Question

5. A farmer has an increase of 12.5% in the output of wheat in his farm every year, he produced 2916 quintals of wheat. What was his annual production of wheat 2 years ago?​

Answer 1

$★ {\pmb{\underline{\sf{Understanding \ Concepts ... }}}}$

As We know that a Farmer has an increase in his Output of Wheat in his Farm Every year by Some amount of Rate.

Here we gonna take Time interval as a Negative (—ve) value because We have to calculate Time Period in Past time.

• Time Period = —2 Years

Now, We consider Present time Production as the Principal of the Calculation to make the desired decision as Follows.

• Principal = 2916 Quintal of wheat

So, We can calculate Rate of increasing in the output of the Production of the wheat.

• Rate = 25/2%

$★ {\pmb{\underline{\sf{Required \ Solution ... }}}}$

As We know that We can apply Amount gaining Formula to Find the Desired results as:

$\circ \ {\underline{\boxed{\sf\gray{ Amount = p \left( 1 + \dfrac{r}{100} \right) ^ {-n} }}}}$

Here's, The Time Period is in negative because we have to find the Past time Production of Wheat as Well.

$\colon\implies{\tt{ 2916 \left( 1 + \dfrac{ \dfrac{25}{2} }{100} \right) ^ {-2} }} \\ \\ \\ \colon\implies{\tt{ 2916 \left( 1 + \dfrac{25 }{200} \right) ^ {-2} }} \\ \\ \\ \colon\implies{\tt{ 2916 \left( \dfrac{225 }{200} \right) ^ {-2} }} \\ \\ \\ \colon\implies{\tt{ 2916 \left( \dfrac{9 }{8} \right) ^ {-2} }} \\ \\ \\ \colon\implies{\tt{ 2916 \left( \dfrac{8 }{9} \right) ^ {2} }} \\ \\ \\ \colon\implies{\tt{ 2916 \times \dfrac{8 }{9} \times \dfrac{8}{9} }} \\ \\ \\ \colon\implies{\tt{ 36 \times 64 }} \\ \\ \colon\implies{\tt{ 2304 \ Quintal }}$

Hence,

The Production of the Wheat 2 years ago in the farm of Farmer was 2304 Quintals.