if C.

    A. 100
    B. 120
    C. 140
    D. 160


Answer: Option B

Explanation :
\(\frac{1}{15}\) + \(\frac{1}{12}\) + \(\frac{1}{c}\) = \(\frac{1}{6}\), we got C = 60 (it means C will take 60 days to complete the work alone)
so ratio of work done by P:Q:R = 4:5:1
so c share = (\(\frac{1}{100}\))*1200 = 120


Q4. P does half as much work as Q in three-fourth of the time. If together they take 24 days to complete the work, how much time shall P take to complete the work?

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


Answer: Option B

Explanation :
Let Q take x days to complete the work, so P will take 2* \(\frac{3}{4}\) of X day to complete the work i.e. 3x/2 days
\(\frac{1}{x}\) + \(\frac{2}{3x}\) = \(\frac{1}{24}\), we get x = 40 days, so P will take = \(\frac{3}{2}\) of 40 = 60 days


Q5. Neha takes 5 hours to type 40 pages while sunil takes 6 hours to type 60 pages. How much time will they take working together on different computer to type an assignment of 180 pages.

    A. 5hr
    B. 7hr
    C. 9hr
    D. 10hr


Answer: Option D

Explanation :
In one hour number of pages type by neha = \(\frac{40}{5}\) = 8 and similarly for sunil it is \(\frac{60}{6}\) = 10.
Now to type 180 pages they will take, (8 + 10)*T = 180, T = 10 hours

1. If the divisor is five times the quotient and six times the remainder, if the remainder is 5 then the dividend is

    A. 225
    B. 300
    C. 185
    D. 412


Answer: Option C

Explanation: a + b = 6
Q = 6, R = 5
Dividend = divisor*Q + R
Dividend = 30*6 + 5 = 180 + 5 = 185


2. If the sum and the product of 2 numbers are 25 and 144 respectively then the difference of the number should be

    A. 11
    B. 9
    C. 7
    D. 5


Answer: Option C

Explanation:
A+B = 25
AB = 24
\({(a-b)}^{2}\) = \({(a+b)}^{2}\)
\({(a-b)}^{2}\) = 252 – 4(144)
\({(a-b)}^{2}\) = 625 – 576
\({(a-b)}^{2}\) = 49
a – b = 7


3. The sum of the two digit number is 11. The number obtained by reversing the digit is 27 less than the original number.Then the original number is

    A. 74
    B. 47
    C. 65
    D. 83


Answer: Option A
4+7 = 11
74-47 = 27


4. The product of 2 consecutive even number is 1224, then one of the number is

    A. 26
    B. 36
    C. 28
    D. 42


Answer: Option B

Explanation:
42


5. A number when divided by 361 gives a remainder 47. If the Same number is divided by 19 ,the remainder is

    A. 18
    B. 11
    C. 9
    D. 13


Answer: Option C

Explanation:
47÷19=> remainder 9
Ex : 769 ÷ 361 => remainder 47
769 ÷ 19 => remainder 9


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