MCQ 11 Mark
Tick $(\checkmark)$ the correct answer in the following: If the compound interest on a sum for $2$ years at $12\frac{1}{2}\%$ per annum is $Rs. 510,$ the simple interest on the same sum at the same rate for the same period of time is.
- A
$Rs. 400$
- B
$Rs. 450$
- C
$Rs. 460$
- ✓
$Rs. 480$
AnswerCorrect option: D. $Rs. 480$
$C.I$ on a sum $= Rs. 510$
Rate $(R)12\frac{1}{2}\%=\frac{25}{2}\%\text{ p.a}$
Period $(n) = 2$ years
$\text{C.I} = \text{A} - \text{P}=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-\text{P}$
$\text{C.I}=\text{P}\Bigg[\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}-1\Bigg]$
$\Rightarrow510=\text{P}\Bigg[\Big(1+\frac{\text{R}}{2\times100}\Big)^2-1\Bigg]$
$\Rightarrow510=\text{P}\Big[\frac{9}{8}\times\frac{9}{8}-1\Big]$
$\Rightarrow510=\text{P}\Big[\frac{81}{64}-1\Big]$
$\Rightarrow510=\text{P}\Big(\frac{17}{64}\Big)=\text{P}=\frac{510\times64}{17}$
$\text{P}=\text{Rs. }1920$
$\therefore\text{S.I}=\frac{\text{PRT}}{100}=\frac{1920\times25\times2}{100\times2}$
$=\text{Rs. }480$
View full question & answer→MCQ 21 Mark
Tick $(\checkmark)$ the correct answer in the following:
A sum of $Rs. 25000$ was given as loan on compound interest for $3$ years compounded annually at $5\%$ per annum during the first year, $6\%$ per annum during the second year and $8\%$ per annum during the third year. The compound interest is.
- A
$Rs. 5035$
- ✓
$Rs. 5051$
- C
$Rs. 5072$
- D
$Rs. 5150$
AnswerCorrect option: B. $Rs. 5051$
Principal $(P) = Rs. 25000$
Rate $(R_1) = 5\%$ for the first year
$R_2= 6\%$ for the second year
$R_3= 8\%$ for the third year
$\therefore$ Amount $(A) =\text{P}\Big(1+\frac{\text{R}_1}{100}\Big)\Big(1+\frac{\text{R}_2}{100}\Big)\Big(1+\frac{\text{R}_3}{100}\Big)$
$=\text{Rs. }25000\Big(1+\frac{5}{100}\Big)\Big(1+\frac{6}{100}\Big)\Big(1+\frac{8}{100}\Big)$
$=\text{Rs. }25000\times\frac{21}{20}\times\frac{53}{50}\times\frac{27}{25}$
$=\text{Rs. }30051$
$\therefore C.I = A - P = Rs. 30051 - 25000$
$= Rs. 5051$
View full question & answer→MCQ 31 Mark
Tick $(\checkmark)$ the correct answer in the following: The compound interest on $Rs. 6250$ at $8\%$ per annum for $1$ year, compounded half yearly, is.
- A
$Rs. 500$
- ✓
$Rs. 510$
- C
$Rs. 550$
- D
$Rs. 512.50$
AnswerCorrect option: B. $Rs. 510$
Principal $(P) = Rs. 6250$
Rate $(R) = 8\% \ p.a$ or $4\%$ half yearly
Period $(n) = 1$ years or $2$ half years
$\therefore$ Amount $(A)=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }6250\Big(1+\frac{4}{100}\Big)^2$
$=\text{Rs. }6250\times\frac{26}{25}\times\frac{26}{25}$
$=\text{Rs. }6760$
$\therefore C.I = A - P $
$= Rs. 6760 - 6250$
$= Rs. 510$
View full question & answer→MCQ 41 Mark
If the compound interest on a certain sum for $2$ years at $10\%$ per annum is $Rs.1050,$ the sum is:
- A
$Rs. 3000$
- B
$Rs. 4000$
- ✓
$Rs. 5000$
- D
$Rs. 6000$
AnswerCorrect option: C. $Rs. 5000$
Here, $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{P}\times\Big(1+\frac{10}{100}\Big)^2$
$=\text{P}\times\Big(\frac{110}{100}\Big)^2$
$=\text{P}\times\Big(\frac{11}{10}\Big)\times\Big(\frac{11}{10}\Big)$
Now, $\text{CI}=\text{A}-\text{P}$
$\Rightarrow\text{Rs. }1050$
$=\frac{121\text{p}}{100}-\text{P}$
$=\frac{121\text{P}-100\text{P}}{100}$
$=\frac{12\text{P}}{100}$
$\therefore\text{P}=\text{Rs. }\frac{1050\times100}{21}$
$=\text{Rs. }5000$
View full question & answer→MCQ 51 Mark
Tick $(\checkmark)$ the correct answer in the following: At what rate percent per annum will a sum of $Rs. 7500$ amount to $Rs. 8427$ in $2$ years, compounded annually?
AnswerHere $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)$
$=\text{Rs. }7500\times\Big(1+\frac{\text{R}}{100}\Big)^2$
$=\text{Rs. }7500\times\Big(1+\frac{\text{R}}{100}\Big)^2$
However, amount $= Rs. 8427$
Now, $Rs. 8427$
$=\text{Rs. }7500\times\Big(1+\frac{\text{R}}{100}\Big)^2$
$\Rightarrow\frac{\text{Rs. }8427}{\text{Rs. }7500}=\Big(1+\frac{\text{R}}{100}\Big)^2$
$\Rightarrow\Big(\frac{53}{50}\Big)^2=\Big(1+\frac{\text{R}}{100}\Big)^2$
$\Rightarrow\Big(1+\frac{\text{R}}{100}\Big)=\Big(\frac{53}{50}\Big)$
$\Rightarrow\frac{\text{R}}{100}=\frac{53}{50}-1$
$\Rightarrow\frac{\text{R}}{100}=\frac{53-50}{50}=\frac{3}{50}$
$\therefore\text{R}=\frac{300}{50}=6\%$
View full question & answer→MCQ 61 Mark
Tick $(\checkmark)$ the correct answer in the following: If the simple interest on a sum of money at $5\%$ per annum for $3$ years is $Rs. 1200$ then the compound interest on the same sum for the same period at the same rate will be.
- A
$Rs. 1225$
- B
$Rs. 1236$
- C
$Rs. 1248$
- ✓
$Rs. 1261$
AnswerCorrect option: D. $Rs. 1261$
$S.I = Rs. 60000$
Rate $(R) = 5\%$
Period $(T) = 3$ years
$\therefore$ Principal $=\frac{\text{S.I}\times100}{\text{R}\times\text{T}}$
$=\frac{1200\times100}{5\times3}$
$=\text{Rs. }8000$
$\therefore$ Amount on $C.I=8000\Big(1+\frac{5}{100}\Big)^3$
$=\text{Rs. }8000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}$
$=\text{Rs. }9261$
$\therefore C.I = A - P $
$= Rs. 9261 - 8000$
$= Rs. 1261$
View full question & answer→MCQ 71 Mark
If the simple interest on a sum of money at $10\%$ per annum for $3$ years is $Rs. 1500,$ then the compound interest on the same sum at the same rate for the same period is:
- ✓
$Rs. 1655$
- B
$Rs. 1155$
- C
$Rs. 1555$
- D
$Rs. 1855$
AnswerCorrect option: A. $Rs. 1655$
Here, $\text{SI}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow\text{Rs. }1500=\frac{\text{P}\times10\times3}{100}$
$\Rightarrow\text{P}=\frac{1500\times100}{10\times3}$
$=\text{Rs. }5000$
Now, $\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs. }5000\times\Big(1+\frac{10}{100}\Big)^3$
$=\text{Rs. }5000\times\Big(\frac{110}{100}\Big)^3$
$=\text{Rs. }5000\times\Big(\frac{11}{10}\Big)\times\Big(\frac{11}{10}\Big)\times\Big(\frac{11}{10}\Big)$
$=\text{Rs. }(5\times11\times11\times11)$
$=\text{Rs. }6655$
$\therefore CI = A - P $
$= Rs. (6655 - 5000) $
$= Rs. 1655$
View full question & answer→MCQ 81 Mark
Tick $(\checkmark)$ the correct answer in the following: The compound interest on $Rs. 40000$ at $6\%$ per annum for $6$ months, compounded quarterly, is.
- ✓
$Rs. 1209$
- B
$Rs. 1902$
- C
$Rs. 1200$
- D
$Rs. 1306$
AnswerCorrect option: A. $Rs. 1209$
Principal $(P) = Rs. 40000$
Rate $(R)=6\%\ \text{p.a}\ \frac{6}{4}=\frac{3}{2}\%$ quarterly
Period $(n) = 6$ months $= 2$ quaters
$\therefore$ Amount $(A)=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }40000\Big(1+\frac{3}{2\times100}\Big)^2$
$=\text{Rs. }40000\times\frac{230}{200}\times\frac{203}{200}$
$=\text{Rs. }41209$
$\therefore C.I = A - P $
$= Rs. 41209 - 40000$
$= Rs. 1209$
View full question & answer→MCQ 91 Mark
Tick $(\checkmark)$ the correct answer in the following: The compound interest on $Rs. 5000$ at $8\%$ per annum for $2$ years, compounded annually, is.
- A
$Rs. 5035$
- B
$Rs. 5051$
- ✓
$Rs. 5072$
- D
$Rs. 5150$
AnswerCorrect option: C. $Rs. 5072$
Principal $(P) = Rs. 5000$
Rate $(r) = 8\% \ p.a$
Period $(n) = 2$ years
Amount $(A)=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }5000\Big(1+\frac{8}{100}\Big)^2$
$=\text{Rs. }5000\times\frac{27}{25}\times\frac{27}{25}$
$=\text{Rs. }5832$
$\therefore C.I = A - P $
$= Rs. 5832 - 5000$
$= Rs. 832$
View full question & answer→MCQ 101 Mark
Tick $(\checkmark)$ the correct answer in the following: The value of a machine depreciates at the rate of $20\%$ per annum. It was purchased $2$ years ago. If its present value is $Rs. 40000,$ for how much was it purchased?
- A
$Rs. 56000$
- ✓
$Rs. 62500$
- C
$Rs. 65200$
- D
$Rs. 56500$
AnswerCorrect option: B. $Rs. 62500$
Principal value $= Rs. 40000$
Rate of depreciation $(R) = 20\% \ p.a$
Value of machine $2$ years ago
$=\text{A}\div\Big(1-\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }40000\div\Big(1-\frac{20}{100}\Big)^2$
$=\text{Rs. }40000\div\Big(\frac{4}{5}\Big)^2$
$=\text{Rs. }40000\times\frac{5}{4}\times\frac{5}{4}$
$=\text{Rs. }62500$
View full question & answer→MCQ 111 Mark
Tick $(\checkmark)$ the correct answer in the following: The sum that amounts to $Rs. 4913$ in $3$ years at $6\frac{1}{4}\%$ per annum compounded annually, is.
- A
$Rs. 3096$
- B
$Rs. 4076$
- C
$Rs. 4085$
- ✓
$Rs. 4096$
AnswerCorrect option: D. $Rs. 4096$
Amount $= Rs. 4913$
Rate $(R)=6\frac{1}{4}=\frac{25}{4}\%$
Period $(n) = 3$ years
$\therefore \text{A}=\text{P}\div\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=4913\div\Big(1\frac{25}{4\times100}\Big)^3$
$=\text{Rs. }4913\div\Big(1\div\frac{1}{16}\Big)^3$
$=4913\div\Big(\frac{17}{16}\Big)^3$
$=4913\times\frac{16}{17}\times\frac{16}{17}\times\frac{16}{17}$
$=\text{Rs. }4096$
$\therefore\text{Sum}=\text{Rs. }4096$
View full question & answer→MCQ 121 Mark
Tick $(\checkmark)$ the correct answer in the following: The compound interest on $Rs. 10000$ at $12\%$ per annum for $112112$ years, compounded annually, is.
- ✓
$Rs. 1872$
- B
$Rs. 1720$
- C
$Rs. 1910.16$
- D
$Rs. 1782$
AnswerCorrect option: A. $Rs. 1872$
Principal $(P) = Rs. 10000$
Rate $(r) = 12\% \ p.a$
Period $(n) =1\frac{1}{2}\text{year}$
$\therefore $ Amount $(A)=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }10000\Big(1+\frac{12}{100}\Big)^1\Big(1+\frac{12}{2\times100}\Big)$
$=\text{Rs. }10000\times\frac{28}{25}\times\frac{53}{50}$
$=\text{Rs. }11872$
$\therefore C.I = A - P $
$= Rs. 11872 - 10000$
$= Rs. 1872$
View full question & answer→MCQ 131 Mark
Tick $(\checkmark)$ the correct answer in the following: The value of a machine depreciates at the rate of $10\%$ per annum. It was purchased $3$ years ago for $Rs. 60000.$ What is the present value of the machine?
- A
$Rs. 53640$
- B
$Rs. 51680$
- ✓
$Rs. 43740$
- D
$Rs. 43470$
AnswerCorrect option: C. $Rs. 43740$
$3$ years ago, the value of machine $= Rs. 60000$
Rate of depreciation $(R) = 10\%$
Period $(n) = 3 $ years
$\therefore$ Present value $=\text{P}\Big(1-\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }60000\Big(1-\frac{10}{100}\Big)^3$
$=\text{Rs. }60000\times\Big(\frac{9}{10}\Big)^3$
$=\text{Rs. }60000\times\frac{9}{10}\times\frac{9}{10}\times\frac{9}{10}$
$=\text{Rs. }43740$
View full question & answer→MCQ 141 Mark
Tick $(\checkmark)$ the correct answer in the following: The compound interest on $Rs. 4000$ at $10\%$ per annum for $2$ years $3$ months, compounded annually, is.
- A
$Rs. 1331$
- ✓
$Rs. 3310$
- C
$Rs. 3130$
- D
$Rs. 13310$
AnswerCorrect option: B. $Rs. 3310$
Principal $(P) = Rs. 10000$
Rate $(r) = 10\% \ p.a$
Period $(n) = 3$ years
$\therefore $ Amount $(A)=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }10000\Big(1+\frac{10}{100}\Big)^3$
$=\text{Rs. }10000\times\frac{11}{10}\times\frac{11}{10}\times\frac{11}{10}$
$=\text{Rs. }13310$
$\therefore C.I = A - P $
$= Rs. 13310 - 10000$
$= Rs. 3310$
View full question & answer→MCQ 151 Mark
At what rate per cent per annum will $Rs. 5000$ amount to $Rs. 5832$ in $2$ years, compounded annually?
- A
$11\%$
- B
$10\%$
- C
$9\%$
- ✓
$8\%$
Answer$\text{A}=\text{Rs}.\ \text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$\Rightarrow\text{Rs}.\ 5932=\text{Rs}.\ 5000\times\Big(1+\frac{\text{R}}{100}\Big)^2$
$\Rightarrow\frac{\text{Rs. }5832}{\text{Rs. }5000}=\Big(1+\frac{\text{R}}{100}\Big)^2$
$\Rightarrow\Big(\frac{27}{25}\Big)^2=\Big(1+\frac{\text{R}}{100}\Big)^2$
$\Rightarrow1+\frac{\text{R}}{100}=\frac{27}{25}$
$\Rightarrow\frac{\text{R}}{100}=\frac{27}{25}-1$
$=\frac{27-25}{25}=\frac{2}{25}$
$\therefore\text{R}=\frac{100\times2}{25}$
$=8\%$
View full question & answer→MCQ 161 Mark
Tick $(\checkmark)$ the correct answer in the following: The compound interest on $Rs. 4000$ at $10\%$ per annum for $2$ years $3$ months, compounded annually, is.
- A
$Rs. 916$
- B
$Rs. 900$
- ✓
$Rs. 961$
- D
$Rs. 896$
AnswerCorrect option: C. $Rs. 961$
Principal $(P) = Rs. 4000$
Rate $(r) = 10\%\ p.a$
Period $(n) = 2$ years $3$ months $=2\frac{1}{4}\text{years}$
$\therefore$ Amount $(A)=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }4000\Big(1+\frac{10}{100}\Big)^2\Big(1+\frac{10\times1}{100\times4}\Big)$
$=\text{Rs. }4000\times\frac{11}{10}\times\frac{11}{10}\times\frac{41}{40}$
$=\text{Rs. }4961$
$\therefore C.I = A - P $
$= Rs. 4961 - 4000$
$= Rs. 961$
View full question & answer→MCQ 171 Mark
Tick $(\checkmark)$ the correct answer in the following: The annual rate of growth in population of a town is $10\%.$ If its present population is $33275,$ what was it $3$ years ago?
- ✓
$Rs. 25000$
- B
$Rs. 27500$
- C
$Rs. 30000$
- D
$Rs. 26000$
AnswerCorrect option: A. $Rs. 25000$
Rate of growth in population $(R) = 20% \ p.a$
persent population $= 33275$
Population $3$ years ago
$=\text{A}\div\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }33275\div\Big(1+\frac{10}{100}\Big)^3$
$=\text{Rs. }33275\div\Big(\frac{11}{10}\Big)^3$
$=\text{Rs. }33275\times\frac{10}{11}\times\frac{10}{11}\times\frac{10}{11}$
$=\text{Rs. }25000$
View full question & answer→MCQ 181 Mark
The annual rate of growth in population of a town is $5\%.$ If its present population is $4000,$ what will be its population after $2$ years?
- A
$4441$
- B
$4400$
- ✓
$4410$
- D
$4800$
AnswerCorrect option: C. $4410$
Present population $= 4000$
Rate of growth $= 5\%$
To find the population of the town after $2$ years, we have:
$\text{A}=\text{P}\times\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=4000\times\Big(1+\frac{5}{100}\Big)^2$
$=4000\times\Big(\frac{5}{100}\Big)^2$
$=4000\times\Big(\frac{21}{20}\Big)\times\Big(\frac{21}{20}\Big)$
$=(10\times21\times21)$
$=4410$
View full question & answer→MCQ 191 Mark
Tick $(\checkmark)$ the correct answer in the following: The present population of a town is $24000.$ If it increases at the rate of $5\%$ per annum, what will be its population after $2$ years?
- A
$26400$
- ✓
$26460$
- C
$24460$
- D
$26640$
AnswerCorrect option: B. $26460$
Present population $(P) = 24000$
Rate of increase $(R) = 5\% \ p.a$
Period $(n) = 2$ years
$\therefore$ Population after $2$ years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }24000\Big(1+\frac{5}{100}\Big)^2$
$=\text{Rs. }24000\times\frac{21}{20}\times\frac{21}{20}$
$=26460$
View full question & answer→