Question

find it compound interest and amount.
18,000 for 2½
years at 10% per annum compounded annually
2.​

Answer 1

Answer:

amount =22869 & interest =4869

Step-by-step explanation:

• firstly calculate compound interest of two years :-A=p(1+r/100)^n.
• A=18000{1+(10/100)}^2
• A=18000{1+(1/10)^2.
• A=18000(11/10)^2. (by L.C.M.).
• A=18000×121/100.
• A=180×121
• A=21780. (amount after two years)
• I = A-P
• I = 21780-18000
• I=3780 (compound interest of two years)
• now calculate simple interest of 1/2 year
• P=21780 ,R= 10% ,T=1/2years
• I = (p×r×t)/100
• I =(21780×10×1/2)/100
• (21780×10)/100×2
• 217800/200
• I =1089
• Interest for 2×1/2 years = 3780+1089 =4869
• Total Amount of 2×1/2years=21780+1089= 22869

Answer 2

$\Large {\underline { \sf \orange{Clarification :}}}$

Here, we are given that the principal of 18,000 for time 2½ years compounded annually at the rate of 10% . We have to find the compound interest and amount.

In order to calculate the compound interest, first we need to calculate the amount. So, there is a formula used to calculate amount when interest is compounded annually. We'll use it to calculate the amount. After that, by using the formula of compound interest i.e, difference of amount and principal, we'll find compound interest.

$\Large {\underline { \sf \orange{Explication \: of \: Steps :}}}$

Given :

• Principal (P) = 18,000

• Time (n years) = 2½ years

• Rate (R) = 10 % p.a

To calculate :

• Amount and C.I

Calculations :

Formula we have to use here,

$\bigstar \: \boxed{\sf { A ={ P \Bigg \lgroup 1+ \dfrac{R}{100} \Bigg \rgroup }^{2}\times \Bigg \lgroup 1 + \dfrac{\cfrac{1}{2} \times R}{100} \Bigg \rgroup }}$

$\longrightarrow \sf{ A = {18,000 \Bigg \lgroup 1+ \dfrac{10}{100} \Bigg \rgroup }^{2}\times \Bigg \lgroup 1 + \dfrac{\cfrac{1}{\cancel{2}} \times \cancel{10}}{100} \Bigg \rgroup}$

$\longrightarrow \sf{ A = {18,000 \Bigg \lgroup \dfrac{100 + 10}{100} \Bigg \rgroup }^{2}\times \Bigg \lgroup 1 + \dfrac{1\times 5}{100} \Bigg \rgroup}$

$\longrightarrow \sf{ A = {18,000 \Bigg \lgroup \dfrac{110}{100} \Bigg \rgroup }^{2}\times \Bigg \lgroup 1 + \dfrac{5}{100} \Bigg \rgroup}$

$\longrightarrow \sf{ A = {18,000 \Bigg \lgroup \dfrac{110}{100} \Bigg \rgroup }^{2}\times \Bigg \lgroup \dfrac{100 + 5}{100} \Bigg \rgroup}$

$\longrightarrow \sf{ A = {18,000 \Bigg \lgroup \dfrac{110}{100} \Bigg \rgroup }^{2}\times \Bigg \lgroup \dfrac{105}{100} \Bigg \rgroup}$

$\longrightarrow \sf{ A = 18,000 \times \dfrac{110}{100} \times \dfrac{110}{100} \times\dfrac{105}{100} }$

$\longrightarrow \sf{ A = 180 \times 110 \times \dfrac{110}{100} \times\dfrac{105}{100} }$

$\longrightarrow \sf{ A = 18 \times 110 \times \dfrac{110}{100} \times\dfrac{105}{10} }$

$\longrightarrow \sf{ A = 18 \times 110 \times \dfrac{11}{10} \times\dfrac{105}{10} }$

$\longrightarrow \sf{ A = 18 \times 11 \times 11 \times \cancel{\dfrac{105}{10}} }$

$\longrightarrow \sf{ A = \cancel{18} \times 11 \times 11 \times \dfrac{21}{\cancel{2}} }$

$\longrightarrow \sf{ A = 9 \times 11 \times 11 \times 21 }$

❝ $\longrightarrow \boxed{ \sf \orange { Amount = Rs. \: 22869 }}$ ❞

Now, we know that :

$\bigstar \: \boxed{\sf { C.I = Amount - Principal}}$

$\longrightarrow \sf{ C.I = Rs. \: ( 22869 - 18000) }$

❝ $\longrightarrow \boxed{ \sf \orange { C.I = Rs. \: 4869 }}$ ❞

❝ Therefore, compound interest is Rs. 4869 and compound interest is Rs. 22869.❞