5) Dinkarrao borrowed 45000 to buy a zerox machine from a company at 9

1.	5) Dinkarrao borrowed 45000 to buy a zerox machine from a company at 9p.c.p.a. fororteyear. Then after one year what amount will he have to pay to the company?
Answer» ➤ Given :- Principle :- ₹45000 Rate of interest :- 9% Time :- 1 year $\:$ ➤ To Find :- The total amount to be returned back after one year. $\:$ ★ How to do :- Here, we are given with the principle amount, the rate of interest and the time taken to return back the money. We are asked to find the total amount to be returned back to the bank after the time period. So, we have an appropriate formula to calculate the same. But, first we should find the interest amount and then we should find the total amount by USING the SIMPLE interest and the principle amount. We ALSO have the formula for simple interest and to the total amount which should be used here. So, LET's solve!! $\:$ ➤ Solution :- Simple Interest :- ${\sf \leadsto \underline{\boxed{\sf \dfrac{P \times R \times T}{100}}}}$ Substitute the given VALUES. ${\tt \leadsto \dfrac{45000 \times 9 \times 1}{100}}$ Cancel the zeros in numerator and denominator. ${\tt \leadsto \dfrac{450 \cancel{00} \times 9 \times 1}{1 \cancel{00}} = \dfrac{450 \times 9 \times 1}{1}}$ Multiply the remaining numbers as they cannot be cancelled further. ${\tt \leadsto 450 \times 9 \times 1 = 450 \times 9}$ ${\tt \leadsto Rs \: \: 4050}$ $\:$ Now, let's find the total amount to be paid. Total Amount :- ${\sf \leadsto \underline{\boxed{\sf Simple \: Interest + Principle}}}$ Substitute the given values. ${\tt \leadsto 4050 + 45000}$ Add the values to get the final answer. ${\tt \leadsto Rs \: \: 49050}$ $\:$ ${\red{\underline{\boxed{\bf So, \: the \: total \: amount \: paid \: is \: Rs.49050}}}}$ ━━━━━━━━━━━━━━━━━━━━━━━ $\dashrightarrow$ Some related formulas :- $\begin{gathered} \small \boxed{\begin{array} {cc} \large \dag \: \sf More \: formulas \\ \\ \sf{\bigstar \: Principle = \dfrac{SI \times 100}{R \times T}} \\ \\ \sf{\bigstar \: Rate \: of \: interest = \dfrac{SI \times 100}{P \times T}} \\ \\ \sf{\bigstar \: Time = \dfrac{SI \times 100}{P \times R}} \end{array}} \end{gathered}$

5) Dinkarrao borrowed 45000 to buy a zerox machine from a company at 9p.c.p.a. fororteyear. Then after one year what amount will he have to pay to the company?

Answer»

➤ Given :-

Principle :- ₹45000

Rate of interest :- 9%

Time :- 1 year

$\:$

➤ To Find :-

The total amount to be returned back after one year.

$\:$

★ How to do :-

Here, we are given with the principle amount, the rate of interest and the time taken to return back the money. We are asked to find the total amount to be returned back to the bank after the time period. So, we have an appropriate formula to calculate the same. But, first we should find the interest amount and then we should find the total amount by USING the SIMPLE interest and the principle amount. We ALSO have the formula for simple interest and to the total amount which should be used here. So, LET's solve!!

$\:$

➤ Solution :-

Simple Interest :-

${\sf \leadsto \underline{\boxed{\sf \dfrac{P \times R \times T}{100}}}}$

Substitute the given VALUES.

${\tt \leadsto \dfrac{45000 \times 9 \times 1}{100}}$

Cancel the zeros in numerator and denominator.

${\tt \leadsto \dfrac{450 \cancel{00} \times 9 \times 1}{1 \cancel{00}} = \dfrac{450 \times 9 \times 1}{1}}$

Multiply the remaining numbers as they cannot be cancelled further.

${\tt \leadsto 450 \times 9 \times 1 = 450 \times 9}$

${\tt \leadsto Rs \: \: 4050}$

$\:$

Now, let's find the total amount to be paid.

Total Amount :-

${\sf \leadsto \underline{\boxed{\sf Simple \: Interest + Principle}}}$

Substitute the given values.

${\tt \leadsto 4050 + 45000}$

Add the values to get the final answer.

${\tt \leadsto Rs \: \: 49050}$

$\:$

${\red{\underline{\boxed{\bf So, \: the \: total \: amount \: paid \: is \: Rs.49050}}}}$

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$\dashrightarrow$ Some related formulas :-

$\begin{gathered} \small \boxed{\begin{array} {cc} \large \dag \: \sf More \: formulas \\ \\ \sf{\bigstar \: Principle = \dfrac{SI \times 100}{R \times T}} \\ \\ \sf{\bigstar \: Rate \: of \: interest = \dfrac{SI \times 100}{P \times T}} \\ \\ \sf{\bigstar \: Time = \dfrac{SI \times 100}{P \times R}} \end{array}} \end{gathered}$