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.
Learn more about fractions here:
#SPJ4
You might be interested in
Answer:
(1, 1) is the only one on the graph
Step-by-step explanation:
To determine if an ordered pair lies on the graph, substitute the x-coordinate of the ordered pair into f(x) and evaluate. If the value equals the y-coordinate of the ordered pair then it lies on the graph
f(1) = (2 × 1) - 1 = 2 - 1 = 1 ⇒ (1, 1) lies on graph
f(- 1) = (2 × - 1) - 1 = - 2 - 1 = - 3 ≠ - 4 ⇔ (- 1, - 4 ) not on graph
f(- 2) = (2 × - 2) - 1 = - 4 - 1 = - 5 ≠ - ⇒ (- 2, -
) not on graph
f(0) = (2 × 0) - 1 = 0 - 1 = - 1 ≠ ⇒ not on graph