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

Word Problem

Problem : A reduction of 20 percent in the price of tea enables a purchaser to obtain 4 kg more for Rs 80. The reduced price per kg of tea is ;

a. Rs 4                b. Rs 3           c. Rs 7                 d. Rs 8

Finite math, business math –


Let the original price be Rs ‘y’ per kg

20 percent of this price;



20percent Of y'= \frac{20}{100}y

 \Rightarrow 20percent Of y'= \frac{y}{5}

So new price per kg can be calculated by subtracting this 20% of y – from original price ‘y’


New Price = Original Price - 20% of Original Price

 \Rightarrow New Price = y - \frac{y}{5}

 \Rightarrow New Price = \frac{4y}{5}

Now According to statement of question – purchaser gets 4 kg more – if  Rs 80 given more!

i-e For new price he gets 1 kg (because this is the new price per kg) – and if he gives Rs 80 more he will get 4 kg more i-e 1+4=5 kg he will get !

i-e   New Price Per KG+Rs 80 = 1Kg +4 kg

\Rightarrow \frac{4y}{5}+80=1+4

\Rightarrow \frac{4y}{5}+80=5

\Rightarrow 4y+400=25

\Rightarrow 4y=375

\Rightarrow y=\frac{375}{4}

Put this value of ‘y’ in New Price above

i-e New Price=\frac{4y}{5}

New Price=\frac{4\times (375/4)}{5}

New Price=\frac{375}{5}

New Price=75

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