Question

A takes 6 days less than the time taken by B to complete a work. If they together can complete a work in 4 days find the time taken by each to finish the work ?​

Answer 1

Suppose B alone takes x days to finish the work. Then A alone can finish it in (x−6) days

Now A's one day's work + B's one day's work =
x
1
​
+
x−6
1
​

and (A+B)'s one day's work =
4
1
​


∴
x
1
​
+
x−6
1
​
=
4
1
​


⇒
x(x−6)
x−6+x
​
=
4
1
​
⇒
x
2
−6x
2x−6
​
=
4
1
​


⇒8x−24=x
2
−6x

⇒x
2
−14x+24=0

⇒x
2
−12x−2x+24=0

⇒(x−12)(x−2)=0

⇒x=12orx=2

But x cannot be less than 6.
So x=12

Hence, B also can finish the work in 12 days.

Answer 2

Answer:

no. of days taken by A to finish that work = x

then, no. of days taken by B to finish that work = x+6

now, work done by A in one day = 1/x and work done by B in one day = 1/x+6

now, work done by both A and B in one day,

1/x + 1/x+6 = 1/4

{(x+6)+x}/x(x+6) = 1/4

8x + 24 = x2 + 6x

x2-2x - 24 = 0

x2 -6x +4x -24 = 0

x(x-6) +4(x-6) = 0

(x+4) (x-6) = 0

so, x= 6  (neglecting x = -4)

so, time taken by B to finish the work = x+6 = 12 days