A, B, & C can do same job in 8, 24, & 48 hours respectively working

Question

A, B, & C can do same job in 8, 24, & 48 hours respectively working individually. It was decided that each of them will work on this job alternatively for an hour. They can start the work in any order.

Select 'Minimum hours', & 'Maximum hours' required to complete the job in the table. Make only two selections, one in each column.

mandyrhtdm · Accepted Answer

An easier way to work will be take the amount of work to be 48 units this allows to do away with fractions.
in one hour A does 48/8 = 6 units 
B does 48/24 = 2 units
C does 48/48 = 1 unit

In three hours no matter in which order they work they will complete 9 units of work 
3 hr = 9 units
15 hrs = 45 units
now three units are left, for minimum time as A to do it he will do 6 units in 1 hr so 3 units in .5 hr answer = 15.5 units
For maximum time let a and b finish the job they will take 1unit + 2unit = 3unit that is 2 hrs to complete it.

EvaJager · Answer

If A, B, and C each work for an hour, in a total of 3 hours they accomplish  \frac{1}{8}+\frac{1}{24}+\frac{1}{48}=\frac{9}{48}=\frac{3}{16}  of the entire job.
After 5 such cycles, a total of 5 x 3 = 15 hours, they accomplish  \frac{15}{16}  of the entire job. For the remaining 1/16th of the job, if it is the slowest worker's turn to work (this is C), another 1/48th is accomplished, then B would do another 1/24th, which gives the missing  \frac{1}{48}+\frac{1}{24}=\frac{3}{48}=\frac{1}{16}  of the job. So, the maximum time is 5 x 3 + 2 =17 hours. Choice (D,B) (row, column)

For the minimum time, after the 5 cycles of 3 hours, the fastest worker, which is A, needs just 1/2 an hour to do the remaining 1/16th of the job.
So, minimum time is 5 x 3 + 0.5 =15.5 hours. Choice (B,A) (row, column)

Kmohan1 · Answer

In one hour, A does 1/8 of work, B does 1/24 of work and C does 1/48 of work. Since the question states that the machines work on every alternative hour, eg A for the first hour, B for the second hour, and C for the third hour. For any multiples of 3, all 3 machines perform equal hours. In case the hours are 3n+1 ie 1 more than the multiple of 3, the one hour can be added to A, B or C. in case of 3n+2 hours, the additional 2 hours are added to pair of machines ie both to A and B or B and C etc.
For the minimum number of hours taken by all 3, most of the time should be assigned to the machine that does the most work. 15 is thus rejected, since for within 15 hours, all the 3 machines will have to work for 5 hours. Considering 15.5 hours, the number of hours can be distribution in the below fashion, 5 complete cycle of A-B-C and an extra 0.5 to A.
5.5/8 +5/24 + 5/48 = 48/48 hence this is the minimum number of hours.
For maximum number of hours, most of the time should be assigned to the machine that does the least amount of work. Because 17 is a whole number I will choose that. 5 cycles of C-B-A and additional 1 hour for machine C and 1 hour for machine B. 6/48+ 6/24 + 5/24= 48/48- this is the maximum number of hours

PSKhore · Answer

Lets take the total work done as the LCM of 8,24,48 = 48
So the total work is 48 units
Individual work in units :
A = 48/8 = 6 units
B = 48/24 = 2 units
C = 48/48 = 1 unit
Total work in 3 hours is 9 units
Now 48/9 = 5.xx X 3 = 15.xx
remaining units = 3 units
if A + B work together in 2 hours we can complete the 3 units work.
therefore maximum will be 17 hours

if we want the minimum time to complete those 3 units, we can ask A
A will complete 3 units in half an hour
So, minimum will be 15.5 hours

xyzKunal · Answer

This is a simple approach that does not need much calculation-

Let, Work = 48 (LCM of 8,24,48)
so, Rate of a = 6 unit of work an hour
rate of b = 2 unit of work an hour
rate of c = 1 unit of work an hour.

In 3 hours (working simultaneously), they complete 9 units of work.
in 15 hours, they complete 45 units.

Remaining 3 units, will give us the answer.
SLOWEST- C complete 1 unit and B complete 2 unit. ( total- 15+2 = 17 hours)
FASTEST - A complete the 3 units in 1/2 hour (total- 15+0.5 = 15.5 hours)

mkeshri185 · Answer

I took almost same approach. The remaining work is 1/16
Now to finish this in minimum time we are assigning it to the fastest worker that is A and getting the min time.
But for the fastest time why are we not considering assigning it to the slowest worker that is C? or to the second slowest worker that is B? Why are we combining B and C?

Bunuel · Answer

Your doubt is not clear. To  minimize the time , you assign the remaining work to the fastest worker, A, because A completes the task the quickest. To  maximize the time , you would assign the remaining work to the slowest worker, C, or a combination of slower workers like B and C.
What is unclear there?

mkeshri185 · Answer

Oh my mistake i got why we are not considering only C to complete the rem work which is 1/16 bcz C would take 3 hrs to do 1/16 work but people are allowed to work for 1 hr at a time only so we can't consider C to do the whole of the rest work. Same with the B which would also take more than 1 hr to complete the rem work. So from this its clear that some part of the remaining part will be done by B and some part by C (bcz they are the slowest so we are combining them)

AditiDeokar · Answer

Same doubt why are we combining B and C, is it because the stem says when the work alternatively?

Bunuel · Answer

Yes, exactly, it is because they work alternately.

You cannot just “give” the whole remaining 1/16 of the job to C alone. After 15 hours, whoever’s turn it is works for 1 hour, then must stop, and the next person in the fixed order takes over.

A, B, & C can do same job in 8, 24, & 48 hours respectively working

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Minimum hours	Maximum hours
		15
		15.5
		16.17
		17
		17.33

GMAT Two-part Analysis (TPA) Questions

A, B, & C can do same job in 8, 24, & 48 hours respectively working

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