d Word Problem 19 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Word Problem 19

Problem 19: A man buys milk at Rs 2 per kg. Mixing it with water he sells the mixture at the same price. If his profit is 11-1/9 %, the quantity of water added to each kg of milk is ;

a) 1/3 kg                b) 1/9 kg                     c) 1/5 kg           d) 1/2 kg


We are given the following information;

Cost of milk = Rs 2 per kg

Sale price of mixture = Rs 2 per kg

Profit of mixture sale = 11-1/9 %

Quantity of water added per kg of milk = ?


We suppose

cost price of mixture = Rs Y per kg

We know Profit is given by the formula;

Profit = SP – CP

Where SP = sale price

CP= cost price

and Profit = 11-1/9 % of CP


\[ Profit = \frac{11 \frac{1}{9} }{100}\times Y \]


\[ Profit = \frac{100Y}{900} \]

Putting value of profit, SP and CP in above formula, we get;

\[ \frac{100Y}{900}=2-Y \]

We get;

\[ \frac{1000Y}{900}=2 \]

and finally CP Y is given as;

\[ Y=\frac{1800}{1000}=\frac{18}{10}=\frac{9}{5}=1.8 Per Kg \]

Now assuming that water is free of cost and use blending formula as below;



We subtract both entities

prices from mixture

price and put result across

crossed lines in blending formula !)

So we got;

Milk : Water = 1.8 : 0.2 = 9 : 1


Milk : Water = 1:1/9


we can say that for each 1 kg of milk 1/9 kg of water is added






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