A submarine after travelling 100km from the shore, descended $1\frac{1}{10}$ Km. Then it ascended about $\frac{3}{5}$km. The next day, it again descended by $\frac{9}{10}$ Km. What is its current position with respect to the sea level?

Given: 

A submarine after travelling 100km from the shore, descended $1\frac{1}{10}$ Km

Then it ascended about $\frac{3}{5}$km

The next day, it again descended By $\frac{9}{10}$ Km

To do: Find its current position with respect to the sea level

Solution:

Let us take ascending of the submarine as positive and descending as negative with respect to the sea level

First its descending  $1\frac{1}{10}$ Km or $\frac{11}{10}$ Km or $\frac{-11}{10}$km

Next its ascending $\frac{3}{5}$

Again descending $\frac{9}{10}$or $\frac{-9}{10}$km

Total movement with respect to sea level $\frac{-11}{10}+ \frac{3}{5} – \frac{-9}{10}$

or $\frac{-11}{10}+ \frac{6}{10} – \frac{-9}{10}$ 

= $\frac{-11+6-9}{10}$

= $\frac{-14}{10}$

=$\frac{ -7}{5}$ km.

So the submarine is $\frac{7}{5}$ km below the sea level.

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Updated on: 10-Oct-2022

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