Narjidn
Answered

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how many liters of 10% alcohol solution must be mixed with 4 liters of 50% solution to get a 20% solution?

Sagot :

SugoiOverloadidn
Let x be the amount (in liters) of the 10% solution that we need

4 (50%) + x (10%) = (4 + x)(20%)
Multiply both sides by 10 to cancel out the % sign

4(50) + x(10) = (4 + x)(20)
200 + 10x = 80 + 20x

Combine like terms
10x = 120
x = 12 Liters

Hope this helps :)