Chocolate costing $\$10$ per pound is mixed with nuts costing $\$4$ per pound to make a mixture costing $\$6$ per pound. What fr
1 answer:
1/3 portion of the mixture, costing $\$6$ contains chocolate.
<h2>Calculation of the weight of the chocolate in the mixture:</h2>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
=
=
= 1/3
Therefore, chocolate makes 1/3 of the mixture's weight.
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