d Word Problem 1 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Word Problem 1

Problem 1: Income of A is 25 % less than B’s income. How much percent B’s income would be more than that of A ?

(word problems on job exam mathematics.)


To solve this problem we will suppose income of B

i-e B's Income = x

then because A’s income is 25% less than that of B, we write;

A's income = x-\frac{25}{100}x

Now we can find how much B is earning extra than A – by subtracting A’s income from B’s income.

i-e B's Extra Earning = B's Income - A's Incom

\Rightarrow B's Extra Earning = x- \frac{75}{100}x

\Rightarrow B's Extra Earning = \frac{25}{100}x

Now we use percentage formula  – to find what percent  this B’s extra income is of A’s income !

%age = \frac{Amount To Be Calculated With %age}{Reference Amount For Calculating %age}\times 100

So, here we will put B’s extra earning as nominator and A’s income as denominator;

\Rightarrow B's Extra Earning %age=\frac{\frac{25}{100}x}{\frac{75}{100}x}\times 100

\Rightarrow B's Extra Earning %age= \frac{25}{75}\times 100

\Rightarrow B's Extra Earning %age= \frac{1}{3}\times 100

\Rightarrow B's Extra Earning %age= 33 % (approx))

So, B is earning approximately 33% more than A



Problem 65: If the price of the coal be raised by 10%, by how much percent a man must reduce its consumption so as not to increase his expenditure ?


This is the problem involving percentage calculation. Here we assume that intially price of caol and expenditure are same.

i-e Let’s suppose;

Price Of Coal = Expenditure By Man = x

Rise in price is 10%, so we will find first 10% of x (price)

10% of Price = \frac{10}{100}\times x

\Rightarrow 10% of Price = \frac{1}{10}\times x


10% of Price = \frac{x}{10}\

New price will be the sum of increment and original price


New Price Of Coal After Rising= x+\frac{x}{10}

or simply

New Price =\frac{11x}{10}

If the man keeps on burning same quantity of coal then again his consumption will be equal to the new price

i-e he will stay with

New Consumption= \frac{11x}{10}

Now suppose to keep his expenditure same (i-e ‘x’) the man now will have to reduce his consumption by y% of ‘new consumption’


Reduction = \frac{y}{100}Of \frac{11x}{10}

\Rightarrow Reduction = \frac{y}{100}\times \frac{11x}{10}

\Rightarrow Reduction = \frac{11xy}{1000}

Now If we subtract from ‘new consumption’ this ‘reduction’ then consumptiom will stay equal to original expenditure ‘x’


New Consumption – Reduction in Consumption = x


cancelling ‘x’ term





 \Rightarrow y =\frac{100}{11}

  \Rightarrow y =9\frac{1}{11}%

So, the man will have to reduce his consumption by 9\frac{1}{11}%





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