Question 14 Marks
The simple interest on a certain sum of money for $3$ years at $5\%$ per annum is $Rs. 1,200.$ Find the amount and the compound interest due on this sum of money at the same rate and after $2$ years. Interest is reckoned annually.
AnswerLet $Rs. X$ be the sum of money.
Rate $=5 \% p.a.$ Simple interest $= Rs. 1,200, n=3$ years.
$1,200=\frac{X \times 5 \times 3}{100}$
$ \Rightarrow X=\frac{12,00,000}{15}$
$ \Rightarrow X=8,000$
The amount due and the compound interest on this sum of money at the same rate and after $2$ years.
$P=\text { Rs. } 8,000 \text {; rate }=5 \% \text { p.a., } n=3$ years
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow A=8,000\left(1+\frac{5}{100}\right)^2$
$\Rightarrow A=8,000(1.1025)$
$\Rightarrow \mathrm{A}=8,820$
$\text { C.I. }=\text { A - P }$
$ \Rightarrow \text { C.I. }=8,820-8,000$
$ \Rightarrow \text { C.I. }=820 .$
The amount due after $2$ years is $Rs. 8,820$ and the compound interest is $Rs. 820 .$
View full question & answer→Question 24 Marks
Anuj and Rajesh each lent the same sum of money for $2$ years at $8\%$ simple interest and compound interest respectively. Rajesh received $Rs. 64$ more than Anuj . Find the money lent by each and interest received.
AnswerLet the sum of money lent by both $Rs. y$
For Anuj
$P= Rs.y ;$ rate $=8 \%$ and time $=2$ years
$\therefore \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{y \times 8 \times 2}{100}=\frac{4 y}{25}$
For Rajesh
$P= Rs. y$; rate $=8 \%$ and time $=2$ years
$\therefore \text { C.I. }=\mathrm{P}\left[\left(1+\frac{r}{100}\right)^n-1\right]=y\left[\left(1+\frac{8}{100}\right)^2-1\right]=\frac{104 y}{625}$
Given $: C.I. - Rs. 64$
$\Rightarrow \frac{104 y}{625}-\frac{4 y}{25}=64$
$ \Rightarrow \frac{4 y}{625}=64$
$ \Rightarrow y=\frac{64 \times 625}{4}=\text { Rs. } 10,000$
Interest received by Anuj $=\frac{4 \times 10,000}{25}=\text { Rs. } 1600$
Interest received by Rajes$h=\frac{104 \times 10,000}{625}=\text { Rs. } 1664 \text {. }$
View full question & answer→Question 34 Marks
Nikita invests $Rs.6,000$ for two years at a certain rate of interest compounded annually. At the end of first year it amounts to $Rs.6,720.$ Calculate:$(a)$ The rate of interest.$(b)$ The amount at the end of the second year.
AnswerLet $\mathrm{X} \%$ be the rate of interest.
$P=\text { Rs. } 6,000, n=2$ years , $A=\text { Rs. } 6,720$
For the first year
$A=P\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow 6,720=6,000\left(1+\frac{\mathrm{X}}{100}\right)^1$
$ \Rightarrow 6,720-6,000=60 \mathrm{X}$
$ \Rightarrow \mathrm{X}=12$
The rate of interest is $\mathrm{X} \%=12 \%$.
The amount at the end of the second year.
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow \mathrm{A}=6,000\left(1+\frac{12}{100}\right)^2$
$ \Rightarrow \mathrm{A}=6,000\left(\frac{112}{100}\right)^2$
$ \Rightarrow \mathrm{A}=7,526.40$
The amount at the end of the second year $= Rs. 7,526.40$
View full question & answer→Question 44 Marks
A sum of money lent out at $C.I.$ at a certain rate per annum becomes three times of itself in $10$ years. Find in how many years will the money become twenty$-$seven times of itself at the same rate of interest $p.a.$
AnswerLet Principal be $Rs. y$ and rate $=r \%$
According to $1^{\text {st }}$ condition
Amount in $10$ years $= Rs.3 y$
$ \therefore A=P\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow 3 y=y\left(1+\frac{r}{100}\right)^{10}$
$\Rightarrow 3=\left(1+\frac{r}{100}\right)^{10} ...(1)$
According to $2^{\text {nd }}$ condition
Let after $n$ years amount will be $Rs. 27 y$
$ \therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow 27 \mathrm{y}=\mathrm{y}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow(3)^{\wedge} 3=\left(1+\frac{r}{100}\right)^n$
Put value from first equation
$\Rightarrow\left[\left(1+\frac{r}{100}\right)^{10}\right]^3=\left(1+\frac{r}{100}\right)^n$
On comparing,
we get $n=10 \times 3=30$ years
View full question & answer→Question 54 Marks
Find the difference between compound interest and simple interest on $Rs. 12,000$ and in $1 \frac{1}{2}$ years at $10 \%$ compounded yearly.
AnswerGiven: $P= Rs. 12,000 ; n=1 \frac{1}{2}$ years and $r=10 \%$
$ \text { S.I. }=\frac{P \times R \times T}{100}=\frac{12,000 \times 10 \times \frac{3}{2}}{100}=R s .1,800 $
To Calculate $C.I.$ annually
$P =\text { Rs. } 12,000 ; n =1$ year and $r =10 \%$
$A = P \left(1+\frac{r}{100}\right)^n\left(1+\frac{r}{200}\right)^{\frac{1}{2} \times 2}$
$=12,000\left(1+\frac{10}{100}\right)\left(1+\frac{10}{200}\right)$
$=12000 \times \frac{11}{10} \times \frac{21}{20}$
$=660 \times 21$
$=\text { Rs. } 13,860^{\circ}$
$\text { C.I }=13860-12000$
$=1860$
So Difference between $C.I$ and $C.I$
$\text { C.I. - S.I }$
$=1860-1800$
$Rs.60 $
View full question & answer→Question 64 Marks
At what rate of interest per annum will a sum of $Rs. 62,500$ earn a compound interest of $Rs. 5,100$ in one year? The interest is to be compounded half yearly.
Answer$ r =? P =62,500; C . I = A - P =5100$
$A=5100+62500$
$=67,600$
$A= P \left(1+\frac{r}{200}\right)^{2 n}$
$67600=62500\left(1+\frac{r}{200}\right)^2$
$=\left(1+\frac{r}{200}\right)^2=\frac{67,600}{62,500}$
$=\left(\frac{26}{25}\right)^2$
$=1+\frac{r}{200}=\frac{26}{25}$
$=\frac{r}{200}=\frac{62}{25}-1$
$=\frac{1}{5}$
$r=\frac{200}{25}$
$=8 \%$
$=r=8$
The rate of interest is $8\%.$
View full question & answer→Question 74 Marks
A man borrowed $Rs.16,000$ for $3$ years under the following terms:
$20\%$ simple interest for the first $2$ years.
$20\% C.I.$ for the remaining one year on the amount due after $2$ years, the interest being compounded half$-$yearly.
Find the total amount to be paid at the end of the three years.
AnswerFor the first $2$ years
$\text { S.I. }=\frac{P \times N \times R}{100}$
$ \Rightarrow \text { S.I. }=\frac{16,000 \times 2 \times 20}{100}$
$ \Rightarrow \text { S.I. }=6,400$
Amount $= S.I. +\mathrm{P}$
$\Rightarrow$ Amount $=6,400+16,000=\text { Rs. } 22,400$
Amount in the account at the end of the two years is $Rs. 22,400 .$
For the remaining one year
$A=P\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}$
$ \Rightarrow A=22,400\left(1+\frac{20}{200}\right)^2$
$ \Rightarrow A=22,400\left(\frac{11}{10}\right)^2$
$ \Rightarrow A=27,104$
The total amount to be paid at the end of the three years is $Rs. 27,104 .$
View full question & answer→Question 84 Marks
A certain sum of money amounts to $Rs. 23,400$ in $3$ years at $10\%$ per annum simple interest. Find the amount of the same sum in $2$ years and at $10\% p.a$. compound interest.
AnswerLet the sum $($principle$) =X$
Given Amount $=23400, R=10 \%$ and $T=3$ years
$\Rightarrow$ Interest $=\frac{\mathrm{X} \times 10 \times 3}{100}=\frac{3 X}{10}$
Amount $=$ Principle $+$ Interest
$23400=X+\frac{3 X}{10}$
$\mathrm{X}=18000$
Principle $=18000$
Now,
Principle $=18000, r=10 \%$ and $n=2$ years
$\mathrm{A}=\mathrm{P}\left[1+\frac{r}{100}\right]^n$
$A=18000\left[1+\frac{10}{100}\right]^2$
$ A=18000\left[\frac{11}{10}\right]^2$
$ A=18000\left[\frac{121}{100}\right]$
$A=21780$
The amount of the same sum in $2$ years and at $10 \% p.a.$ compound interest is $21780.$
View full question & answer→Question 94 Marks
Simple interest on a certain sum of money for $4$ years at $4\%$ per annum exceeds the compound interest on the same sum for $3$ years at $5$ per cent per annum by $Rs. 228.$ Find the sum.
AnswerLet principal $(P),$ $R=4 \%, T=4$ years
$\mathrm{SI}=\frac{\mathrm{P} \times 4 \times 4}{100}=\frac{4 \mathrm{P}}{25}$
$ \mathrm{CI}=\mathrm{P}\left[1+\frac{5}{100}\right]-\mathrm{P}$
$ =\mathrm{P}\left|\left(\frac{21}{20}\right)^3-1\right|$
$ =\mathrm{P}\left[\frac{9261}{8000}-1\right]$
$ =\frac{1261}{8000} \mathrm{P}$
Given:
$\mathrm{SI}=\mathrm{Cl}= Rs. 228$
$\frac{4 \mathrm{P}}{25}-\frac{1261}{8000} \mathrm{P}$
$ \frac{4 \times 320 \mathrm{P}-1261 \mathrm{P}}{8000}=228$
$19 \mathrm{P}=228 \times 8000$
$\mathrm{P}=\frac{228 \times 8000}{19}$
$P=\text { Rs. } 96000$
Thus, Principal $= Rs. 96000$
View full question & answer→Question 104 Marks
A sum of money lent out at $C.I.$ at a certain rate per annum becomes three times of itself in $8$ years. Find in how many years will the money becomes twenty$-$seven times of itself at the same rate of interest $p.a.$
AnswerLet principal $=\mathrm{X}, \mathrm{A}=3 \mathrm{X}, \mathrm{T}=8$ years, $\mathrm{R}=$ ?
Case $ I,$
$A=P\left[1+\frac{R}{100}\right]^T$
$ 3 X=X\left[1+\frac{R}{100}\right]^8$
$ 3^{\frac{1}{8}}=1+\frac{R}{100}$
Case $ II,$
$P=X, A=27 X, T=?$
$ 27 X=X\left[1+\frac{R}{100}\right]^T$
$ 27^{\frac{1}{T}}=1+\frac{R}{100}$
From $(1)$ and $(2)$
$3^{\frac{1}{8}}=27^{\frac{1}{T}}$
$3^{\frac{1}{8}}=3^{\frac{1}{8}}=\frac{3}{\mathrm{~T}}$
$\mathrm{T}=24$
Time $=24$ years.
View full question & answer→Question 114 Marks
The difference between simple interest and compound interest on a certain sum is $Rs. 54.40$ for $2$ years at $8$ per cent per annum. Find the sum.
Answer Let principal $(P)=x$
$ \mathrm{R}=8 \%$
$ \mathrm{~T}=2$ years
$ \mathrm{SI}=\frac{\mathrm{X} \times 8 \times 2}{100}=\frac{4 X}{25}$
$ \mathrm{Cl}=\mathrm{A}-\mathrm{P}=\mathrm{X}\left(1+\frac{8}{100}\right)^2-\mathrm{X}$
$ =\mathrm{X}\left|\left[1+\frac{2}{25}\right]^2-1\right|$
$ =\mathrm{X}\left|\left[\frac{27}{25}\right]^2-1\right|$
$ =\frac{104}{625} X$
Given,
$\mathrm{Cl}=\mathrm{SI}=54.40$
$ \frac{104 X}{625}-\frac{4 X}{25}=\text { Rs. } 54.40$
$ X\left(\frac{104}{625}-\frac{4}{25} \times \frac{25}{25}\right)=54.40$
$ X\left(\frac{4}{625}\right)=54.40$
$ X=\frac{54.40 \times 625}{4}$
$X=\text { Rs. } 8500$
Thus, principal sum $= Rs. 8500$
View full question & answer→Question 124 Marks
Find the difference between compound interest and simple interest on $Rs. 8,000$ in $2$ years and at $5\%$ per annum.
AnswerGiven : $P= Rs. 8,000, R=5 \%, T=2$ years
For simple interest,
$\text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =\frac{8,000 \times 5 \times 2}{100}$
$ =\text { Rs. } 800$
For compound interest,
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ A=8,000\left(1+\frac{5}{100}\right)^2$
$ =8,000 \times \frac{21}{20} \times \frac{21}{20}$
$ =\text { Rs. } 8,820$
$ \text { C.I. }=\mathrm{A}-\mathrm{P}$
$ =\text { Rs. }(8,820-8,000)$
$ =\text { Rs. } 820$
Now,
$ C.I. - S.I. = Rs. (820-800)= Rs. 20.$
Thus, the difference between the compound interest and the simple interest is $Rs. 20.$
View full question & answer→Question 134 Marks
Divide $Rs. 28,730$ between $A$ and $B$ so that when their shares are lent out at $10\%$ compound interest compounded per year, the amount that $A$ receives in $3$ years is the same as what $B$ receives in $5$ years.
AnswerLet share of $A= Rs. y$
share of $B=Rs(28,730-y)$
rate of interest $=10 \%$
According to question,
Amount of $A$ in $3$ years $=$ Amount of $B$ in $5$ years
$\Rightarrow y\left(1+\frac{10}{100}\right)^3=(28,730-y)\left(1+\frac{10}{100}\right)^5$
$ \Rightarrow y=(28,730-y)\left(1+\frac{10}{100}\right)^2$
$ \Rightarrow y=(28,730-y)\left(\frac{121}{100}\right)$
$ \Rightarrow 100 y=121(28,730-y)$
$ \Rightarrow 100 y+121 y=121 \times 28,730$
$ \Rightarrow 221 y=121 \times 28,730$
$ \Rightarrow y=\frac{121 \times 28,730}{221}=\text { Rs. } 15,730$
Therefore share of $A= Rs. 15,730$
Share of $B= Rs. 28,730 - Rs. 15,730$
$= Rs. 13,000$
View full question & answer→Question 144 Marks
A person invests $Rs.5,000$ for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to $Rs.6,272.$ Calculate $:(i) $the rate of interest per annum$.(ii)$ the amount at the end of the third year.
AnswerGiven: $\mathrm{P}= Rs. 5,000 ; \mathrm{A}= Rs. 6,272$ and $\mathrm{n}=2$ years.
$\text { (i) } \therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow 6,272=5,000\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow 6, \frac{272}{5}, 000=\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow \frac{784}{625}=\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow\left(\frac{28}{25}\right)^2=\left(1+\frac{r}{100}\right)^2$
On comparing,
$\frac{28}{25}=1+\frac{r}{100}$
On solving,
we get
$r=12 \%$
(ii) Amount at the third year
$=5,000\left(1+\frac{12}{100}\right)^3$
$ =5000\left(\frac{28}{25}\right)^3$
$ =\text { Rs. } 7,024.64$
View full question & answer→Question 154 Marks
A sum of money was invested for $3$ years, interest being compounded annually. The rates for successive years were $10\%, 15\%$ and $18\%$ respectively. If the compound interest for the second year amounted to $Rs. 4,950,$ find the sum invested.
AnswerGiven : $C.I.$ for the $2^{\text {nd }}$ year $= Rs. 4,950$ and rate $=15 \%$
Then, $C.I.=P\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$ \Rightarrow 4,950=P\left[\left(1+\frac{15}{100}\right)^1-1\right]$
$ \Rightarrow 4,950=P\left[\frac{3}{20}\right]$
$ \Rightarrow P=\frac{4,950 \times 20}{3}$
$ \Rightarrow P=\text { Rs. } 33,000 .$
Then amount at the end of $2^{\text {nd }}$ year$= Rs. 33,000$
For first $2$ years
$A=\text { Rs. } 33,000 ; r_1=10 \%$
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r_1}{100}\right)$
$\Rightarrow 33,000=P\left(1+\frac{10}{100}\right)$
$ \Rightarrow 33,000=P\left(\frac{11}{10}\right)$
$\Rightarrow \mathrm{P}=\frac{33,000 \times 10}{11}= Rs. 30,000$
The $\sum$ invested is $Rs. 30,000 .$
View full question & answer→Question 164 Marks
The value of a machine, purchased two years ago, depreciates at the annual rate of $10\%$. If its present value is $Rs.97,200$, find:
- Its value after 2 years.
- Its value when it was purchased.
Answer$(i)$ Present value of machine $(\mathrm{P})=\mathrm{Rs} .97,200$
Depreciation rate $=10 \%$
$\therefore$ Value of machine after $2$ years $=\mathrm{P}\left(1-\frac{r}{100}\right)^n$
$=97,200\left(1-\frac{10}{100}\right)^2$
$ =97,200\left(\frac{9}{10}\right)^2$
$ =\text { Rs. } 78732 .$
$(ii)$ Present value of machine $(A)= Rs. 97,200$
Depreciation rate $=10 \%$ and time $=2$ years
To calculate the cost $2$ years ago
$\therefore A=P\left(1-\frac{r}{100}\right)^n$
$ \Rightarrow 97,200=P\left(1-\frac{10}{100}\right)^2$
$ \Rightarrow 97,200=P\left(\frac{9}{10}\right)^2$
$ \Rightarrow P=\text { Rs. } 97,200 \times\left(\frac{10}{9}\right)^2$
$=1,20,000$
View full question & answer→Question 174 Marks
A $\sum$ of money, invested at compound interest, amounts to $Rs. 16,500$ in $1$ year and to $Rs. 19,96$5 in $3$ years. Find the rate per cent and the original sum of money invested.
AnswerLet $\sum $of money be $Rs. \mathrm{P}$ and rate of interest $=\mathrm{r} \%$
Money after $1$ year $= Rs. 16,500$
Money after $3$ years $= Rs. 19,965$
For $1$ year
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 16,500=\mathrm{P}\left(1+\frac{r}{100}\right)^1 \dots...(1)$
For $3$ years
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow 19,965=\mathrm{P}\left(1+\frac{r}{100}\right)^3 \dots...(2)$
Divide eq ${ }^{\mathrm{n}}(2)$ by eq ${ }^{\mathrm{n}}(1)$
$ \frac{19,965}{16,500}=\frac{P\left(1+\frac{r}{100}\right)^3}{P\left(1+\frac{r}{100}\right)^1}$
$ \Rightarrow \frac{121}{100}=\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow\left(\frac{11}{10}\right)^2=\left(1+\frac{r}{100}\right)^2$
On comparing, we get
$\Rightarrow \frac{11}{10}=1+\frac{r}{100}$
$ \Rightarrow r=10 \%$
Put value of $r$ in eq ${ }^n(1)$
$16,500=P\left(1+\frac{10}{100}\right)$
$ \Rightarrow P=\frac{16,500 \times 10}{11}$
$=\text { Rs. } 15,000 .$
View full question & answer→Question 184 Marks
Ashok invests a certain sum of money at $20\%$ per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half$-$yearly. If Geeta gets $Rs. 33$ more than Ashok in $18$ months, calculate the money invested.
Answer$(i)$ For Ashok$($interest is compounded yearly$) $:
Let $P= Rs. y ; n=18$ months $=1 \frac{1}{2}$ year and $r=20 \% p.a.$
For $1$ year
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n=y\left(1+\frac{20}{100}\right)^1=\left(\frac{6}{5}\right) y$
For $\frac{1}{2}$ year
$\mathrm{P}=$ Rs. $\left(\frac{6}{5}\right) y ; n=\frac{1}{2}$ year and $\mathrm{r}=20 \%$
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}=R s .\left(\frac{6}{5}\right) y\left(1+\frac{20}{2 \times 100}\right)^{\frac{1}{2} \times 2}=R s .\left(\frac{66}{50}\right) y$
$(ii)$ For Geeta $($ interest is compounded half$-$yearly $)$
$P= Rs. y ; n=1 \frac{1}{2}$ year and $r=20 \% p.a.$
$\mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}=y\left(1+\frac{20}{2 \times 100}\right)^{\frac{3}{2} \times 2}=y\left(\frac{11}{10}\right)^3$
$ =\text { Rs. } \frac{1,331}{1,000} y$
According to question
$\therefore \frac{1,331}{1,000} y-\left(\frac{66}{50}\right) y=R s .33$
$ =\left(\frac{11}{1,000}\right) y=R s .33$
$ =\mathrm{y}=\text { Rs. } \frac{33 \times 1,000}{11}=R s .3,000$
Money invested by each person$=Rs. 3,000.$
View full question & answer→Question 194 Marks
Find the difference between the compound interest compounded yearly and half$-$yearly on $Rs. 10,000$ for $18$ months at $10\%$ per annum.
Answer$(i)$ When interest is compounded yearly:
Given : $P=R s .10,000 ; n=18$ months $=1 \frac{1}{2}$ year and $r=10 \% p.a.$
For $1$year
$A=P\left(1+\frac{r}{100}\right)^n\left(1+\frac{r}{200}\right)^{\frac{1}{ Z }} \times Z$
$=10,000 \times \frac{11}{10} \times \frac{21}{20}$
$=1020 \times 11$
$=11,550$
When interest is compounded half$-$yearly
$A=P\left(1+\frac{r}{200}\right)^{n \times 2}$
$=10,000\left(1+\frac{10}{200}\right)^{\frac{3}{\nexists} \times \not 2}$
$=10,000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$
$=\frac{105 \times 441}{4}$
$=\frac{46305}{4}$
$=11576.25$
$=\frac{46305}{4}$
$=11576.25$
Therefore, the difference between both $C.I$
$₹11,576.25 - ₹11,550$
$= ₹26.25$
Therefore, the difference between both $C.I$
View full question & answer→Question 204 Marks
Kamal and Anand each lent the same sum of money for $2$ years at $5\%$ at simple interest and compound interst respectively. Anand recived $Rs. 15$ more than Kamal. Find the amount of money lent by each and the interest received.
AnswerLet principal $= Rs. 100, R=5 \% T=2$ years
For Kamal, $SI =\frac{100 \times 5 \times 2}{100}=Rs. 10$
For Anand, $ A=P\left[1+\frac{R}{100}\right]^T$
$=100\left[1+\frac{5}{100}\right]^2$
$ =100 \times \frac{21}{20} \times \frac{21}{20}$
$=\frac{441}{4}$
$\mathrm{Cl}=\frac{441}{4}-100=\frac{41}{4}$
Difference of $\mathrm{Cl}$ and $\mathrm{SI}=\frac{41}{4}-10$
$=\frac{41-40}{4}$
$ =\text { Rs. } \frac{1}{4}$
When difference is $Rs. \frac{1}{4}$,
then principal $= Rs. 100 .$
If difference is $1 $, principal $=100 \times 4$
If difference is $Rs. 15 ,$ principal $=100 \times 4 \times 15= Rs. 6000$
For kamal, interest $=\frac{6000 \times 5 \times 2}{100}= Rs. 600 .$
For Anand, interest $=6000\left[1+\frac{5}{100}\right]^2-6000$
$=6000\left[\left(\frac{21}{20}\right)^2-1\right]$
$ =6000\left[\frac{441}{400}-1\right]$
$ =6000 \times \frac{41}{400}$
$ =\text { Rs. } 615$
View full question & answer→Question 214 Marks
A sum of money, invested at compound interest, amounts to $Rs. 19,360$ in $2$ years and to $Rs. 23,425.60$ in $4$ years. Find the rate per cent and the original sum of money.
Answer$($for $2$ years$) A = Rs. 19360$
$T=2$ years
Let $P=X$
$\mathrm{X}\left(1+\frac{\mathrm{R}}{100}\right)^2=19,360\dots ......(1)$
$A($ for $4$ years$) = Rs. 23425.60$
$X\left(1+\frac{R}{100}\right)^4=23425.60 \dots......(2)$
$ 2 \div 1$
$ {\left[1+\frac{\mathrm{R}}{100}\right]^2=\frac{23425.60}{19360}}$
$ {\left[1+\frac{\mathrm{R}}{100}\right]^2=\frac{2342560}{1936000}}$
$ {\left[1+\frac{\mathrm{R}}{100}\right]^2=\frac{14641}{12100}}$
$ {\left[1+\frac{\mathrm{R}}{100}\right]^2=\left[\frac{121}{110}\right]^2}$
$ 1+\frac{\mathrm{R}}{100}=\frac{121}{110}$
$ \frac{\mathrm{R}}{100}=\frac{121}{110}-1$
$R=10 \%$
From $(1) \times\left[1+\frac{10}{100}\right]^2=19360$
$X=\frac{19360 \times 10 \times 10}{11 \times 11}$
$X= Rs. 16,000$
Thus,$ \sum = Rs. 16000$
View full question & answer→Question 224 Marks
Mohit borrowed a certain sum at $5\%$ per annum compound interest and cleared this loan by paying $Rs. 12,600$ at the end of the first year and $Rs. 17,640$ at the end of the second year. Find the sum borrowed.
AnswerFor the payment of $Rs. 12,600$ at the end of first year : $A= Rs. 12,600 ; n=1$ year and $r=5 \%$
Now, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 12,600=\mathrm{P}\left(1+\frac{5}{100}\right)^1$
$ \Rightarrow 12,600=\mathrm{P}\left(\frac{21}{20}\right)$
$ \Rightarrow \mathrm{P}=\frac{20}{21} \times 12,600=\text { Rs. } 12,000$
For the payment of $Rs. 17,640$ at the end of second year : $A=Rs. 17,640, n=2$ years and $r=5 \%$
Now, $\mathrm{A}=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 17,640=\mathrm{P}\left(1+\frac{5}{100}\right)^2$
$ \Rightarrow 17,640=\mathrm{P}\left(\frac{21}{20}\right)^2$
$ \Rightarrow \mathrm{P}=\frac{20}{21} \times \frac{20}{21} \times 17,640=\text { Rs. } 16,000$
$\therefore$ Sum borrowed $= Rs. (12,000+16,000)= Rs. 28,000$.
View full question & answer→Question 234 Marks
A sum of $Rs. 44,200$ is divided between John and Smith, $12$ years and $14$ years old respectively, in such a way that if their portions be invested at $10\%$ per annum compound interest, they will receive equal amounts on reaching $16$ years of age.$(i)$ What is the share of each out of $Rs.44,200$ ?$(ii)$ What will each receive, when $16$ years old ?
Answer$(i)$Let share of John $=\text { Rs y }$
share of Smith =$\text { Rs }(44,200-y)$
rate of interest$=10 \%$
According to question,
Amount of John in $4$ years $=$ Amount of Smith in $2$ years
$\Rightarrow y\left(1+\frac{10}{100}\right)^4=(44,200-y)\left(1+\frac{10}{100}\right)^2$
$ \Rightarrow y\left(1+\frac{10}{100}\right)^2=(44,200-y)$
$ \Rightarrow y\left(\frac{11}{10}\right)^2=(44,200-y)$
$ \Rightarrow 121 y=100(44,200-y)$
$ \Rightarrow 121 y=100 \times 44,200-100 y$
$ \Rightarrow 121 y+100 y=100 \times 44,200$
$ \Rightarrow 221 y=100 \times 44,200$
$ \Rightarrow y=100 \times 44, \frac{200}{221}=\text { Rs. } 20,000 .$
Therefore share of John $=Rs. 20,000$
Share of Smith $= Rs. 44,200- Rs. 20,000= Rs. 24,200$
$(ii)$ Amount that each will receive
$=20,000\left(1+\frac{10}{100}\right)^4$
$ =20,000\left(\frac{11}{10}\right)^4$
$= Rs. 29,282$
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