Difficult Percentage Problems Solved in 10 Seconds

Published on Wednesday, March 01, 2017

Ques 1. B as a % of A is equal to A as a % of (A+B). Find B as a % of A.

Solution:

→B/A = A/ (A+B)
→Assume, B = Ax (B is some percent of A. So consider that some percent as ‘x’.)
→Ax/A = A/(A+Ax)
→X=1/(1+x)
→X^2+ X -1 = 0
→Solving the above equation, we get 0.62(consider only positive terms),

    So the answer is 62%.

Ques 2. If the price of petrol increased by 25% and Raj intends to spend only an additional 15% on petrol, by how much will he reduce the quantity of petrol purchased?

Solution:

→Reduction = (% increased - % additional)/ (100 + % increased)
→Reduction in litres = (25-15)/ (100+25) = 8 litres

Ques 3. A candidate who got 20% marks fails by 10 marks but another candidate who gets 42% marks get 12% more than the passing marks. Find the maximum.

Solution:

→42% - 12% = 30% (Actual pass mark)
→30% of x – 20% of x = 10
→10% of x = 10
→X = (10*100)/10 = 100 Maximum mark 

Ques 4. Peter got 30% of the maximum marks in the examination and failed by 10 marks. However, Paul who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination? 

Solution:

→Difference of % of marks obtained = difference of the mark obtained
→40% of x - 30% of x = 15 – (-10)
→10% of x = 25
→X = 25*100/10 = 250 (maximum marks)
→Passing marks = 30% of 250 + 10 = 85%

Ques 5. In an election between two candidates, a person who got 58% of total votes won the election by a majority of 960. Find the total number of votes.

Solution:

→Assume the population be 100.
→58% of 100 = 58 votes got by winning party.
→42% of 100 = 42 votes got by loser.
→Difference between them = 58% – 42% = 16% (majority)
→Now in the given question, majority = 960 = 16%
→If 16% = 960 then 100% is 6000.
→So the total number of votes is 6000.

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