Question

$\begin{gathered}\large{\color{cyan}{\bigstar}} \: \: {\underline{\boxed{\red{\sf{ question \: \div \: }}}}} \end{gathered} \\ \\ \large \ \sf find \: the \: ammount \: and \: the \\ \large \ \sf \: compound \: intrest \: on \: ₹10000 \\ \large \ \sf \: for 1\frac{1}{2} \: years \: at \: 10percent \: annum \\ \large \ \sf \: compounded \: half \: yearly \: .would \\ \large \ \sf \: this \: intrest \: be \: more \: than \: the \\ \large \ \sf \: intrest \: he \: would \: get \: if \: it \: was \\ \large \ \sf \: compounded \: annually \: \: ??$



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Answer 1

$\large\underline{\sf{Solution-}}$

Case :- 1 Interest is compounded half-yearly

Principal, p = ₹ 10000

Rate of interest, r = 10 % per annum compounded half - yearly.

Time, n = 3/2 years

We know,

Amount received on a certain sum of money of ₹ p invested at the rate of r % per annum compounded half - yearly for n years is given by

$\boxed{\tt{ \: \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{200} \bigg]}^{2n} \: \: }}$

So, on substituting the values, we get

$\rm \: Amount \: = \: 10000 {\bigg[1 + \dfrac{10}{200} \bigg]}^{3}$

$\rm \: Amount \: = \: 10000 {\bigg[1 + \dfrac{1}{20} \bigg]}^{3}$

$\rm \: Amount \: = \: 10000 {\bigg[\dfrac{20 + 1}{20} \bigg]}^{3}$

$\rm \: Amount \: = \: 10000 {\bigg[\dfrac{21}{20} \bigg]}^{3}$

$\rm \: Amount \: = \: 10000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20}$

$\rm\implies \:Amount \: = \: 11576.25$

Now, We know that,

$\rm \: Compound \: interest \: = \: Amount \: - \: Principal$

$\rm \: Compound \: interest \: = \: 11576.25 \: - \: 10000$

$\rm\implies \:Compound \: interest \: = \: 1576.25$

Hence,

Amount = ₹ 11576. 25

Compound interest = ₹ 1576. 25

Case :- 2 Interest is compounded annually.

Principal, p = ₹ 10000

Rate of interest, r = 10 % per annum compounded anually.

Time, n = 3/2 years

We know,

Amount received on a certain sum of money of ₹ p invested at the rate of r % per annum compounded annually for n q/s years is given by

$\boxed{\tt{ \: \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n}\bigg[1 + \dfrac{r}{100} \times \frac{q}{s} \bigg] \: \: }}$

$\rm \: \: Amount \: = \: 10000 \: {\bigg[1 + \dfrac{10}{100} \bigg]}^{1}\bigg[1 + \dfrac{10}{100} \times \dfrac{1}{2} \bigg] \: \:$

$\rm \: \: Amount \: = \: 10000 \: {\bigg[1 + \dfrac{1}{10} \bigg]}\bigg[1 + \dfrac{1}{20}\bigg] \: \:$

$\rm \: \: Amount \: = \: 10000 \: {\bigg[\dfrac{10 + 1}{10} \bigg]}\bigg[\dfrac{20 + 1}{20}\bigg] \: \:$

$\rm \: \: Amount \: = \: 10000 \: {\bigg[\dfrac{11}{10} \bigg]}\bigg[\dfrac{21}{20}\bigg] \: \:$

$\rm \: Amount = 50 \times 11 \times 21$

$\rm\implies \:Amount = 11550$

So,

$\rm \: Compound \: interest \: = \: Amount \: - \: Principal$

$\rm \: Compound \: interest \: = \: 11550 \: - \: 10000$

$\rm \: Compound \: interest \: = \: 1550 \:$

So, from both the cases, we concluded that

Compound interest in first case is greater than compound interest in second case.

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ADDITIONAL INFORMATION

1. Amount received on a certain sum of money of ₹ p invested at the rate of r % per annum compounded yearly for n years is given by

$\boxed{\tt{ \: \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{100} \bigg]}^{n} \: \: }}$

2. Amount received on a certain sum of money of ₹ p invested at the rate of r % per annum compounded quarterly for n years is given by

$\boxed{\tt{ \: \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{400} \bigg]}^{4n} \: \: }}$

3. Amount received on a certain sum of money of ₹ p invested at the rate of r % per annum compounded monthly for n years is given by

$\boxed{\tt{ \: \: Amount \: = \: p \: {\bigg[1 + \dfrac{r}{1200} \bigg]}^{12n} \: \: }}$

Answer 2

Step By Step Explanation:

Interest when compounded half yearly

Principal = ₹10,000.

Time = 1 ½ years = 3/2 years.

Rate = 10% p.a.

Compounded - Half yearly.

Here, interest is compounded half yearly so we will multiply the time by 2 and divide the rate by 2.

Let's do :-

Time = 3/2 × 2 = 3years.

Rate = 10/2 = 5% p.a.

Now, by using formula of C.I, we have

C.I. = P(1+ R/100)^n - P

⇒C.I. = 10000(1+ 5/100)³ - P

⇒C.I. = 10000(1+ 1/20)³ - P

⇒C.I. = 10000(21/20)³ - P

⇒C.I. = 10000 × 21/20 × 21/20 × 21/20 - P

⇒C.I. = 10 × 9261/8 - 10000

⇒C.I. = 92610/8 - 10000

⇒C.I. = 11576.25 - 10000

⇒C.I. = ₹1,576.25.

A = P + C.I.

⇒A = 10000 + 1576.25

⇒A = ₹11,576.25.

Therefore,

• Amount = ₹11,576.25.
• Compound Interest = ₹1,576.25.

Interest when compounded annually

Principal = ₹10,000.

Time = 1 ½ years.

Rate = 10% p.a.

Compound - Annually.

We will first take out interest for 1year by C.I. then for next ½ year by S.I.

For 1year.

C.I. = P(1+ R/100)^n - P

⇒C.I. = 10000(1+ 10/100)¹ - P

⇒C.I. = 10000(1+ 1/10)¹ - P

⇒C.I. = 10000 × 11/10 - 10000

⇒C.I. = 1000 × 11 - 10000

⇒C.I. = 11000 - 10000

⇒C.I. = ₹1,000.

A = P + C.I.

⇒A = 10000 + 1000

⇒A = ₹11,000.

∴ Principal for next ½ year = ₹11000

Rate = 10/2 = p.a.

Time = ½

Principal = ₹10000.

S.I. = P × R × T/100

⇒S.I. = 11000 × 10 × ½/100

⇒S.I. = 11000 × 10 × 1/100 × 2

⇒S.I. = ₹550.

Therefore,

• Total C.I. = 1000 + 500 = ₹1,500.

Now, we have taken out the interest of both. We can see that the interest compounded half yearly is more by ₹76.25(1576.25-1500).

Used Abbreviations

A= Amount.

P = Principal.

R = Rate.

T = Time.

C.I. = Compound Interest.

S.I. = Simple Interest.

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