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Algebraic Word Problems and Modeling

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Question 61Question →

A coffee merchant creates a 5050-pound custom mixture combining Grade A coffee beans, which cost $8\$8 per pound, and Grade B coffee beans, which cost $14\$14 per pound. If the total cost of the mixture must be at least $460\$460 and at most $520\$520, which of the following could be the weight, in pounds, of Grade A coffee beans used in the mixture? Select all such weights.

Select all that apply

Show answer & explanation

Answer: 3232; 3535; 4040

Answer

The weight of Grade A coffee beans can be 32 pounds, 35 pounds, or 40 pounds.
Let xx represent the number of pounds of Grade A coffee beans. The remaining 50−x50 - x pounds consist of Grade B beans. The total cost of the mixture is given by 8x+14(50−x)=700−6x8x + 14(50 - x) = 700 - 6x. Setting up the given inequality constraints, we have 460≤700−6x≤520460 \le 700 - 6x \le 520. Subtracting 700 yields −240≤−6x≤−180-240 \le -6x \le -180. Dividing by −6-6 and reversing the inequality signs gives 30≤x≤4030 \le x \le 40. Therefore, any weight of Grade A beans between 30 and 40 pounds inclusive is valid. The values 32, 35, and 40 satisfy this condition.

Step-by-Step Solution

1
Define variables for the quantities of Grade A and Grade B beans.
Let xx be the weight in pounds of Grade A beans. Then 50−x50 - x is the weight in pounds of Grade B beans.
The total weight of the mixture is fixed at 50 pounds.
2
Set up an expression for the total cost of the mixture in terms of xx.
Total Cost = 8x+14(50−x)=700−6x8x + 14(50 - x) = 700 - 6x.
Multiply the weight of each component by its price per pound.
3
Set up the compound inequality representing the given cost constraints.
460≤700−6x≤520460 \le 700 - 6x \le 520.
The total cost must be at least $460\$460 and at most $520\$520.
4
Solve the compound inequality for xx.
Subtracting 700 from all parts gives −240≤−6x≤−180-240 \le -6x \le -180. Dividing by −6-6 and reversing the inequality signs yields 30≤x≤4030 \le x \le 40.
Dividing by a negative number reverses the direction of the inequality signs.
5
Identify which options fall within the valid range [30,40][30, 40].
The values 32, 35, and 40 fall within the range 30≤x≤4030 \le x \le 40.
Any weight between 30 and 40 pounds inclusive satisfies the cost constraint.

Key Concept

Linear Modeling and Inequality Constraints in Mixture Problems
Estimated Time:1m 40s
Question 62Question →

A delivery truck traveled from Warehouse X to Warehouse Y at a constant speed of 5050 miles per hour. On the return trip along the exact same route, heavy traffic reduced the truck's constant speed to 3030 miles per hour. If the total driving time for the entire round trip was 88 hours, what was the distance, in miles, between Warehouse X and Warehouse Y?

Show answer & explanation

Answer: 150150

Answer

The distance between Warehouse X and Warehouse Y is 150150 miles.
The correct distance between the two warehouses is 150150 miles. Since time equals distance divided by rate, the time taken going to Warehouse Y is d50\frac{d}{50} hours and returning is d30\frac{d}{30} hours. Summing these expressions to equal the total 88 hours gives d50+d30=8\frac{d}{50} + \frac{d}{30} = 8. Solving for dd yields 8d150=8\frac{8d}{150} = 8, which simplifies to d=150d = 150 miles.

Step-by-Step Solution

1
Define variables and write time expressions for each leg of the trip
Let dd be the distance in miles between Warehouse X and Warehouse Y. Time outbound is t1=d50t_1 = \frac{d}{50} hours, and time inbound is t2=d30t_2 = \frac{d}{30} hours.
Distance divided by rate gives the time taken for each individual leg of the journey.
2
Set up the equation using total elapsed time
d50+d30=8\frac{d}{50} + \frac{d}{30} = 8
The total driving time for both legs combined is given as 88 hours.
3
Solve the algebraic equation for dd
\frac{3d + 5d}{150} = 8 \implies \frac{8d}{150} = 8 \implies 8d = 1200 \implies d = 150
Finding a common denominator of 150150 allows combining the fractions to solve for the unknown distance dd.

Key Concept

Distance-Rate-Time Relationship & Harmonic Mean Rate Modeling
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