Question

A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 45 % and the third contains 90 %. He wants to use all three solutions to obtain a mixture of 50 liters containing 70 % acid, using 2 times as much of the 90 % solution as the 45 % solution. How many liters of each solution should be used?
the chemist should use:​

Answer 1

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ALGEBRA WORD PROBLEM

Dorian S. asked • 06/28/17

THREE LINEAR EQUATIONS WITH THREE VARIABLES

A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 25%,

and the third contains 70%. He wants to use all three solutions to obtain a mixture of 40 liters containing 45%

acid, using 2 times as much of the 70% solution as the 25% solution. How many liters of each solution should be used?

How to set up as three linear equations to find the answer?

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Arthur D. answered • 06/28/17

TUTOR 4.9 (67)

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set up 2 equations

the first equation...

15%*x+25%*y+70%*2y=45%*40

0.15x+0.25y+1.4y=0.45*40

0.15x+1.65y=18

multiply all terms by 100

15x+165y=1800

divide all terms by 15

x+11y=120

the second equation...

x+y+2y=40

x+3y=40

x=40-3y

substitute into the first equation x+11y=120

40-3y+11y=120

40+8y=120

8y=120-40

8y=80

y=80/8

y=10 liters

x+3*10=40

x+30=40

x=40-30

x=10 liters

10 liters of 15%, 10 liters of 25%, and 20 liters of 70%

check:

10*15%=1.5

10*25%=2.5

20*70%=14

1.5+2.5+14=18

18/40=9/20=45/100=45%

45%*40=18

hope it helps

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