Question

Raju borrowed Rs 6,500 from Rohan at the rate of
10% S.I. He returned it after 2 years. What was the
amount returned?​

Answer 1

Step-by-step explanation:

so,

P= Rs6,500

R= 10%

T=2years

S.I.=P×R×T\100

= Rs6,500×10%×2/100

= Rs1,300

A= S.I.+ P

= Rs1,300 + Rs6,500

= Rs7,800

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Answer 2

☆ To Find :

The Amount returned after 2 years.

☆ We Know :

Amount Formula :

$\implies \purple{\sf{\underline{\boxed{A = P\left(1 + \dfrac{R}{100}\right)^{n}}}}}$

➝ Where ,

• A = Amount
• P = Principal
• R = Rate of Interest
• 0n = time period

Interest :

$\purple{\sf{\underline{\boxed{I = \dfrac{P \times R \times t}{100}}}}}$

➝ Where ,

• SI = Simple Interest
• P = Principal
• R = Rate of Interest
• t = time period

☆ Solution :

Given :

• Principal = ₹ 6500

• Rate of interest = 10 % p.a,

• Time = 2 years

Using the formula and putting the value in it , we get :

$\purple{\sf{A = P\left(1 + \dfrac{R}{100}\right)^{n}}} \\ \\ \\ \\ \:\:\:\:\:\:\implies \sf{A = 6500\left(1 + \dfrac{10}{100}\right)^{2}} \\ \\ \\ \\ \:\:\:\:\:\:\implies \sf{A = 6500\left(\dfrac{100 + 10}{100}\right)^{2}} \\ \\ \\ \\ \:\:\:\:\:\:\implies \sf{A = 6500 \times \dfrac{10}{100} \times \dfrac{10}{100}} \\ \\ \\ \\ \:\:\:\:\:\:\implies \sf{A = 65 \times \dfrac{10}{100} \times 10} \\ \\ \\ \\ \:\:\:\:\:\:\implies \sf{A = 65 \times 11 \times 11} \\ \\ \\ \\ \:\:\:\:\:\:\implies \sf{A = 65 \times 121} \\ \\ \\ \\ \implies \sf{A = 7865}$

Hence , the amount after 2 years is ₹ 7865.

☆ Alternative Method :

Principal for first year = ₹ 6500

Interest for the first year = $\sf{I = \dfrac{P \times R \times t}{100}}$

Putting the value in the formula, we get :

$\implies \sf{I = \dfrac{6500 \times 10 \times 1}{100}} \\ \\ \implies \sf{I = \dfrac{650\not{0} \times 1\not{0} \times 1}{1\not{0}\not{0}}} \\ \\ \implies \sf{I = 650} \\ \\ \therefore \purple{I = 650}$

Amount at the end of Year = Principal + Interest

$\implies \sf{A = 6500 + 650}$

$\implies \sf{A = 7150}$

Hence , the amount at the end of first year is ₹ 7150.

Principal for Second year = ₹ 7150

Interest for the first year = $\sf{I = \dfrac{P \times R \times t}{100}}$

Putting the value in the formula, we get :

$\implies \sf{I = \dfrac{7150 \times 10 \times 1}{100}} \\ \\ \implies \sf{I = \dfrac{715\not{0} \times 1\not{0} \times 1}{1\not{0}\not{0}}} \\ \\ \implies \sf{I = 715} \\ \\ \therefore \purple{I = 715}$

Amount at the end of Year = Principal + Interest

$\implies \sf{A = 7150 + 715}$

$\implies \sf{A = 7865}$

Hence , the amount at the end of first year is ₹ 7865.

Thus , the amount returned is ₹ 7865.

☆ Additional information :

• Simple Interest = $\sf{SI = \dfrac{P \times R \times t}{100}}$

• Principal = $\sf{P = \dfrac{100 \times SI}{R \times t}}$

• Rate = $\sf{R = \dfrac{100 \times SI}{P \times t}}$

• Time = Principal = $\sf{t = \dfrac{100 \times SI}{R \times P}}$