Question 12 Marks
Rohit borrows $Rs. 86,000$ from Arun for two years at $5\%$ per annum simple interest. He immediately lends out this money to Akshay at $5\%$ compound interest compounded annually for the same period. Calculate Rohit's profit in the transaction at the end of two years.
Answer$P=Rs. 86,000 ;$ $
$ time $=2$ years and rate $=5 \%\ p.a.$
To calculate $S.I.$
$\therefore \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{86,000 \times 5 \times 2}{100}=\text { Rs. } 8,600$
To calculate $C.I.$
$\therefore \text { C.I. }=\mathrm{P}\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$ =86,000\left[\left(1+\frac{5}{100}\right)^2-1\right]$
$ =86,000\left(\frac{41}{400}\right)=\text { Rs. } 8,815$
Profit $=C.I. - S.I.$
$= Rs.8,815-Rs. 8,600$
$= Rs. 215$
View full question & answer→Question 22 Marks
Find the compound interest to the nearest rupee on $Rs. 10,800$ for $2 \frac{1}{2}$ years at $10 \%$ per annum.
AnswerGiven : $P= Rs. 10,800$ ; Time $=2 \frac{1}{2}$ years and Rate $=10 \% \ p.a$
For $2 $ years
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n=10,800(1+10 / 100)^2=\text { Rs. } 13,068$
For $\frac{1}{2}$ year
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}=13068\left(1+\frac{10}{2 \times 100}\right)^{\frac{1}{2} \times 2}$
$ =13068 \times \frac{21}{20}=\text { Rs. } 13721.40$
$=\text { Rs. } 13721 \text { ( nearest rupee) }$
$ \therefore \text { Rs. } 13,721-\text { Rs. } 10,800$
$=\text { Rs. } 2,921$
View full question & answer→Question 32 Marks
The population of a town decreased by $12\%$ during $1998 $ and then increased by $8\%$ during $1999. $ Find the population of the town $, $ at the beginning of $1998,$ if at the end of $1999$ its population was $2,85,120.$
AnswerLet the population in the beginning of $1998=P$
The population at the end of $1999=2,85,120(\mathrm{~A})$
$ \mathrm{r}_1=-12 \% \text { and } \mathrm{r}_2=+8 \%$
$ \therefore \mathrm{A}=\mathrm{P}\left(1-\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)$
$ \Rightarrow 2,85,120=\mathrm{P}\left(1-\frac{12}{100}\right)\left(1+\frac{8}{100}\right)$
$ \Rightarrow 2,85,120=\mathrm{P}\left(\frac{22}{25}\right)\left(\frac{27}{25}\right)$
$ \Rightarrow \mathrm{P}=\frac{2,85,120 \times 25 \times 25}{22 \times 27}=3,00,000 .$
View full question & answer→Question 42 Marks
According to a census taken towards the end of the year $ 2009, $ the population of a rural town was found to be $64,000. $The census authority also found that the population of this particular town had a growth of $5\%$ per annum. In how many years after $2009$ did the population of this town reach $74,088 $ ?
AnswerPopulation in $2009(P)=64,000$
Let after $n$ years its population be $74,088(A)$ Growth rate $=5 \%$ per annum
$
\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow \frac{74,088}{64,000}=\left(\frac{21}{20}\right)^n$
$ \Rightarrow \frac{9,261}{8,000}=\left(\frac{21}{20}\right)^n$
$ \Rightarrow\left(\frac{21}{20}\right)^3=\left(\frac{21}{20}\right)^n
$
On comparing $,$ we get $,$
$
n=3 \text { years }
$
View full question & answer→Question 52 Marks
What sum of money will amount to $Rs. 27,783 $ in one and a half years at $10\%$ per annum compounded half yearly ?
Answer$A=P\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}$
$ \Rightarrow 27,783=P\left(1+\frac{10}{200}\right)^{\frac{3}{2} \times 2}$
$ \Rightarrow 27,783=P\left(\frac{21}{20}\right)^3$
$ \Rightarrow P=27,783\left(\frac{20}{21}\right)^3$
$ \Rightarrow P=24,000$
The sum of $Rs. 24,000$ amount $ Rs. 27,783$ in one and a half years at $10 \%$ per annum compounded half yearly.
View full question & answer→Question 62 Marks
If the interest is compounded half $-$ yearly $, $ calculate the amount when principal is $Rs. 7,400;$ the rate of interest is $5\%$ per annum and the duration is one year.
AnswerGiven: $P=$ $Rs. 7,400 ; r=5 \%$ p.a. and $n=1$ year
Since the interest is compounded half$-$yearly,
Then, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}$
$=7,400\left(1+\frac{5}{2 \times 100}\right)^{1 \times 2}$
$ =7,400\left(\frac{41}{40}\right)^2$
$ =\text { Rs. } 7,774.63$
View full question & answer→Question 72 Marks
Compound interest on a certain sum of money at $5\%$ per annum for two years is $Rs. 246. $ Calculate simple interest on the same sum for $3$ years at $ 6\%$ per annum.
Answer$\mathrm{Cl}=\text { Rs. } 246, \mathrm{R}=5 \%, \mathrm{~T}=2 \text { years }$
$ \mathrm{Cl}=\mathrm{A}-\mathrm{P}$
$ 246=\mathrm{P}\left[1+\frac{5}{100}\right]^2-\mathrm{P}$
$ 246=\mathrm{P}\left|\left(\frac{21}{20}\right)^2-1\right|$
$ 246=\mathrm{P}\left[\frac{41}{400}\right]$
$ \mathrm{P}=\frac{246 \times 400}{41}$
$ =\text { Rs. } 2400$
Now, $P=Rs.$ $2400, R=6 \%, T=3$ years$
\text { SI }=\frac{2400 \times 6 \times 3}{100}$
$ =\text { Rs. } 432 .$
View full question & answer→Question 82 Marks
At what per cent per annum will $Rs.6,000$ amount to $ Rs.6,615$ in $2$ years when interest is compounded annually?
Answer$\text { Amount }=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow 6,615=6,000\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow\left(1+\frac{r}{100}\right)^2=\frac{6,615}{6,000}$
$ \Rightarrow 1+\frac{r}{100}=\frac{21}{20}$
$ =r=5 \%$
At $5\%$ per annum the sum of $Rs. 6,000$ amounts to $Rs. 6,615 $ in $ 2 $ years when the interest is compounded annually.
View full question & answer→Question 92 Marks
What principal will amount to $Rs. 9,856$ in two years, if the rates of interest for successive years are $10\%$ and $12\%$ respectively ?
AnswerGiven : $A=$ Rs. 9,$856 ; n=2$ years; $r_1=10 \%$ and $r_2=12 \%$
$\text { Amount }=\mathrm{P}\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)$
$ \Rightarrow 9,856=\mathrm{P}\left(1+\frac{10}{100}\right)\left(1+\frac{12}{100}\right)$
$ \Rightarrow 9,856=P\left(\frac{11}{10}\right)\left(\frac{28}{25}\right)$
$ \Rightarrow \mathrm{P}=\text { Rs. } \frac{9,856 \times 10 \times 25}{11 \times 28}=\text { Rs. } 8,000$
View full question & answer→Question 102 Marks
Find the sum on which the compound interest for $3$ years at $10\%$ per annum amounts to $Rs. 1,655.$
AnswerGiven $: C.I. = Rs. 1,655 $;$ n=3$ years and $r=10 \%$
$\begin{aligned}
& \therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n \\
& \Rightarrow \mathrm{A}=\mathrm{P}\left(1+\frac{10}{100}\right)^3 \\
& \Rightarrow \mathrm{A}=\mathrm{P}\left(\frac{11}{10}\right)^3=\frac{1,331}{1,000} \mathrm{P} \\
& \therefore \mathrm{A}-\mathrm{P}=\text { C.I. } \\
& \Rightarrow \frac{1,331}{1,000} \mathrm{P}-\mathrm{P}=\text { Rs. } 1,655 \\
& \Rightarrow \frac{331}{1,000} \mathrm{P}=\text { Rs. } 1,665 \\
& \Rightarrow \mathrm{P}=\text { Rs. } \frac{1,655 \times 1,000}{331}=\text { Rs. } 5,000 .\end{aligned}$
View full question & answer→Question 112 Marks
What sum of money will amount to $Rs. 5,445$ in $2$ years at $10\%$ per annum compound interest ?
AnswerGiven : $P= Rs. 5,445 ; n=2$ years and $r=10 \%$
$\text { Amount }=\mathrm{P}\left(1+\frac{\mathrm{r}}{100}\right)^n$
$ \Rightarrow 5,445=\mathrm{P}\left(1+\frac{10}{100}\right)^2$
$ \Rightarrow 5,445=\mathrm{P}\left(\frac{11}{10}\right)^2$
$ \Rightarrow \mathrm{P}=5,445\left(\frac{10}{11}\right)^2$
$ \Rightarrow \text { Rs. } 4,500$
View full question & answer→Question 122 Marks
Calculate the compound interest accrued on $Rs. 6,000 $ in $3 $ years $,$ compounded yearly $,$ if the rates for the successive years are $5\%, 8\%$ and $10\% $ respectively.
AnswerGiven : $P= Rs. 6,000 ; n=3$ years; $r_1=5 \% ; r_2=8 \%$ and $r_3=10 \%$$
\text { Amount } =\mathrm{P}\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)\left(1+\frac{r_3}{100}\right)$
$ =6000\left(1+\frac{5}{100}\right)\left(1+\frac{8}{100}\right)\left(1+\frac{10}{100}\right)$
$ =6000\left(\frac{21}{20}\right)\left(\frac{27}{25}\right)\left(\frac{11}{10}\right)$
$ \quad=\text { Rs. } 7,484.40$
$ \therefore \text { C.I. }=\text { Rs. } 7,484.40-\text { Rs. } 6,000$
$=\text { Rs. } 1,484.40 .$
View full question & answer→Question 132 Marks
Calculate the amount of $Rs. 15,000$ is lent at compound interest for $2$ years and the rates for the successive years are $8\%$ and $10\%$ respectively.
AnswerGiven : $P= Rs. 15,000 ; n=2$ years; $r_1=8 \%$ and $r_2=10 \%$
$\text { Amount } =\mathrm{P}\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)$
$ =15000\left(1+\frac{8}{100}\right)\left(1+\frac{10}{100}\right)$
$ =15,000\left(\frac{27}{25}\right)\left(\frac{11}{10}\right)$
$ =\text { Rs. } 17,820 .$
View full question & answer→Question 142 Marks
Find the amount and the compound interest on $Rs. 12,000$ in $3$ years at $ 5\%$ compounded annually.
AnswerGiven : $P= Rs. 12,000 ; n=3$ years and $r=5 \%$
$\text { Amount } =\mathrm{P}\left(1+\frac{\mathrm{r}}{100}\right)^n$
$ =12000\left(1+\frac{5}{100}\right)^3$
$ =12000\left(\frac{21}{20}\right)^3$
$ =\text { Rs. } 13,891.50$
$\text { C.I. }=\text { Rs. } 13,891.50 \text { - Rs. } 12,000$
$=\text { Rs. } 1,891.50 \text {. }$
View full question & answer→Question 152 Marks
In how many years will $Rs. 7,000$ amount to $Rs. 9,317$ at $10\%$ per annum compound interest ?
AnswerGiven : $\mathrm{P}= Rs. 7,000 ; \mathrm{A}= Rs. 9,317$ and $\mathrm{r}=10 \%$.
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{r}}{100}\right)^{\mathrm{n}}$
$ \Rightarrow 9,317=7,000\left(1+\frac{10}{100}\right)^{\mathrm{n}}$
$ \Rightarrow \frac{9,317}{7,000}=\left(\frac{11}{10}\right)^{\mathrm{n}}$
$ \Rightarrow \frac{1,331}{1,000}=\left(\frac{11}{10}\right)^{\mathrm{n}}$
$ \Rightarrow\left(\frac{11}{10}\right)^3=\left(\frac{11}{10}\right)^{\mathrm{n}}$
On comparing,
$n=3 \text { years }$
View full question & answer→