I went by options for this one!

1800 = number of days(let it be x) * units of work per day(let it be u)

For 1/3rd of the total days, the rate is u - 20

For the remaining 2/3rd of the days is u + 20

Since, the finish the work in 1 day less, u units of work must be reduced from the total amount of work

We are given the answer options for the initial amount of units, u.

If u=50, x=1800/50 = 36

First 1/3rd days = x/3*(u-20) = 12*30 = 360

Next 2/3rd days - 1day(since work completes before time) = (2x/3 - 1)*(u+20) = 25*70 = 1750

Sum is greater than 1800

If u=200, x=1800/200 = 9

First 1/3rd days = x/3*(u-20) = 3*180 = 540

Next 2/3rd days - 1day(since work completes before time) = (2x/3 - 1)*(u+20) = 8*220 = 1780

Sum is greater than 1800

If u=150, x=1800/150 = 12

First 1/3rd days = x/3*(u-20) = 4*130 = 520

Next 2/3rd days - 1day(since work completes before time) = (2x/3 - 1)*(u+20) = 7*170 = 1190

Sum is lesser than 1800

If u=100, x=1800/100 = 18

First 1/3rd days = x/3*(u-20) = 6*80 = 480

Next 2/3rd days - 1day(since work completes before time) = (2x/3 - 1)*(u+20) = 11*120 = 1320

Sum is equal to 1800. Hence, that's our answer.

(Option D)

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