Math, asked by third42, 6 months ago

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 ?​

Answers

Answered by Anonymous
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.

Answered by Anonymous
1

Answer:

Step-by-step explanation:

(A+B)'s one day's work =

30

1

(B+C)'s one day's work =

24

1

..........(1)

(C+A)'s one day's work =

20

1

∴(A+B+C)'s one day's work =

2

1

[

30

1

+

24

1

+

20

1

]=

2

1

×[

600

20+25+30

]=

1200

75

=

16

1

...........(2)

(A+B+C)'s 10 day's work =

16

10

=

8

5

Remaining work,  1−

8

5

=

8

3

 (which is done by A)

From (1) and (2), A's one day's work =

16

1

24

1

=

48

1

∴  Remaining  

8

3

 of the work is done by A alone in  

8

3

×48=18  days.

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