DAV Class 8 Maths Chapter 6 Worksheet 4
Compound Interest Worksheet 4
1. The population of a town is increasing at the rate of 8% per annum. What will be the population of the town after two years if the present population is 12,500?
Solution
\[ \begin{aligned} \text{Present population } (P) &= 12500 \\[6pt] R &= 8\% \text{ per annum} \\[6pt] n &= 2\ \text{years} \end{aligned} \]\[ \begin{aligned} \color{magenta}\textbf{Population after 2 years} &= \color{magenta} P\left(1 + \frac{R}{100}\right)^n \\[10pt] &= 12500\left(1 + \frac{8}{100}\right)^2 \\[10pt] &= 12500\left(\frac{108}{100}\right)^2 \\[10pt] &= 12500\left(\frac{27}{25}\right)^2 \\[10pt] &= 12500 \times \frac{27}{25} \times \frac{27}{25} \\[10pt] &= {\cancel{12500}^{\cancel{500}}}^{ \ 20} \times \frac{27}{\cancel{25}_1} \times \frac{27}{\cancel{25}_1} \\[6pt] &= 20 \times 27 \times 27 \\[6pt] &= 20 \times 729 \\[6pt] &= 14580 \end{aligned} \]Answer Population after 2 years \(= \color{red}{14{,}580}\)
2. Three years ago, the population of a town was 50,000. If the annual increase during three successive years was at the rate 4%, 5% and 4% per annum respectively, find the present population.
Solution
\[ \begin{aligned} \text{Population 3 years ago }(P) &= 50000 \\[6pt] a &= 4\% \\[4pt] b &= 5\% \\[4pt] c &= 4\% \end{aligned} \] \[ \begin{aligned} \color{magenta}\textbf{Population after 3 years} &= \color{magenta} P\left(1 + \frac{a}{100}\right) \left(1 + \frac{b}{100}\right)\left(1 + \frac{c}{100}\right) \\[10pt]\end{aligned} \]\[ \begin{aligned} &= 50000\left(1 + \frac{4}{100}\right) \left(1 + \frac{5}{100}\right)\left(1 + \frac{4}{100}\right) \\[10pt] &= 50000\left(1 + \frac{1}{25}\right) \left(1 + \frac{1}{20}\right)\left(1 + \frac{1}{25}\right) \\[10pt] &= 50000\left(\frac{25 + 1}{25}\right) \left(\frac{20 + 1}{20}\right)\left(\frac{25 + 1}{25}\right) \\[10pt] &= 50000 \times \frac{26}{25} \times \frac{21}{20} \times \frac{26}{25} \\[10pt] &= {{\cancel{50000}^{\cancel{2000}}}^{ \ \cancel{80}}}^{4} \times \frac{26}{\cancel{25}_1} \times \frac{21}{\cancel{20}_1} \times \frac{26}{\cancel{25}_1} \\[6pt] &= 4 \times 26 \times 21 \times 26 \\[6pt] &= 56784 \end{aligned} \]Answer Present population \(= \color{red}{56{,}784}\)
3. Madhu bought a house for ₹ 1,31,25,000. If its value depreciates at the rate of 10% per annum, what will be its sale price after three years?
Solution
\[ \begin{aligned} \text{Cost price }(P) &= \text{₹ }1{,}31{,}25{,}000 \\[6pt] \text{Depreciation rate }(R) &= 10\% \\[6pt] n &= 3\ \text{years} \end{aligned} \]\[ \begin{aligned} \color{magenta}\textbf{Value after 3 years} &= \color{magenta} P\left(1 - \frac{R}{100}\right)^n \\[6pt] &= 13125000 \left(1 - \frac{10}{100}\right)^3 \\[8pt] &= 13125000 \left(1 - \frac{1}{10}\right)^3 \\[8pt] &= 13125000 \left(\frac{10 - 1}{10}\right)^3 \\[8pt] &= 13125000 \left(\frac{9}{10}\right)^3 \\[8pt] &= 13125000 \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} \\[8pt] &= 13125 \times 729 \\[8pt] &= 9568125 \end{aligned} \]Answer Sale price after 3 years \(=\; \color{red}{\text{₹ }95{,}68{,}125}\)
4. The profits of a firm were ₹ 72,000 in the year 2014. During the next year, it increased by 7% and it decreased by 5% in the following year. What are the profits of the firm after two years?
Solution
\[ \begin{aligned} \text{Profit in 2014 }(P) &= \text{₹ }72000 \\[6pt] \text{Increase in 1st year }(a) &= 7\% \\[4pt] \text{Decrease in 2nd year }(b) &= 5\% \end{aligned} \]\[ \begin{aligned} \color{magenta}\textbf{Profit after 2 years} &= \color{magenta} P\left(1 + \frac{a}{100}\right)\left(1 - \frac{b}{100}\right) \\[8pt] &= 72000\left(1 + \frac{7}{100}\right)\left(1 - \frac{5}{100}\right) \\[8pt] &= 72000\left(\frac{107}{100}\right)\left(\frac{95}{100}\right) \\[8pt] &= 72000 \times \frac{107 \times 95}{100 \times 100} \\[8pt] &= \frac{\cancel{72}^{36} \times 107 \times \cancel{95}^{19}}{\cancel{10}_{\cancel 2_1}} \\[8pt] &= 36 \times 107 \times 19 \\[8pt] &= 73188 \end{aligned} \]Answer Profits after 2 years \(= \color{red}{\text{₹ }73{,}188}\)
5. The population of a town is 64,000. If the annual birth rate is 10.7% and the annual death rate is 3.2%, calculate the population after three years. (Hint: Net growth rate = (Birth rate − Death rate)%)
Solution
\[ \begin{aligned} \text{Present population }(P) &= 64000 \\[6pt] \text{Net growth rate} (R) &= 10.7\% - 3.2\% \implies 7.5\% \\[6pt] n &= 3\ \text{years} \end{aligned} \]\[ \begin{aligned} \color{magenta}\textbf{Population after 3 years} &= \color{magenta} P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 64000 \left(1 + \frac{7.5}{100}\right)^3 \\[8pt] &= 64000 \left( \frac{100 + 7.5}{100}\right)^3 \\[8pt] &= 64000 \left(\frac{107.5}{100}\right)^3 \\[8pt] &= 64000 \left(\frac{1075}{1000}\right)^3 \\[8pt] &= 64000 \left(\frac{43}{40}\right)^3 \\[8pt] &= 64000 \times \frac{43}{40} \times \frac{43}{40} \times \frac{43}{40} \\[8pt] &= 64 \times \frac{79507}{64} \\[6pt] &= 79507 \end{aligned} \]Answer Population after 3 years \(= \color{red}{79{,}507}\)
6. In a factory, the production of motor bikes was 40,000 in a particular year, which rose to 48,400 in two years. Find the rate of growth per annum, if it was uniform during two years.
Solution
\[ \begin{aligned} \text{Initial production }(P) &= 40000 \\[6pt] \text{Production after 2 years }(A) &= 48400 \\[6pt] n &= 2 \text{ years} \\[6pt] \color{magenta} P\left(1 + \frac{R}{100}\right)^n &= \color{magenta} A \\[8pt] 40000 \left(1 + \frac{R}{100}\right)^2 &= 48400 \\[8pt] \left(\frac{100 + R}{100}\right)^2 &= \frac{48400}{40000} \\[8pt] \left(\frac{100 + R}{100}\right)^2 &= \frac{484}{400} \\[8pt] \left(\frac{100 + R}{100}\right)^2 &= \frac{121}{100} \\[8pt] \frac{100 + R}{100} &= \sqrt{\frac{121}{100}} \\[8pt] \frac{100 + R}{100} &= \frac{11}{10} \\[8pt] 100 + R &= \frac{11}{10} \times 100 \\[8pt] 100 + R &= 110 \\[8pt] R &= 110 - 100 \\[8pt] R &= 10\% \end{aligned} \]Answer Rate of growth \(= \color{red}{10\%}\)