
The total number of industries in a particular portion of the country is approximately 1600. If the government has decided to increase the number of industries in the area by 20% every year. Find the approximate number of industries after 2 years
Answer
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Hint: From the question we can get the present number of industries which increases by 20% for next two years, so using the below mentioned formula we can easily find the number of industries after 2 years.
Formula. $ {P_n} = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} $ ;
where $ {P_n} $ is the number of industries after n years
P is the no. of industries presently
r is the increment every year
n is the no of years after which the new value to be found.
Complete step-by-step answer:
Let the Present number of industries in a particular portion of the country be P.
So, from the given question P=1600;
Now, the government has decided to increase the number of industries in the area by 20% every year.
So, let the number of industries after 2 years be $ {P_n} $ .
Now, we need to use the formula $ {P_n} = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} $ to find the value of $ {P_n} $
Hence, the number of industries after 2 years $ ({P_n}) $ = $ P{\left( {1 + \dfrac{r}{{100}}} \right)^n} $
$\Rightarrow {P_n} = 1600{\left( {1 + \dfrac{{20}}{{100}}} \right)^2} $
$\Rightarrow {P_n} = 1600 \times \dfrac{{120}}{{100}} \times \dfrac{{120}}{{100}} $
$\Rightarrow {P_n} = 1600 \times \dfrac{{120}}{{100}} \times \dfrac{{120}}{{100}} $
$\Rightarrow {P_n} = 2304 $
Hence, the number of industries after 2 years will be 2304.
So, the correct answer is “2304”.
Additional Information.
If the present value P of a quantity decreases at the rate of r% per unit of time then the value Q of the quantity after n units of time is given by
$\Rightarrow Q = P{\left( {1 - \dfrac{r}{{100}}} \right)^n} $
Note: In the above case the number of industries increased uniformly for both the years so we used the above formula. But if the number of industries would have increased un-uniformly for both the years then we need to take $ {r_1} $ & $ {r_2} $ in the same formula and need to solve accordingly.
Formula. $ {P_n} = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} $ ;
where $ {P_n} $ is the number of industries after n years
P is the no. of industries presently
r is the increment every year
n is the no of years after which the new value to be found.
Complete step-by-step answer:
Let the Present number of industries in a particular portion of the country be P.
So, from the given question P=1600;
Now, the government has decided to increase the number of industries in the area by 20% every year.
So, let the number of industries after 2 years be $ {P_n} $ .
Now, we need to use the formula $ {P_n} = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} $ to find the value of $ {P_n} $
Hence, the number of industries after 2 years $ ({P_n}) $ = $ P{\left( {1 + \dfrac{r}{{100}}} \right)^n} $
$\Rightarrow {P_n} = 1600{\left( {1 + \dfrac{{20}}{{100}}} \right)^2} $
$\Rightarrow {P_n} = 1600 \times \dfrac{{120}}{{100}} \times \dfrac{{120}}{{100}} $
$\Rightarrow {P_n} = 1600 \times \dfrac{{120}}{{100}} \times \dfrac{{120}}{{100}} $
$\Rightarrow {P_n} = 2304 $
Hence, the number of industries after 2 years will be 2304.
So, the correct answer is “2304”.
Additional Information.
If the present value P of a quantity decreases at the rate of r% per unit of time then the value Q of the quantity after n units of time is given by
$\Rightarrow Q = P{\left( {1 - \dfrac{r}{{100}}} \right)^n} $
Note: In the above case the number of industries increased uniformly for both the years so we used the above formula. But if the number of industries would have increased un-uniformly for both the years then we need to take $ {r_1} $ & $ {r_2} $ in the same formula and need to solve accordingly.
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