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Difficulty: Very hardRatios, Rates, and Proportions

A manufacturing plant operates two types of machines, Type A and Type B. The ratio of the number of Type A machines to Type B machines is 3:53:5. Each Type A machine produces widgets at a constant rate that is 60%60\% faster than the constant rate of a Type B machine. When all machines of both types are operating simultaneously, they produce a total of 9898 widgets per hour. If the plant increases the number of Type A machines by 50%50\% and decreases the number of Type B machines by 20%20\%, what is the total number of widgets the new setup will produce in 33 hours?

  1. A
    112
  2. B
    246
  3. C
    312
  4. 336Answer
  5. E
    382

Answer

336
The correct answer is 336 because setting up the initial production equation based on the ratios gives a compound constant of xr=10xr = 10. Applying the percentage adjustments yields a new hourly production rate of 112 widgets. Multiplying this hourly rate by the specified 3 hours gives a total of 336 widgets.

Step-by-Step Solution

1
Represent the machine counts and rates using variables based on the given ratios.
Let the number of Type A machines be 3x3x and the number of Type B machines be 5x5x. Let the rate of a Type B machine be rr widgets/hour. The rate of a Type A machine is 1.6r1.6r widgets/hour.
This establishes algebraic expressions for the quantities in terms of common multipliers xx and rr.
2
Set up an equation for the initial total hourly production and solve for xrxr.
(3x)(1.6r)+(5x)(r)=98  ⟹  4.8xr+5xr=98  ⟹  9.8xr=98  ⟹  xr=10(3x)(1.6r) + (5x)(r) = 98 \implies 4.8xr + 5xr = 98 \implies 9.8xr = 98 \implies xr = 10.
This allows us to find the value of the joint constant factor xrxr needed for subsequent calculations.
3
Determine the new machine counts after the percentage changes.
New Type A count: 3x×1.50=4.5x3x \times 1.50 = 4.5x. New Type B count: 5x×0.80=4x5x \times 0.80 = 4x.
This applies the 50% increase to Type A and the 20% decrease to Type B machine counts.
4
Calculate the new total hourly production rate using the value of xrxr.
New hourly rate: (4.5x)(1.6r)+(4x)(r)=7.2xr+4xr=11.2xr(4.5x)(1.6r) + (4x)(r) = 7.2xr + 4xr = 11.2xr. Substituting xr=10xr = 10 yields 11.2×10=11211.2 \times 10 = 112 widgets/hour.
This finds the rate at which widgets are produced in the new configuration.
5
Multiply the new hourly rate by the specified time of 3 hours.
Total production: 112 widgets/hour×3 hours=336112 \text{ widgets/hour} \times 3 \text{ hours} = 336 widgets.
This yields the final total quantity of widgets produced over the 3-hour duration.

Key Concept

Solving compound ratio and rate problems with percentage changes

Alternative Method

Instead of variables, you can plug in convenient numbers that satisfy the ratios. Suppose there are initially 3030 machines of Type A and 5050 machines of Type B. Let a Type B machine produce at a rate of 11 widget/hour, so a Type A machine produces 1.61.6 widgets/hour. The total initial hourly rate is 30(1.6)+50(1)=48+50=9830(1.6) + 50(1) = 48 + 50 = 98 widgets/hour, which matches the prompt. Increasing Type A by 50% gives 4545 machines, and decreasing Type B by 20% gives 4040 machines. The new hourly rate is 45(1.6)+40(1)=72+40=11245(1.6) + 40(1) = 72 + 40 = 112 widgets/hour. For 3 hours, this gives 112×3=336112 \times 3 = 336 widgets.
Estimated Time:3m 0s
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Topics

Ratios, Rates, and Proportions

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