SOLUTION: metallurgist has one alloy containing 23% titanium and another containing 51% titanium. How many pounds of each alloy must he use to make 56 pounds of a third alloy containing 47%

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Question 1153659: metallurgist has one alloy containing 23% titanium and another containing 51% titanium. How many pounds of each alloy must he use to make 56 pounds of a third alloy containing 47% titanium? (Round to two decimal places if necessary.)

Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
H=51 percent
L=23 percent
T=47 percent
M=56 pounds
v=56 pounds, (unknown)
M-v, pounds unknown also for L concentration material

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
Just substitute your given values and evaluate v and M-v.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


A quick and easy non-algebraic path to the answer to this problem and a wide variety of similar "mixture" problems....

Look at the percentages 23, 47 and 51 on a number line.
47 is 24/28 = 6/7 of the way from 23 to 51.
That means 6/7 of the mixture is the 51% alloy.

With a total of 56 pounds, that means 48 pounds of the 51% alloy an 8 pounds of the 23% alloy.