There are 4 card-processing machines in an office. The fastest of thes

slowest rate = 7
Fastest rate= 8.

Now , we can find out the range of average time taken by 4 machines .

minimum average time ; 7 + 7 + 7 + 8 = 29 / 4 = 7.25. but we know the actual average time will be greater than 7.25.

Maximum Average Time ; 8 + 8 + 8 + 7 = 31 / 4 = 7.45. but actual average must be lower than 7.45.

Average time must be in between 7.25 and 7.45.

So, 7.2 can be the average time at any means.

Thus the best answer is A.

The extremes of the other 2 machines are either all 7 per hour in which case the average will be
(7+7+7+8)/4= 7.25

Or

all 8per hour making the overall average to be
(7+8+8+8)/4= 31/4=7.75

Hence the answer cannot be lower than 7.25 per hour making A the answer

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Here we can test the two extremes.

The maximum possible is 8*3 + 7 = 24 + 7 = 31/4 = 28/4 + 3/4 = 7 3/4 = 7.75

The minimum possible is 7*3 + 8 = 21 + 8 = 29/4 = 28/4 + 1/4 = 7 1/4 = 7.25.

It cannot be less than 7.25

Answer choice A

VeritasKarishma chetan2u niks18 gmatbusters Gladiator59 GMATPrepNow EMPOWERgmatRichC

I understood the above solutions; my query is:
Is there a way to skip long division? The answers seems so close by to approximate.

adkikani,

I think looking at the options and otherwise ( four numbers between 7 and 8) we know that the average will be 7. Something..

To minimize average, three machines have to be at max output and one at min .. so the average will be closer to 7.

So the only shortcut I can think of here would be (3*7+7+1)/4 = 7 + 1/4 = 7.25 ... So 7.2 is not possible.

Still 1/4 needs to be done but one can calculation of 29/4.

Best,
Gladi

Posted from my mobile device

Hi adkikani,

Since the answer choices are so 'close' to one another, we have to do enough 'precision' math to make sure that we have the correct answer (if you approximate the answer incorrectly, then you might convince yourself that one of the wrong answers is correct). Thankfully, the work here isn't too bad - and you really just have to think about no more than two possibilities:

We know the exact rates of 2 of the 4 machines. The FASTEST processes 7 cards in an hour and the SLOWEST processes 8 cards in an hour. We don't know the exact speeds of the other two machines, but we know that they are some rate BETWEEN 7 cards and 8 cards per hour. Thus, there will be a limit to how fast or how slow the overall average could be. By extension, the correct answer will either be TOO FAST or TOO SLOW for what that group is capable of averaging (meaning that the correct answer will be Answer A or Answer E).

Even though we can't have the other two machines working at 7 cards/hour, IF they did work at that rate, then the average would be (7+7+7+8)/4 = 29/4 = 7.25 cards/hour. There is NO POSSIBLE OPTION for the average to be a lower number than that (those other two numbers can only get bigger than 7). At this point, we can stop working; 7.2 is NOT a possibility, so it has to be the answer to this question.

GMAT assassins aren't born, they're made,
Rich

Best case (max fastest machine, min time)-> (7+7+7+8)/4 = 7.2
So, the minimum time has to be equal to or greater than 7.2
So, option A -> average 7 hours - which is lesser than 7.2

Hence, option A

What I did, is looked at the answer choices. I do believe this approach is only applicable to this question.

But I thought, if 7.3 is answer choice and 7.7 is answer choice those are both net .3 around from an integer, so only 7.2 is the odd one out.

Hi Karishma,
Can you Elaborate more about this approach ,
1. how we get to 1/4 th part away from 7?
2. how can I use this in other questions.

Hi guys,
I did not average out the speeds of these machines fr max and min thinking that we cannot simply average out the speeds (as we do with SDT questions, suppose when we have to find the avg speed of 2 trains at 60 m/s and 40 m/s we don't simply avg it out to 50m/s we take individual D/T and then find it). Please can someone explain to me why there's a difference between those questions and this question? They are both talking about rates?

The average of the two machines that we know is 7.5.
We know we need an extreme value, so B, C, and D are out.
7.2 deviates from 7.5 by 0.3.
7.7 deviates from 7.5 by 0.2.
How would it be possible to get to 7.7 but not be possible to get to 7.2? E is out.

Answer choice A.

let us assume the 2 machines that have the rate between the slowest and fastest machines have a rate of x and y.
The fastest machine has a rate of 8 and the slowest one has a rate of 7. so both x and y must lie between 7 and 8.

so avg rate = 8+7+x+y/4
now substitute each of the values, only A will not fit this equation as x+y = 13.8 if avg rate were to be 7.2 and x and y must be greater than 7. Please let me know if my understanding is incorrect!

This is exactly how I solved it too, makes sense.

Just look at the answer choices really. There are the same restrictions for max and minimum in terms of distance from both. 7.5 is easy to tell, then 7.6 is easy too. If you cannot have 7.7 then FOR SURE you won't have 7.2 and 7.3. If you can have 7.7, then you can have 7.3.

Hi KarishmaB, what if the other limit would have been 9 or 10 - how would you approach this question in that case?

It's no different. If 1 is at 7 and other at 9, we know that all 4 together cannot take less than 7.5 hrs, a quarter away from 7.