DAV Class 8 Maths Chapter 6 Brain Teasers
Compound Interest Brain Teasers
1. A. Tick (✓) the correct option.
(a)
For a given rate of interest and time, what is more profitable to the depositor?
\(
\begin{aligned}
(i)\ &\ \text{compound interest} \\[5pt]
(ii)\ &\ \text{simple interest} \\[5pt]
(iii)\ &\ \text{both are equally profitable} \\[5pt]
(iv)\ &\ \text{cannot be determined}
\end{aligned}
\)
Answer \( {\color{orange}(i)}\ \color{red}{\text{compound interest}} \)
(b)
A sum of ₹ 10,000 at 8% per annum for six months compounded quarterly amounts to—
\(
\begin{aligned}
(i)\ &\ \text{₹}\,400 \\[5pt]
(ii)\ &\ \text{₹}\,404 \\[5pt]
(iii)\ &\ \text{₹}\,408 \\[5pt]
(iv)\ &\ \text{₹}\,10{,}404
\end{aligned}
\)
Answer \( {\color{orange}(iv)}\ \color{red}{\text{₹}\,10{,}404} \)
(c)
In \(A = P\left(1 + \dfrac{R}{100}\right)^n\), A stands for—
\(
\begin{aligned}
(i)\ &\ \text{time period} \\[5pt]
(ii)\ &\ \text{principal} \\[5pt]
(iii)\ &\ \text{principal + interest} \\[5pt]
(iv)\ &\ \text{interest}
\end{aligned}
\)
Answer \( {\color{orange}(iii)}\ \color{red}{\text{principal + interest}} \)
(d)
If the number of conversion periods is greater than or equal to 2, then compound interest is—
\(
\begin{aligned}
(i)\ &\ \text{less than simple interest} \\[5pt]
(ii)\ &\ \text{greater than simple interest} \\[5pt]
(iii)\ &\ \text{less than or equal to simple interest} \\[5pt]
(iv)\ &\ \text{greater than or equal to simple interest}
\end{aligned}
\)
Answer \( {\color{orange}(ii)}\ \color{red}{\text{greater than simple interest}} \)
(e)
Reema wants to do a one-year deposit. She should opt for—
\(
\begin{aligned}
(i)\ &\ \text{a simple interest of 10%} \\[5pt]
(ii)\ &\ \text{a compound interest of 10% compounded quarterly} \\[5pt]
(iii)\ &\ \text{a compound interest of 10% compounded annually} \\[5pt]
(iv)\ &\ \text{a compound interest of 10% compounded half-yearly}
\end{aligned}
\)
Answer \( {\color{orange}(ii)}\ \color{red}{\text{a compound interest of 10% compounded quarterly}} \)
B. Answer the following questions.
(a) Find the compound interest on ₹ 1,000 at 10% per annum for two years.
Solution
\[ \begin{aligned} P &= \text{₹}\,1000 \\ R &= 10\% \\ n &= 2 \text{ years} \\[6pt] \color{green}A &= \color{green}P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 1000\left(1 + \frac{10}{100}\right)^2 \\[6pt] &= 1000\left(1 + \frac{1}{10}\right)^2 \\[6pt] &= 1000\left(\frac{11}{10}\right)^2 \\[6pt] &= 1000 \times \frac{11}{10} \times \frac{11}{10} \\[6pt] &= 10 \times 121 \\[6pt] A &= \text{₹}\,1210 \\[6pt] \color{green}\text{C.I.} &= \color{green}A - P \\[4pt] &= 1210 - 1000 \\[4pt] &= \text{₹}\,210 \end{aligned} \]Answer Compound Interest \(=\ \color{red}{\text{₹}\,210}\)
(b) If ₹ 6,000 is deposited for two years at 4% per annum compounded quarterly, then find the time period and rate to compute compound interest.
Solution
\[ \begin{aligned} P &= \text{₹}\,6000 \\ R &= 4\% \text{ per annum} \\ n &= 2 \text{ years} \\[6pt] \text{Compounded quarterly} &\implies 4 \text{ times per year} \\[6pt] \color{green}\text{Time period for CI} &= \color{green}2 \times 4 \implies 8 \text{ quarters} \\[6pt] \color{green}\text{Rate per quarter} &= \color{green}\frac{4\%}{4} \implies 1\% \\[6pt] \end{aligned} \]Answer Time period \(= \color{red}{8 \text{ quarters}}\), Rate \(= \color{red}{1\% \text{ per quarter}}\)
(c) The value of a machine worth ₹ 5,00,000 depreciates at the rate of 10% every year. In how many years will its value be ₹ 3,64,500?
Solution
\[ \begin{aligned} \text{Initial value }(P) &= \text{₹}\,5{,}00{,}000 \\ \text{Depreciation rate }(R) &= 10\% \\ \text{Final value }(A) &= \text{₹}\,3{,}64{,}500 \\ n &= n \text{ years} \\[6pt] \color{green}\text{Depreciation formula: } A &= \color{green}P\left(1 - \frac{R}{100}\right)^n \\[6pt] 3{,}64{,}500 &= 5{,}00{,}000\left(1 - \frac{10}{100}\right)^n \\[6pt] 3{,}64{,}500 &= 5{,}00{,}000\left(1 - \frac{1}{10}\right)^n \\[6pt] \frac{364500}{500000} &= \left(\frac{9}{10}\right)^n \\[6pt] \frac{\cancel{3645}^{\ 729}}{\cancel{5000}_{1000}} &= \left(\frac{9}{10}\right)^n \\[6pt] \frac{729}{1000} &= \left(\frac{9}{10}\right)^n \\[6pt] \left(\frac{9}{10}\right)^3 &= \left(\frac{9}{10}\right)^n \\[6pt] \therefore\ n &= 3 \text{ years} \end{aligned} \]Answer The value will be ₹ 3,64,500 after \(\color{red}{3 \text{ years}}\)
(d) Find the difference between the compound interest and simple interest on ₹ 5,000 for two years at 5% per annum.
Solution
\[ \begin{aligned} P &= \text{₹}\,5000 \\ R &= 5\% \\ n &= 2 \text{ years} \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Compound Interest} \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 5000\left(1 + \frac{5}{100}\right)^2 \\[6pt] &= 5000\left(1 + \frac{1}{20}\right)^2 \\[6pt] &= 5000\left(\frac{21}{20}\right)^2 \\[6pt] &= 5000 \times \frac{21}{20} \times \frac{21}{20} \\[6pt] &= \frac{50 \times 21 \times 21}{4} \\[6pt] &= \frac{22050}{4} \\[6pt] A &= \text{₹}\,5512.50 \\[6pt] \text{C.I.} &= A - P \\ & = 5512.50 - 5000 \\ &= \text{₹}\,512.50 \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Simple Interest} \\[6pt] \text{S.I.} &= \frac{P \times R \times T}{100} \\[6pt] &= \frac{5000 \times 5 \times 2}{100} \\[6pt] &= 50 \times 10 \\[6pt] &= \text{₹}\,500 \end{aligned} \] \[ \begin{aligned} \text{Difference} &= \text{C.I.} - \text{S.I.} \\[4pt] &= 512.50 - 500 \\[4pt] &= \text{₹}\,12.50 \end{aligned} \]Answer Difference between C.I. and S.I. \(= \color{red}{\text{₹}\,12.50}\)
(e) If ₹ 20,000 is deposited for three years at 5% compounded annually, then what will be the principal for the second year?
Solution
\[ \begin{aligned} P &= \text{₹}\,20{,}000 \\ R &= 5\% \\ T &= 1 \ year \\[6pt] \color{green}\text{S.I} &= \color{green}\frac{P \times R \times T}{100} \\[6pt] &= \frac{20000 \times 5 \times 1}{100} \\[6pt] &= 200 \times 5\\[6pt] \text{S.I}&= 1000 \\[6pt] A &= P + SI \\[6pt] &= 20000 + 1000 \\[6pt] A&= \text{₹}\,21{,}000 \\[6pt] & \text{Amount will be principal for the second year} \end{aligned} \]Answer Principal for the second year \(= \color{red}{\text{₹}\,21{,}000}\)
2.
Preeti invested ₹ 50,000 at 8% per annum for three years and the interest is compounded annually.
Calculate:
(a) the amount standing to her credit at the end of the second year.
(b) the interest for the third year.
Solution
\[ \begin{aligned} P &= \text{₹}\,50{,}000 \\ R &= 8\% \\ n &= 2 \text{ years} \end{aligned} \] \[ \begin{aligned} & \color{magenta} \textbf{(a) Amount at the end of 2nd year} \\[6pt] P &= \text{₹}\,50{,}000 \\ R &= 8\% \\ n &= 2 \text{ years} \\ \color{green}A &= \color{green}P\left(1 + \frac{R}{100}\right)^2 \\[6pt] &= 50000\left(1 + \frac{8}{100}\right)^2 \\[6pt] &= 50000\left(1 + \frac{2}{25}\right)^2 \\[6pt] &= 50000\left(\frac{27}{25}\right)^2 \\[6pt] &= {\cancel{50000}^{\cancel{\ 2000}}}^{80} \times \frac{27}{\cancel{25}_1} \times \frac{27}{\cancel{25}_1} \\[6pt] &= 80 \times 729 \\[6pt] A &= \text{₹}\,58{,}320 \ \color{green} \textbf{ (at the end of 2nd year)} \\ \end{aligned} \] \[ \begin{aligned} & \color{magenta} \textbf{(b) Interest for the 3rd year} \\[6pt] P &= \text{₹}\,58{,}320 \\ R &= 8\% \\ n &= 1 \text{ years} \\[6pt] \color{green} \text{S.I.} &= \color{green} \frac{P \times R \times T}{100} \\[6pt] &= \frac{58320 \times 8 \times 1}{100} \\[6pt] &= 583.2 \times 8 \\[6pt] &= \text{₹}\,4{,}665.60 \ \color{green} \textbf{ (Interest for the 3rd year)} \end{aligned} \]Answer (a) Amount at end of 2nd year \(=
\color{red}{\text{₹}\,58{,}320}\)
(b) Interest for 3rd year \(= \color{red}{\text{₹}\,4{,}665.60}\)
3. A man had ₹ 75,000. He invested ₹ 35,000 in a company which pays him 9% interest per annum and he invested rest of the money in another company which pays him 9.5% interest per annum. Find the total compound interest received by him after two years.
Solution
\[ \begin{aligned} \text{Total money} &= \text{₹}\,75{,}000 \\ n &= 2 \text{ years} \\[6pt] \text{First investment }(P_1) &= \text{₹}\,35{,}000 \\ R_1 &= 9\% \\[6pt] \text{Second investment }(P_2) &= 75{,}000 - 35{,}000 \\ & \implies \text{₹}\,40{,}000 \\ R_2 &= 9.5\% \\ \end{aligned} \]\[ \begin{aligned} & \color{magenta} \textbf{Compound Interest from first investment} \\[6pt] A_1 &= P_1\left(1 + \frac{R_1}{100}\right)^2 \\[6pt] &= 35000\left(1 + \frac{9}{100}\right)^2 \\[6pt] &= 35000\left(\frac{109}{100}\right)^2 \\[6pt] &= 35000 \times \frac{109}{100} \times \frac{109}{100} \\[6pt] &= \frac{35 \times 109 \times 109}{10} \\[6pt] &= \frac{415835}{10} \\[6pt] A_1 &= \text{₹}\,41{,}583.50 \\[6pt] \text{C.I.}_1 &= 41{,}583.50 - 35{,}000 \\ &= \text{₹}\,6{,}583.50 \end{aligned} \]\[ \begin{aligned} & \color{magenta} \textbf{Compound Interest from second investment} \\[6pt] A_2 &= P_2\left(1 + \frac{R_2}{100}\right)^2 \\[6pt] &= 40000\left(1 + \frac{9.5}{100}\right)^2 \\[6pt] &= 40000\left(1 + \frac{95}{1000}\right)^2 \\[6pt] &= 40000\left(1 + \frac{19}{200}\right)^2 \\[6pt] &= 40000\left(\frac{219}{200}\right)^2 \\[6pt] &= 40000 \times \frac{219}{200} \times \frac{219}{200} \\[6pt] &= 219 \times 219 \\[6pt] A_2 &= \text{₹}\,47961 \\[6pt] \text{C.I.}_2 &= 47961 - 40000 \\ &= \text{₹}\,7{,}961 \\[8pt] \color{green}\text{Total C.I.} &= \color{green}\text{C.I.}_1 + \text{C.I.}_2 \\[6pt] &= 6583.50 + 7961 \\[6pt] \color{green}\text{Total C.I.}&= \text{₹}\,14544.50 \end{aligned} \]Answer Total compound interest \(= \color{red}{\text{₹}\,14{,}544.50}\)
4. A certain sum of money at compound interest becomes ₹ 7,396 in two years and ₹ 7,950.70 in three years. Find the rate of interest.
Solution
\[ \begin{aligned} \text{Amount after 2 years} &= \text{₹}\,7396 \\[6pt] \text{Principal for 3rd year} &= \text{₹}\,7396 \\[6pt] \text{Amount after 3 years} &= \text{₹}\,7950.70 \\[6pt] S.I \text{ for 3rd year} &= A - P \\[6pt] &= 7950.70 - 7396 \\[6pt] S.I &= \text{₹}\,554.70 \\[6pt] \frac{P \times R \times T}{100} & = 554.70 \\[6pt] \frac{7396 \times R \times 1}{100} & = 554.70 \\[6pt] R &=\frac{554.70 \times 100}{7396}\\[6pt] R &= \frac{55470}{7396} \\[6pt] R &= 7.5\% \end{aligned} \]Answer Rate of interest \(= \color{red}{7.5\%}\)
5. Pooja started a business by investing ₹ 2,00,000. During the first three successive years, she earned a profit of 5%, 8% and 12% per annum respectively. If in each year, the profit was added on the capital at the end of the previous year, calculate her total profit after three years.
Solution
\[ \begin{aligned} \text{Initial capital }(P) &= \text{₹ }2{,}00{,}000 \\[6pt] a &= 5\% \\[4pt] b &= 8\% \\[4pt] c &= 12\% \end{aligned} \]\[ \begin{aligned} \color{magenta}\textbf{Capital after 3 years} &= \color{magenta} P\left(1 + \frac{a}{100}\right)\left(1 + \frac{b}{100}\right)\left(1 + \frac{c}{100}\right) \\[10pt] &= 200000\left(1 + \frac{5}{100}\right)\left(1 + \frac{8}{100}\right)\left(1 + \frac{12}{100}\right) \\[10pt] &= 200000\left(1 + \frac{1}{20}\right)\left(1 + \frac{2}{25}\right)\left(1 + \frac{3}{25}\right) \\[10pt] &= 200000\left(\frac{20+1}{20}\right)\left(\frac{25+2}{25}\right)\left(\frac{25+3}{25}\right) \\[10pt] &= \frac{200000 \times 21 \times 27 \times 28}{20 \times 25 \times 25} \\[10pt] &= 16 \times 21 \times 27 \times 28 \\[8pt] &= 16 \times 15876 \\[6pt] &= 254016 \end{aligned} \]\[ \begin{aligned} \color{green}\text{Total profit} &= \color{green}\text{A} - \text{P} \\[6pt] &= 254016 - 200000 \\[6pt] &= \text{₹ }54016 \end{aligned} \]Answer Total profit after 3 years \(= \color{red}{\text{₹ }54{,}016}\)
6. Mahesh borrowed a certain sum for two years at simple interest from Bhim. Mahesh lent this sum to Vishnu at the same rate for two years compound interest. At the end of two years, Mahesh received ₹ 410 as compound interest but paid ₹ 400 as simple interest. Find the sum and rate of interest.
Solution
\[ \begin{aligned} \text{Let the principal }(P) &= P \\[4pt] \text{Rate of interest }(R) &= R\% \\[4pt] n &= 2\ \text{years} \\[6pt] \text{S.I. paid by Mahesh} &= \text{₹ }400 \\[4pt] \text{C.I. received by Mahesh} &= \text{₹ }410 \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{Simple Interest (paid by Mahesh)} \\[6pt] \end{aligned} \] \[ \begin{aligned} \text{S.I.} &= \text{₹ }400 \\[6pt] \frac{P \times R \times T}{100} &= 400 \\[6pt] \frac{P \times R \times 2}{100} & = 400 \\[6pt] \frac{PR}{50} & = 400 \\[6pt] PR &= 400 \times 50 \\[6pt] PR &= 20000 \qquad \implies \text{(i)} \end{aligned} \]\[ \begin{aligned} &\color{magenta}\textbf{Compound Interest (received by Mahesh)} \\[6pt] \end{aligned} \] \[ \begin{aligned} \text{C.I.} &= \text{₹ }410 \\[6pt] P\left[\left(1 + \frac{R}{100}\right)^2 - 1\right] & = 410\\[6pt] P\left[\left(\frac{100 + R}{100}\right)^2 - 1\right] & = 410\\[6pt] P\left[\frac{(100 + R)^2 }{10000} - 1 \right ] & = 410\\[6pt] P\left[\frac{(100 + R)^2 - 10000}{10000}\right] & = 410\\[6pt] P\left[\frac{10000 + 200R + R^2 - 10000}{10000}\right] & = 410\\[6pt] P\left[\frac{200R + R^2}{10000}\right] & = 410\\[6pt] P\left[\frac{R(200 + R)}{10000}\right] & = 410\\[6pt] \frac{PR(200 + R)}{10000} & = 410 \\[6pt] PR(200 + R) & = 410 \times 10000 \\[6pt] PR(200 + R) & = 4100000 \qquad \implies \text{(ii)} \end{aligned} \]Substituting \(PR = 20000\) from (i) in (ii):
\[ \begin{aligned} PR(200 + R) & = 4100000 \\[6pt] 20000(200 + R) & = 4100000 \\[6pt] 200 + R &= \frac{4100000}{20000} \\[6pt] 200 + R &= 205 \\[6pt] R &= 205 - 200 \\[4pt] \color{green} R &= \color{green} 5\% \end{aligned} \]\[ \begin{aligned} \therefore PR &= 20000 \\[6pt] P &= \frac{20000}{R} \\[6pt] P &= \frac{20000}{5} \\[6pt] P &= \text{₹ }4000 \end{aligned} \]Answer Sum \(=\; \color{red}{\text{₹ }4{,}000}\), Rate \(=\; \color{red}{5\% \text{ per annum}}\)
7. A man invested ₹ 1,000 for three years at 11% simple interest per annum and ₹ 1,000 at 10% compound interest per annum compounded annually for three years. Find which investment is better.
Solution
\[ \begin{aligned} & \color{magenta} \textbf{Investment 1: Simple Interest} \\[6pt] P &= \text{₹}\,1{,}000 \\ R &= 11\% \\ T &= 3 \text{ years} \\[6pt] \color{green}\text{S.I.} &= \color{green}\frac{P \times R \times T}{100} \\[6pt] &= \frac{1000 \times 11 \times 3}{100} \\[6pt] &= 10 \times 33 \\[6pt] \color{green}\text{S.I.} &= \text{₹}\,330 \end{aligned} \]\[ \begin{aligned} & \color{magenta} \textbf{Investment 2: Compound Interest} \\[6pt] P &= \text{₹}\,1{,}000 \\ R &= 10\% \\ n &= 3 \text{ years} \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 1000\left(1 + \frac{10}{100}\right)^3 \\[6pt] &= 1000\left(\frac{11}{10}\right)^3 \\[6pt] &= 1000 \times \frac{1331}{1000} \\[6pt] A &= \text{₹}\,1{,}331 \\[6pt] \color{green}\text{C.I.} &= \color{green}1{,}331 - 1{,}000 \\ &= \text{₹}\,331 \end{aligned} \]Answer \(\color{red}{\text{Compound Interest investment is better.}}\)
8. M/s Heera Associates let out ₹ 4,00,000 for one year at 16% per annum compounded annually. How much they could earn if the interest is compounded half-yearly?
Solution
\[ \begin{aligned} P &= \text{₹}\,4{,}00{,}000 \\ R &= 16\% \text{ per annum} \\ n &= 1 \text{ year} \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{CI - half yearly} \\[6pt] A &= P\left(1 + \frac{R}{200}\right)^{2n} \\[6pt] &= 400000\left(1 + \frac{16}{200}\right)^2 \\[6pt] &= 400000\left(\frac{216}{200}\right)^2 \\[6pt] &= 400000 \times \frac{216}{200} \times \frac{216}{200} \\[6pt] &= 40 \times \frac{216}{2} \times \frac{216}{2} \\[6pt] &= 10 \times 46656 \\[6pt] A &= \text{₹}\,466560 \\[6pt] \text{C.I} &= 466560 - 400000 \\ &= \text{₹}\,66560 \end{aligned} \]Answer If compounded half-yearly, they earn \( = \color{red}{\text{₹}\,66{,}560}\)
9. Sirish borrowed a sum of ₹ 1,63,840 at 12.5% per annum compounded annually. On the same day, he lent out the. same amount to Sahej at the same rate of interest but compounded half-yearly. Find his gain after two years.
Solution
\[ \begin{aligned} P &= \text{₹}\,1{,}63{,}840 \\ R &= 12.5\% \implies \frac{25}{2}\% \\ n &= 2 \text{ years} \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Sirish borrowed (CI)} \\[6pt] A &= P\left(1 + \frac{R}{100}\right)^n \\[6pt] &= 163840\left(1 + \frac{12.5}{100}\right)^2 \\[6pt] &= 163840\left(\frac{112.5}{100}\right)^2 \\[6pt] &= 163840\left(\frac{1125}{1000}\right)^2 \\[6pt] &= 163840\left(\frac{9}{8}\right)^2 \\[6pt] &= 163840 \times \frac{81}{64} \\[6pt] &= 2560 \times 81 \\[6pt] A &= \text{₹}\,207360 \\[6pt] \text{C.I} &= 207360 - 163840 \\ \text{Interest paid} &= \text{₹}\,43{,}520 \end{aligned} \] \[ \begin{aligned} &\color{magenta}\textbf{Sirish lent (CI - half yearly)} \\[6pt] A &= P\left(1 + \frac{R}{200}\right)^{2n} \\[6pt] &= 163840\left(1 + \frac{12.5}{200}\right)^4 \\[6pt] &= 163840\left(1 + \frac{125}{2000}\right)^4 \\[6pt] &= 163840\left(1 + \frac{1}{16}\right)^4 \\[6pt] &= 163840\left(\frac{17}{16}\right)^4 \\[6pt] &= 163840 \times \frac{17 \times 17 \times 17 \times 17}{16 \times 16 \times 16 \times 16} \\[6pt] &= \frac{5 \times 83521}{2} \\[6pt] &= \frac{417605}{2} \\[6pt] A&= \text{₹}\,208802.50 \\[6pt] \text{C.I} &= 208802.50 - 163840 \\ \text{Interest received} &= \text{₹}\,44{,}962.50 \end{aligned} \] \[ \begin{aligned} \color{green}\text{Gain} &= \color{green}\text{Interest received} - \text{Interest paid} \\ &= 44962.50 - 43520 \\ &= \text{₹}\,1{,}442.50 \end{aligned} \]Answer Sirish's gain after 2 years \(= \color{red}{\text{₹}\,1{,}442.50}\)
10. The annual rate of growth in population of a certain city is 8%. If its present population is 1,96,830, what was the population three years ago?
Solution
\[ \begin{aligned} P &= P \\[6pt] R &= 8\% \text{ per annum} \\[6pt] n &= 3 \text{ years} \\[6pt] \text{Present population }(A) &= 1,96,830 \\[6pt] \color{magenta} A & = \color{magenta}P\left(1 + \frac{R}{100}\right)^n \\[8pt] P\left(1 + \frac{8}{100}\right)^3 & = 196830 \\[8pt] P\left(1 + \frac{2}{25}\right)^3 & = 196830 \\[8pt] P \left(\frac{27}{25}\right)^3 & = 196830 \\[8pt] P & = 196830 \times \left(\frac{25}{27}\right)^3 \\[8pt] P &= \frac{196830 \times 15625}{19683} \\[8pt] P &= 10 \times 15625 \\[8pt] P &= 156250 \end{aligned} \]Answer Population three years ago \(=\; \color{red}{1{,}56{,}250}\)