Question

109. A and B together can do a piece of work in 30 days. B and
C together can do it in 20 days. A starts the work and
works on it for 5 days, then B takes up and works for 15
days. Finally C finishes the work in 18 days. The number
of days in which C alone can do the work when doing it
separately is
(2) 40 (6) 120 (C) 60
(d) 24​

Answer 1

Answer:

Option d) 24

Step-by-step explanation:

According to question

A and B together do a piece of work in 30 days.

Means, A + B = 30 days

Similarly,

B and C together do a piece of work in 20 days.

B + C = 20 days

LCM of 20 and 30 is 60.

Work done by A and B in 1 day = 60/30

= 2 units

Work done by B and C in 1 day = 60/20

= 3 units

So,

A + B = 2 and B + C = 3 ____ (eq 1)

A starts the work and work on it for 5 days.

A = 5 days

Similarly, B and C starts the work and work on it for 15 and 18 days.

B = 15 days and C = 18 days

We have to find the number of days in which the C alone do the work.

A starts the work and works on it for 5 days, then B takes up and works for 15 days. Finally C finishes the work in 18 days.

Also, work done by both A and B is 60 units.

So,

• A = 5 days

• B = (5 + 10) days

• C = (10 + 8) days

Remaining work = 5A + 15B = 60

= 5(A + B) + 10(B + C) = 60

= 5(2) + 10(3) = 60 [From (eq 1)]

= 10 + 30 = 60

= 60 - 40

= 20 units

So, work done by C in 8 days

= 20/8

= 5/2 unit/day

Total work = 60

Time taken by C alone to complete the work = (60 × 2)/5

24 days

Answer 2

Answer:

24 days

Step-by-step explanation:

A and B together can do the work in

30 days

and

B and c together can do it in 20 days.

LCM of 30 and 20 is 60

So total work is 60 unit's

( A + B ) per day work is 60/30 = 2 unit's

and

( B + C ) per day work is 60/20 = 3 unit's

now,

A starts the work on it 5 day's

then B takes up and work per 15 days

C finished remaining work in 18 days

this means

A = 5 days

B = ( 5 + 10 ) days

C= ( 10 + 8 ) days

Now work done by ( A + B ) in 5 days

is = 2 × 5 = 10 unit's

and work done by ( B + C) in 5 days

is = 3 × 10 = 30 unit's

total work finished till now

10 + 30 = 40 unit's

remaining work is 60 - 40 = 20 unit's

Here we know C does remaining 20 unit's

in 8 days

C efficiency is work/day = 20/8 = 5/2

Hence time taken by C alone to

complete the whole work is,

( 60/5 ) × 2 = 12 × 2 = 24 days.