d Word Problem 8 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Word Problem 8

Problem 8: A reduction of 20% in the price of sugar enables a purchaser to buy 4 kgs more for Rs 80. The original price of 1 kg of sugar was;

a) Rs 4                          b) Rs 4.50                   c) Rs 8                         d) Rs 5.50

(word problems on job exam mathematics.)


We suppose;

Original price of sugar per kg = y Rs 

It means y rupees is equal to 1 kg of sugar i-e

\Rightarrow yRs = 1 kg

Now we calculated quantity of sugar in 80 Rs according to original price, so, from above we get;

1 Rs = \frac{1 kg}{y}

Multiply both sides by 80

\Rightarrow80 Rs = \frac{80 kg}{y}

Now we come to new price per kg,  after reduction of 20 % in original price. i-e after reduction of \frac{20}{100}y per kg. But new price per kg is calculated by subtracting reduction from original price i-e;

New Price

=y - \frac{20}{100}y

= \frac{100 y - 20 y }{100}

= \frac{80y }{100}

This simply means now 1 kg of sugar can be bought in Rs  \frac{80y }{100}.  We can write;

 \frac{80y }{100} = 1 kg

\Rightarrow y Rs = \frac{100}{80}

\Rightarrow 1 Rs = \frac{100}{80y}

Now we will calculate quantity of sugar in 80 Rs according to New Price ( After Reduction);, So multiply both sides by 80;

\Rightarrow 80 Rs = \frac{100}{80y}\times 80

\Rightarrow 80 Rs = \frac{100}y}

But according to condition of question, this quantity of sugar in Rs 80 as per new price should be 4 kg more than quantity of sugar in 80 Rs as per original price ! i-e we will add 4 kg more to \frac{80 kg}{y} , then both quantities of Rs 80 will be equal!

i-e \frac{100}{y}= \frac{80}{y}+4

Shift ‘y’ terms to one side;

\Rightarrow \frac{100}{y}- \frac{80}{y}= 4

\Rightarrow \frac{20}{y}= 4

\Rightarrow {20}= 4y

\Rightarrow y= 5

So original price of sugar per kg was Rs 5


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