Machines A and B each produce tablets at their respective co

I wouldn't call it an assumption.. I think the better term would be "inference". We are told that B does catch up to A. If B is slower than A, then that can never happen... So B should be faster than A...

As for the solution :

1) A = 2(B-A), B=1.5A = 45, A =30.. Answer can be found.. Sufficient...

2)A+B = 5(B-A), 6A=4B, For same reasons as 1, the statement is sufficient.

Answer is D

But don't you think this question is poorly worded and it should actually mention this.

I am no expert but IMHO as long as there is no ambiguity I think the question is good... In fact, I think the question you ask can probably be converted to a Critical Reasoning Inference question..

How did you get that the rate of A is 30?
Where is this fact stated?

Sorry for the lack of clarity.
The equation was supposed to mean that in the time A produces 30 units, B will produce 45 units. And using this equation the answer can be determined.

Hi MacFauz,

I see how you come up with the two equation but how that leads to the solution is still rather unclear. The question ask's How many tablets does Machine A produce in the time that it takes Machine B to catch up?

Don't you need to know the individual rates to come to the conclusion? These statements are giving us the relative rates, which is fine but how can you tell how long it'll take machine B to make 30 additional tablets without knowing the difference between the two and the rate of A?

Hi Russ9 ,
This one is tricky and more you practice,the more hang you should get for such q's.
Nevertheless,to make this concept clear :

Suppose the production rate for
m/c A is "a" tablets per unit time
and m/c B is "b" tablets per unit time

Now, as per problem statement,if for time t,B produces (during which A is also producing);then
b*t-a*t = 30
=> a*t (b/a-1) = 30
=> a*t = 30 / (b/a-1) ------ (1)
where no. of tablets m/c A produces in the time when m/c B catches up that is during t = a*t
Now, finding this value boils down to finding the ratio b:a
for which each option could be checked:

option 1:
a = 2*(b-a) ------ as B is faster so as to catch up
=> 3*a = 2*b
=> b/a = 3/2 which gives a*t =60 using (1)
option 2:
a+b = 5*(b-a)
=> 6*a = 4*b
=> b/a = 3/2 which again gives a*t =60 using (1)

So, either option is independently sufficient.

(B – A)t = 30 or Bt – At = 30

(1) SUFFICIENT: Translate the statement: A = 2(B − A)

Simplify:
A = 2(B − A)
A = 2B − 2A
3A = 2B

Combine with the given equation Bt – At = 30 and try to solve for At. Solve the simpler equation, above, for B (3A/2 = B) and then substitute into the more complicated equation.
(3A/2)t – At = 30
(3At)/2 – (2At)/2 = 30
(At)/2 = 30
At = 60


**** Could anyone explain, please, why the red colored part is -(2At)/2 in stead of just At.

IMO the equation should be like this;

(3At/2)-At=30
2At/2=30
2At=60
At=30

Hello JeroenReunis

before you make adding or subtraction fractions you should equalize their denominators.
First number \(\frac{3AT}{2}\) has a denominator 2
Second number \(\frac{At}{1}\) has denominator 1
For making common denominator we multiple both part of fraction \(\frac{At}{1}\) on \(2\) and received \(\frac{2At}{2}\)
Just in case: multiplying both part of fraction on some number doesn't change this fraction
\(\frac{At}{1} = \frac{2At}{2}\)

Another way of solving it could be by assuming a and b to be the units of time it takes A and B to make 1 tablet, respectively. Then, if t is the total time starting after machine B joins and ending when machine B has made as many tablets as A, t/a and t/b are the number of tablets machine A and machine B have made in time t, respectively.

Hence, according to info in the quest, we have: 30 + t/a = t/b . Let this be equation (1)

1) This gives us (1/a)=2(1/b - 1/a) [Machine b's rate is higher according to info in the question]
=> 3b=2a
Substituting this in equation (1), we get t=90b
Hence, t/b = 90 and t/a=60. This means a makes 60 tablets AFTER b has joined. Together they make 90 tablets each. INFORMATION 1 IS SUFFICIENT.

2) This gives us 1/a + 1/b = 5(1/b - 1/a)
=> 6b=4a or 3b=2a, which is same as above. INFORMATION 2 IS SUFFICIENT.

Thus, answer is D i.e. both INFORMATION 1 as well as INFORMATION 2 are sufficient to answer the question.

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.

Question
Rate A = Ra
Rate B = Rb
\(Time=t\), for both mashines, because after Mashine A has produced 30 tablets, they have worked the same amount of time
Rb x t = Ra x t+30

(1) Ra = 2 x (Rb-Ra)
Rb = 1.5Ra insert this information in our initial equation:

1.5Ra x t = Ra x t+30
Ra x t = 60, Sufficient

(2) Ra+Rb = 5(Rb-Ra)
6Ra = 4Ra -> 3Ra = 2Ra same expression as in statement 1, hence Sufficient.

Answer D

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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.


In the original condition and the question, when producing 30 tablets, B can catch up A once you find out the ratio between A and B. That is, there are 2 variables(Ra, Rb). You need 2 equations as well in order to match with the number of equations. For 1) 1 equation, for 2) 1equation, which is likely to make C the answer.
When 1) & 2), for 1), Ra=2(Rb-Ra)=2Rb-2Ra becomes 3Ra=2Rb and you can find out the relation between A and B, which is sufficient. For 2), Ra+Rb=5(Rb-Ra) becomes 3Ra=2Rb as well, which is sufficient. When 1)=2), D is 95% to be the answer. Therefore, the answer is D.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Key idea here, each statement says the same thing

Each is going to give you A's rate in terms of B's rate

And since all you need to know to say how long it will take A to catch B is how much faster A is than B, each statement works

Or sorry, how long it takes B to catch A, I got the machines backwards

So once we know that B = 2A, or whatever, we can say that B will do twice as much as A over ANY span of time

You actually can't get the time, you can only get the # of tablets produced

All we know is that B is twice as fast

So it could do all the work in 2 minutes or in 2 days or in 2 years. We don't know.

We just know that B will do 60 in the time it takes A to do 30 more, that is at which point the machines will have produced equal numbers of tablets

Answer : D

Answer: Option D

Check the attachment for explanation

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Bunuel - How can we interpret - The sum of Machine A’s rate and Machine B’s rate is five times the difference between the two rates. as A+B = 5(B-A) and not A + B 5 (A-B)?

I believe this has been mentioned previously, but to reiterate: the problem states that "Both machines continue to run until Machine B’s total production CATCHES UP to Machine A’s total production". Consequently, Machine B's production rate must be greater than that of Machine A; otherwise, Machine B will never be able to catch up.

thanks a lot Bunuel