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Salsk061 [2.6K]

3 years ago

10

There are two peice pf gold and silver alloy. The ratio between the gold and silver in the first piece is 2:3, in the second pei

ce 3:7. If we want to have 8 gram of gold and silver alloy with gold-silver ratio 5:11 how much of each peice of alloy is needed.

1 answer:

STALIN [3.7K]3 years ago

3 0

Answer:

1 grams of one of the alloy; and
7 grams of the other corresponding alloy.

Step-by-step explanation:

The ratio between the gold and silver in the first piece = 2:3

The ratio between the gold and silver in the second piece =3:7

The ratio in the mixture = 5:11

We want to have 8 gram of the new mixture.

Let the gram of alloy taken from the first piece=x

Therefore: gram of alloy would be taken from the second piece=(8-x)

This gives:

$\dfrac{2}{5}x+ \dfrac{3}{10}(8-x)=\dfrac{5}{16}*8$

We simplify the equation above for the value of x.

$\dfrac{2x}{5}+ \dfrac{3(8-x)}{10}=\dfrac{5}{2} \\\dfrac{4x+24-3x}{10}=\dfrac{5}{2}\\\dfrac{x+24}{10}=\dfrac{5}{2}\\2x+48=50\\2x=50-48\\2x=2\\x=1$

Therefore to create 8 gram of gold and silver alloy with gold-silver ratio 5:11, we take 1 grams of one of the alloy and 7 grams of the other alloy.

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