1.

How many gallons of each of a 4% and an 11% acid solutions should be mixed to obtain 35 gallons of a 7% solution?1). 20 gallons, 15 gallons2). 22 gallons, 13 gallons3). 13 gallons, 22 gallons4). 15 gallons, 20 gallons

Answer»

Let US denote by y gallons the AMOUNT of 4% solution.

⇒ The amount of solution of 11% acid solution = 35 - y gallons

 

Amount of solution (gallons)

Percentage

Amount of SOLVENT (gallons)

Component 1

y

0.04

0.04y

Component 2

35 - y

0.11

0.11 (35 - y)

Mixture

35

0.07

0.07 (35) = 0.04y + 0.11 (35 - y)

 

SOLVING 0.07 (35) = 0.04y + 0.11 (35 - y),we get

y = 20 gallons

If y = 20 gallons; then the other amount, denoted by 35 - y must be 35 - 20 = 15 gallons.


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