A specialized coffee roastery produces three custom blends—Roast Alpha, Roast Beta, and Ro… — GMAT

A specialized coffee roastery produces three custom blends—Roast Alpha, Roast Beta, and Roast Gamma—using three varieties of single-origin beans: Grade A, Grade B, and Grade C.

- One batch of Roast Alpha requires 3 kg of Grade A, 1 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of $64. - One batch of Roast Beta requires 1 kg of Grade A, 4 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of$ 64.
- One batch of Roast Gamma requires 2 kg of Grade A, 2 kg of Grade B, and 5 kg of Grade C beans, and has a total raw material cost of $90.

What is the cost, in dollars, of 1 kg of Grade A beans?

Cevap: 12 dollars

Cevap

The cost of 1 kg of Grade A beans is 12 dollars.

Setting up the 3-variable linear system

3x + y + 2z = 64

x + 4y + 2z = 64

, and

2x + 2y + 5z = 90

allows us to eliminate

z

by subtracting the second equation from the first, yielding

2x - 3y = 0

, or

x = 1.5y

. Substituting this relationship back into the system leads to

y = 8

and

x = 12

. Thus, 1 kg of Grade A beans costs 12 dollars.

Adım Adım Çözüm

Formulate linear equations representing the total cost of each coffee blend batch.

Let

x

be the cost per kg of Grade A beans,

y

be the cost per kg of Grade B beans, and

z

be the cost per kg of Grade C beans:

\begin{aligned} (1)\quad 3x + y + 2z &= 64 \\ (2)\quad x + 4y + 2z &= 64 \\ (3)\quad 2x + 2y + 5z &= 90 \end{aligned}

Translating the word problem into a system of 3 linear equations with 3 variables.

Eliminate variable

z

by subtracting Equation (2) from Equation (1).

(3x + y + 2z) - (x + 4y + 2z) = 64 - 64 \implies 2x - 3y = 0 \implies x = 1.5y

Since both equations (1) and (2) contain the term

+2z

, subtracting them removes

z

directly and provides a simple relation between

x

and

y

Substitute

x = 1.5y

into Equation (1) and Equation (3) to obtain a system in terms of

y

and

z

From Equation (1):

3(1.5y) + y + 2z = 64 \implies 5.5y + 2z = 64 \implies 11y + 4z = 128 \quad (4)

From Equation (3):

2(1.5y) + 2y + 5z = 90 \implies 5y + 5z = 90 \implies y + z = 18 \implies z = 18 - y

Reducing the system from 3 variables down to 2 variables.

Substitute

z = 18 - y

into Equation (4) to solve for

y

, and subsequently calculate

x

11y + 4(18 - y) = 128 \implies 7y + 72 = 128 \implies 7y = 56 \implies y = 8

Using

x = 1.5y

x = 1.5 \times 8 = 12

Solving the single-variable linear equation for

y

, then substituting back to find the required cost

x

for Grade A beans.

Anahtar Kavram

Solving a 3-Variable System of Linear Equations via Variable Elimination

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Cevap

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