Question

A and B can do a work in 30 days and 10 days respectively. If they work on alternate days starting with A, then in how many days will the work be completed ?

• A. 16 1/3 days
• B. 15 1/3 days
• C. 14 1/3 days
• D. 12 1/3 days

Answer 1

Answer:

B. 15 1/3 days

A’s 1-day work =$( \frac{1}{30} )$
B’s 1-day work = $( \frac{1}{10} )$

They work alternately starting with A.

Work done in 2 days

Day 1 (A):$( \frac{1}{30} )$
Day 2 (B): $( \frac{1}{10} )$

Total in 2 days:
$\frac{1}{30} + \frac{1}{10}$
$= \frac{1}{30} + \frac{3}{30}$
$= \frac{4}{30}$
$= \frac{2}{15}$

Work done in 14 days (7 cycles)

$7 \times \frac{2}{15} = \frac{14}{15}$

Remaining work:
$1 - \frac{14}{15} = \frac{1}{15}$

Day 15 (A works)

$A’s 1-day work = ( \frac{1}{30} )$

Remaining after Day 15:
$\frac{1}{15} - \frac{1}{30} = \frac{1}{30}$

Day 16 (B works)

Time needed for $( \frac{1}{30} )$ work by B:
$\frac{\frac{1}{30}}{\frac{1}{10}} = \frac{1}{3}\text{ day}$

Total time

$15 + \frac{1}{3} = 15\frac{1}{3} \text{ days}$
$=(15\frac{1}{3}) days$