Machines A and B each produce tablets at their respective constant rates. Machine A has produced 30 tablets when Machine B is turned on. Both machines continue to run until Machine B's total production catches up to Machine A's total production. How many tablets does Machine A produce in the time that it takes Machine B to catch up?

(1) Machine A's rate is twice the difference between the rates of the two machines.

(2) The sum of Machine A's rate and Machine B's rate is five times the difference between the two rates.

The answer provided is D; Its posted in BTG itself (answered by GMATGuruNY also):

https://www.beatthegmat.com/machines-a-a ... 70314.html

If you approach this question with the concept of relative speed, I get this way:

Tablets = Distance (miles)

Rate (tablets per hour) = Speed (miles per hour)

A: Rate (speed) of machine A.

B: Rate (speed) of machine B.

B > A;

Relative Speed = (Speed of B) - (Speed of A) = (0.5*A);

(this can be derived from (1) and (2) independently)

Time (B catches up with A)

= (Extra Distance/Relative Speed)

= (30+x)/(0.5* A)

I think that the solution provided ignores the number of tablets that A produces after B is turned on. In that case, the numerator of the time to meet will be: (30+x) where x is the number of tablets that A will produce after B is turned on. So, for B to catch up A, B will have to produce 30 (that A already produced) plus x (that A will produce after B is turned on).

I think the answer should be E. Please let me know if I am wrong.

## MGMAT DS Question - Time,Speed,Work(MachinesA&B tablets)

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If B can catch up to A, then B must be faster than A.

Using that information, translate the statements exactly.

(1) Machine A's rate is twice the difference between the rates of the two machines.

A = 2(B - A)

A = 2B - 2A

3A = 2B

A = (2/3)B

Since we know that A = (2/3)B, we can calculate how many tablets each will make when the other makes a certain number.

For instance, in the time it takes for B to make 30 tablets, A will make 20.

Every time A makes a 2 tablets, B will make 3, which is 1 more than A makes. So they need to do that 30 times for B to make 30 more tablets than A makes and to thus catch up.

So in catching up B will make 90 tablets while A makes 60.

You don't actually have to do the math though. You just need to realize that the production of one can be determined given the production of the other.

Sufficient.

(2) The sum of Machine A's rate and Machine B's rate is five times the difference between the two rates.

A + B = 5(B - A)

A + B = 5B - 5A

6A = 4B

A = (2/3)B

Sufficient.

The correct answer is D.

Using that information, translate the statements exactly.

(1) Machine A's rate is twice the difference between the rates of the two machines.

A = 2(B - A)

A = 2B - 2A

3A = 2B

A = (2/3)B

Since we know that A = (2/3)B, we can calculate how many tablets each will make when the other makes a certain number.

For instance, in the time it takes for B to make 30 tablets, A will make 20.

Every time A makes a 2 tablets, B will make 3, which is 1 more than A makes. So they need to do that 30 times for B to make 30 more tablets than A makes and to thus catch up.

So in catching up B will make 90 tablets while A makes 60.

You don't actually have to do the math though. You just need to realize that the production of one can be determined given the production of the other.

Sufficient.

(2) The sum of Machine A's rate and Machine B's rate is five times the difference between the two rates.

A + B = 5(B - A)

A + B = 5B - 5A

6A = 4B

A = (2/3)B

Sufficient.

The correct answer is D.

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The expression in red is incorrect.s777 wrote:Relative Speed = (Speed of B) - (Speed of A) = (0.5*A)

B - A â‰ 0.5A.

An algebraic approach to Statement 1:

*Machine A's rate is twice the difference between the rates of the two machines*.

A = 2(B-A)

A = 2B - 2A

3A = 2B

B = (3/2)A.

Let t = the time for B to catch up.

Work produced by A in t hours = At.

Since B = (3/2)A, work produced by B in t hours = Bt = (3/2)(At).

At the end of the t hours, A's work = B's work.

But BEFORE the t hours begin, A has already produced 30 tablets.

Implication:

During the t hours, A produces 30 fewer tablets than B.

Since A's work in t hours is equal to 30 less than B's work in t hours, we get:

At = (3/2)At - 30

30 = (1/2)(At)

60 = At.

Thus, the work produced by A in t hours = 60 tablets.

SUFFICIENT.

Last edited by GMATGuruNY on Mon Feb 29, 2016 3:37 am, edited 1 time in total.

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Thanks very much for the clarity on this topic. I now understand that I cannot use Relative Speed concept in Work-Time questions. I really appreciate your help.

Just wanted to clarify something in your post; there are a couple of typos:

At the end of t hours, A's work = B's work.

But BEFORE the t hours begin, A (not B) has already produced 30 tablets.

Implication:

During the t hours that A (and B ofcourse) works, A (not B) produces 30 fewer tablets than B (not A).

Since A's (not B's) work in t hours is equal to 30 less than B's (not A's) work in t hours, we get:

(3/2)(At) = At + 30 (no minus sign)

30 = (1/2)(At)

60 = At.

But the concept is absolutely fine. Thanks very much.

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Good catch.s777 wrote:Hello Guru,

Thanks very much for the clarity on this topic. I now understand that I cannot use Relative Speed concept in Work-Time questions. I really appreciate your help.

Just wanted to clarify something in your post; there are a couple of typos:

I've amended my solution as follows:

Let t = the time for B to catch up.

Work produced by A in t hours = At.

Since B = (3/2)A, work produced by B in t hours = Bt = (3/2)(At).

At the end of the t hours, A's work = B's work.

But BEFORE the t hours begin, A has already produced 30 tablets.

Implication:

During the t hours, A produces 30 fewer tablets than B.

Since A's work in t hours is equal to 30 less than B's work in t hours, we get:

At = (3/2)At - 30

30 = (1/2)(At)

60 = At.

Thus, the work produced by A in t hours = 60 tablets.

SUFFICIENT.

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