Question

A dishonest milkman claims to sell milk at cost price, but the milk he sells is adulterated with 20% of water. His profit per cent is (a) 20% (b) 25% (C) 80% (d) 0%​

Answer 1

According to the question,

Let's assume milkman has 100 liter of milk, If he added x liter of water, the percentage of water in the mixture

$\bf= \frac{x} {(100 + x)} \times 100$

The milk gain 25%, he must have added 25 liter water in 100 liter of milk. Then the percentage of water in the mixture

$\bf = \frac{25}{(100 + 25)} \times 100 - 20\%$

In the new mixture, If the milk is 80% then 80% of total mixture should be 100 liter.

$\bf(100 + x) \times \frac{80}{100} - 100$

$\bf = 8x = 200$

$\bf = x - 25$

Then, the percentage of water in the mixture,

$\bf= \frac{ 25}{ (100 + 25)} \times 100 - 20\%$

Answer 2

Step-by-step explanation:

According to the question,

Let's assume milkman has 100 liter of milk, If he added x liter of water, the percentage of water in the mixture

$\bf= \frac{x} {(100 + x)} \times 100=$

The milk gain 25%, he must have added 25 liter water in 100 liter of milk. Then the percentage of water in the mixture

$\bf = \frac{25}{(100 + 25)} \times 100 - 20\%=$

In the new mixture, If the milk is 80% then 80% of total mixture should be 100 liter.

$\bf(100 + x) \times \frac{80}{100} - 100(100+x)$

$\bf = 8x = 200$

$\bf = x - 25$

Then, the percentage of water in the mixture,

$\bf= \frac{ 25}{ (100 + 25)} \times 100 - 20\%=$

REGARDS.

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