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Machines A and B

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Machines A and B

by jain2016 » Mon Apr 04, 2016 9:45 pm
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.

OAD

Hi Experts ,

Please explain.

Thanks,

SJ
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Source: — Data Sufficiency |
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by MartyMurray » Mon Apr 04, 2016 10:42 pm
Translate the statements into mathematical expressions, using A to denote the rate of machine A and B to denote the rate of machine B.

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

A = (2/3)B

So Statement 1 gives us the ratio of their rates. We don't care about the time, only how much work A does when B does a certain amount of work. So the ratio of their rates is sufficient. We don't have to do any math at this point, but I will to prove the concept.

For every two tablets that A makes B makes 3, which is 1 greater than 2.

For B to catch up to A, B has to make 30 more tablets than A does.

So repeat the 2:3 ratio 30 times. B will make one more tablet each time, catching up to A after 30 times.

A makes 30 x 2 = 60 more tablets.

B makes 30 x 3 = 90 tablets.

The proof can also be done algebraically.

We want to know how many tablets A makes while B makes 30 more during the same time, T.

(B x T) - (A x T) = 30

We know from Statement 1 that B = 3/2A.

3/2AT - AT = 30

1/2AT = 30

AT = 60

You can also just plug in for A and B any numbers with a 2:3 ratio.

Let A be 30 and B be 45.

45T - 30T = 30

15T = 30

T = 2

30T = 60

In every case we can show that A produces 60 tablets while B is catching up.

Sufficient.

Statement 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

3A = 2B

That's a repeat of Statement 1.

Sufficient.

The correct answer is D.
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by GMATGuruNY » Tue Apr 05, 2016 4:43 am
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.
For B to catch up to A, B's rate MUST be faster than A's rate.
The rate at which B will CATCH UP to A is equal to the DIFFERENCE between the two rates: B-A.

Statement 1: Machine A's rate is twice the difference between the rates of the two machines.
Test TWO cases.

Case 1: A's rate = 2 tablets per hour
Since A's rate is twice the difference between the two rates, B's rate = 3 tablets per hour, so that B-A = 3-2 = 1.
Time for B to catch up = w/(rate difference) = 30/1 = 30 hours.
Work produced by A in 30 hours = r*t = 2*30 = 60 tablets.

Case 2: A's rate = 4 tablets per hour
Since A's rate is twice the difference between the two rates, B's rate = 6 tablets per hour, so that B-A = 6-4 = 2.
Time for B to catch up = w/(rate difference) = 30/2 = 15 hours.
Work produced by A in 15 hours = r*t = 4*15 = 60 tablets.

Since Machine A produces the same number of tablets in each case, SUFFICIENT.

Statement 2: The sum of Machine A's rate and Machine B's rate is five times the difference between the two rates.
Both Case 1 and Case 2 also satisfy statement 2.
Case 1: (B+A)/(B-A) = (3+2)/(3-2) = 5.
Case 2: (B+A)/(B-A) = (6+4)/(6-4) = 5.
Thus -- like statement 1 -- statement 2 is SUFFICIENT.

The correct answer is D.
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by GMATGuruNY » Tue Apr 05, 2016 4:44 am
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.
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