​P and Q alone can complete a piece of work in 9 days and 12 days, respectively. In how many days will the work be completed if they work on alternate days starting with Q ?​

Correct option is A

Given:

P  can finish task in 9 days

Q can finish task in 12days

Formula Used:

Let total work be 1.

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

Solution:

P’s one day’s work = $$\frac{1}{9}$$​​

Q’s one day’s work = $$\frac{1}{12}$$​​

P and Q’s one pair of day’s work =  $$\frac{1}{9} + \frac{1}{12} = \frac{4+3}{36} = \frac{7}{36}$$​​

Total work = 36 and one pair of day is 7

Pair of days required to complete the work = $$\frac{7}{36} \times 5 = \frac{35}{36}$$​​

Total days required to complete $$\frac{35}{36}$$​ of work = $$5 \times 2$$ = 10 days

Work remaining = $$1 - \frac{35}{36} = \frac{1}{36}$$​​

Work was started with Q hence when Q started working on 11^th day he worked for:

$$= \frac{\frac{1}{36}}{\frac{1}{12}} = \frac{1}{3}$$​​

Hence total time required = $$10 + \frac{1}{3} =10\frac{1}{3} days$$​​

Alternative Method:

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

Total Work = LCM of 9 and 12  =36

Efficiency of P =$$\frac{36}{9}$$ = 4 units per day.

Efficiency of Q = $$\frac{36}{12}$$ = 3 units per day

Efficiency of P and Q together in one pair of days working alternatively = 4 + 3 =7 units per day

Time taken by P and Q together to complete the work = $$\frac{36}{7} = 5 + \frac{1}{7}days$$​​

Work left = 1

Time required by Q  = $$\frac{1}{3}$$​

Hence, total time required = $$5 \times 2 + \frac{1}{3} =10\frac{1}{3} days$$​​