Question

A and b together can do a piece of work in 40 days. A having worked for 20 days, b finishes the remaining work alone in 60 days. In how many days shall b finish the  whole work alone

Answer 1

Answer:

Step-by-step explanation:

a+b=40

A+b in one day=1/40

A in 1day=1/40-b=1-40b/40

A in 20 days=1-40b/40*20=1-40b/2

remaining work=1-(1-40b)/2=(1+40b)/2

B did remain work in 60days

1+40b/2 work in 60 days

wOrk done in 1 day=1+40b/120

=>b=1+40b/120

=>120b=40b+1

=>80b=1

b=1/80 work in 1 day so he will finish in 80 days

mArk it brainliest

Answer 2

HEY Buddy...!! here is ur answer

Answer : 80 days

Let, work of A of 1 day = x and work of B of 1 day = y

Also Given that A worked for 20 days and B finished the remaining work in 60 days.

According to the question :

$x + y = \frac{1}{40} ....(1) \\ \\ 20x + 60y = 1....(2)$

On solving these equations :

On Multiplying by 20 in equation (1)

$20x + 20y = \frac{1}{2} ....(3)$

On subtracting equation (2) from (3)

$- 40y = - \frac{1}{2} \\ \\ = > y = \frac{1}{80}$

On putting the value of y in equation (1)

$x = \frac{1}{40} - \frac{1}{80} \\ \\ = > x = \frac{1}{80}$

So, the work of A and work of B of 1 day = 1/80

It mean they both can complete the whole works in 80 days.

I hope it will be helpful for you...!!

THANK YOU ✌️✌️