Question

find the sum of money when
(i) final amount is 11300 at 4% p.a. for 3 years 3 months.
​

Answer 1

↝To Find :

The Sum of Money or the Principal .

↝We Know :

Amount Formula :

(When n years and n(m) months are given)

$\purple{\sf{\underline{\boxed{A = P\left(1 + \dfrac{R}{100}\right)^{n}\left(1 + \dfrac{R \times \dfrac{n_{m}}{12}}{100}\right)}}}}$

Where ,

• A = Amount
• P = Principal
• R = Rate of interest per annum
• n(m) = months
• n = time period

↝Solution :

→ Given :

• Amount = ₹ 11300
• Rate = 4 % p.a
• Time = 3 years 3 months

→ Taken :

Let the Principal or sum of money be P .

Using the formula and by substituting the values in it , we get :

$\purple{\sf{A = P\left(1 + \dfrac{R}{100}\right)^{n}\left(1 + \dfrac{R \times \dfrac{n_{m}}{12}}{100}\right)}}$

$\sf{\Rightarrow 11300 = P\left(1 + \dfrac{4}{100}\right)^{3}\left(1 + \dfrac{4 \times \dfrac{\cancel{3}}{\cancel{12}}}{100}\right)}$

$\sf{\Rightarrow 11300 = P\left(\dfrac{100 + 4}{100}\right)^{3}\left(1 + \dfrac{\cancel{4} \times \dfrac{1}{\cancel{4}}}{100}\right)}$

$\sf{\Rightarrow 11300 = P\left(\dfrac{104}{100}\right)^{3}\left(1 + \dfrac{1}{100}\right)}$

$\sf{\Rightarrow 11300 = P\left(\dfrac{104}{100}\right)^{3}\left(\dfrac{100 + 1}{100}\right)}$

$\sf{\Rightarrow 11300 = P\left(\dfrac{104}{100}\right) \times \left(\dfrac{104}{100}\right) \times \left(\dfrac{104}{100}\right) \times \left(\dfrac{100 + 1}{100}\right)}$

$\sf{\Rightarrow \dfrac{11300}{P} = \left(\dfrac{104}{100}\right) \times \left(\dfrac{104}{100}\right) \times \left(\dfrac{104}{100}\right) \times \left(\dfrac{101}{100}\right)}$

$\sf{\Rightarrow \dfrac{1}{P} = \dfrac{\left(\dfrac{\cancel{104}}{\cancel{100}}\right) \times \left(\dfrac{\cancel{104}}{\cancel{100}}\right) \times \left(\dfrac{\cancel{104}}{\cancel{100}}\right) \times \left(\dfrac{101}{\cancel{100}}\right)}{11300}}$

$\sf{\Rightarrow \dfrac{1}{P} = \dfrac{\left(\dfrac{13}{25}\right) \times \left(\dfrac{26}{25}\right) \times \left(\dfrac{13}{25}\right) \times \left(\dfrac{101}{25}\right)}{11300}}$

$\sf{\Rightarrow \dfrac{1}{P} = \dfrac{\left(\dfrac{13 \times 26 \times 13 \times 101}{25 \times 25 \times 25 \times 25}\right)}{11300}}$

$\sf{\Rightarrow \dfrac{1}{P} = \dfrac{\left(\dfrac{443794}{390625}\right)}{11300}}$

$\sf{\Rightarrow \dfrac{1}{P} = \dfrac{1.134}{11300}}$

By cross-multiplication , we get :

$\sf{\Rightarrow 11300 = 1.134P}$

$\sf{\Rightarrow \dfrac{11300}{1.134} = P}$

$\sf{\Rightarrow 8432.85 = P}$

$\purple{\sf{\therefore P = 8432.85}}$

Hence ,the sum of money or Principal is ₹ 8432.85.

» Additional Information :

• Simple Interest :

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf{SI = \dfrac{P \times R \times T}{100}}$

• Rate :

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf{R = \dfrac{100 \times t}{P \times T}}$

• Time :

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf{T = \dfrac{100 \times t}{P \times R}}$

• Principal :

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf{P = \dfrac{100 \times t}{R \times T}}$

Answer 2

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