Robert, Chris and Jeremy can finish a certain piece of work in 10, 14 and 21 days, respectively. All three of them started the work together. Robert left the work after 5 days and Chris left just 2 days before the work was completed. Find the total number of days taken for the work to be completed.​

Correct option is D

Given:

Robert finishes the work in 10 days, Chris in 14 days, and Jeremy in 21 days.

Robert works for 5 days, Chris stops 2 days before completion.

Formula Used:

Work = Rate × Time

Solution: 

Let T be the total time taken to compete the work 

The LCM of 10, 14, and 21 is 210 work units.

Robert's work rate = $$\frac{210}{10}$$​= 21 units/day

Chris's work rate = $$\frac{210}{14}$$​ = 15 units/day

Jeremy's work rate = $$\frac{210}{21}$$​ = 10 units/day

Robert works for 5 days → 21 × 5 = 105 units

Chris works T - 2 days → $$15 \times (T - 2)$$​ units

Jeremy works T days $$10 \times T$$​ units

Total work:

105 + 15(T - 2) + 10T = 210

105 + 15T - 30 + 10T = 210

25T + 75 = 210

25T = 135

T =$$\frac{135}{25}$$​ = 5.4 days