Question

A pharmacist has 20 L of a 10% drug solution. How many liters of a 5% solution must be added to get a mixture that is 8%?

Answer 1

13.33 L of 5% solution must be added to get a mixture that is 8%.

Step-by-step explanation:

• Let 'x' be the required liters of 5% solution to be added  to get 8% mixture.
• Already we have 10% of 20 L, with this 5% of x liters must be added.

⇒ 0.05x + 0.10(20) = 0.08(20+x)

⇒ 0.05x + 2 = 1.6 + 0.08x

⇒ 2 - 1.6 = 0.08x - 0.05x

⇒ 0.4 = 0.03x

x = 0.4/0.03

x = 13.33 liters.

∴ 13.33 L of 5% solution is required.

Answer 2

Answer:

13.33 L

Step-by-step explanation:

y represents the required liters of 5% solution to be added  to get 8%mixture.

Already we have 10% of 20 L, with this 5% of x liters must be added.

0.05x + 0.10(20) = 0.08(20 + x)

0.05x + 2 = 1.6 + 0.08x

2 - 1.6 = 0.08x - 0.05x

0.4 = 0.08x - 0.05x

0.4 = 0.03x

0.4 ÷ 0.03 = 0.03x ÷ 0.03

x = 0.4 ÷ 0.03

$x=13\frac{1}{3}$

x = 13.33 liters.

13.33 L