A milkman sells his mixture of milk and water at the cost price of mil...
Solution:
Let's assume that the milkman has x litres of mixture, out of which y litres is milk and (x-y) litres is water.
As per the given condition, the cost price of the mixture is equal to the cost price of milk. Let's assume the cost price of 1 litre of milk is Rs. 10.
So, the cost price of y litres of milk is Rs. 10y and the cost price of x litres of mixture is Rs. 10x.
As per the question, the milkman gains a profit of 20%, which means he sells the mixture at 120% of its cost price.
Selling price of x litres of mixture = 120% of cost price of x litres of mixture
Selling price of x litres of mixture = 120/100 * 10x
Selling price of x litres of mixture = 12x
Now, we know that the milkman sells x litres of mixture, which contains y litres of milk. So, the selling price of y litres of milk is equal to the selling price of x litres of mixture.
Selling price of y litres of milk = Selling price of x litres of mixture
10y = 12x
y/x = 12/10
y/x = 6/5
Therefore, the ratio of milk and water in the mixture is 6:5, which is option (b) 5:1.
Explanation:
- We assume the quantity of mixture and the quantity of milk in it.
- We use the cost price of milk to calculate the cost price of milk in the mixture and the selling price of the mixture.
- We equate the selling price of milk with the selling price of the mixture to get the ratio of milk and water in the mixture.
- We choose the option which matches the ratio obtained.
A milkman sells his mixture of milk and water at the cost price of mil...
It can be solved by the allegation and mixture .20% profit- assume cp of milk and water is 100 then sp is 120,and if actual sp is 1 rs. then cp is 5/6.
1 0 5/6. =5/6 : 1/6 =5:1
0-5/6 1-5/6
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