A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 35% and the third contains 55%. He wants to use all three solutions to obtain a mixture of 252 liters containing 25% acid, using 3 times as much of the 55% solution as the 35% solution. How many liters of each solution should be used?
The chemist should use _________ liters of 15% solution, ________ liters of 35%
solution, and _______ liters of 55% solution.

Respuesta :

Answer Below:

Sorry for the late notice, but here you go:

Let +a+ = liters of 15% solution needed

Let +b+ = liters of 35% solution needed

Let +c+ = liters of 80% solution needed

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(1) +a+%2B+b+%2B+c+=+100+

(2) +c+=+2b+

(3) +%28+.15a+%2B+.35b+%2B+.8c+%29+%2F+100+=+.3+

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(3) +.15a+%2B+.35b+%2B+.8c+=+30+

(3) +15a+%2B+35b+%2B+80c++3000+

(3) +3a+%2B+7b+%2B+16c+=+600+

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Multiply both sides of (1) by +3+ and

subtract (1) from (3)

(3) +3a+%2B+7b+%2B+16c+=+600+

(1) +-3a+-+3b+-+3c+=+-300+

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+4b+%2B+13c+=+300+

Substitute (2) into this result

+4b+%2B+13%2A%282b%29+=+300+

+4b+%2B+26b+=+300+

 

+30b+=+300+

+b+=+10+

and

(2) +c+=+2b+

(2) +c+=+2%2A10+

(2) +c+=+20+

and

(1) +a+%2B+10+%2B+20+=+100+

(1) +a+=+100+-+30+

(1) +a+=+70+

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70 liters of 15% solution is needed

10 liters of 35% solution is needed

20 liters of 80% solution is needed

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check:

(3) +%28+.15a+%2B+.35b+%2B+.8c+%29+%2F+100+=+.3+

(3) +%28+.15%2A70+%2B+.35%2A10+%2B+.8%2A20+%29+%2F+100+=+.3+

(3) +%28+10.5+%2B+3.5+%2B+16+%29+%2F+100+=+.3+

(3) +10.5+%2B+3.5+%2B+16+=+30+

(3) +30+=+30+

Hope this helped!