DAV Class 8 Maths Chapter 6 Worksheet 1
Compound Interest Worksheet 1
1. Find the compound interest on ₹ 25,000 at the rate of 12% per annum for three years (compounded annually).
Solution
\[ \begin{aligned} & \color{magenta} \textbf{First Year} \\[6pt] P &= \text{₹ } 25000 \\ R &= 12 \% \\ T &= 1 \ year \\[8pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] &= \frac{25000 \times 12 \times 1}{100} \\[6pt] &= 250 \times 12 \\[6pt] \text{S.I.} &= \text{₹ }3000 \\[6pt] \text{Amount} &= P + SI \\[6pt] &= 25000 + 3000 \\[6pt] &= \text{₹ }28000 \\ \\ & \color{magenta} \textbf{Second Year} \\[6pt] P &= \text{₹ } 28000 \\ R &= 12 \% \\ T &= 1 \ year \\[8pt] \text{S.I.} &= \frac{28000 \times 12 \times 1}{100} \\[6pt] &= 280 \times 12 \\[6pt] \text{S.I.} &= \text{₹ } 3360 \\[6pt] \text{Amount} &= 28000 + 3360 \\[6pt] &= \text{₹ }31360 \\ \\ & \color{magenta} \textbf{Third Year} \\[6pt] P &= \text{₹ } 31360 \\ R &= 12 \% \\ T &= 1 \ year \\[8pt] \text{S.I.} &= \frac{31360 \times 12 \times 1}{100}\\[6pt] &= 313.60 \times 12 \\[6pt] \text{S.I.} &= \text{₹ }3763.20 \\[6pt] \text{Amount} &= 31360 + 3763.20 \\[6pt] &= \text{₹ }35123.20 \end{aligned} \]\[ \begin{aligned} & \textbf{Compound Interest for 3 yrs} \\ &= \text{Amount} - \text{Principal} \\ &= 35123.20 - \ 25000 \\ &= \text{₹ }10123.20 \end{aligned} \]\[ \begin{aligned} & \color{brown} \boxed{\textbf{Method - } 2} \\[6pt] P &= \text{₹ } 25000 \\ R &= 12 \% \\ n &= 3 \ years \\[8pt] \text{A} &= P \left( 1 + \frac{R}{100} \right)^n \\[8pt] &= 25000 \left( 1 + \frac{\cancel{12}^3}{\cancel{100}_{25}} \right)^3 \\[8pt] &= 25000 \left( 1 + \frac{3}{25} \right)^3 \\[8pt] &= 25000 \left( \frac{25 + 3}{25} \right)^3 \\[8pt] &= 25000 \left( \frac{28}{25} \right)^3 \\[8pt] &= {\cancel{25000}^{\cancel{1000}}}^{\ 40} \times \frac{28}{\cancel{25}_1} \times \frac{28}{\cancel{25}_1} \times \frac{28}{25} \\[8pt] &= \frac{\cancel{40}^8 \times 21952}{\cancel{25}_5} \\[8pt] &= \frac{175616}{5} \\[6pt] \text{A} &= \text{₹ }35123.20 \\[6pt] \text{C.I} &= A - P \\ &= 35123.20 - \ 25000 \\ &= \text{₹ }10123.20 \end{aligned} \]
Answer Compound Interest \(=\; \color{red}{\text{₹ }10{,}123.20}\)
2. Find the compound interest on ₹ 6,500 for two years at 9% per annum (compounded annually).
Solution
\[ \begin{aligned} & \color{magenta} \textbf{First Year} \\[6pt] P &= \text{₹ } 6500 \\ R &= 9 \% \\ T &= 1 \ year \\[8pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] &= \frac{6500 \times 9 \times 1}{100} \\[6pt] &= 65 \times 9 \\[6pt] \text{S.I.} &= \text{₹ }585 \\[6pt] \text{Amount} &= P + SI \\[6pt] &= 6500 + 585 \\[6pt] &= \text{₹ }7085 \\ \\ & \color{magenta} \textbf{Second Year} \\[6pt] P &= \text{₹ } 7085 \\ R &= 9 \% \\ T &= 1 \ year \\[8pt] \text{S.I.} &= \frac{7085 \times 9 \times 1}{100} \\[6pt] &= 70.85 \times 9 \\[6pt] \text{S.I.} &= \text{₹ }637.65 \\[6pt] \text{Amount} &= 7085 + 637.65 \\[6pt] &= \text{₹ }7722.65 \end{aligned} \]\[ \begin{aligned} & \textbf{Compound Interest for 2 yrs} \\ &= \text{Amount} - \text{Principal} \\ &= 7722.65 - \ 6500 \\ &= \text{₹ }1222.65 \end{aligned} \]\[ \begin{aligned} & \color{brown} \boxed{\textbf{Method - } 2} \\[6pt] P &= \text{₹ } 6500 \\ R &= 9 \% \\ n &= 2 \ years \\[8pt] \text{A} &= P \left( 1 + \frac{R}{100} \right)^n \\[8pt] &= 6500 \left( 1 + \frac{9}{100} \right)^2 \\[8pt] &= 6500 \left( \frac{100 + 9}{100} \right)^2 \\[8pt] &= 6500 \left( \frac{109}{100} \right)^2 \\[8pt] &= \cancel{6500}^{65} \times \frac{109}{\cancel{100}_1} \times \frac{109}{100} \\[8pt] &= \frac{65 \times 109 \times 109}{100} \\[8pt] &= \frac{772265}{100} \\[8pt] \text{A} &= \text{₹ }7722.65 \\[6pt] \text{C.I} &= A - P \\ &= 7722.65 - \ 6500 \\ &= \text{₹ }1222.65 \end{aligned} \]
Answer Compound Interest \(= \color{red}{\text{₹ }1{,}222.65}\)
3. Find the amount and compound interest on a sum of ₹ 8,000 at 5% per annum for three years (compounded annually).
Solution
\[ \begin{aligned} & \color{magenta} \textbf{First Year} \\[6pt] P &= \text{₹ } 8000 \\ R &= 5 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] &= \frac{8000 \times 5 \times 1}{100} \\[6pt] &= 80 \times 5 \\[6pt] \text{S.I.} &= \text{₹ }400 \\[6pt] \text{Amount} &= P + \text{S.I.} \\[6pt] &= 8000 + 400 \\[6pt] &= \text{₹ }8400 \\ \\ & \color{magenta} \textbf{Second Year} \\[6pt] P &= \text{₹ } 8400 \\ R &= 5 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{8400 \times 5 \times 1}{100} \\[6pt] &= 84 \times 5 \\[6pt] \text{S.I.} &= \text{₹ }420 \\[6pt] \text{Amount} &= 8400 + 420 \\[6pt] &= \text{₹ }8820 \\ \\ & \color{magenta} \textbf{Third Year} \\[6pt] P &= \text{₹ } 8820 \\ R &= 5 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{8820 \times 5 \times 1}{100} \\[6pt] &= 88.20 \times 5 \\[6pt] \text{S.I.} &= \text{₹ }441 \\[6pt] \text{Amount} &= 8820 + 441 \\[6pt] &= \text{₹ }9261 \end{aligned} \]\[ \begin{aligned} &\textbf{Compound Interest for 3 yrs} \\ &= \text{Amount} - \text{Principal} \\ &= 9261 - 8000 \\ &= \text{₹ }1261 \end{aligned} \]\[ \begin{aligned} & \color{brown} \boxed{\textbf{Method - } 2} \\[6pt] P &= \text{₹ } 8000 \\ R &= 5 \% \\ n &= 3 \ years \\[8pt] \text{A} &= P \left( 1 + \frac{R}{100} \right)^n \\[8pt] &= 8000 \left( 1 + \frac{\cancel{5}^1}{\cancel{100}_{20}} \right)^3 \\[8pt] &= 8000 \left( 1 + \frac{1}{20} \right)^3 \\[8pt] &= 8000 \left( \frac{20 + 1}{20} \right)^3 \\[8pt] &= 8000 \left( \frac{21}{20} \right)^3 \\[8pt] &= \cancel{8000}^1 \times \frac{9261}{\cancel{8000}_1} \\[8pt] \text{A} &= \text{₹ }9261 \\[6pt] \text{C.I} &= A - P \\ &= 9261 - \ 8000 \\ &= \text{₹ }1261 \end{aligned} \]
Answer Amount \(= \color{red}{\text{₹ }9{,}261}\) , Compound Interest \(=\; \color{red}{\text{₹ }1{,}261}\)
4. Harvinder deposited ₹ 40,000 in a post office for a period of three years. The post office credits the interest yearly in his account at 7% p.a., compounded annually. Find the balance in his account after three years.
Solution
\[ \begin{aligned} & \color{magenta} \textbf{First Year} \\[6pt] P &= \text{₹ } 40000 \\ R &= 7 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] &= \frac{40000 \times 7 \times 1}{100} \\[6pt] &= 400 \times 7 \\[6pt] \text{S.I.} &= \text{₹ }2800 \\[6pt] \text{Amount} &= P + \text{S.I.} \\[6pt] &= 40000 + 2800 \\[6pt] &= \text{₹ }42800 \\ \\ & \color{magenta} \textbf{Second Year} \\[6pt] P &= \text{₹ } 42800 \\ R &= 7 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{42800 \times 7 \times 1}{100} \\[6pt] &= 428 \times 7 \\[6pt] \text{S.I.} &= \text{₹ }2996 \\[6pt] \text{Amount} &= 42800 + 2996 \\[6pt] &= \text{₹ }45796 \\ \\ & \color{magenta} \textbf{Third Year} \\[6pt] P &= \text{₹ } 45796 \\ R &= 7 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{45796 \times 7 \times 1}{100} \\[6pt] &= 457.96 \times 7 \\[6pt] \text{S.I.} &= \text{₹ }3205.72 \\[6pt] \text{Amount} &= 45796 + 3205.72 \\[6pt] &= \text{₹ }49001.72 \end{aligned} \] \[ \begin{aligned} \text{Balance after 3 years} &= \text{₹ }49001.72 \end{aligned} \]\[ \begin{aligned} & \color{brown} \boxed{\textbf{Method - } 2} \\[6pt] P &= \text{₹ } 40000 \\ R &= 7 \% \\ n &= 3 \ years \\[8pt] \text{A} &= P \left( 1 + \frac{R}{100} \right)^n \\[8pt] &= 40000 \left( 1 + \frac{7}{100} \right)^3 \\[8pt] &= 40000 \left( \frac{100 + 7}{100} \right)^3 \\[8pt] &= 40000 \left( \frac{107}{100} \right)^3 \\[8pt] &= \cancel{40000}^4 \times \frac{107 \times 107 \times 107}{\cancel{100} \times \cancel{100} \times 100} \\[8pt] &= \frac{4 \times 107 \times 107 \times 107}{100} \\[8pt] &= \frac{4900172}{100} \\[8pt] \text{A} &= \text{₹ }49001.72 \end{aligned} \]
Answer Balance after 3 years \(=\; \color{red}{\text{₹ }49{,}001.72}\)
5. Monika borrowed ₹ 4,096 from Shalini for three years at \(6\dfrac{1}{4}\%\) per annum. Find the amount and the compound interest after three years if the interest is compounded annually.
Solution
\[ \begin{aligned} & \color{magenta} \textbf{First Year} \\[6pt] P &= \text{₹ } 4096 \\[6pt] R &= 6\frac{1}{4} \implies \frac{25}{4} \% \\[6pt] T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{\cancel{4096}^{1024} \times \cancel{25}^1 \times 1}{ \cancel4_1 \times \cancel{100}_4} \\[6pt] &= \frac{1024}{4} \\[6pt] \text{S.I.} &= \text{₹ }256 \\[6pt] \text{Amount} &= 4096 + 256 \\[6pt] &= \text{₹ }4352 \\ \\ & \color{magenta} \textbf{Second Year} \\[6pt] P &= \text{₹ } 4352 \\[6pt] R &= 6\frac{1}{4} \implies \frac{25}{4} \% \\[6pt] T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{\cancel{4352}^{1088} \times \cancel{25}^1 \times 1}{\cancel4_1 \times \cancel{100}_4} \\[6pt] &= \frac{1088}{4} \\[6pt] \text{S.I.} &= \text{₹ }272 \\[6pt] \text{Amount} &= 4352 + 272 \\[6pt] &= \text{₹ }4624 \\ \\ & \color{magenta} \textbf{Third Year} \\[6pt] P &= \text{₹ } 4624 \\[6pt] R &= 6\frac{1}{4} \implies \frac{25}{4} \% \\[6pt] T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{\cancel{4624}^{1156} \times \cancel{25}^1 \times 1}{\cancel4_1 \times \cancel{100}_4} \\[6pt] &= \frac{1156}{4} \\[6pt] \text{S.I.} &= \text{₹ }289 \\[6pt] \text{Amount} &= 4624 + 289 \\ &= \text{₹ }4913 \end{aligned} \] \[ \begin{aligned} & \text{Amount paid after 3 years} \implies \text{₹ }4913 \\[6pt] &\textbf{Compound Interest} \\ &= \text{Amount} - \text{Principal} \\ &= 4913 - 4096 \\ &= \text{₹ }817 \end{aligned} \]\[ \begin{aligned} & \color{brown} \boxed{\textbf{Method - } 2} \\[6pt] P &= \text{₹ } 4096 \\ R &= 6\frac{1}{4} \implies \frac{25}{4} \% \\[6pt] n &= 3 \ years \\[8pt] \text{A} &= P \left( 1 + \frac{R}{100} \right)^n \\[8pt] &= 4096 \left( 1 + \frac{\cancel{25}^1}{4 \times \cancel{100}_4} \right)^3 \\[8pt] &= 4096 \left( 1 + \frac{1}{16} \right)^3 \\[8pt] &= 4096 \left( \frac{16 + 1}{16} \right)^3 \\[8pt] &= 4096 \left( \frac{17}{16} \right)^3 \\[8pt] &= \cancel{4096} \times \frac{4913}{\cancel{4096}} \\[8pt] \text{A} &= \text{₹ }4913 \\[6pt] \text{C.I} &= A - P \\ &= 4913 - \ 4096 \\ &= \text{₹ }817 \end{aligned} \]
Answer Amount \(=\; \color{red}{\text{₹ }4{,}913}\) Compound Interest \(=\; \color{red}{\text{₹ }817}\)
6. Ravi purchased a house from DDA on credit. If the cost of the house is ₹ 7,50,000 and DDA charges interest at 6% per annum compounded annually, find the interest paid by Ravi if he makes payment to DDA after three years.
Solution
\[ \begin{aligned} & \color{magenta} \textbf{First Year} \\[6pt] P &= \text{₹ } 750000 \\ R &= 6 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{750000 \times 6 \times 1}{100} \\[6pt] &= 7500 \times 6 \\[6pt] \text{S.I.} &= \text{₹ }45000 \\[6pt] \text{Amount} &= 750000 + 45000 \\[6pt] &= \text{₹ }795000 \\ \\ & \color{magenta} \textbf{Second Year} \\[6pt] P &= \text{₹ } 795000 \\ R &= 6 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{795000 \times 6 \times 1}{100} \\[6pt] &= 7950 \times 6 \\[6pt] \text{S.I.} &= \text{₹ }47700 \\[6pt] \text{Amount} &= 795000 + 47700 \\[6pt] &= \text{₹ }842700 \\ \\ & \color{magenta} \textbf{Third Year} \\[6pt] P &= \text{₹ } 842700 \\ R &= 6 \% \\ T &= 1 \ \text{year} \\[8pt] \text{S.I.} &= \frac{842700 \times 6 \times 1}{100} \\[6pt] &= 8427 \times 6 \\[6pt] \text{S.I.} &= \text{₹ }50562 \\[6pt] \text{Amount} &= 842700 + 50562 \\[6pt] &= \text{₹ }893262 \\ \\ \end{aligned} \] \[ \begin{aligned} & \color{magenta} \textbf{C.I. paid by Ravi after 3 years} \\ &= \text{Amount} - \text{Principal} \\ &= 893262 - 750000 \\ &= \text{₹ }1{,}43{,}262 \end{aligned} \]\[ \begin{aligned} & \color{brown} \boxed{\textbf{Method - } 2} \\[6pt] P &= \text{₹ } 750000 \\ R &= 6 \% \\ n &= 3 \ years \\[8pt] \text{A} &= P \left( 1 + \frac{R}{100} \right)^n \\[8pt] &= 750000 \left( 1 + \frac{\cancel6^3}{\cancel{100}_{50}} \right)^3 \\[8pt] &= 750000 \left( \frac{50 + 3}{50} \right)^3 \\[8pt] &= 750000 \left( \frac{53}{50} \right)^3 \\[8pt] &= \cancel{750000}^6 \times \frac{148877}{\cancel{125000}} \\[8pt] &= 6 \times 148877 \\[8pt] \text{A} &= \text{₹ }893262 \\[6pt] \text{C.I} &= A - P \\ &= 893262 - \ 750000 \\ &= \text{₹ }1{,}43{,}262 \end{aligned} \]
Answer Compound Interest paid by Ravi after 3 years \(=\; \color{red}{\text{₹ }1{,}43{,}262}\)