Question

A and B can do a piece of work in 30 days. While B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work ?​

Answer 1

Answer:

$\huge\underline\bold {Answer:}$

(A + B)’s one days’ work = 1/30

(B + C)’s one days’ work = 1/24 .....(1)

(C + A)’s one days’ work = 1/20

Therefore (A + B + C)’s one days’ work

$\frac{1}{2} ( \frac{1}{30} + \frac{1}{24} + \frac{1}{20} ) \\ = \frac{1}{2} \times ( \frac{20 + 25 + 30}{600} ) \\ = \frac{75}{1200} \\ = \frac{1}{16} \: \: \: \: \: \: \: ...(2)$

(A + B + C)’s 10 days’ work

= 10/6 = 5/8

From (1) and (2), A's one days’ work

= 1/16 – 24

= 1/48

Therefore remaining 3/8 of the work is done by A alone in 3/8 × 48

= 18 days.

Answer 2

Answer:

Thus A will take 18 days to finish the work

Step-by-step explanation:

A+B=1/30

B+C=1/24.......(.1)

C+A=1/20

adding we get

2(A+B+C)=1/30+1/24+1/20

=1/120( 4+5+6)=15/120=1/8

So A+B+C=1/16

From (1)

A+B+C-(B+C)=1/16-1/24=1/48( 3-2)=1/48

So A=1/48.............(2)

Work done by all 3 in 10 days

=10*(A+B+C)=10*1/16=10/16=5/8

Work left=1-5/8=3/8

Thus no of days taken by  A to finish the work

=(3/8) / (1/48)

=3/8*(48)

=3*6=18 days

Thus A will take 18 days to finish the work