Dwight and Mose decide to build a new barn on their farm, and they beg

Dwight = First Shift

Remaining Work = Total Work (Let us represent the total work as 1) = 1

Diwght completes 1/4th of the remaining work

Remaining Work = 1 - 1/4 = 3/4


Mose = First Shift

Remaining Work = 3/4

Mose completes 1/6th of the remaining work = 3/4 * 1/6 = 1/8

Remaining work = 3/4 - 1/8 = 5/8


Dwight = Second Shift

Remaining Work = 5/8

Diwght completes 1/4th of the remaining work = 5/8 * 1/4 = 5/32

Remaining work = 5/8 - 5/32 = 15/32


Mose = Second Shift

Remaining Work = 15/32

Mose completes 1/6th of the remaining work = 15/32 * 1/6 = 5/64

Remaining work = 15/32 - 5/64 = 25/64


Dwight = Third Shift

Remaining Work = 25/64

Diwght completes 1/4th of the remaining work = 25/64 * 1/4 = 25/256

Remaining work = 25/64 - 25/256 = 75/256

Work on the barn that is left to finish when Mose begins his third shift = 75/256

Answer is D. 75/256

This seems like it would take too long

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It wouldn't if you assume the work to be of 256 units.

Why do you have to multiply the remaining amount of work by the amount being completed (as opposed to subtract)?

This is something that we have to keep dividing by 4 and 6 (since 3/4 work is left, and 5/6 work is left), so better take a number which has atleast three 4's and three 6's. So lets take the initial barn work to be 4^3 * 6^3.

First shift:
Dwight does = 1/4 and work remaining = 3/4 * 4^3 * 6^3 = 3 * 4^2 * 6^3
Then Mose does = 1/6 and work remaining = 5/6 * 3 * 4^2 * 6^3 = 3 * 5 * 4^2 * 6^2

Second shift
Dwight does= 1/4 and remaining = 3/4 * 3 * 5 * 4^2 * 6^2 = 3^2 * 4 * 5 * 6^2
Mose does = 1/6 and work remaining = 5/6 * 3^2 * 4 * 5 * 6^2 = 3^2 * 5^2 * 4 * 6

Third Shift
Dwight does = 1/4 and work remaining for Mose = 3/4 * 3^2 * 5^2 * 4 * 6 = 3^3 * 5^2 * 6

So the fraction of work left = (3^3 * 5^2 * 6)/(4^3 * 6^3) = (27*25)/(64*36) = 75/256

Thus D is answer

Let the work to be completed be 1.
D completes 1/4 of the remaining work. Therefore work remaining is \((1-1/4)\).
M completes 1/6 of the remaining work. Hence \([(1-1/4)-(1-1/4)(1/6)]\). Therefore work remaining is \((1-1/4)(1-1/6)\).

Hence work remaining when M starts work 3rd time is \((1-1/4)^3*(1-1/6)^2 = 75/256\)

Can you explain why we assume the work to be of 256 units please?
Thank a lot.

Hi,
Lets say total work is 256 .

D does 1/4 and M does 1/ of the remaining Unit .

Ist Attempt:

D = 1/4*256 = 64 , M = (256-64)*1/6 = 32

2nd Attempt:

D = (256 - 96)*1/4 = 40 , M = (256 - 136) *1/6= 20

3rd Attempt:

D = (256-156)*1/4= 25

At this point work remaining for M is 75 out of 256 . Hence 75/256 is answer

The best way to solve this problem is to keep track of the fraction of the barn that still needs to be completed after each of their shifts:

Dwight’s 1st shift: 1/4 → Fraction of the barn still needs to be completed: 3/4

Mose’s 1st shift: 1/6(3/4) = 1/8 → Fraction of the barn still needs to be completed: 5/6(3/4) = 5/8

Dwight’s 2nd shift: 1/4(5/8) = 5/32 → Fraction of the barn still needs to be completed: 3/4(5/8) = 15/32

Mose’s 2nd shift: 1/6(15/32) = 5/64 → Fraction of the barn still needs to be completed: 5/6(15/32) = 25/64

Dwight’s 3rd shift: 1/4(25/64) = 25/256 → Fraction of the barn still needs to be completed: 3/4(25/64) = 75/256

We can see that before Mose begins his 3rd shift, 75/256 of the barn still needs to be completed.

Answer: D

I dont understand this pattern. Could you please clarify how you got to your solution step by step?

If (1/4) is taken away from the whole ——- (3/4) remains

If (1/6) is taken away from whatever is left——- (5/6) remains


D 1st shift: 3/4 remains

M 1st shift: (5/6) (3/4) remains

D 2nd shift: (3/4) (5/6) (3/4) remains

M 2nd shift: (5/6) (3/4) (5/6) (3/4) remains

D 3rd shift: (3/4)^3 * (5/6)^2 remains for Moses beginning his 3rd shift

(3 * 3 * 3 * 5 * 5)
______________
(4 * 4 * 4 * 6 * 6)

Cancel two of the 3’s in the numerator

75 / 256

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