A milkman purchases 10 litres of milk at Rs. 7 per litre and forms a mixture by adding freely available water which constitutes 16.66 % of the mixture. Later on he replaced the mixture by some freely available water and thus the ratio of milk is to water is 2 : 1. He then sold the new mixture at cost price of milk and replaced amount of mixture at twice the cost of the milk then what is the profit percentage?
Answers
Solution :-
→ CP of milk = 10 * 7 = Rs.70
now,
→ water in the mixture = 16.66% = (1/6)
so,
→ Milk : water in mixture = 5 : 1 [ 10 litre milk and 2 litre water .]
now, let us assume that, he replaced x litre of mixture with water .
then,
→ milk in x litre mixture = x * (5/6) = (5x/6) litre .
→ water in x litre mixture = x * (1/6) = (x/6) litre .
A/q, ratio becomes,
→ Milk : water = 2 : 1
→ [{10 - (5x/6)} / {2 - (x/6) + x} = 2/1
→ (60 - 5x/6) / (12 - x + 6x/6) = 2
→ 60 - 5x = 2(12 + 5x)
→ 60 - 5x = 24 + 10x
→ 60 - 24 = 10x + 5x
→ 15x = 36
→ x = 2.4 litre .
now,
→ Total mixture sold = 10 + 2 + 2.4 = 14.4 litre .
→ SP (mixture sold at cost price) = 7 * 12 = Rs.84
→ SP of Replaced amount (Twice of CP) = 2.4 * 14 = Rs.33
6
therefore,
→ Total SP = 84 + 33.6 = Rs.117.6
hence,
→ Profit % = {(SP - CP) * 100} / CP
→ Profit % = {(117.6 - 70) * 100} / 70
→ Profit % = (47.6 * 10)/7
→ Profit % = 476/7
→ Profit % = 68% (Ans.)
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