Question

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?

• A. 5 3/7
• B. 5 4/7
• C. 5 5/7
• D. 5 6/7

Answer 1

Answer:

C. 5 5/7

Let the total work = LCM of (12, 6, 15) = 60 units

Daily work rates

• Rajeshwar: (60/12 = 5) units/day

• Vikram: (60/6 = 10) units/day

• Tiger: (60/15 = 4) units/day

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Work done in different periods

First 2 days (all three work):
$(5 + 10 + 4) \times 2 = 19 \times 2 = 38 \text{ units}$

Last 3 days (only Tiger works):
$4 \times 3 = 12 \text{ units}$

Let the total time = x days

So, the middle period (after Rajeshwar leaves and before Vikram leaves) is:
$x - 5 \text{ days}$
$During this time, <b>Vikram + Tiger</b> work:$
$(10 + 4)(x - 5) = 14(x - 5)$

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Total work equation

$38 + 14(x - 5) + 12 = 60$
$38 + 14x - 70 + 12 = 60$
$14x - 20 = 60$
$14x = 80$
$x = \frac{80}{14} = \frac{40}{7}$
]

The work is completed in $5\frac{5}{7}$