A coffee roaster creates a custom blend by combining Grade X beans costing \p per pound wi… — GRE General Test

A coffee roaster creates a custom blend by combining Grade X beans costing $\$p$ per pound with Grade Y beans costing $\$q$ per pound, where $0 < p < q$ . The batch contains a total of $M$ pounds, consisting of $x$ pounds of Grade X and $y$ pounds of Grade Y. The total cost of the batch is $C$ dollars, and the average cost per pound of the blend is $A$ dollars. Which of the following algebraic relationships MUST be true? Indicate all such statements.

The weight of Grade X beans in pounds is given by $x = \frac{qM - C}{q - p}$ .Answer
B
The weight of Grade X beans in pounds is given by $x = \frac{C - pM}{q - p}$ .
The average cost per pound of the blend satisfies $A = q - \frac{x(q - p)}{M}$ .Answer
D
If the weight of Grade X in the blend is increased by $20\%$ while keeping the weight of Grade Y constant, the total cost of the batch increases by $0.20C$ dollars.
The ratio of the weight of Grade X to the weight of Grade Y in the blend is equal to $\frac{q - A}{A - p}$ .Answer

Answer

The true statements are that the weight of Grade X beans is

x = \frac{qM - C}{q - p}

, the average cost per pound satisfies

A = q - \frac{x(q - p)}{M}

, and the ratio of Grade X weight to Grade Y weight is

\frac{q - A}{A - p}

The correct statements correctly model the system of equations

x + y = M

and

C = px + qy

. Solving for

x

in terms of total cost gives

x = \frac{qM - C}{q - p}

. Dividing the expression for total cost by total mass

M

yields

A = q - \frac{x(q - p)}{M}

. Finally, setting total cost

C = A(x+y) = px + qy

and rearranging gives the ratio

\frac{x}{y} = \frac{q - A}{A - p}

Step-by-Step Solution

Set up equations for total mass

M

and total cost

C

using individual weights

x

and

y

x + y = M \implies y = M - x

and

C = px + qy = px + q(M - x)

Relating the two variables through total weight eliminates

y

to express cost purely in terms of

x

Solve the total cost equation for

x

C = qM - (q - p)x \implies (q - p)x = qM - C \implies x = \frac{qM - C}{q - p}

Isolating

x

confirms the valid algebraic formula for the weight of Grade X.

Calculate average cost

A = \frac{C}{M}

by substituting the simplified expression for

C

A = \frac{qM - (q - p)x}{M} = q - \frac{x(q - p)}{M}

Dividing total cost by total weight

M

yields the weighted average cost per pound.

Determine the ratio of component weights

\frac{x}{y}

in terms of unit costs

p, q

and average cost

A

A(x + y) = px + qy \implies Ax + Ay = px + qy \implies y(q - A) = x(A - p) \implies \frac{x}{y} = \frac{q - A}{A - p}

Rearranging the weighted average equation isolates the ratio of the quantities of the two components.

Key Concept

Linear weighted averages and multi-variable system modeling in mixture word problems.

Estimated Time:2m 30s

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