How many kg of the first alloy and of the second alloy should be alloyed together to obtain 28 kgs of new alloy with equal contents of copper and zinc if in two alloys, ratios of copper to zinc are 5:2 and 3:4?

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How many kg of the first alloy and of the second alloy should be alloyed together to obtain 28 kgs of new alloy with equal contents of copper and zinc if in two alloys, ratios of copper to zinc are 5:2 and 3:4?

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Answer 1 · 2015-03-31T23:46:00

Taking $X$ $X$ kg of alloy 1 (with 5:2 ratio of copper and zinc) and $Y$ $Y$ kg of alloy 2 (with 3:4 ratio) we will get the following quantities of copper and zinc:

1st alloy: $\frac{5}{7} \cdot X$ $5/7*X$ kg of cooper and $\frac{2}{7} \cdot X$ $2/7*X$ kg of zinc,
2nd alloy: $\frac{3}{7} \cdot Y$ $3/7*Y$ kg of copper and $\frac{4}{3} \cdot Y$ $4/3*Y$ kg of zinc.

Together we will get $\frac{5 X + 3 Y}{7}$ $(5X+3Y)/7$ kg of copper and $\frac{2 X + 4 Y}{7}$ $(2X+4Y)/7$ kg of zinc.

There are two conditions in this problem:
(a) combined weight should be $28$ $28$ kg and
(b) amount copper and zinc must equal (that is, $14$ $14$ kg each).

Therefore, we have to solve the following system of two linear equations with two unknown:
$\frac{5 X + 3 Y}{7} = 14$ $(5X+3Y)/7=14$
$\frac{2 X + 4 Y}{7} = 14$ $(2X+4Y)/7=14$

Multiplying by 7 both equations:
$5 X + 3 Y = 98$ $5X+3Y=98$
$2 X + 4 Y = 98$ $2X+4Y=98$

Multiplying the first by 4 and the second by 3:
$20 X + 12 Y = 392$ $20X+12Y=392$
$6 X + 12 Y = 294$ $6X+12Y=294$

Subtract the second from the first:
$14 X = 98$ $14X=98$

Divide by 14:
$X = 7$ $X=7$

Using the first equation in its original form and substituting $X$ $X$ :
$5 \cdot 7 + 3 Y = 98$ $5*7+3Y=98$
$35 + 3 Y = 98$ $35+3Y=98$
$3 Y = 63$ $3Y=63$
$Y = 21$ $Y=21$

The answer: we have to take $7$ $7$ kg of the first and $21$ $21$ kg of the second alloy.

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How many kg of the first alloy and of the second alloy should be alloyed together to obtain 28 kgs of new alloy with equal contents of copper and zinc if in two alloys, ratios of copper to zinc are 5:2 and 3:4?

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