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Coffee R 24 pounds

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Coffee R 24 pounds

by CrackGMAC » Thu Aug 27, 2009 8:06 am
101. Coffee R 24 pounds, coffee P 25 pounds. Mix these two kinds of coffee to produce mixture X, in which the rate of R/P is 4/1; and mixture Y, in which the rate of R/P is 1/3. What is the amount of mixture Y?

Don't have a penny clue about this one as in how to proceed.
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by CrackGMAC » Thu Sep 17, 2009 6:30 am
Any update???
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by m&m » Thu Sep 17, 2009 6:41 am
since all 24 and 25 pounds are used you can setup 2 eqns
call x the multiple used in coffee 1 and call y the multiple used in coffee 2, so that

4x + x = some amnt
y + 3y = some other amnt

some amnt + some other amnt = 24 + 25 = 49

so
4x + y = 24 and
x + 3y = 25

solving we get

4(25-3y) + y = 24
100 - 12y + y = 24
76 = 11y ~ 7 (less than 7)

so (~7) + 3(~7) = ~28 (less than 28)
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by caspermonday » Wed Sep 23, 2009 11:47 am
CrackGMAC: Is the solution of m&m right?
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by grockit_jake » Thu Sep 24, 2009 3:15 pm
Right. I would find it easier to subtract equations, instead of solving for a variable. See below.

4x + x = Total in mixture X

y + 3y = Total in mixture Y

Tx + Ty = 49 = 5x + 4y

AND we know that:

4x + y = 25 lbs
x + 3y = 24 lbs

Since the question asks for Ty, we can cancel out the x's in the system of equations:

4x + y = 25 lbs -
4x + 12y = 96 lbs

11y = 71 lbs, and we are looking for 4y. See above.
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