Question

A and B can do a work in 20 days. B and C can do it in 30 days. A worked on it for 10 days and B worked on it for 15 days. C finished the remaining work in 17 days. How many days would it have taken for C to complete the entire work all alone?

Answer 1

ANSWER

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If A and B take 20days to complete a work they will complete 1

20 of the work in a day

so,

=> A + B = 1/20

like,

=> B + C = 1/30

like, => 10A + 15B + 17C = 1

=> 10A + 10B + 5B + 5C + 12C = 1

=> 10( 1 ) 5( 1 ) + 12C = 1

20 30

=> 12C = 1 - 1 - 1 = 12 - 6 - 2 = ( 12 - 6 -2 )/12

2 6 12 12 12

= 4 = 1

12 3

hence,

=> C= 1 Units per day.

36

: It takes 36 days to complete the work alone.

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Answer 2

$\large\underline\mathfrak\pink{Answer-}$

36 days.

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$\large\underline\mathfrak\pink{Explanation-}$

A and B can do a work in 20 days.

So, A and B done $\dfrac{1}{20}$ of the work in one day.

$\therefore$ A + B = $\dfrac{1}{20}$ _______(1)

Similarly, B and C can do it in 30 days.

So, B and C done $\dfrac{1}{30}$ of the work in one day.

$\therefore$ B + C = done $\dfrac{1}{30}$ _______(2)

According to question,

10A + 15B + 17C = 1

$\rightarrow$ 10A + 10B + 5B + 5C + 12C = 1

Taking 10 and 5 as common,

$\rightarrow$ 10(A+B) + 5(B+C) + 12C = 1______(3)

Put the value of eq (1) and (2) in eq (3) :

$\rightarrow$ 10 ( $\dfrac{1}{20}$ ) + 5 ( $\dfrac{1}{30}$ ) + 12C = 1

$\rightarrow$ $\dfrac{1}{2}$ + $\dfrac{1}{6}$ + 12C = 1

$\rightarrow$ 12C = 1 - $\dfrac{1}{2}$ - $\dfrac{1}{6}$

$\rightarrow$ 12C = $\dfrac{4}{12}$

$\rightarrow$ 12C = $\dfrac{1}{3}$

$\rightarrow$ C = $\dfrac{1}{3×12}$

$\rightarrow$ C = $\dfrac{1}{36}$

$\therefore$ It takes 36 days for C to complete the entire work all alone.