Question 13 Marks
A sum of money is invested at $10\%$ per annum compounded half yearly. If the difference of amounts at the end of $6 $months and $12$ months is $Rs.189,$ find the sum of money invested.
AnswerLet the sum of money be $Rs. y$
and rate $=10 \%$ p.a. compounded half yearly
For first $6$ months
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}=y\left(1+\frac{10}{2} \times 100\right)^{\frac{1}{2} \times 2}=\left(\frac{21}{20}\right) y$
For first $12$ months
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}=y\left(1+\frac{10}{2} \times 100\right)^{1 \times2}=\left(\frac{441}{400}\right) y$
Given: The difference between the above amounts $= Rs. 189$
$\Rightarrow\left(\frac{441}{400}\right) y-\left(\frac{21}{20}\right) y=189$
$ \Rightarrow\left(\frac{21}{400}\right) y=189$
$ \Rightarrow y=\frac{189 \times 400}{21}$
$ y=3600$
View full question & answer→Question 23 Marks
Calculate the sum of money on which the compound interest $($payable annually$)$ for $2$ years be four times the simple interest on $Rs. 4,715$ for $5$ years, both at the rate of $5\%$ per annum.
AnswerGiven : Principal $=Rs. 4,715 ;$ time $=5$ years and rate $=5 \%$ p.a.
$\therefore \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{4715 \times 5 \times 5}{100}=R s .1,178.75$
Then C.I. $= Rs. 1,178.75 \times 4= Rs. 4,715$
Time $=2$ years and rate $=5 \%$
$\therefore \text { C.I. }=\mathrm{P}\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$ \Rightarrow 4,715=\mathrm{P}\left[\left(1+\frac{5}{100}\right)^2-1\right]$
$ \Rightarrow 4,715=\mathrm{P}\left[\frac{41}{400}\right]$
$ \Rightarrow \mathrm{P}=\text { Rs. } \frac{4,715 \times 400}{41}=\text { Rs. } 46,000 .$
View full question & answer→Question 33 Marks
Simple interest on a sum of money for $2$ years at $4\%$ is $Rs .450.$ Find compound interest on the same sum and at the same rate for $1$ year, if the interest is reckoned half yearly.
Answer$1^{\text {st }}$ case
Given $:\text{ S.I.}={\text{Rs.} 450} ;$ Time $=2$ years and Rate $=4 \%$
$\therefore$ Principle $=\frac{\mathrm{I} \times 100}{\mathrm{R} \times \mathrm{T}}=\frac{450 \times 100}{4 \times 2}= Rs. 5625 .$
$2^{\text {nd }}$ case $($ compounded half$-$yearly $)$
$P= Rs. 5,625 ; n=1$ year and $r=4 \%$
$\therefore \mathrm{A}=\mathrm{P}\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}=5,625\left(1+\frac{4}{2 \times 100}\right)^{1 \times 2}$
$ =5625\left(\frac{51}{50}\right)^2$
$ =\text { Rs. } 5852.25$
$\therefore \text { C.I. }=5,852 \cdot 25-5,625$
$=\text { Rs. } 227.25$
View full question & answer→Question 43 Marks
The difference between compound interest for a year payable half$-$yearly and simple interest on a certain sum of money lent out at $10\%$ for a year is $Rs. 15.$ Find the sum of money lent out.
AnswerLet sum of money be $Rs. y$
To calculate $ \text{S.I.}$
$\text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{y \times 10 \times 1}{100}=R s . \frac{y}{10}$
To calculate $ \text{C.I}.($compounded half$-$yearly$)$
$\therefore \text { C.I. }=\mathrm{P}$
$ {\left[\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}-1\right]=y\left[\left(1+\frac{10}{2 x 100}\right)^{1 \times 2}-1\right]}$
$ =y\left[\left(\frac{21}{20}\right)^2-1\right]=\left(\frac{41}{400}\right) y$
Given $:\text {C.I.} - \text{S.I.} = Rs. 15$
$\Rightarrow\left(\frac{41}{400}\right) y-\frac{y}{10}=15$
$ \Rightarrow \frac{y}{400}=15$
$\Rightarrow y=\text { Rs. } 6,000 .$
View full question & answer→Question 53 Marks
Mr. Sharma borrowed a certain sum of money at $10\%$ per annum compounded annually. If by paying $Rs.19,360$ at the end of the second year and $Rs. 31,944$ at the end of the third year he clears the debt; find the sum borrowed by him.
AnswerAt the end of the two years the amount is
$A_1=P\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow A_1=P\left(1+\frac{10}{100}\right)^2$
Mr. Sharma paid $Rs.19,360$ at the end of the second year.
So for the third year the principal is $\mathrm{A}_1-19,360$.
Also he cleared the debt by paying $Rs.31,944$ at the end of the third year.
$\mathrm{A}_2=\mathrm{P}\left(1+\frac{r}{100}\right)^n$
$ \Rightarrow 31,944=\left(\mathrm{P}\left(1+\frac{10}{100}\right)^2-19,360\right)\left(1+\frac{10}{100}\right)^1$
$ \Rightarrow 29040=\left(\mathrm{P}\left(1+\frac{10}{100}\right)^2-19,360\right)$
$ \Rightarrow\left(\mathrm{P}\left(1+\frac{10}{100}\right)^2=48,400\right.$
$ \Rightarrow \mathrm{P}=\text { Rs. } 40,000$
Mr. Sharma borrowed $Rs. 40,000.$
View full question & answer→Question 63 Marks
The difference between $\text{C.I.}$ and $\text{S.I.}$ on $Rs. 7,500$ for two years is $Rs. 12$ at the same rate of interest per annum. Find the rate of interest.
AnswerGiven $: P= Rs. 7,50$0 and Time $(n)=2$ years
Let rate of interest $=y \%$
$\therefore \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\frac{7500 \times y \times 2}{100}=R s .150 y$
$ \therefore \text { C.I. }=\mathrm{P}$
$ \left(1+\frac{r}{100}\right)^n-\mathrm{P}=\text { Rs. } 7,500\left(1+\frac{y}{100}\right)^2-\text { Rs. } 7,500$
Given $: C.I. : S.I. = Rs. 12$
$\Rightarrow 7,500\left[1+\frac{y}{100}\right]^2-7,500-15 y=12$
$ \Rightarrow 7,500\left[1+\frac{y^2}{10000}+\frac{2 y}{100}\right]-7,500-150 y=12$
$ \Rightarrow 7,500+\frac{7500 y^2}{10000}+150 y-7,500-150 y=12$
$ \Rightarrow \frac{3 y^2}{4}=12$
$ \Rightarrow y^2=16$
$ \Rightarrow y=4 \%$
View full question & answer→Question 73 Marks
The value of an article decreases for two years at the rate of $10\%$ per year and then in the third year it increases by $10\%.$ Find the original value of the article, if its value at the end of $3$ years is $Rs. 40,095.$
AnswerLet $X$ be the value of the article.
The value of an article decreases for two years at the rate of $10 \%$ per year.
The value of the article at the end of the $1^{\text {st }}$ year is
$X-10 \%$ of $X=0.90 X$
The value of the article at the end of the $2^{\text {nd }}$ year is
$0.90 X-10 \%$ of $(0.90 X)=0.81 X$
The value of the article increases in the $3^{\text {rd }}$ year by $10 \%$.
The value of the article at the end of $3^{\text {rd }}$ year is
$0.81 x+10 \%$ of $(0.81 x)=0.891 x$
The value of the article at the end of $3$ years is $Rs. 40,095$ .
$0.891 X=40,095$
$\Rightarrow X=45,000$
The original value of the article is $Rs. 45,000 .$
View full question & answer→Question 83 Marks
The cost of a machine is supposed to depreciate each year at $12\%$ of its value at the beginning of the year. If the machine is valued at $Rs. 44,000$ at the beginning of $2008,$ find its value :
$(i)$ at the end of $2009.$
$(ii)$ at the beginning of $2007.$
AnswerCost of machine in $2008=Rs. 44,000$
Depreciation rate $=12 \%$
$(i)\therefore$ Cost of machine at the end of $2009$
$=\mathrm{P}\left(1-\frac{r}{100}\right)^n$
$ =44,000\left(1-\frac{12}{100}\right)^2$
$ =44,000 \times\left(\frac{88}{100}\right)^2$
$ =\text { Rs. } 34,073.60$
$(ii)$ Cost of machine at the beginning of $2007(P)$
$A=P\left(1-\frac{r}{100}\right)^n$
$ \Rightarrow 44,000=\mathrm{P}\left(1-\frac{12}{100}\right)^1$
$ \Rightarrow 44,000=\mathrm{P}\left(\frac{88}{100}\right)^1$
$ \Rightarrow \mathrm{P}=\frac{44,000 \times 100}{88}$
$=\text { Rs. } 50,000$
View full question & answer→Question 93 Marks
The ages of Pramod and Rohitare $16$ years and $18$ years respectively. In what ratio must they invest money at $5\%$ p.a. compounded yearly so that both get the same sum on attaining the age of $25$ years?
AnswerLet $Rs.X$ and $Rs.Y$ be the money invested by Pramod and Rohit respectively such that they will get the same sum on attaining the age of $25$ years.
Pramod will attain the age of $25$ years after $25-16=9$ years Rohit will attain the age of $25$ years after $25-18=7$ years
$X\left(1+\frac{5}{100}\right)^9=y\left(1+\frac{5}{100}\right)^7$
$ \Rightarrow \frac{X}{Y}=\frac{1}{\left(1+\frac{5}{100}\right)^2}$
$ \Rightarrow \frac{X}{Y}=\frac{400}{441}$
Pramod and Rohit should invest in $400: 441$ ratio respectively such that they will get the same sum on attaining the age of $25$ years.
View full question & answer→Question 103 Marks
Calculate the $\text{C.I.}$ on $Rs. 3,500$ at $6\%$ per annum for $3$ years, the interest being compounded half$-$yearly.
Do not use mathematical tables. Use the necessary information from the following:
$(1.06)^3=1.191016; (1.03)^3= 1.092727$
$(1.06)^6=1.418519; (1.03)^6= 1.194052$
AnswerGiven $: P= Rs. 3,500 ; r=6 \%$ and $n=3$ years
Since interest is being compounded half$-$yearly
Then $, \text{C.I.}=\mathrm{P}\left[\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}-100\right]$
$ =3,500\left[\left(1+\frac{6}{2 \times 100}\right)^{n \times 2}-1\right] $
$ =3,500\left[\left(\frac{103}{100}\right)^6-1\right] $
$ =3,500\left[(1.03)^6-1\right] $
$ =3,500[1.194052-1] $
$ =3,500 \times 0.194052 $
$ =\text { Rs. } 679.18$
View full question & answer→Question 113 Marks
In what time will $Rs. 1,500$ yield $Rs. 496.50$ as compound interest at $20\%$ per year compounded half$-$yearly ?
AnswerGiven: $P= Rs. 1,500 ; \text{C.I. }= Rs. 496.50$ and $r=20 \%$
Since interest is compounded semi$-$annually
Then, $\text{C.I.} =\mathrm{P}\left[\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}-1\right]$
$\Rightarrow 496.50=1,500\left[\left(1+\frac{20}{2 \times 100}\right)^{n \times 2}-1\right]$
$\Rightarrow \frac{496.50}{1500}=\left(\frac{11}{10}\right)^{2 n}-1$
$\Rightarrow \frac{331}{1000}+1=\left(\frac{11}{10}\right)^{2 n}$
$\Rightarrow \frac{1331}{1000}=\left(\frac{11}{10}\right)^{2 n}$
$\Rightarrow\left(\frac{11}{10}\right)^3=\left(\frac{11}{10}\right)^{2 n}$
On comparing, we get,
$2 n=3$
$\Rightarrow n=1 \frac{1}{2}$ years
View full question & answer→Question 123 Marks
Find the difference between compound interest and simple interest on $Rs. 12,000$ and in $1 \frac{1}{2}$ years at $10 \%$ compounded half$-$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 $\text{C.I.} ($ Compound half $-$ yearly$ ) :$
$P= Rs. 12,000 ; n=1 \frac{1}{2}$ years and $r=10 \%$
$A=P\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}$
$ A=12,000\left(1+\frac{10}{2 \times 100}\right)^{\frac{3}{2} \times 2}$
$ A=12,000\left(\frac{21}{20}\right)^3$
$A=\text { Rs. } 13,891.50$
$ \text{ C.I. }=\text { Rs. } 13,891.50 \text { - Rs. } 12,000=\text { Rs. } 1,891.50$
$\therefore$ Difference between $\text{C.I.}$ and $\text{S.I} = Rs. 1,891.50- Rs. 1,800$
$= Rs. 91.50 .$
View full question & answer→Question 133 Marks
Simple interest on a sum of money for $2$ years at $4\%$ is $Rs. 450.$ Find compound interest of the same sum and at the same rate for $2$ years.
Answer$\text{ S I}=\text { Rs. } 450$
$ R=4 \%$
$ T=2 \text { years }$
$ P=?$
$ P=\frac{S I \times 100}{R \times T}=\frac{450 \times 100}{4 \times 2}=R s .5625$
$ A=5625\left[1+\frac{4}{100}\right]^2$
$ =5625\left(\frac{26}{25}\right)^2$
$ =\frac{3802500}{625}$
$ =\text { Rs. } 6084$
$\text{C I}=A-P$
$=6084-5625$
$= Rs. 459$
View full question & answer→Question 143 Marks
On what sum of money will compound interest $($payable annually$)$ for $2$ years be the same as simple interest on $Rs. 9,430$ for $10$ years, both at the rate of $5$ per cent per annum ?
Answer$P=\text { Rs. } 9430$
$ R=5 \%$
$ T=10$ years
$ \text{SI}=\frac{9430 \times 5 \times 10}{100}=\text { Rs. } 4715$
Let sum $=\mathrm{X}$
$\mathrm{Cl}=4715, \mathrm{~T}=2$ years,$\mathrm{Rs}=5 \%$
$\mathrm{Cl}=\mathrm{AP}$
$4715=\mathrm{X}\left[1+\frac{\mathrm{R}}{100}\right]^{\mathrm{T}}-\mathrm{X}$
$ 4715=\mathrm{X}\left[1+\frac{5}{100}\right]^2-\mathrm{X}$
$4715=X\left|\left(\frac{21}{20}\right)^2-1\right|$
$4715=X\left(\frac{441-400}{400}\right)$
$X=\frac{4715 \times 400}{41}=\text { Rs. } 46,000$
Thus principal from $= Rs. 46,000$
View full question & answer→Question 153 Marks
On a certain sum, the compound interest in $2$ years amounts to$\text{ Rs.} 4,240.$ If the rate of interest for the successive years is $10\%$ and $15\%$ respectively, find the sum.
Answer$\text { C.I }=\mathrm{A}-\mathrm{P}$
$ 4240=\mathrm{P}\left(1+\frac{\mathrm{r}_1}{100}\right)\left(1+\frac{\mathrm{r}_2}{100}\right)-\mathrm{P}$
$ 4240=\mathrm{P}\left[\left(1+\frac{10}{100}\right)\left(1+\frac{15}{100}\right)-1\right]$
$ 4240=\mathrm{P}\left[\frac{11}{10} \times \frac{115}{100}-1\right]$
$ 4240=\mathrm{P}\left[\frac{1265}{1000}-1\right]$
$ 4240=\mathrm{P}\left[\frac{1265-1000}{1000}\right]$
$ 4240=\mathrm{P}\left[\frac{265}{1000}\right]$
$ \frac{4240 \times 1000}{265}=\mathrm{P}$
$ 16000=\mathrm{P}$
The sum is $\text{Rs.} 16,000 .$
View full question & answer→Question 163 Marks
On what sum of money will the compound interest for $2$ years at $5\%$ per annum amount to $\text{Rs.} 768.75$?
AnswerGiven $: \text{C.I.} = Rs. 768.75;$
$n=2$ years and $r=5 \%$
$\therefore \text { Amt. }= P \left(1+\frac{r}{100}\right)^n$
$\Rightarrow A = P \left(1+\frac{5}{100}\right)^2$
$\Rightarrow A =P\left(\frac{21}{20}\right)^2=\frac{441}{400} P$
$Cl =\text { Amt }- P$
$=P \times\left(\frac{21}{20} \times \frac{21}{20}\right)-P=768.75$
$=P\left(\frac{21}{20} \times \frac{21}{20}-1\right)=768.75$
$=P \times\left(\frac{441-400}{400}\right)=768.75$
$=P \times \frac{41}{400}=768.75$
$=\frac{768.75 \times 400}{41 \times 100}$
$P=\text{Rs.} 7500$
View full question & answer→Question 173 Marks
The simple interest on a certain sum of money and at $10\%$ per annum is $Rs. 6,000$ in $2$ years, Find$:$
- thesum.
- theamount due to the end of 3 years and at the same rate of interest compounded annually.
- thecompound interest earned in 3 years.
Answer$(i) I = Rs. 6,000, T=2$ years and $R=10 \%$
$\therefore \mathrm{P}=\frac{\mathrm{I} \times 100}{\mathrm{R} \times \mathrm{T}}=\frac{6,000 \times 100}{10 \times 2}=\text { Rs. } 30,000$
$(ii) P= Rs. 30,000, n=3$ years and $r=10 \%$
$A=P\left(1+\frac{r}{100}\right)^n$
$ =30,000\left(1+\frac{10}{100}\right)^3$
$ =30,000\left(\frac{11}{10}\right)^3$
$ =30,000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$
$ =\text { Rs. } 39,930$
$(iii) C.I.$ earned in $3$ years $=A-P= Rs. (39,930-30,000)$
$= Rs. 9,930.$
View full question & answer→Question 183 Marks
Find the time, in years, in which $\text{Rs.} 4,000$ will produce $\text{Rs.} 630.50$ as compound interest at $5\%$ compounded annually.
AnswerGiven $: P= Rs. 4,000 ; C.I. = Rs. 630.50$ and $r=5 \%$
$\therefore \text { C.I. }=P\left[\left(1+\frac{r}{100}\right)^n-1\right]$
$ \Rightarrow 630.50=4,000\left[\left(1+\frac{5}{100}\right)^n-1\right]$
$ \Rightarrow \frac{630.50}{4,000}=\left[\left(\frac{21}{20}\right)^n-1\right]$
$ \Rightarrow \frac{1,261}{8,000}=\left[\left(\frac{21}{20}\right)^n-1\right]$
$ \Rightarrow \frac{1,261}{8,000}+1=\left[\left(\frac{21}{20}\right)^n-1\right]$
$ \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,
$n=3$ years
View full question & answer→Question 193 Marks
At what rate percent will a sum of $\text{Rs.} 4,000$ yield $\text{Rs.}1,324$ as compound interest in $3$ years?
AnswerGiven $: P= Rs. 4,000, C.I.= Rs. 1,324$ and $n=3$ years
Now, $A=P+I$
$\Rightarrow \mathrm{A}=\text { Rs. }(4,000+1,324)=\text { Rs. } 5,324$
$ A=P\left(1+\frac{r}{100}\right)^3$
$ \Rightarrow 5324=4000\left(1+\frac{r}{100}\right)^3$
$ \Rightarrow \frac{5324}{4000}=\left(1+\frac{r}{100}\right)^3$
$ \Rightarrow \frac{1331}{1000}=\left(1+\frac{r}{100}\right)^3$
$ \Rightarrow\left(1+\frac{R}{100}\right)^3=\frac{1331}{1000}=\left(\frac{11}{10}\right)^3$
$ \Rightarrow 1+\frac{r}{100}=\frac{11}{10}$
$ \Rightarrow \frac{r}{100}=\frac{11}{10}-1=\frac{1}{10}$
$ \Rightarrow r=\frac{100}{10}=10 \%$
Thus, the rate of interest is $10 \%$.
View full question & answer→Question 203 Marks
At what rate per cent compound interest, does a sum of money become $1.44$ times of itself in $2$ years ?
AnswerLet Principal $= \text{Rs.} y$
Then Amount$= \text{Rs.}1.44 \mathrm{y}$
$\mathrm{n}=2$ years
$\therefore$ Amount $=\mathrm{P}\left(1+\frac{\mathrm{r}}{100}\right)^n$
$ \Rightarrow 1.44 \mathrm{y}=y\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow \frac{1.44 y}{y}=\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow \frac{36}{25}=\left(1+\frac{r}{100}\right)^2$
$ \Rightarrow\left(\frac{6}{5}\right)^2=\left(1+\frac{r}{100}\right)^2$
On comparing,
$\frac{6}{5}=1+\frac{r}{100}$
On solving, we get
$r=20 \%$
View full question & answer→