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
Solution:
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
or
\[ Profit = \frac{11 \frac{1}{9} }{100}\times Y \]
or
\[ Profit = \frac{100Y}{900} \]
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;
(Note:
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
or
Milk : Water = 1:1/9
or
we can say that for each 1 kg of milk 1/9 kg of water is added
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