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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42 GPA: 3.82
U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 81% (02:13) correct 19% (02:48) wrong based on 277 sessions

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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 in 4x/3 hours alone at the constant rate, in how many hours can T produce 10,000 units alone at the constant rate, in terms of x?
A. 5x/2 B. 4x C. 2x D. x E. x/2

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Re: U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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MathRevolution wrote:
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 in 4x/3 hours alone at the constant rate, in how many hours can T produce 10,000 units alone at the constant rate, in terms of x?
A. 5x/2
B. 4x
C. 2x
D. x
E. x/2

For work questions, there are two useful rules to consider:

Rule #1: If a person can complete an entire job in k hours, then in one hour, the person can complete 1/k of the job
Example: If it takes Sue 5 hours to complete a job, then in one hour, she can complete 1/5 of the job. In other words, her work rate is 1/5 of the job per hour

Rule #2: If a person completes a/b of the job in one hour, then it will take b/a hours to complete the entire job
Example: If Sam can complete 1/8 of the job in one hour, then it will take him 8/1 hours to complete the job.
Likewise, if Joe can complete 2/3 of the job in one hour, then it will take him 3/2 hours to complete the job.

Let’s use these rules to solve the question. . . . . . . .

For convenience, let's just say that The Entire Job = producing 10,000 units.

U and T can produce 10,000 units in x hours when working together at their constant rates.
In other words, machines U and T (working together) can complete the entire job in x hours.
When we apply Rule #1, we see that, IN ONE HOUR, the machines can complete 1/x of the entire job

U can produce 10,000 units in 4x/3 hours alone at the constant rate
In other words, machine U (working alone) can complete the entire job in 4x/3 hours.
When we apply Rule #1, we see that, IN ONE HOUR, machine U can complete 3/4x of the entire job

We'll use the fact that: (Machine U's output in ONE HOUR) + (Machine T's output in ONE HOUR) = (The combined machines' output in ONE HOUR)
In other words, 3/4x + (Machine T's output in ONE HOUR) = 1/x
So, Machine T's output in ONE HOUR = 1/x - 3/4x
= 4/4x - 3/4x
= 1/4x
In other words, in ONE HOUR machine T can complete 1/4x of the entire job
When we apply Rule #2, we see that it takes machine T a total of 4x/1 hours to complete the entire job
Answer:

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Re: U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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MathRevolution wrote:
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 in 4x/3 hours alone at the constant rate, in how many hours can T produce 10,000 units alone at the constant rate, in terms of x?
A. 5x/2 B. 4x C. 2x D. x E. x/2

Assume x=3

Since work done is same.

1/4 + 1/y = 1/3

1/y = 1/12

y has to be written in terms of x.

Therefore y = 12 = 4*3 = 4x

B

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Re: U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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MathRevolution wrote:
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 in 4x/3 hours alone at the constant rate, in how many hours can T produce 10,000 units alone at the constant rate, in terms of x?

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

$$Efficiency of U + T = \frac{10000}{x}$$

$$Efficiency of U = \frac{30000}{4x}$$

So, Efficiency of T = ( Efficiency of U + T ) - ( Efficiency of U )

Or, Efficiency of T = $$\frac{10000}{x} - \frac{30000}{4x}$$

Or, Efficiency of T = $$\frac{10000}{4x}$$

Time required to produce 10000 units = $$10000/\frac{10000}{4x}$$ = $$4x$$

Hence, correct answer will be (B) 4x

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Re: U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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Abhishek009 wrote:
MathRevolution wrote:
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 in 4x/3 hours alone at the constant rate, in how many hours can T produce 10,000 units alone at the constant rate, in terms of x?

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

$$Efficiency of U + T = \frac{10000}{x}$$

$$Efficiency of U = \frac{30000}{4x}$$

So, Efficiency of T = ( Efficiency of U + T ) - ( Efficiency of U )

Or, Efficiency of T = $$\frac{10000}{x} - \frac{30000}{4x}$$

Or, Efficiency of T = $$\frac{10000}{4x}$$

Time required to produce 10000 units = $$10000/\frac{10000}{4x}$$ = $$4x$$

Hence, correct answer will be (B) 4x

a

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Re: U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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==> In case of work rate questions, if it is “together and alone”, you solve it reciprocally. In other words, if you assume the time it takes for T to produce 10,000 units alone as t hrs, you get t=4x from 1/(4x/3)+1/t=1/x. Therefore, B is the answer.

Answer: B
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Re: U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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MathRevolution wrote:
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 in 4x/3 hours alone at the constant rate, in how many hours can T produce 10,000 units alone at the constant rate, in terms of x?
A. 5x/2 B. 4x C. 2x D. x E. x/2

We are given that U and T can produce 10,000 units in x hours when working together at their constant rates and that U can produce 10,000 units in 4x/3 hours alone at the constant rate.

Thus, we can say the following:

Time for machines U and T = x hours.

Rate for machine U = 10,000/(4x/3) = 30,000/(4x) = 7,500/x

Rate for machine T = 10,000/t (where t = the number of hours machine T takes to complete the job alone).

Since work = rate x time, we can determine the amount of work done by both machines when working together.

Work done by machine U = (7,500/x)(x) = 7,500

Work done by machine T = (10,000/t)(x) = 10,000x/t

We can now use the combined work equation to determine a value for t.

Work (machine U) + Work (machine T) = total work completed

7,500 + 10,000x/t = 10,000

10,000x/t = 2,500

10,000x = 2,500t

10,000x/2,500 = t

4x = t

Answer: B
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Re: U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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Total work = rate of U + rate of T

$$\frac{10,000}{x}=\frac{10,000}{4x/3}+\frac{10,000}{T}$$
where we must find value of T in terms of x
On simplifying we get T=4x

Answer is B
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Re: U and T can produce 10,000 units in x hours when working together at t  [#permalink]

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Abhishek009 wrote:
MathRevolution wrote:
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 in 4x/3 hours alone at the constant rate, in how many hours can T produce 10,000 units alone at the constant rate, in terms of x?

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

$$Efficiency of U + T = \frac{10000}{x}$$

$$Efficiency of U = \frac{30000}{4x}$$

So, Efficiency of T = ( Efficiency of U + T ) - ( Efficiency of U )

Or, Efficiency of T = $$\frac{10000}{x} - \frac{30000}{4x}$$

Or, Efficiency of T = $$\frac{10000}{4x}$$

Time required to produce 10000 units = $$10000/\frac{10000}{4x}$$ = $$4x$$

Hence, correct answer will be (B) 4x

Hi...quick question:

Efficiency and rate are the same?

regards Re: U and T can produce 10,000 units in x hours when working together at t   [#permalink] 28 Sep 2018, 04:19
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