Question

A can finish a work in 24 hours, B in 40 hours and C in 60 hours. They all begin
together but A alone continues to work till the end, while B leaves 2 hours and C
leave 7 hours before completion. In what time is the work finished?​

Answer 1

Answer:

the total work is finished in 14 hours

Step-by-step explanation:

Given,

• A can finish a work in 24 hours, B in 40 hours and C in 60 hours.
• They all begin  together but A alone continues to work till the end.
• B leaves 2 hours and C  leaves 7 hours before completion.

To find,

• In what time is the work finished ?

Solution,

➙ A can finish work in 24 hours

      In one hour, A completes = (1/24) part of work

➙ B can finish the work in 40 hours

     In one hour, B completes = (1/40) part of work

➙ C can finish the work in 60 hours

    In one hour, C completes = (1/60) part of work

A words till the end , B leaves 2 hours and C  leaves 7 hours before completion.

let the work be finished in "x hours"

So, to finish the work

 ➙ A completes (x/24) part of work

 ➙ B completes (x - 2)/40 part of work

 ➙ C completes (x - 7)/40 part of work

    $\sf \dfrac{x}{24}+\dfrac{x-2}{40}+\dfrac{x-7}{60}=1 \\\\\\ \dfrac{1}{4} (\dfrac{x}{6}+\dfrac{x-2}{10}+\dfrac{x-7}{15})=1 \\\\\\ \dfrac{x}{6}+\dfrac{x-2}{10}+\dfrac{x-7}{15}= 1 \times 4 \\\\\\ \dfrac{x}{6}+\dfrac{x-2}{10}+\dfrac{x-7}{15}=4 \\\\\\ LCM=30 \\\\\\ \dfrac{x \times 5}{6 \times 5}+\dfrac{(x-2) \times 3}{10 \times 3}+\dfrac{(x-7) \times 2}{15 \times 2}=4 \\\\\\ \dfrac{5x}{30}+\dfrac{3x-6}{30}+\dfrac{2x-14}{30}=4 \\\\\\ \dfrac{5x+3x-6+2x-14}{30}=4 \\\\\\ \dfrac{10x-20}{30}=4$

    10x - 20 = 4 × 30

    10x - 20 = 120

     10x = 120 + 20

    10x = 140

      x = 140/10

      x = 14 hours

Therefore, the total work is finished in 14 hours