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

Question

There are 4 card-processing machines in an office. The fastest of these machines processes x cards in 7 hours and the slowest processes x cards in 8 hours. Which of the following could NOT be the average time per machine for each of the 4 machines to process x cards?

(A) 7.2
(B) 7.3
(C) 7.5
(D) 7.6
(E) 7.7

(A) 7.2
(B) 7.3
(C) 7.5
(D) 7.6
(E) 7.7

u1983 · Accepted Answer

By, condition the machines those are not fastest or the slowest , can take any time between 7 & 8 hrs to process x cards.
The minimum possible timimg : 7,7,7,8 ; total time = 29 ; Avg time = 7.25 >7.2 ( min of the options) --Thus Option 1 is not possible , all others possible.
The maximum possible timimg : 7,8,8,8 ; total time = 31 ; Avg time = 7.75 >7.7( the max of the options)- Thus all cases possible.
 Ans A .

KarishmaB · Answer

Hey adkikani,

There is no calculation involved in this question.

Out of 4 values, we know two are 7 (least) and 8 (greatest).
The average must be between 7 and 8. All options are within the range 7 to 8. 
Then what would be the constraint? Th problem has to be the least or the greatest value since one of those would lie outside the possible values range.

Think of weighted averages. If 3 machines take 7 hrs each and 1 machine takes 8 hrs, the average would be 1/4 away from 7 i.e. at 7.25.
Average time cannot be less than 7.25 hrs so 7.2 hrs is not possible.

Answer (A)

KSBGC · Answer

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.

mayokanmi · Answer

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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Salsanousi · Answer

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

adkikani · Answer

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.

Gladiator59 · Answer

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

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EMPOWERgmatRichC · Answer

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

aa1090 · Answer

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

kapstone1996 · Answer

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.

Ahmed9955 · Answer

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.

kop18 · Answer

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?

ThatDudeKnows · Answer

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.

Anupama_1090 · Answer

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!

oscarpiastri81 · Answer

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

akwenok · Answer

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.

akshitab2912 · Answer

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

KarishmaB · Answer

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.

aditya khanna · Answer

31/4 = 7.75 
so the maximum avg can be of 7.75 not 7.45

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

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