Question

Anil has deposited certain amount in the bank to earn compoundbinterest at the rate of 10% p.a. The difference in the interest on the amount in the 3rd and 2nd years is Rs. 1100. What amount has Anil deposited ?
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Answer 1

Let the P be x

Therefore c.i for 2nd year= x{(1 + 10/100)^2 -1}

= x{( 11/10)^2 - 1}

= x{ 121/100 - 1}

= x × 21/ 100

= 21x/100

Again,c.i for the 3rd year= x{( 1 + 10/100)^3 -1}

= x{( 11/ 10)^3 -1}

= x { 1331/1000 -1}

= x × 331/ 1000

= 331x / 1000

From the condition we get,

331x/1000 - 21x/100 = 1100

=> 121x/ 1000 = 1100

=> x = 1100000/ 121 = 100000/ 11 = ₹9090.91 (approx)

Therefore the required principal amount is ₹9090.91 (approx)

Answer 2

Let the principle be $P$

Given the rate of interest $r=10\% p.a.$

We know, $A=P(1+\frac{r}{n})^{nt}$,

where $A$ = final amount

$P$ = initial principal balance

$r$ = interest rate

$n$ = number of times interest applied per time period

$t$ = number of time periods elapsed

Therefore $C.I$ for 2nd year= $P((1 + \frac{10}{100})^2 -1})$

=$P((\frac{11}{10}^{2})-1)=P(\frac{121}{100}-1)=\frac{21P}{100}$

Similarly $C.I.$ for the 3rd year =$P((1 + \frac{10}{100})^3 -1})$

$P((\frac{11}{10}^{3})-1)=P(\frac{1331}{1000}-1)=\frac{331P}{1000}$

Now, Given $\frac{331P}{1000}-\frac{21P}{100}=1100\\\frac{121P}{1000}=1100\\P=\frac{1100\times 1000}{121}=9090.90$

$\therefore$ Anil deposited Rs. 9090

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