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Difficulty: Very hardAlgebraic Word Problems and Modeling

Two automated assembly lines, Line A and Line B, produce components at constant individual rates. Under normal operating conditions, Line A operating for 33 hours and Line B operating for 44 hours together produce a combined total of 1,4001,400 units. Under adjusted operating conditions, Line A operates at a rate 20%20\% higher than its normal rate, while Line B operates at a rate 10%10\% lower than its normal rate. Operating together under these adjusted conditions for 55 hours, the two lines produce a total of 2,1002,100 units. What is the normal rate of Line A, in units per hour?

  1. A
    160160
  2. 200200Answer
  3. C
    240240
  4. D
    250250
  5. E
    300300

Answer

The normal rate of Line A is 200200 units per hour.
Let rAr_A and rBr_B represent the normal production rates in units per hour for Line A and Line B, respectively. From the first condition, 3rA+4rB=14003r_A + 4r_B = 1400. From the second condition, operating for 55 hours at rates 1.20rA1.20r_A and 0.90rB0.90r_B yields 5(1.20rA+0.90rB)=21005(1.20r_A + 0.90r_B) = 2100, which simplifies to 1.20rA+0.90rB=4201.20r_A + 0.90r_B = 420, or 4rA+3rB=14004r_A + 3r_B = 1400. Subtracting 3rA+4rB=14003r_A + 4r_B = 1400 from 4rA+3rB=14004r_A + 3r_B = 1400 gives rA−rB=0r_A - r_B = 0, meaning rA=rBr_A = r_B. Substituting rB=rAr_B = r_A into 3rA+4rA=14003r_A + 4r_A = 1400 gives 7rA=14007r_A = 1400, so rA=200r_A = 200 units per hour.

Step-by-Step Solution

1
Define variables and set up the equation for normal operating conditions.
3rA+4rB=14003r_A + 4r_B = 1400
Line A operates for 33 hours at rate rAr_A and Line B operates for 44 hours at rate rBr_B to produce 1,4001,400 units.
2
Set up the equation for adjusted operating conditions.
5(1.20rA+0.90rB)=2100  ⟹  1.20rA+0.90rB=4205(1.20r_A + 0.90r_B) = 2100 \implies 1.20r_A + 0.90r_B = 420
Line A's rate increases by 20%20\% (1.20rA1.20r_A) and Line B's rate decreases by 10%10\% (0.90rB0.90r_B). Divided by 55 hours, their combined hourly adjusted rate is 420420 units per hour.
3
Multiply the simplified adjusted equation by 1010 to clear decimals.
12rA+9rB=4200  ⟹  4rA+3rB=140012r_A + 9r_B = 4200 \implies 4r_A + 3r_B = 1400
Dividing all terms by 33 simplifies the linear equation for easier elimination.
4
Solve the system of equations for rAr_A.
rA=200r_A = 200
From Step 1, 4rB=1400−3rA  ⟹  rB=350−0.75rA4r_B = 1400 - 3r_A \implies r_B = 350 - 0.75r_A. Substituting into 4rA+3(350−0.75rA)=14004r_A + 3(350 - 0.75r_A) = 1400 gives 4rA+1050−2.25rA=1400  ⟹  1.75rA=350  ⟹  rA=2004r_A + 1050 - 2.25r_A = 1400 \implies 1.75r_A = 350 \implies r_A = 200.

Key Concept

Linear Modeling of Combined Work and Rates
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Topics

Algebraic Word Problems and Modeling

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