Question

X stars a certain work and after he completes exactly half the work, he leaves and Y takes up the remaining work and it takes a total of 45 days for the work to be completed this way. If X and Y working together can complete the work in 20 days, and Y is slower than X. Then, find the number of days in which the work will be completed, if X alone first works on exactly one-third, after which Y alone completes the rest of the work?
1. 40 days
2. 50 days
3. 60 days
4. 45 days
5. 35 days

Answer 1

Correct Answer - Option 2 : 50 days

Suppose the number of days taken by X alone and Y alone be x and y, respectively. Since each X and Y worked on exactly half the work and finished it in 45 days.

⇒ x/2 + y/2 = 45

⇒ x + y = 90       ----- (1)

Also, together they take 20 days.

⇒ xy/(x + y) = 20

⇒ xy = 20(x + y)

⇒ xy = 20(90)

⇒ xy = 1800

⇒ x = 1800/y       --- (2)

Put the value of x in equation (1), we have –

⇒ 1800/y + y = 90

⇒ y² – 90y + 1800 = 0

⇒ (y -30) (y – 60) = 0

⇒ y = 30 or 60

Put this value in equation (1), we have –

⇒ x = 60 or 30

Given, Y is slower than X.

∴ y = 60 days and x = 30 days

Suppose total work is LCM of (60 and 30), i.e., 60 units

One day work by X = 60/30 = 2 units

One day work by Y = 60/60 = 1 unit

According to question –

⇒ Required Time to complete work = (1/3 × 60)/2 + (2/3 × 60)/1

⇒ 20/2 + 40/1

⇒ 10 + 40

⇒ 50 days

∴ the required number of days = 50 days