Question

find the compound interest on rs 4000 for 2.5 years at 10% per annum

Answer 1

Answer: Compound interest for 2.5 years is Rs. 1082.

Step-by-step explanation:

Since we have given that

Principal = Rs. 4000

Rate of interest = 10% p.a.

Time = 2.5 years

First we calculate the amount for 2 years using the Compound interest formula:

$A=P(1+\dfrac{r}{100})^n\\\\A=4000(1+\dfrac{10}{100})^2\\\\A=4000(1+0.1)^2\\\\A=4000\times (1.1)^2\\\\A=4840$

Now, Rs. 4840 becomes the principal at the end of 2nd year.

So, we will calculate the interest on Rs. 4840 for 0.5 year @ 10%.

So, it becomes,

$I=\dfrac{4840\times 0.5\times 10}{100}\\\\I=Rs.\ 242$

So, interest on RS. 4000 @10% for the first year = Rs. 400

Interest on Rs. 4000+400=Rs. 4400 for the second year = Rs. 440

Interest on Rs. 4840 for the 0.5 more year = Rs. 242

So,Total compound interest = 400+440+242 = Rs. 1082

Hence, compound interest for 2.5 years is Rs. 1082.

Answer 2

Answer:

$\red {Compound \: Interest} \green {= Rs \: 1082 }$

Step-by-step explanation:

$i) Principal (P) = Rs \: 4000 ,\\Time (T) = 2 \: years \\Rate \: of \: interest (r) = 10\% \:per \: annum$

$Conversation \: period (n) = 2$

$\boxed {\pink { Amount (A) = P\left( 1 + \frac{r}{100}\right)^{n} }}$

$A_{1} = 4000\left( 1 + \frac{10}{100}\right)^{2}\\= 4000 \times (1.1)^{2}\\= 4000\times 1.1 \times 1.1 \\= Rs\:4840$

$Compound \: Interest (C.I) = A_{1} - P\\= Rs\:4840 - Rs\: 4000 \\= Rs\: 840 \: ---(1)$

$ii ) For \: 6 \: months \: we \:have \: to \: find \\simple \: Interest$

$Principal = A_{1} = Rs \: 4840 ,\\R = 10\% \\Time (T) = \frac{1}{2} \: year$

$\boxed { \pink { Simple \: Interest (I) = \frac{PTR}{100} }}$

$= \frac{ 4840 \times \frac{1}{2} \times 10}{100}\\= \frac{ 4840 \times 5}{100} \\= Rs\: 242 \: ---(2)$

Therefore,.

$Compound \: Interest = (1) + (2) \\= Rs \: 840 + Rs\: 242 \\= Rs\:1082$

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