Math, asked by svetajayaprakash, 9 months ago

A works 3 times as fast as B and is able to complete a task in 24 days less than the days taken by B. Find the time in which they can complete the work together please explain step by step and strictly no spam ​

Answers

Answered by Anonymous
27

Answer:

9 days

Step-by-step explanation:

Let the no. of days taken by A and B be A and B days respectively

Work done by A in one day = 1 / A

Work done by B in one day = 1 / B

A works 3 times as fast as B

⇒ 1 / A = 3 / B

A takes 24 days less than the days taken by B

⇒ A = B - 24

Substituting A = B - 24 in 1 / A = 3 / B

⇒ 1 / ( B - 24 ) = 3 / B

⇒ B = 3( B - 24 )

⇒ B = 3B - 72

⇒ 72 = 3B - B

⇒ 72 = 2B

⇒ 72 / 2 = B

⇒ 36 = B

⇒ B = 36

⇒ Work done by B in 1 day = 1 / B = 1 / 36

Substituting B = 36 in 1 / A = 3 / B

⇒ 1 / A = 3 / 36

⇒Work done by A in one day =  1 / A = 1 / 12

Work done by A and B in one day = 1 / 12 + 1 / 36 = ( 3 + 1 ) / 36 = 4 / 36 = 1 / 9

∴  A and B can finish the task in 9 days.

Answered by madhushree3
25

Answer:

Ans :9days

Step-by-step explanation:

If B does the work in 3 days,then A will do it in 1 day.that is,the difference is 2 days.here given that the difference between A and B in completing the works is 24 days.therefore,A will take 24/2=12days and B will take 3×12=36days to complete the work separately.

hence the time taken by A and B together to complete the work =ab/a+b days.

=12×36/12+36

=12×36/48=9 days

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