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Bunuel
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Machines U and T can produce 10,000 units in x hours when working 26 Aug 2026, 19:00
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91% (01:51) correct 9% (03:10) wrong based on 46 sessions

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Machines U and T can produce 10,000 units in x hours when working together at their constant rates. If U can produce 10,000 units alone in 4x/3 hours, in terms of x, how many hours will it take T to produce 10,000 units alone?

A. 5x/2
B. 4x
C. 2x
D. x
E. x/2


Source: Math Revolution

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B
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paragw
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Machines U and T can produce 10,000 units in x hours when working 27 Aug 2026, 00:02
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1/U + 1/T = 1/Both
3/4x + 1/t = 1/x
t = 4x
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LucidGhost
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Machines U and T can produce 10,000 units in x hours when working 27 Aug 2026, 17:13
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U rate= 10000/4x/3 = 30000/4x
T rate = 10000/T
Combined Rate = 1/T+ 1/U
(10000/T)+(30000/4x)= 10000/x
(1/T)+(3/4x)=1/x
1/T = (4/4x)-(3/4x)= 1/4x
T = 4x

B
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Machines U and T can produce 10,000 units in x hours when working 27 Aug 2026, 18:57
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Rate at which both machine work together = 10,000/x
Rate at which machine U work alone = (10,000 * 3) / 4x = 30,000/4x
Then, rate at which machine T work alone = 10,000/T

30,000/4x + 10,000/T = 10,000/x
3/4x + 1/T = 1/x
T = 4x

Answer is option B.
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Machines U and T can produce 10,000 units in x hours when working 27 Aug 2026, 19:09
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Bunuel
Machines U and T can produce 10,000 units in x hours when working together at their constant rates. If U can produce 10,000 units alone in 4x/3 hours, in terms of x, how many hours will it take T to produce 10,000 units alone?

A. 5x/2
B. 4x
C. 2x
D. x
E. x/2
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On combined work / rate problems, you can often sometimes simplify your work by picking numbers for unknowns or awkward quantities.

Remember also that you can add rates -- not times.

Here, U's rate + T's rate = combined rate, meaning that:
T's rate = combined rate - U's rate

The amount of work is 10,000 units in all 3 cases, so to simplify things, we can instead say that the amount of work is 1 production lot.

Similarly, let's use x = 1 hour to simplify our work.

Then the combined rate is 1 lot / hour.

U's rate is 1 lot per 4/3 hours, or 3/4 lot / hour.

So, T's rate = 1 - 3/4 = 1/4 lot / hour.

That means it takes T 4 hours to produce 1 lot.

The answer is B. Confirm your answer using common sense. If it took 1 hour together, and it took U 4/3 hours alone (not too much longer), it makes sense that T alone is considerably slower.

While you could also solve using 10,000 and x, picking 1 hour and 1 lot makes your calculations very straightforward. You can use this technique in work / rate problems when the actual values don't matter as long as they are consistent.
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Machines U and T can produce 10,000 units in x hours when working 27 Aug 2026, 19:27
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Combined rate of U & T= 10,000/x
10,000/T + 30,000/4x = 10,000/x
1/T = 1/x -3/4x
1/T= 4-3/4x =1/4x
T=4x
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Bunuel
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