Question

A and B together can complete a task in 12 days. However, if A works alone, completes half the job and leaves and then B works alone and completes the rest of the work, it takes 25 days in all to complete the work. If B is more efficient than A, how many days would it have taken B to do the work by herself?​

Answer 1

Answer:

TIME TAKEN BY A+B TO FINISH WORK=12days

time taken by B alone to finish the half work =25days

A+B)'s 1day's work=1/12

and, B 1day's work=1/25

A's day's work =1/12-1/25=(25-12)300

=13/300

time taken to complete half work=(25/2)/(13/300)=1/24*24=24 days

Answer 2

Step-by-step explanation:

Given that A and B can complete a work in 12 days.

Therefore, 1/A + 1/B =  (1/12)   ---- (i)

According to the given condition,

=> A/2 + B/2 = 25

=> A + B = 50

=> A = 50 - B    ----- (ii)

Substitute (ii) in (i), we get

→ 1/(50 - B) + 1/B = 1/12

→ 12B + 12(-B + 50) = B(-B + 50)

→ -B² + 50B - 600 = 0

→ B² - 50B + 600 = 0

→ B² - 20B - 30B + 600 = 0

→ B(B - 20) - 30(B - 20) = 0

→ (B - 20)(B - 30) = 0

→ B = 20,30

If B = 20:

A = 30

If B = 30:

A = 20.

Given that B is more efficient that A,he will take lesser time to do the job alone. Hence A will take 20 days and A will take 30 days.

Therefore,B will do the work in 20 days.

Hope it helps you !