Question

A factory kept increasing its output by the same percentage every year. Find the percentage it is known that the output has doubled in the last two years.

Answer 1

Hint: First find production after one year then after two year taking initial investment as P and rate as x. Then equate finding one with 2P so hence one can find the value of x.

Complete step by step answer:

In the question we were given information about the factory that its output is increasing by some percentage every year. So, now we have to find a percentage of growth so that output has doubled in the last two years.
Now let say that the initial production be considered as P which is the time of 2 years ago.
After one year production will be,
\[\text{P}\ \text{+}\ \dfrac{\text{P}x}{100}\ =\ \text{P}\left( 1+\dfrac{x}{100} \right)\]
where, x is percent increase of production.
After one more year means altogether after to year we get,
\[\text{P}\left( 1+\dfrac{x}{100} \right)\ \text{+}\ \dfrac{x}{100}\left( \text{P}\left( 1+\dfrac{x}{100} \right) \right)\ =\ \text{P}\left( 1+\dfrac{x}{100} \right)\left( 1+\dfrac{x}{100} \right)\]
Which is equal to \[\ \text{P}{{\left( 1+\dfrac{x}{100} \right)}^{2}}\]
Now, in the equation it is said that the production doubled in the last two years so we can say that product became 2P.
Hence, we can equate,
\[\text{P}{{\left( 1+\dfrac{x}{100} \right)}^{2}}\ =\ 2\text{P}\]
So, by cancelling we can write,
\[{{\left( 1+\dfrac{x}{100} \right)}^{2}}\ =\ 2\]
Now, we know that \[{{\left( 1+\dfrac{x}{100} \right)}^{2}}\] is always positive so,
\[\left( 1+\dfrac{x}{100} \right)\ =\ \sqrt{2}\] and \[\left( -\sqrt{2} \right)\]
So, \[\dfrac{x}{100}\ =\ \sqrt{2}-1\]
Now, we know that value of \[\sqrt{2}\ =\ 1.4142\] so, \[\sqrt{2}-1\ =\ 1.4142-1\ =\ 0.4142\]
Hence, \[\dfrac{x}{100}\ =\ 0.4142\]
So, \[x=\ 0.4142\times 100\]
Hence \[x=\ 41.42%\]

Hence, the percent of increase is \[41.42%\].

Note: We can directly find amount if initial production P and rate x is given with given time ‘t’ using formula, \[\text{P}{{\left( 1+\dfrac{x}{100} \right)}^{t}}\] .