SOLUTION: How many liters of 10% acid solution must be added to a 40% acid solution to produce an 8 liters volume 30% acid solution

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Question 1154192: How many liters of 10% acid solution must be added to a 40% acid solution to produce an 8 liters volume 30% acid solution
Found 3 solutions by josmiceli, josgarithmetic, greenestamps:
Answer by josmiceli(19441) About Me  (Show Source):
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Let +x+ = liters of 10% acid solution needed
+8+-+x+ = liters of 40% acid solution needed
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+%28+.1x+%2B+.4%2A%28+8+-+x+%29+%29+%2F+8+=+.3+
+.1x+%2B+.4%2A%28+8+-+x+%29+=+2.4+
+.1x+%2B+3.2+-+.4x+=+2.4+
+.3x+=+.8+
+x+=+8%2F3+
+2.667+ liters of 10% solution are needed
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check:
+%28+.1x+%2B+.4%2A%28+8+-+x+%29+%29+%2F+8+=+.3+
+%28+.1%2A%288%2F3%29+%2B+.4%2A%28+8+-+8%2F3+%29+%29+%2F+8+=+.3+
++.1%2A%288%2F3%29+%2B+.4%2A%28+8+-+8%2F3+%29+=+2.4+
++.1%2A%288%2F3%29+%2B+.4%2A%28+24%2F3+-+8%2F3+%29+=+2.4+
++.1%2A%288%2F3%29+%2B+.4%2A%28+16%2F3+%29+=+2.4+
+.1%2A8+%2B+.4%2A16+=+7.2+
+.8+%2B+6.4+=+7.2+
+7.2+=+7.2+
OK

Answer by josgarithmetic(39618) About Me  (Show Source):
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Two-part mixture can generally be solved as:
How many liters of L% solution be added to H% solution to produce M liters of T% solution?

v, volume of the H% solution
M-v volume of the L% solution

highlight_green%28Hv%2BL%28M-v%29=TM%29
-
Hv%2BLM-Lv=TM
Hv-Lv=TM-LM
v%28H-L%29=M%28T-L%29
highlight%28v=M%28%28T-L%29%2F%28H-L%29%29%29

Substitute your given system%28M=8%2CT=30%2CL=10%2CH=40%29, and evaluate v.

Answer by greenestamps(13200) About Me  (Show Source):
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Here is a quick and easy way (if you understand it!) to solve two-part mixture problems like this.

(1) 30% is "twice as close" to 40% as it is to 10%. (The difference between 40 and 30 is half the difference between 30 and 10.)

(2) That means the mixture needs to use twice as much of the 40% acid as the 10% acid.

(3) Twice as much means 2/3 of the mixture needs to be the 40% acid, so 1/3 is the 10% acid.

ANSWER: 2/3 of 8 liters, or 16/3 liters, of 40% acid; 1/3 of 8 liters, or 8/3 liters, of 10% acid.