A pole is painted as \(\frac{1}{10}th\) red, \(\frac{1}{20}th\) white, \(\frac{1}{30}th\) blue, \(\frac{1}{40}th\) black, \(\frac{1}{50}th\) violet, \(\frac{1}{60}th\) yellow and the rest green. If the portion in green has length 12.08 metre, length of the pole is

Given:

Red portion = \(\frac{1}{10}\)

White portion = \(\frac{1}{20}\)

Blue portion = \(\frac{1}{30}\)

Black portion = \(\frac{1}{40}\)

Violet portion = \(\frac{1}{50}\)

Yellow portion = \(\frac{1}{60}\)

Green portion length = 12.08 metre

Formula Used:

Sum of fractions of painted portions = Total painted length / Total length of pole

Remaining portion = 1 - Sum of fractions of painted portions

Calculations:

Sum of the fractions of the pole painted in different colors (excluding green):

⇒ \(\frac{1}{10} + \frac{1}{20} + \frac{1}{30} + \frac{1}{40} + \frac{1}{50} + \frac{1}{60}\)

⇒ \(\frac{60 + 30 + 20 + 15 + 12 + 10}{600}\)

⇒ \(\frac{147}{600}\) 

⇒ \(\frac{49}{200}\)

The portion painted green is the rest of the pole:

⇒ Green portion (fraction) = 1 - \(\frac{49}{200}\)

⇒ Green portion (fraction) = \(\frac{200 - 49}{200}\) = \(\frac{151}{200}\)

We are given that the length of the green portion is 12.08 metre.

Let L be the total length of the pole.

⇒ \(\frac{151}{200}\) × L = 12.08

⇒ L = \(\frac{12.08 \times 200}{151}\)

⇒ L = 16 metres

The correct answer is Option (3).