Question

If a number is 42 more than the average of its half, one-third and one-
fifteenth, then the number is​

Answer 1

Given:

• If a number is 42 more than the average of its half, one-third and one- fifteenth, then the the number is.

Answer:

• 60

Solution:

Let the number be x

Then,

• ${ \boldsymbol { Half \: of \: x \: = \: \dfrac{1}{2} x}}$
• ${ \boldsymbol { one -third \: of \: x \: = \: \dfrac{1}{3} x}}$
• ${ \boldsymbol { one - fifteenth \: of \: x \: = \: \dfrac{1}{15} x}}$

Now,

• Average of half

• One-third of x

• One-fifteenth of x

$\\ \bf \: = \: \: \: \: \: \frac{ \bigg( \dfrac{1}{2} x \: + \: \dfrac{1}{3}x \: + \: \dfrac{1}{15} x \bigg) }{3} \\ \\ \bf \: = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{ 3} \bigg( \frac{x}{2 } + \frac{x}{3} + \frac{x}{15} \bigg )$

${ \boldsymbol{ \ \: As \: per \: question, }} \\ \\ \: \: \: \: \: \bf \: x \: = \: \frac{x}{6} + \frac {x}{9} + \frac{x}{45} + 42$

Transposing variables on one side and constant terms on other side, we get:

$\\ \bf \: x \: - \: \dfrac{x}{6} - \dfrac {x}{9} - \dfrac{x}{45} = 42$

Multiplying both sides by 90 (L.C.M of 6, 9 and 15), we get:

$\\ \bf \: 90x \: - 15x - 10x - 2x = 40 \times 90 \\ \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: \: \: \: \: \: \: 63x = 40 \times 90 \\ \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: \: \: \: \: \: \: x = \frac{40 \times 90}{63} \\ \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = { \underline{ \boxed{ \bigstar{ \pink{ \boldsymbol{60}}}}}}$

• Hence, The number is 60