Work-time related math.

Question

Working alone at their respective constant rates, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?

Answer 1

We know that A can complete work in "a" days

So 1/a part completed in 1 day

Similarly

We know that B can complete work in "b" days

So 1/b part completed in 1 day

Now let assume "1/a = x." And "1/b = y"

If They work 2 day at a time and

A starts working they complete work in 10 days

So they work like

AA BB AA BB AA

Means "A 6 days " and " B 4 days"

So 6x + 4y = 1..................Eq1

Similarly if B starts working then

BB AA BB AA BB and 1/2 day extra by a

Then 4x + 6y + 0.5x = 1

4.5x + 6y = 1.........Eq 2

Now we will solve both equation

Multiplying both sides of "Eq1 by 3" and "Eq2 by 2"

We get

18x + 12y = 3...........Eq3

9x + 12y = 2 ............Eq4

Eq3 - Eq4

9x = 1

x = 1/9

1/a = 1/9

"a = 9 "

Substituting x in Eq 1

6/9 + 4y = 1

4y = 1 - 6/9

4y = 3/9

y = 1/12

1/b = 1/12

"b = 12"

If they both work together then

No of days d will be

d(x+y) = 1

d(1/12 + 1/9) = 1

d(3/36 + 4/36) = 1

d(7/36) = 1

d = 36/7 = 5.14

So it will take both "5.14" days to complete work if thry work together