“A” finishes his work in 15 days while “B” takes 10 days. How many days will the same
work be done if they work together? get

2 thoughts on ““A” finishes his work in 15 days while “B” takes 10 days. How many days will the same<br />work be done if they work together? get”

1. Given :-

   • A can do a piece of work in 15 days
   • B can do the work in 10 days

   Aim :-

   • To find the total number of days taken to complete the work when A and B both work together

   If B can do the work in 10 days, then in one day B, can do :-

   [tex]\sf \dfrac{1}{10}[/tex] th part of the work.

   If A can do a piece of work in 15 days, then in one day A did :-

   [tex]\sf \dfrac{1}{15}[/tex]th part of the work.

   Therefore, for 1 day, if they work together, then work done will be :-

   [tex]\implies \sf \dfrac{1}{15} + \dfrac{1}{10}[/tex]

   Taking LCM = 30,

   [tex]\implies \sf \dfrac{1 \times 2}{15 \times 2} = \dfrac{2}{30}[/tex]

   [tex]\implies \sf \dfrac{1 \times 3}{10 \times 3} = \dfrac{3}{30}[/tex]

   Putting the values in the equation,

   [tex]\implies \sf \dfrac{2}{30} + \dfrac{3}{30}[/tex]

   Adding,

   [tex]\implies \sf \dfrac{5}{30}[/tex]

   Reducing to the lowest terms, (dividing the numerator and the denominator by 5)

   [tex]\implies \sf \dfrac{1}{6}[/tex]

   This is the work done in 1 day. Therefore the total number of days taken to finish the work :-

   [tex]\implies \sf \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{6}{6} \longrightarrow 1[/tex]

   [tex]\implies \sf 6 \times \bigg(\dfrac{1}{6} \bigg) = 1[/tex]

   Hence, if A and B both work together, then the work can be done in 6 days

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2. Answer:

   Your correct answer is…

   Step-by-step explanation:

   6 Days…

   Please mark me as Brainliest...

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