Question 13 Marks
Three years ago, Beeru purchased a buffalo from Surjeet for $Rs. 11000.$ What payment will discharge his debt now, the rate of interest being $10\%$ per annum, compounded annually$?$
AnswerPrice of a buffalo $(P) = Rs. 11000$
Rate of interest $(r) = 8\%$ p.a
Period $(n) = 3$ years Price of buffalo at present $(A)$
$=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }11000\times\Big(1+\frac{10}{100}\Big)^3$
$=\text{Rs. }11000\times\Big(\frac{11}{10}\Big)^3$
$=\text{Rs. }11000\times\frac{11}{10}\times\frac{11}{10}\times\frac{11}{10}$
$=\text{Rs. }14641$
View full question & answer→Question 23 Marks
A sum of money amounts to $Rs. 10240$ in $2$ years at $6\frac{2}{3}\%$ per annum, compounded annually. Find the sum.
AnswerAmount $(A) = Rs. 10240$
Rate $ (r) =6\frac{2}{3}\%=\frac{20}{3}\%\text{ p.a}$
Period $(n) = 2$ years
Let sum $-\ P,$ then $\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow10240=\text{p}\Big(1+\frac{20}{3\times100}\Big)^2$
$\Rightarrow10240=\text{p}\Big(1+\frac{1}{15}\Big)^2$
$\Rightarrow10240=\text{p}\times\Big(\frac{16}{15}\Big)^2$
$\therefore\text{P }=\text{Rs. }10240\times\Big(\frac{15}{16}\Big)^2$
$=\text{Rs. }10240\times\frac{15}{16}\times\frac{15}{16}$
$=\text{Rs. }9000$
$\therefore\text{Sum}=-\text{Rs. }9000$
View full question & answer→Question 33 Marks
Three years ago, the population of a town was $50000.$ If the annual increase during three successive years be at the rate of $5\%, 4\%$ and $3\%$ respectively, what is its present population$?$
Answer$3$ years ago, the population was $=50000$
Rate of increase successively $\left(r_1, r_2, r_3\right)=4 \%, 5 \%$ and $3 \%$ p.a.
Period $(n)=3$ years
Present Population
$=P\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)$
$=50000\left(1+\frac{5}{100}\right)\left(1+\frac{4}{100}\right)\left(1+\frac{3}{100}\right)$
$=50000 \times \frac{21}{20} \times \frac{26}{25} \times \frac{103}{100}$
$=56238$
View full question & answer→Question 43 Marks
A car is purchased for $Rs. 348000.$ Its value depreciates at $10\%$ per annum during the first year and at $20\%$ per annum during the second year. What will be its value after $2$ years$?$
AnswerCost of car $= Rs. 34800$
Rate of depreciation $(R1) = 10\%$ p.a. for first year
$\therefore(\text{R}_2)=20\%$ p.a. for second year
$\therefore$ Value after $2$ years $=\text{P}\Big(1-\frac{\text{R}_1}{100}\Big)\Big(1-\frac{\text{R}_2}{100}\Big)$
$=\text{Rs. }348000\Big(1-\frac{10}{100}\Big)\Big(1-\frac{20}{100}\Big)$
$=\text{Rs. }348000\times\frac{9}{10}\times\frac{4}{5}$
$=\text{Rs. }250560$
View full question & answer→Question 53 Marks
By using the formula, find the amount and compound interest on. $Rs. 62500$ for $2$ years $6$ months at $12\%$ per annum compounded annually.
AnswerPrincipal $(P) = Rs. 62500$
Rate $(R) = 12\%$ p.a
Period $(n) = 2$ years $6$ months
$\therefore$ Amount $(A)$
$=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }62500\Big(1+\frac{12}{100}\Big)^2\Big(1+\frac{12\times1}{100\times2}\Big)$
$=\text{Rs. }62500\times\Big(\frac{28}{25}\Big)^2\times\frac{53}{50}$
$=\text{Rs. }62500\times\frac{28}{25}\times\frac{28}{25}\times\frac{53}{50}$
$=\text{Rs. }83104$
$\therefore C.I = A - P = Rs. 83104 - Rs. 62500= Rs. 20604.$
View full question & answer→Question 63 Marks
Find the amount and the compound interest on $Rs. 2500$ for $2$ years at $10\%$ per annum, compounded annually.
AnswerPrincipal $(p) = Rs. 2500$
Rate $(r) = 10\%$ p.a
Period $(t) = 2$ years Interest for the first year
$=\frac{\text{prt}}{100}=\frac{2500\times10\times1}{100}$
$=\text{Rs. }250$
Amount at the end of the first year $= Rs. (2500 + 250 ) = Rs. 2750$
Principal for the second year $= Rs. 2750$
Interest for the second year $=\text{Rs. }=\frac{2500\times10\times1}{100}=\text{Rs. }2750$
Amount at the end of the second year $= Rs. 2750 + 275 = Rs. 3025$ And compound interest for the $2$ years $= Rs. 3025 - 2500 = Rs. 525$
View full question & answer→Question 73 Marks
The population of a city was $120000$ in the year $2009.$ During next year it increased by $6\%$ but due to an epidemic it decreased by $5\%$ in the following year. what is its population in the year $2011?$
AnswerPopulation of a city in $2013 = 120000$
Increase in next year $= 6\%$ and decrease in the following year $= 5\%$
Population in $ 2015=\text{P}\Big(1+\frac{\text{R}_1}{100}\Big)^\text{1}\Big(1+\frac{\text{R}_2}{100}\Big)^\text{1}$
$=120000\Big(1+\frac{6}{100}\Big)\Big(1-\frac{5}{100}\Big)$
$=120000\times\frac{53}{50}\times\frac{19}{20}$$=120840$
View full question & answer→Question 83 Marks
In how many years will $Rs. 6250$ amount to $Rs. 7290$ at $8\%$ per annum, compounded annually$?$
AnswerPrincipal $(P) = Rs. 6250$
Amount $(A) = Rs. 7290$
Rate $(R) = 8\%\ p.a.$
Let $n$ be the time, then
$\frac{\text{A}}{\text{P}}=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$\frac{7290}{6250}=\Big(1+\frac{8}{100}\Big)^\text{n}$
$\Rightarrow\frac{729}{625}=\Big(\frac{27}{25}\Big)^\text{n}$
$\Rightarrow\Big(\frac{27}{25}\Big)^\text{n}=\Big(\frac{27}{25}\Big)^\text{n}$
Comparing, we get $n = 2$
$\therefore$ Period $= 2$ years.
View full question & answer→Question 93 Marks
At what rate per cent per annum will $Rs. 640$ amount to $Rs. 774.40$ in $2$ years when compounded annually$?$
AnswerPrincipal $(P) = Rs. 640$
Amount $(A) = Rs. 774.40$
Period $(n) = 2$ years
Let $r$ be the rate percent annum
We know that,
$\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\frac{\text{A}}{\text{P}}=\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\frac{774.40}{640.00}=\Big(1+\frac{\text{r}}{100}\Big)^2$
$\Rightarrow\frac{121}{100}=\Big(1+\frac{\text{r}}{100}\Big)^2 ($Dividing by $64)$
$\Rightarrow\Big(\frac{11}{10}\Big)^2=\Big(1+\frac{\text{r}}{100}\Big)^2$
$\Rightarrow1+\frac{\text{r}}{100}=\frac{11}{10}$
$\Rightarrow1+\frac{1}{100}\times100=10$
$\therefore$ Rate percent is $10\%$ p.a
View full question & answer→Question 103 Marks
A scooter is bought at $Rs. 56000.$ Its value depreciates at the rate of $10\%$ per annum. What will be its value after $3$ years$?$
AnswerValue of scooter $(P) = Rs. 56000$
Rate of depreciation $(R) = 10\%\ p.a.$
Period $(n) = 3$ years
$\therefore$ Value of scooter after $3$ years
$=\text{P}\Big(1-\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs. }56000\Big(1-\frac{10}{100}\Big)^\text{3}$
$=\text{Rs. }56000\times\frac{9}{10}\times\frac{9}{10}\times\frac{9}{10}$
$=\text{Rs. }40824$
View full question & answer→Question 113 Marks
The simple interest on a sum of money for $2$ years at $8\%$ per annum is $Rs. 2400$. What will be the compound interest on that sum at the same rate and for the same period$?$
AnswerSimple interest $(S.I) = Rs. 2400$
Rate $(R) = 8\%$ p.a
Period $(T) = 2$ years
$\therefore\text{Sum}(\text{P})=\frac{\text{S.I}\times100}{\text{R}\times\text{T}}$
$=\frac{2400\times100}{8\times2}=\text{Rs. }15000$
$\therefore$ Amount on compound interest, $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^4=15000\Big(1+\frac{8}{100}\Big)^2$
$=\text{Rs. }15000\times\Big(\frac{27}{25}\Big)^2$
$=\text{Rs. }15000\times\frac{27}{25}\times\frac{27}{25}$
$=\text{Rs. }17496$
$\therefore C.I = A - P = Rs. 17496 - 15000 = Rs. 2496$
View full question & answer→Question 123 Marks
In how many years will $Rs. 1800$ amount to $Rs. 2178$ at $10\%$ per annum when compounded annually$?$
AnswerPrincipal $(P) = Rs. 1800$
Amount $(A) = Rs. 2178$
Rate $(r) = 10\%\ p.a.$
Let $n$ be the number of years.
We know that
$\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}=\frac{\text{A}}{\text{P}}$
$\Rightarrow\Big(1+\frac{10}{100}\Big)^\text{n}=\frac{1}{1800}$
$\Rightarrow\Big(\frac{11}{10}\Big)^\text{n}=\frac{121}{100}=\Big(\frac{11}{10}\Big)($Dividing by $18)$
$\therefore$ While comparing, we get $n = 2$
$\therefore$ Period $= 2$ years.
View full question & answer→Question 133 Marks
By using the formula, find the amount and compound interest on.
$Rs. 31250$ for $3$ years at $8\%$ per annum compounded annually.
AnswerPrincipal $(P) = Rs. 31250$
Rate $(R) = 8\%$ p.a
Period $(n) = 3$ years
$\therefore$ Amount $(A)$
$=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }31250\Big(1+\frac{8}{100}\Big)^3$
$=\text{Rs. }31250\times\Big(\frac{27}{25}\Big)^3$
$=\text{Rs. }31250\times\frac{27}{25}\times\frac{27}{25}\times\frac{27}{25}$
$=\text{Rs. }39366$
$C.I = A - P = Rs. 39366 - Rs. 31250= Rs. 8116$
View full question & answer→Question 143 Marks
By using the formula, find the amount and compound interest on. $Rs. 6000$ for $2$ years at $9\%$ per annum compounded annually.
AnswerPrincipal $(P) = Rs. 6000$
Rate $(R) = 9\%$ p.a
Period $(n) = 2$ years
Amount $(A)n=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }6000\Big(1+\frac{9}{100}\Big)^2$
$=\text{Rs. }6000\times\frac{109}{100}\times\frac{109}{100}$
$=\text{Rs. }\frac{71286}{10}=\text{ Rs. }7128.60$
$C.I = A - P = Rs. 7128.60 - Rs. 6000 = Rs. 1128.60$
View full question & answer→Question 153 Marks
The count of bacteria in a certain experiment was increasing at the rate of $2\%$ per hour. Find the bacteria at the end of $2$ hours if the count was initially $500000.$
AnswerInitially bacteria $= 500000$
Increase in bacteria $= 2\%$ per hour
Period $(2 n) =$ hours
$\therefore$ Bacteria after $2$ hours $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=500000\Big(1+\frac{2}{100}\Big)^2$
$=500000\times\Big(\frac{51}{50}\Big)^2$
$=500000\times\frac{51}{50}\times\frac{51}{50}$
$=520200$
View full question & answer→Question 163 Marks
Find the amount of $Rs.8000$ for $2$ years compounded annually and the rates being $9\%$ per annum during the first year and $10\%$ per annum during the second year.
AnswerPrincipal $(P) = Rs. 8000$
Period $(n) = 2$ years
Rate $(R_1) = 9\%$ for the first years
$R_2= 10\%$ the second years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}_1}{100}\Big)^1$
$=\Big(1+\frac{\text{R}_2}{100}\Big)^1$
$=8000\Big(1+\frac{9}{100}\Big)\Big(1+\frac{10}{100}\Big)$
$=\text{Rs. }8000\times\frac{109}{100}\times\frac{110}{100}$
$=\text{Rs. }11253$
View full question & answer→Question 173 Marks
By using the formula, find the amount and compound interest on.
$Rs. 10240$ for $3$ years at $12\frac{1}{2}\%$ per annum compounded annually.
AnswerPrincipal $(P) = Rs. 10240$
Rate $(R) 12\frac{1}{2}\%=\frac{25}{2}\%\text{ p.a}$
Period $(n) = 3$ years
$\therefore$ Amount (A) $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }10240\Big(1+\frac{25}{2\times100}\Big)^3$
$=\text{Rs. }10240\times\Big(\frac{9}{8}\Big)^3$
$=\text{Rs. }10240\times\frac{9}{8}\times\frac{9}{8}\times\frac{9}{8}$
$=\text{Rs. }14580$
$\therefore C.I = A - P = Rs. 14580 - Rs. 10240= Rs. 4340.$
View full question & answer→Question 183 Marks
The bacteria in a culture grows by $10\%$ in the first hour, decreases by $10\%$ in the second hour and again increases by $10\%$ in the third hour. Find the bacteria at the end of $3$ hours if the count was initially $20000.$
AnswerGrowth of bacteria in a culture $(R_1) = 10%$ in first hour
Decrease in next hour $(R_2) = 10%$
Increase in the third hour $(R_3) = 10%$
Bacteria in the beginning $= 20000$
Bacteria after $3$ hours
$=\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)$
$20000\Big(1+\frac{10}{100}\Big)\Big(1-\frac{10}{100}\Big)\Big(1+\frac{10}{100}\Big)$
$20000\times\frac{11}{10}\times\frac{9}{10}\times\frac{11}{10}$
$=21780$
View full question & answer→Question 193 Marks
What sum of money will amount to $Rs. 21296$ in $3$ years at $10\%$ per annum, compounded annually$?$
AnswerAmount $(A) = Rs. 21296$
Rate $(r) = 10\%\ p.a$
Period $(n) = 3$ years
Let $P$ be the sum, then $\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow21296=\text{P}\Big(1+\frac{10}{100}\Big)^3$
$\Rightarrow21296=\text{P}\Big(\frac{11}{10}\Big)^3$
$\text{P}=\text{Rs. }21296\times\Big(\frac{10}{11}\Big)^3$
$=21296\times\frac{10}{11}\times\frac{10}{11}\times\frac{10}{11}$
$=\text{Rs. }16000$
$\therefore\text{Sum}=\text{Rs. }16000$
View full question & answer→Question 203 Marks
Find the amount and the compound interest on $Rs. 15625$ for $3$ years at $12\%$ per annum, compounded annually.
AnswerPrincipal $(P) = Rs. 15625$
Rate $(R) = 12\%\ p.a$
Period $(n) =$ years
$\therefore$ Amount $ (A) =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }15625\Big(1+\frac{12}{100}\Big)^3$
$=\text{Rs. }15625\times\Big(\frac{28}{25}\Big)^3$
$=\text{Rs. }15625\times\frac{28}{25}\times\frac{28}{25}\times\frac{28}{25}$
$=\text{Rs. }21952$
$C.I = A - P = Rs. 21952 - 15625 = Rs. 6327$
View full question & answer→Question 213 Marks
Anand obtained a loan of $Rs. 125000$ from the Allahabad Bank for buying computers. The bank charges compound interest at $8\%$ per annum, compounded annually. What amount wil he have to pay after $3$ years to clear the debt$?$
AnswerPrincipal $(P) = Rs. 1,25,000$
Rate of interest $(r) = 8\%\ p.a$
Period $(n) = 3$ years
$\therefore$ Amount $(A) =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }125000\times\Big(1+\frac{8}{100}\Big)^3$
$=\text{Rs. }125000\times\Big(\frac{27}{25}\Big)^3$
$=\text{Rs. }125000\times\frac{27}{25}\times\frac{27}{25}\times\frac{27}{25}$
$=\text{Rs. }157464$
View full question & answer→Question 223 Marks
The population of a town is $125000.$ It is increasing at the rate of $2\%$ per annum. What will be its population after $3$ years$?$
AnswerPresent population $(P) = 125000$
Rate of increasing $(R) = 2\%\ p.a.$
Period $(n) = 3$ years
$\therefore$ Population after $3$ years $=\text{P}\Big(1+\frac{\text{R}}{100}\Big)^\text{n}$
$=125000\Big(1+\frac{2}{100}\Big)^3$
$=125000\times\Big(\frac{51}{50}\Big)^3$
$=125000\times\frac{51}{50}\times\frac{51}{50}\times\frac{51}{50}$
$=132651$
View full question & answer→Question 233 Marks
At what rate percent per annum will $Rs. 4000$ amount to $Rs. 4410$ in $2$ years when compounded annually$?$
AnswerAmount $(A) = Rs. 4000$
Principal $(P) = Rs. 4410$
Period $(n) = 2$ years
Let $r$ be the rate percent per annum We know that, $\text{A}=\text{P}\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\frac{\text{A}}{\text{P}}=\Big(1+\frac{\text{r}}{100}\Big)^{\text{n}}$
$\Rightarrow\frac{4410}{4000}=\Big(1+\frac{\text{r}}{100}\Big)^2$
$\Rightarrow\Big(\frac{21}{20}\Big)^2=\Big(1+\frac{\text{r}}{100}\Big)^2$
$\Rightarrow1+\frac{\text{r}}{100}=\frac{21}{20}$
$\Rightarrow\frac{\text{r}}{100}=\frac{21}{20}-1=\frac{1}{20}$
$\Rightarrow\text{r}=\frac{1}{20}\times100=5$
$\therefore\text{Rate}=5\%\text{ p.a}$
View full question & answer→Question 243 Marks
By using the formula, find the amount and compound interest on. $Rs. 9000$ for $2$ years $4$ months at $10\%$ per annum compounded annually.
AnswerPrincipal $(P) = Rs. 9000$
Rate $(R) = 10\%\ p.a$
Period $(n) = 2$ years $4$ months $=2\frac{1}{3}\text{ years}$
$\therefore$ Amount $(A) =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }9000\Big(1+\frac{10}{100}\Big)^2\Big(1+\frac{10\times1}{100\times3}\Big)$
$=\text{Rs. }9000\times\frac{11}{10}\times\frac{11}{10}\times\frac{31}{30}$
$=\text{Rs. }11253$
$\therefore C.I = A - P = Rs. 11253 - Rs. 9000= Rs. 2253$
View full question & answer→Question 253 Marks
Shubhalaxmi took a loan of $Rs. 18000$ from surya Finance to purchase a $TV$ set. If the company charges compound interest at $12\%$ per annum during the first year and $12\frac{1}{2}\%$ per annum during the second year, how much will she have to pay after $2$ years$?$
AnswerAmount of loan taken $(p) = Rs. 18000$
Rate $(R_1) = 12%$ p.a during first year
$R_2=$ $12\frac{1}{2}\%=\frac{25}{2}\%\text{ p.a}$ during second year
Period $(n) = 2$ years
$\therefore$ Total amount $(A)$
$=\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{p}\Big(1+\frac{\text{R}_1}{100}\Big)^1\Big(1+\frac{\text{R}_2}{100}\Big)^1$
$=\text{Rs. }18000\times\Big(1+\frac{12}{100}\Big)\Big(1+\frac{25}{2\times100}\Big)$
$=\text{Rs. }18000\times\frac{28}{25}\times\frac{9}{8}$
$=\text{Rs. }22680$
View full question & answer→Question 263 Marks
Neha borrowed $Rs. 24000$ from the State Bank of India to buy a scooter. If the rate of interest be $10\%$ per annum compounded annually, what payment will she have to make after $2$ years $3$ months$?$
AnswerAmount borrowed from Bank $(p) = Rs. 24000$
Rate $(R) = 10\%$ p.a Period $(n) = 2$ years $3$ mounths $=2\frac{1}{4}\text{years}$
$\therefore$ Amount after the period $ (A) =\text{p}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=\text{Rs. }24000\Big(1+\frac{10}{100}\Big)^2\Big(14+\frac{10\times1}{100\times4}\Big)$
$=\text{Rs. }24000\times\Big(\frac{11}{10}\Big)^2\times\Big(\frac{41}{40}\Big)$
$=\text{Rs. }29766\times\frac{11}{10}\times\frac{11}{10}\times\frac{41}{40}$
$=\text{Rs. }29766$
View full question & answer→Question 273 Marks
By using the formula, find the amount and compound interest on.
$Rs. 10000$ for $2$ years at $11\%$ per annum compounded annually.
AnswerPrincipal $(P) = Rs. 10000$
Rate $(R) = 11\%$ p.a
Period $(n) = 2$ years
$\therefore$ Amount $(A) =\text{P}\Big(1+\frac{\text{R}}{100}\Big)^{\text{n}}$
$=10000\Big(1+\frac{11}{100}\Big)^2$$=\text{Rs. }10000\times\Big(\frac{111}{100}\Big)^2$
$=\text{Rs. }10000\times\frac{111}{100}\times\frac{111}{100}$
$=\text{Rs. }12321$
$C.I = A - P = Rs. 12321 - Rs. 10000 = Rs. 2321.$
View full question & answer→Question 283 Marks
The value of a machine depreciates at the rate of $10\%$ per annum. It was purchased $3$ years ago. If its present value is $Rs.291600,$ for how much was it purchased$?$
AnswerRate of depreciation $(R) = 10\%$ p.a.
Period $(n) = 3$ years
Present value $(A) = Rs. 291600$
Value of machine $3$ years ago
$(\text{P})=\text{A}\div\Big(1-\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs. }291600\div\Big(1-\frac{10}{100}\Big)^\text{3}$
$=\text{Rs. }291600\div\Big(\frac{9}{10}\Big)^3$
$=\text{Rs. }291600\times\frac{10}{9}\times\frac{10}{9}\times\frac{10}{9}$
$=\text{Rs. }400000$
View full question & answer→Question 293 Marks
A machine is purchased for $Rs. 625000.$ Its value depreciates at the rate of $8\%$ per annum. What will be its value after $2$ years$?$
AnswerValue of machine $(P) = Rs. 625000$
Rate of depreciation $(R) = 8\%$ p.a. Period $(n) = 2$ years
$\therefore$ Value after $2$ years $=\text{P}\Big(1-\frac{\text{R}}{100}\Big)^\text{n}$
$=\text{Rs. }625000\Big(1-\frac{8}{100}\Big)^\text{2}$
$=\text{Rs. }625000\Big(\frac{23}{25}\Big)^\text{2}$
$=\text{Rs. }625000\times\frac{23}{25}\times\frac{23}{25}$
$=\text{Rs. }529000$
View full question & answer→