At what rate per cent per annum will * 625 amount to ₹ 676 in 2 years compounded annually?

(1) Given Principal =P= Rs. 625/-, Amount=A = Rs. 676/-, Time period = n= 2yearsWe know for interest "r%" p.a compounding yearly we have

A = P * ( 1+ r/100)nPutting the values we have

676 = 625* ( 1+r/100)2
or ( 1+r/100)2= 676 / 625 = 262 / 252= (26/25)2or (1+r/100) = 26/25or r/100 = 26/25 - 1 = (26-25) /25 = 1/ 25

or r = 100/ 25 = 4% per annum

Therefore with 4% p.a compounding interst Rs. 625/- will become Rs. 676/- in 2 years (Answer)(2) Given Principal =P= Rs. 1,25,000/-, Amount=A =Rs. 1,48,877/-, interest rate= r =6% p. aIf no. of years = n we have

A =P * ( 1+ r/100)nNow Putting the values we have

148877 = 125000 ( 1+ 6/100)n
or148877/ 12500 =( 1+ 3/50)n
or 533/ 503 = (53/50)n
As bases are equal we have n= 3
Therefore in 3 years with 6% p.a compounding interst Rs.1, 25, 000/- will become Rs.1, 48, 877/- (Answer)

Answer

Verified

Hint: We have to use the formula for amount when the interest is compounded annually which is given by $A=P{{\left( 1+\dfrac{r}{100} \right)}^{t}}$ , where P is the principal amount, r is the rate of interest and t is the time in years. Now, we have to substitute the given values in this equation such that P will be Rs. 62500, A will be Rs. 67600 and t will be 2 years. We have to solve the equation for r which will be the required answer.

Complete step by step solution:

We have to find the rate percent. We know that when the interest is compounded annually, the amount is given by the formula$A=P{{\left( 1+\dfrac{r}{100} \right)}^{t}}...\left( i \right)$where P is the principal amount, r is the rate of interest and t is the time in years.We are given that principal amount, $P=Rs.62500$ ,the amount, $A=Rs.67600$ at the time, $t=2\text{ years}$ . Let us substitute these values in the formula (i) and find the rate, r.$\Rightarrow 67600=62500{{\left( 1+\dfrac{r}{100} \right)}^{2}}$Let us take 62500 to the LHS.$\Rightarrow \dfrac{67600}{62500}={{\left( 1+\dfrac{r}{100} \right)}^{2}}$We have to cancel the zeroes from the numerator and denominator.\[\Rightarrow \dfrac{676\require{cancel}\cancel{0}\require{cancel}\cancel{0}}{625\require{cancel}\cancel{0}\require{cancel}\cancel{0}}={{\left( 1+\dfrac{r}{100} \right)}^{2}}\]We can write the result of the above cancellation as\[\Rightarrow \dfrac{676}{625}={{\left( 1+\dfrac{r}{100} \right)}^{2}}\]Now, we have to take the square root on both sides.\[\Rightarrow \sqrt{\dfrac{676}{625}}=\left( 1+\dfrac{r}{100} \right)\]We know that $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ . Therefore, we can write the above equation as\[\Rightarrow \dfrac{\sqrt{676}}{\sqrt{625}}=\left( 1+\dfrac{r}{100} \right)\]We know that $\sqrt{676}=26$ and $\sqrt{625}=25$ . Let us substitute these values in the above equation.\[\Rightarrow \dfrac{26}{25}=\left( 1+\dfrac{r}{100} \right)\]Let us take 1 from the RHS to the LHS.\[\Rightarrow \dfrac{26}{25}-1=\dfrac{r}{100}\]We have to take LCM and simplify.\[\begin{align} & \Rightarrow \dfrac{26}{25}-\dfrac{1\times 25}{1\times 25}=\dfrac{r}{100} \\ & \Rightarrow \dfrac{26}{25}-\dfrac{25}{25}=\dfrac{r}{100} \\ & \Rightarrow \dfrac{26-25}{25}=\dfrac{r}{100} \\ & \Rightarrow \dfrac{1}{25}=\dfrac{r}{100} \\ \end{align}\]Let us take 100 from the denominator of the RHS to the LHS.\[\begin{align} & \Rightarrow \dfrac{100}{25}=r \\ & \Rightarrow r=\dfrac{100}{25} \\ \end{align}\]Let us divide 100 by 25.\[\Rightarrow r=4\]Hence, the rate percent will be 4%.

Note: Students must be thorough with the formulas of compound interest and simple interest. They can get confused with these formulas. For simple interest, the amount is given by the formula $A=P\left( 1+rt \right)$ . Students must note that in this question, the interest is compounded annually. If the interest was compounded half yearly, we will substitute $n=2$ in the formula $A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}$ , where n is the number of times the interest is compounded in a year and r is the rate of the interest (not in percentage).. We can also find the solution to this question using this formula.

Let us substitute the values in the above formula. Here, $n=1$ since the question specifies ‘compounded annually’.\[\begin{align} & \Rightarrow 67600=62500{{\left( 1+\dfrac{r}{1} \right)}^{1\times 2}} \\ & \Rightarrow \dfrac{67600}{62500}={{\left( 1+r \right)}^{2}} \\ \end{align}\]On simplifying the LHS, we will get$\Rightarrow \dfrac{676}{625}={{\left( 1+r \right)}^{2}}$Now, we have to take the square root on both sides.$\begin{align} & \Rightarrow \sqrt{\dfrac{676}{625}}=\left( 1+r \right) \\ & \Rightarrow \dfrac{26}{25}=\left( 1+r \right) \\ \end{align}$Let us take 1 to the LHS.\[\Rightarrow \dfrac{26}{25}-1=r\]We have to take LCM and simplify.\[\begin{align} & \Rightarrow \dfrac{26-25}{25}=r \\ & \Rightarrow \dfrac{1}{25}=r \\ & \Rightarrow r=0.04 \\ \end{align}\]We have to convert r into percentage by multiplying r by 100.$\Rightarrow r=0.04\times 100\%=4\%$Students may have noticed why the term 100 came on the denominator of the formula (i). This is because we have taken r as a percentage in that equation.

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