Rajeshwar can do a piece of work in 12 days, Vikram in 6 days and Tiger in 15 days. They all start the work together, but Rajeshwar leaves after 2 days and Vikram leaves 3 days before the work is completed. In how many days is the work completed?

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  1. \(5 \frac{4}{7}\)
  2. \(5 \frac{5}{7}\)
  3. \(5 \frac{3}{7}\)
  4. \(5 \frac{6}{7}\)

Answer (Detailed Solution Below)

Option 2 : \(5 \frac{5}{7}\)
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Detailed Solution

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Given:

Rajeshwar's work time = 12 days

Vikram's work time = 6 days

Tiger's work time = 15 days

Formula used:

Total work = LCM of individual work times

Work done = Efficiency × Time

Calculations:

LCM of 12, 6, and 15 = 60 (Total work)

Rajeshwar's efficiency = 60/12 = 5 units/day

Vikram's efficiency = 60/6 = 10 units/day

Tiger's efficiency = 60/15 = 4 units/day

Rajeshwar worked for 2 days, so work done by him = 5 × 2 = 10 units

Let the total number of days the work was completed be 'x' days.

Vikram worked for (x - 3) days.

Tiger worked for x days.

Total work done = Rajeshwar's work + Vikram's work + Tiger's work

60 = 10 + 10(x - 3) + 4x

⇒ 60 = 10 + 10x - 30 + 4x

⇒ 60 = 14x - 20

⇒ 80 = 14x

⇒ x = 80/14 = 40/7  = \(5 \frac{5}{7}\) days. 

∴ The work is completed in 40/7 days, or \(5 \frac{5}{7}\) days.

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