Question

Chocolate costing $\$10$ per pound is mixed with nuts costing $\$4$ per pound to make a mixture costing $\$6$ per pound. What fraction of the mixture's weight is chocolate

Answer 1

1/3 portion of the mixture, costing $\$6$ contains chocolate.

Calculation of the weight of the chocolate in the mixture:

Assume that, in a mixture:

chocolate = x pound

nuts = y pound

mixture= x+y

The desired value is found by dividing the weight of the chocolate by the overall weight, or x/(x+y).

From the given data,

cost of x pound chocolate at $\$10$per pound = 10x

cost of y pound nuts at $\$4$ per pound = 4y

cost of x+y pound mixture at $\$6$ = 6(x+y)

As per condition,

⇒10x + 4y = 6(x+y)

⇒10x + 4y = 6x+ 6y

⇒4x = 2y

⇒y = 2x

The fraction of the chocolate in the mixture is

$x/(x+y)$ = $x/(x+2x)$ = $x/3x$ = 1/3

Therefore, chocolate makes  1/3 of the mixture's weight.

Learn more about fractions here:

https://brainly.com/question/10354322

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