Question

Difficulty: HardTranslating and Solving Algebraic Word Problems

An agricultural cooperative packages a premium seed mixture containing rye grass, fescue, and bluegrass. The weight of the fescue in the mixture is 1010 pounds less than twice the weight of the rye grass. The weight of the bluegrass is 1515 pounds more than half the weight of the fescue. If the total weight of the mixture is 120120 pounds, how many pounds of bluegrass are in the mixture?

Answer: 40 pounds

Answer

The weight of the bluegrass in the mixture is 40 pounds.
By translating the given relationships into equations (f=2r10f = 2r - 10, b=12f+15b = \frac{1}{2}f + 15, and r+f+b=120r + f + b = 120), we can express all variables in terms of rr, yielding r+(2r10)+(r+10)=120r + (2r - 10) + (r + 10) = 120. Solving this gives r=30r = 30. Substituting this back gives the weight of bluegrass as 30+10=4030 + 10 = 40 pounds.

Step-by-Step Solution

1
Define variables for each type of grass in the mixture.
Let rr represent the weight of rye grass, ff represent the weight of fescue, and bb represent the weight of bluegrass.
Establishing variables is necessary to translate the verbal descriptions into algebraic terms.
2
Translate the given relationships into equations.
f=2r10f = 2r - 10 and b=12f+15b = \frac{1}{2}f + 15
The problem states the fescue is 10 pounds less than twice the rye grass, and the bluegrass is 15 pounds more than half the fescue.
3
Substitute the expression for f into the equation for b to express b solely in terms of r.
b=12(2r10)+15=r5+15=r+10b = \frac{1}{2}(2r - 10) + 15 = r - 5 + 15 = r + 10
Reducing the number of variables simplifies the system of equations.
4
Set up the total weight equation and solve for r.
r+(2r10)+(r+10)=120    4r=120    r=30r + (2r - 10) + (r + 10) = 120 \implies 4r = 120 \implies r = 30
The sum of the three grass weights is given as 120 pounds.
5
Calculate the weight of the bluegrass using the value of r.
b=30+10=40b = 30 + 10 = 40
The question asks specifically for the weight of the bluegrass.

Key Concept

Translating verbal relationships into linear equations and solving a system of equations
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