Question

Find the amount on ₹ 12,000 for two years,if the interest is compounded annually and the rate of interest is 10% of the first year and 5% for the second year.​

Answer 1

Given

• Principal = ₹ 12,000
• Time = 2 years
• Rate for the first year = 10 %
• Rate for the second year = 5 %

To find

• Amount

Knowledge required :-

$\boldsymbol{Amount = P \bigg[1 + \dfrac{r}{100} \bigg]^{n}}$

where,

• P = Principal
• r = rate
• n = time

Solution

$\\ : \implies\boldsymbol{Amount = P \bigg[1 + \dfrac{r}{100} \bigg]^{n}}$

$\\ : \implies\boldsymbol{Amount = 12000 \bigg[1 + \dfrac{10}{100} \bigg]\bigg[1 + \dfrac{5}{100} \bigg]}$

$\\ : \implies\boldsymbol{Amount = 12000 \bigg[1 + \dfrac{1 \not0}{10 \not0} \bigg]\bigg[1 + \dfrac{ \not5}{ \cancel{100}} \bigg]}$

$\\ : \implies\boldsymbol{Amount = 12000 \bigg[1 + \dfrac{1}{10} \bigg]\bigg[1 + \dfrac{1}{20} \bigg]}$

$\\ : \implies\boldsymbol{Amount = 12000 \bigg[ \dfrac{10 + 1}{10} \bigg]\bigg[ \dfrac{20 + 1}{20} \bigg]}$

$\\ : \implies\boldsymbol{Amount = 12000 \bigg[ \dfrac{11}{10} \bigg]\bigg[ \dfrac{21}{20} \bigg]}$

$\\ : \implies\boldsymbol{Amount = 12000 \times \dfrac{11}{10} \times \dfrac{21}{20}}$

$\\ : \implies\boldsymbol{Amount = 120 \not0 \not0 \times \dfrac{11}{1 \not0} \times \dfrac{21}{2 \not0}}$

$\\ : \implies\boldsymbol{Amount = 120 \times 11\times \dfrac{21}{2}}$

$\\ : \implies\boldsymbol{Amount = \not120 \times 11\times \dfrac{21}{ \not2}}$

$\\ : \implies\boldsymbol{Amount = 60 \times 11\times 21}$

$\\ : \implies\boldsymbol{Amount = 13,860}$

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Some related formulae :-

$\bullet \: \: \boldsymbol{Simple~Interest = \dfrac{P \times R \times T}{100}}$

$\bullet \: \: \boldsymbol{C.I. = P \bigg[1 + \dfrac{r}{100} \bigg]^{n} - P}$

$\bullet\: \: \boldsymbol{Compound~Interest = Amount - Principal}$

$\bullet\: \: \boldsymbol{Amount = Compound~Interest + Principal}$

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: answer:-}}}}}$

★GIVEN:-

• Principle - Rs. 12,000

• Time - 2 Years

• Rate - 10% and 5%

★TO FIND:-

• AMOUNT

★FORMULA USED:-

${ \boxed{ \underline {\bf{A = P {(1+ \frac{R }{100} )}^{T}}}}}$

★SOLUTION:-

$:\implies \sf{12,000 (1 + \frac{10}{100} )(1+ \frac{5}{100} )}$

$:\implies \sf{12,000( \frac{100 + 10}{100}) ( \frac{100 + 5}{100} })$

$:\implies \sf{12,000( \frac{110}{100}) ( \frac{105}{100} })$

$:\implies \sf{12,0 \cancel{00}( \frac{11}{1 \cancel{0}}) ( \frac{21}{2 \cancel{0}} })$

$: \implies \sf{ \cancel{ 120} (11)( \frac{21}{ \cancel{2}} })$

$: \implies \sf{ \: 60 \times 11 \times 21}$

$\: \: \: \: \: \: \: \: \: \: { \boxed{ \underline{ \sf{ \huge{ \purple{rs. \: 13680}}}}}}$

So, the Amount is Rs. 13,680

★MORE TO KNOW:-

• A = P (1+ r/n)^nt

• C.I. = P(1 + R/100)^n -P

• I = Prt

• C = P [(1+r)^n -1]

• S.I. = P × R × T /100

• Amount = Principal + Compound Interest.

★HERE,

C -- Compound Interest.

P -- Principal (original balance)

r -- rate per period

n -- number of periods

P -- principal sum

T -- time

A -- Future value