A father can complete a task in 8 days, while the son can do it in 7 days. If they work on alternate days, with the father starting, then in how many days will the task be completed?

Correct option is B

Given:

Time taken by father to complete the work = 8 days

Time taken by son to complete the work = 7 days

Formula Used:

Let total work be 1.

Then work done 1 day =$$\frac{1}{Total\ time\ to\ work}$$​​

Solution:

Father’s one day’s work = $$\frac{1}{8}$$​​

Son’s one day’s work =$$\frac{1}{7}$$​​

Father and Son’s one pair of  day work together =$$\frac{1}{8}+\frac{1}{7}$$​​

​$$= \frac{7+8}{56}=\frac{15}{56}$$​​

Total pair of days required to work =$$\frac{15}{56} \times 3 = \frac{45}{56}$$​​

Total days =$$3\times 2$$​ = 6 days

Work left =$$1 - \frac{45}{56} =\frac{11}{56}$$​​

Work on 7^th day $$= \frac{1}{8}$$​​

Work left =$$\frac{11}{56} - \frac{1}{8} =\frac{4}{56} = \frac{1}{14}$$​​

Work done on 8^th day = $$\frac{\frac{1}{14}}{\frac{1}{7}} =\frac{1}{2}$$​​

Total time taken to complete the work = $$7\frac{1}{2}days$$​​

Alternative Method:

Total work = Time $$\times$$​ Efficiency

Total work =LCM of 8 and 7 = 56

Efficiency of Father =$$\frac{56}{8}$$​ = 7 units

Efficiency of Son=$$\frac{56}{7}$$​= 8 units

Total efficiency of Father and Son for 2 days working on alternate days = 7+8  =15

Time required to completer the work = $$\frac{56}{15} =3 \frac{11}{15}$$​​

Work left = 11 units

Work done on 7^th day by father = 7 units

Work left = 11-7 = 4 units

Work done on 8^th day =$$\frac{4}{8} =\frac{1}{2}day$$​​

Total time required by father and son working on alternate days = $$7 + \frac{1}{2} =7\frac{1}{2} days$$​​