Question

A takes 6 days less than the time taken by B to finish a peice of work . If both A and B toghether can finish it in 4 days ,find the time taken by B to finish the work​

Answer 1

Answer:-

• B can alone finish the work in 12 days

Given:-

• A takes 6 days less than B to complete the work.
• Both A and B can finish the same work in 4 days.

To find:-

• Time taken by B to complete the work=?

Solution:-

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

B's 1 day's work = $\frac{1}{x}$

A's 1 day's work = $\frac{1}{x - 6}$

(A+B)'s 1 day's work = $\frac{1}{4}$

:• $\frac{1}{x} + \frac{1}{x - 6} = \frac{1}{4}$

$= > \frac{(x - 6) + x}{x(x - 6)} = \frac{1}{4} \\ = > \frac{(2x - 6)}{( {x}^{2} - 6x) } = \frac{1}{4} \\ = > {x}^{2} - 6x = 8x - 24 \\ = > {x}^{2} - 14x + 24 = 0 \\ = > {x}^{2} - 12x - 2x + 24 = 0 \\ = > x(x - 12) - 2(x - 12) = 0 \\ = > (x - 12)(x - 2) = 0 \\ = > x - 12 = 0 \: or \: x - 2 = 0 \\ = > x = 12 \: or \: x = 2 \\ = > x = 12$

So,[x=2 => (x-6)<0]

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

Hope its help uh❤

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