Question

In a 20-liter solution of water and milk, the percentage of pure milk is 50%. how many liters of pure milk is required to increase the percentage of pure milk to 80%?

Answer 1

$amount \: of \: solution = 20 \\ amount \: of \: milk = 20 \times \frac{50}{100} = 10 \\ \\ let \: amount \: of \: milk \: required \: be \: x \\ then \: 10 + x = (20 + x) \times \frac{80}{100 } \\ 10 + x = 16 + 0.8x \\ 0.2x = 6 \\ x = 30 \: liter$

Answer 2

Answer:

30 liters of pure milk is required to increase the percentage of pure milk to 80%.

Step-by-step explanation:

Total amount of solution = 20 liters

The percentage of pure milk is 50%.

Amount of milk in the solution =$50\% \times 20 = \frac{50}{100} \times 20 = 10$

Let x be the amount of pure milk is required to increase the percentage of pure milk to 80%

So, new Amount of milk = 10+x

ATQ

$10+x=80\% \times (20+x)$

$10+x=\frac{80}{100}\times (20+x)$

$10+x=0.8(20+x)$

$10+x=16+0.8x$

$0.2x=6$

$x=30$

Hence 30 liters of pure milk is required to increase the percentage of pure milk to 80%.