Question

“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​

Answer 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 :-

$\sf \dfrac{1}{10}$ th part of the work.

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

$\sf \dfrac{1}{15}$th part of the work.

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

$\implies \sf \dfrac{1}{15} + \dfrac{1}{10}$

Taking LCM = 30,

$\implies \sf \dfrac{1 \times 2}{15 \times 2} = \dfrac{2}{30}$

$\implies \sf \dfrac{1 \times 3}{10 \times 3} = \dfrac{3}{30}$

Putting the values in the equation,

$\implies \sf \dfrac{2}{30} + \dfrac{3}{30}$

Adding,

$\implies \sf \dfrac{5}{30}$

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

$\implies \sf \dfrac{1}{6}$

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

$\implies \sf \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{6}{6} \longrightarrow 1$

$\implies \sf 6 \times \bigg(\dfrac{1}{6} \bigg) = 1$

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

Answer 2

Answer:

Your correct answer is...

Step-by-step explanation:

6 Days...

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