Time to Catch = Gap/Difference in Speed
Speed of A = A
Speed of B = B
GAP = 30

So, Time to Catch = 30/(B-A)

TO find --> Time to catch * A's rate = 30A/(B-A)

Statement 1:

A = 2(B-A)
A = 2B - 2A
B = 3A/2
A = 2B/3
So,

30A/(B-A) = 30 A / (A/2) = 60
SUFFICIENT


Statement 2:
A + B = 5 ( B - A )
A + B = 5B - 5A
B = 6A/4
SUFFICIENT

Answer [spoiler]{D}[/spoiler]
What is the oA?????

The OA is D Thanks

[email protected] wrote:The OA is D Thanks

Are the Solution Steps clear?

Yes! They are. It's so simple. I was so good at these questions and suddenly they have started confusing me

[email protected] wrote:Yes! They are. It's so simple. I was so good at these questions and suddenly they have started confusing me

It happens.. many a times I also fall in same pitfall..

Just one questions, isn't B's rate more than A based on the info given in the question?

[email protected] wrote:Just one questions, isn't B's rate more than A based on the info given in the question?

Hi Deepika,

I was about to type this error( according to me). I am sure whether it is correct or not. We can't assume anything in the question.
here in this question we are assuming that B's rate is more than A but I don't think it is a correct assumption.( if B's rate is less than A then B will never be able to catch up with A as A is already ahead of B)

Rahul solution should be like this based on the assumption A=2(B-A)
3A=2B
But out point is not to solve the question. yes there is an error but why would we assume that B's rate is more than A.
All the question says is the difference. Now the difference can be A-B or B-A....in those two cases we will have two different outcome. And if its A-B means A's rate is bigger than B than B will never be able to catch up with A.

So I think question is poorly worded.
However, I would like to see what experts say.

[email protected] wrote:Just one questions, isn't B's rate more than A based on the info given in the question?

Deepika, rightly pointed out.. I have corrected my mistake it should have been B-A and not A-B.

However, since the question is asking something different than Rates.. so it won't have any effect on Answer.. because i was considering Modulus
And |5-4| = |4-5|

Is there any other way to solve this question? I think i need more explanation here

[email protected] wrote:Is there any other way to solve this question? I think i need more explanation here

I will think of other solutions.. but, by then, you can share your doubt.. I will try my best to rectify it

rakeshd347 wrote:

[email protected] wrote:Just one questions, isn't B's rate more than A based on the info given in the question?

Hi Deepika,

I was about to type this error( according to me). I am sure whether it is correct or not. We can't assume anything in the question.
here in this question we are assuming that B's rate is more than A but I don't think it is a correct assumption.( if B's rate is less than A then B will never be able to catch up with A as A is already ahead of B)

Rahul solution should be like this based on the assumption A=2(B-A)
3A=2B
But out point is not to solve the question. yes there is an error but why would we assume that B's rate is more than A.
All the question says is the difference. Now the difference can be A-B or B-A....in those two cases we will have two different outcome. And if its A-B means A's rate is bigger than B than B will never be able to catch up with A.

So I think question is poorly worded.
However, I would like to see what experts say.

Rakesh, it hardly matters whether it's A-B or B-A.. the important part is .. you consider Modulus..
While solving the question i used |A-B| so.. it doesn't have any impact on What question is asking..
Also, B > A .. or else it's never possible for B to catch A.

Earlier, my equation was : Time to catch * A's rate = 30A/|A-B|

Now suppose A = 60 & B = 30 .. |60-30| = 30 & |30-60| = 30..

[color=red]Time to Catch = Gap/Difference in Speed [/color]
[b]
What does gap mean here?[/b]

Speed of A = A
Speed of B = B
[color=red]GAP = 3
[/color]

So, Time to Catch = 30/|A-B|[color=red] ( So this means time = Work / Difference in Rate?)[/color]

TO find --> Time to catch * A's rate = 30A/|A-B|
[b][color=red]So we need to find the work done by Machine A in the time period that it takes Machine B to catch up?
I want to understand how to solve this better.
So basically, A's Rate * Time to catchup = 30 / (A-B)
Why is it 30A?[/color]
[/b]
Statement 1:

A = 2(A-B)
A = 2A - 2B
A = 2B
So,

30A/|A-B| = 30 A / (A - A/2) = 60
SUFFICIENT


Statement 2:
A + B = 5 ( A - B )
A + B = 5A - 5B
4A = 6B
B = 2A/3
SUFFICIENT [color=red][/color]

I have solved this question using Speed,Distance & Time Method..

There's logic of CATCH & MEET
MEET: If two objects are moving towards each other then Time taken = (Difference of Distance)/SUM OF SPEED

CATCH: If two objects are moving in same direction then Time taken = (Difference of Distance)/DIFFERENCE OF SPEED

Since, A & B are moving towards target... So CATCH has to be used

What does gap mean here?
--> GAP is the Difference of Distance = 30 [number of tablets]

I want to understand how to solve this better.
--> Question is asking for HOW MANY TABS DOES MACHINE A mAKE WHILE B WAS CATCHING UP.
basically Time Taken to catch x Rate of A will give us that figure..
So, Time taken to Catch = 30/(B-A)
we assumed A's Rate as A
So, total production in that time = 30A/(B-A)

[email protected] wrote: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.