Question

a sum of money amounts to Rs. 17680 when invested for 3years at a certain rate of interest per annum and amounts to Rs.17600 when invested for 2years. at the rate pf 5% per annum simple interest. find the rate of interest in the first investment and the sum invested in both the cases.​

Answer 1

Let :-

• The sum of money be P

To Find :-

• The rate of interest = ?
• The sum invested = ?

Solution :-

To calculate rate of interest and sum invested at first we have to set up equation as per the given clue in question. Then solve the equation.

Calculation for Case - (i)

• Amount = 17680 Rate = R%. Time = 3 years

⟼ A = P(1 + RT/100)

⟼ 17680 = P(1 + R × 3/100)

⟼ 17680 = P( 100 + 3R/100)--------(i)

Calculation for Case - (ii)

• Amount = 17600. Rate = 5%. Time = 2 years

⟼ A = P(1 + RT/100)

⟼ 17600 = P(1 + 5×2/100)

⟼17600 = P(1 + 10/100)

⟼ 17600 = P(110/100)

⟼ 110P = 17600 × 100

⟼ P = 160 × 100

⟼ P = 16000

Now calculate Rate for Case - (i)

• P = 16000. T = 3 years. Amount = 17680

⟼ A = P(1 + RT/100)

⟼ 17680 = 16000(1 + R × 3/100)

⟼ 17680/16000 = (100 + 3R)/100---------(i)

⟼ 16000(100 + 3R) = 17680 × 100

⟼ 16(100 + 3R) = 1768

⟼ 1600 + 48R = 1768

⟼ 48R = 1768 - 1600

⟼ 48R = 168

⟼ R = 3.5%

Hence,

• The rate of interest (Case - (i)) = 3.5%
• The sum invested in (Case - (i) and (ii)) = 16000

Answer 2

Given :-

A sum of money amounts to Rs. 17680 when invested for 3years at a certain rate of interest per annum and amounts to Rs.17600 when invested for 2years. at the rate pf 5% per annum simple interest

To Find :-

Rate of interest in the first investment and the sum invested in both cases.​

Solution :-

We know that

A = P(1 + R/100)ⁿ

But here, we may write it as

A = P(1 + TR/100)

Now

①

$\sf A=P\bigg\{1+\dfrac{TR}{100}\bigg\}$

$\sf 17680=P\bigg\{1+\dfrac{3\times R}{100}\bigg\}$

$\sf 17680=P\bigg\{1+\dfrac{3R}{100}\bigg\}$

$\sf 17680=P\bigg\{\dfrac{100+3R}{100}\bigg\}$

②

$\sf A=P\bigg\{1+\dfrac{TR}{100}\bigg\}$

$\sf 17600=P\bigg\{1+\dfrac{2\times5}{100}\bigg\}$

$\sf 17600=P\bigg\{1+\dfrac{10}{100}\bigg\}$

$\sf 17600=P+\bigg\{1+\dfrac{1}{10}\bigg\}$

$\sf 17600=P\bigg\{\dfrac{10+1}{10}\bigg\}$

$\sf 17600=P\bigg\{\dfrac{11}{10}\bigg\}$

$\sf 17600\times10=P\times 11$

$\sf \dfrac{17600\times10}{11}=P$

$\sf 1600\times 10=P$

$\sf 16000=P$

Here

P(Case ①) = P(Case ②)

16000 = 16000

R a t E :-

$\sf 17680=P\bigg\{\dfrac{100+3R}{100}\bigg\} \bigg\lgroup From \; 1\bigg\rgroup$

$\sf 17680=16000\bigg\{\dfrac{100+3R}{100}\bigg\}$

$\sf \dfrac{17680}{16000}=\dfrac{100+3R}{100}$

$\sf 1.105=\dfrac{100+3R}{100}$

$\sf 1.105\times100=100+3R$

$\sf 110.5=100+3R$

$\sf 110.5-100=3R$

$\sf 10.5=3R$

$\sf\dfrac{10.5}{3}=R$

$\sf\dfrac{105}{30}=R$

$\sf\dfrac{7}{2}=R$

$\sf 3.5=R$