X does half as much work as Y in (1/3)rd of the time. If together they take 20 days to finish a job, how much days shall Y take to do it?

## X does half as much work as Y in (1/3)rd of the time. If together they take 20 days to finish a job, how much days shall Y take to do it?

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# 3 Answers

Shiva:

Let R

_{x}= The rate of work being performed by X and R_{y}= The rate of work being performed by Y. The work accomplished by each can be expressed as the rate * time of each.The total work accomplished by

**both**X and Y in 20 days can be expressed as the sum: W=R_{x}*20 + R_{y}*20Since we know that X accomplishes 1/2 the work in 1/3 the time of Y, Rx would correspond to a rate of ((1/2)/(1/3))Ry or (3/2)Ry

Substituting for Rx in the above equation for work, the total work accomplished by both X and Y in 20 days is: W=20(3/2) Ry+20 Ry = 30Ry+20 Ry = 50Ry

Let T

_{solo}represent the time to accomplish this same amount of work by Y alone. The total work (50Ry) is then equal to R_{y }times T_{solo:}_{ }50Ry=Ry*T_{solo}.Dividing both sides by R

_{y, }we obtain**T**_{solo}=50 days.Hope this helps!

George T.

X does 1/2 as much work as Y, let's break it right here. So that means X does equal amount of work as Y in twice the amount of time mentioned (for half the work done) which is 2 * 1/3 of Y = 2/3 of Y.

So now we know X will do same amount of work as Y in 2/3 the time. if it takes Y 60 minutes to do something, it takes X 60*2/3 = 40 minutes to do the same job.

now we know they're doing equal amounts of work in different amounts of time, so lets see what proportion of work each can do working together for a given amount of time:

let's say the time is 1 hour (the number of hours/days here doesn't matter, we could've just started with 20 days, but goes easier if it's unit time). out of 1 hour, if Y finishes 1 job, X will have already finished it in 40 minutes, as mentioned above,
nonetheless X keeps working for the full hour. How much work will X accumulate in another 20 minutes needed to complete the hour?

well, 40 minutes for a job means 20 minutes for half a job, so that's a total of 1.5 jobs per hour for X, while it is 1 job per hour for Y. How much total work was done by both X and Y combined? 1 + 1.5 = 2.5 jobs in 1 hour(or unit time).

now break down each person's share of jobs accomplished, as a fraction of total jobs accomplished by both X and Y combined.

X's proportion: 1.5/2.5 = 3/5 = 60% of the work

Y's proportion: 1/2.5 = 2/5 = 40% of the work

so Y can accomplish 40% of the work in 20 days, while X puts in the other 60%.

we want to know how long Y takes working alone. it should be intuitive that if both X and Y work together and accomplish the job in 20 days, it should take X or Y individually longer to accomplish the same task working alone.

Y completes 40% of the job in 20 days, so how many days does is take Y to complete a 100% of the task?

so we have 40%/20days = 100%/?days

cross multiply to get:

40 * ? = 100 * 20

which is 40 * ? = 2000

now divide both sides by 40 to find the question mark

2000/40 = 50

it takes Y 50 days to complete the task alone.

X does the same work as Y in 2/3 of the time. It means that his speed of doing work is 3/2 of the speed of his colleague, Y. If the speed of his colleague is

**S**, then their joint speed is:S

_{joint}=3/2S+S=(5/2)S;The time it will take the Y alone to finish is 1/S; We also know that if they work together, they finish it in 20 days. Thus,

5/2*S*20=1; From this we can immediately find 1/S, by dividing both parts by S;

1/S=5/2*20=50;

Answer: it will take 50 days for the Y to finish the job.