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A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
- a)25%
- b)20%
- c)22%
- d)24%
Correct answer is option 'B'. Can you explain this answer?
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A dishonest milkman sells his milk at cost price but he mixes it with ...
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A dishonest milkman sells his milk at cost price but he mixes it with ...
Solution:
Let us assume that the milkman has 1 liter of milk.
As per the question, he sells it at the cost price, i.e., he does not make any profit or loss.
Then he mixes water in it and gains 25%.
This means that he now sells 1.25 liters of the mixture at the cost price of 1 liter of milk.
Let us assume that the percentage of water in the mixture is x%.
This means that the percentage of milk in the mixture is (100-x)%.
Now, let us apply the rule of mixtures:
Quantity of Milk in 1 liter of Milk = Quantity of Milk in 1.25 liters of Mixture
=> (100-x)% of 1 = (100/125)*(100-x)% of 1.25
=> 100-x = (4/5)*(100-x)
=> x/5 = 20
=> x = 20*5
=> x = 100
Therefore, the percentage of water in the mixture is 100- (100-x) = 100-100 = 0%
Hence, the correct answer is option (B) 20%.
Let us assume that the milkman has 1 liter of milk.
As per the question, he sells it at the cost price, i.e., he does not make any profit or loss.
Then he mixes water in it and gains 25%.
This means that he now sells 1.25 liters of the mixture at the cost price of 1 liter of milk.
Let us assume that the percentage of water in the mixture is x%.
This means that the percentage of milk in the mixture is (100-x)%.
Now, let us apply the rule of mixtures:
Quantity of Milk in 1 liter of Milk = Quantity of Milk in 1.25 liters of Mixture
=> (100-x)% of 1 = (100/125)*(100-x)% of 1.25
=> 100-x = (4/5)*(100-x)
=> x/5 = 20
=> x = 20*5
=> x = 100
Therefore, the percentage of water in the mixture is 100- (100-x) = 100-100 = 0%
Hence, the correct answer is option (B) 20%.
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A dishonest milkman sells his milk at cost price but he mixes it with ...
To gain 25% suppose initially milk was 100 mL, now his gain depends on the sp which is defining the new cp i.e. Suppose water content be 'x' mL. .. Suppose Re 1 is cp for every 1 mL then 100 rs = 100mL .. Now, after adding x water new mixture will have 100-x milk in 100 mL mixture. Hence rs = 100-x. Now his gain is 25% hence . . 25/100(100-x) = x( as x amount of milk is saved by him) Solve it and you will get x='20' so resulting mixture will have 80 mL milk, of which he is saving 20mL i.e. 25%
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A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
a)25%b)20%c)22%d)24%Correct answer is option 'B'. Can you explain this answer?
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