Question

A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x ?

Answer 1

The value of x would be
`(3)/(5)y` or
`(3)/(8)z`

Explanation:

Given,

x tons of 10% gravel G mixture is mixed with y ton of 2% gravel G to obtain 5% gravel mixture,

Thus, gravel G in x ton + gravel in y ton = gravel G in the resultant mixture,

⇒ 0.1x + 0.02y = 0.05(x+y),

0.1x - 0.05x = 0.05y - 0.02y

0.05x = 0.03y

`\implies x = (0.03)/(0.05)y=(3)/(5)y----(1)`

Now, resultant mixture is z ton,

i.e.
`z=x+y`

`\implies y = z - x`

From equation (1),

`x=(3)/(5)(z-x) = (3)/(5)z-(3)/(5)x`

`x+(3)/(5)x=(3)/(5)z`

`(8)/(5)x=(3)/(5)z`

`x=(3)/(8)z`