Question 15 Marks
Calculate the rate percent when $Rs. 28000$ amount to $Rs. 30870$ in $2$ years at compounded annually.
AnswerHere $P=R s .28000, A=30870, t=2$ years
Now, $P \left(1+\frac{ r }{100}\right)^{ t }= A$
$\Rightarrow 28000\left(1+\frac{ r }{100}\right)^2 $
$=30870 $
$\Rightarrow\left(1+\frac{ r }{100}\right)^2 $
$=\frac{30870}{28000} $
$=\frac{441}{400} $
$=\left(\frac{21}{20}\right)^2 $
$\Rightarrow 1+\frac{ r }{100}=\frac{21}{20} $
$\Rightarrow \frac{ r }{100}=\frac{21}{20}-1 $
$=\frac{1}{20} $
$\Rightarrow R =\frac{100}{20} $
$=5$
Hence rate of interest is $5 \%$.
View full question & answer→Question 25 Marks
Find the amount and compounded interest on $Rs.15000$ in $2 \frac{1}{2}$years at $10\%p.a.$ compounded annually.
AnswerHere $P=Rs .15000, t=2 \frac{1}{2}, r=10 \%$
Now, Amount after $2$ year
$ =P\left(1+\frac{r}{100}\right)^{ t }$
$=15000\left(1+\frac{10}{100}\right)^2$
$=15000\left(\frac{11}{10}\right)^2$
$=18150 $
Now interest for the next half year
$=\frac{18150 \times 10}{100 \times 2}$
$=907.5 $
Hence, Amount
$ =\text { Rs. } 18150+\text { Rs. } 907.50$
$=\text { Rs. } 19057.50$
Also, $C.I.$
$ =A \cdot P$
$=\text { Rs. } 19057.50-R s .15000$
$=\text { Rs. } 4057.50 . $
View full question & answer→Question 35 Marks
A sum of $Rs.16820$ is to be divided between two girls $A$ and $B, 27$ and $25$ years old respectively, in such a way that, if their portions be invested at $5\%$ per annum compound interest payable annually, they will receive equal amounts on reaching $40$ years of age. What is the share of each in the original sum of money?
AnswerLet the share of $A$ be $Rs. x$.
Then, the share of $B=Rs.(16820 - x) $
For $A: P=R s . x, r=5 \%$ and $n=(40-27)$ years $=13$ years
$ \therefore A=P\left(1+\frac{ r }{100}\right)^{ n }$
$=\operatorname{Rs\cdot x}\left(1+\frac{5}{100}\right)^{13}$
$=\operatorname{Rs} x\left(\frac{21}{20}\right)^{13} $
For $B: P= Rs. (16820-x), r=5 \%$ and $n=(40-25)$ years $=15$ years
$ \therefore A=P\left(1+\frac{r}{100}\right)^{ n }$
$=\operatorname{Rs} \cdot(16820-x)\left(1+\frac{5}{100}\right)^{15}$
$=\operatorname{Rs} \cdot(16820-x)\left(\frac{21}{20}\right)^{15} $
Given; both receive equal sums on reaching the age of $40$ years.
$ \therefore x\left(\frac{21}{20}\right)^{13}$
$=(16820-x)\left(\frac{21}{20}\right)^{15}$
$\Rightarrow x =(16820-x) \times\left(\frac{21}{20}\right)^2$
$\Rightarrow x = Rs.8820$
$\Rightarrow 16820- x$
$=16820-8820$
$=8000 $
$\therefore$ Share of $A=R s .8820$ and Share of $B=R s .8000$.
View full question & answer→Question 45 Marks
Find the amount and the compound interest on the following:$Rs. 12000$ for $1 \frac{1}{2}$ years at $5 \%$ per annum compounded annually.
Answer$Rs. 12000$ for $1 \frac{1}{2}$ years at $5 \%$ per annum compounded annually.
Here $P= Rs. 12000, t=1 \frac{1}{2}$ years, $r=5 \%$
Now, Amount after $1$ year
$= P \left(1+\frac{ r }{100}\right)^{ t }$
$=12000\left(1+\frac{5}{100}\right)$
$=12000\left(\frac{105}{100}\right)$
$=12600$
Now interest for the next half year
$=\frac{12600 \times 5}{100 \times 2} $
$=315$
Hence, Amount
$=\text { Rs. } 12600+\text { Rs. } 315$
$=\text { Rs. } 12915$
Also, $C.I.$
$=A-P . $
$=\text { Rs. } 12915-\text { Rs. } 12000 $
$=\text { Rs. } 915 .$
View full question & answer→Question 55 Marks
The difference between simple interest and compound interest compounded annually on a certain sum is $Rs.448$ for $2$ years at $8$ percent per annum. Find the sum.
AnswerSince, $C.I.$
$=A - P$
$\text { C.I. }=\left(1+\frac{8}{100}\right)^2-P$
$=P\left(\frac{108}{100}\right)^2-P$
$=\frac{11664 P }{10000}- P$
$=\frac{11664 P -10000 P }{10000}$
$=\frac{1664 P }{10000}$
$\text { S.L. }=\frac{ P \times 8 \times 2}{100}$
$=\frac{16 P }{100} $
Now, $C.I. - S.I = Rs. 448$
$\Rightarrow \frac{1664 P }{10000}-\frac{16 P }{100}=\text { Rs. } 448$
$\Rightarrow \frac{1664 P -1600 P }{10000}= Rs .448$
$\Rightarrow 64 P = Rs .4480000$
$\Rightarrow P = Rs {70000}$
Hence, the sum is $Rs. 70000 .$
View full question & answer→Question 65 Marks
The compound interest payable annually on a certain sum for $2$ years is $Rs .40.80$ and the simple interest is $Rs. 40$. Find the sum and the rate percent.
AnswerLet sum be $Rs. P$ and $r \%$ be the rate of interest.
We have $t=2$ years, $C.I. = Rs. 40.80$ and $S.I. = Rs. 40$
Since, Simple interest
$ =\frac{ P \times r \times t }{100}$
$\Rightarrow 40=\frac{ P \times r \times 2}{100}$
$\Rightarrow Pr =\frac{4000}{2}$
$=2000$
Now,
$ \text { C.I. }=A-P$
$=P\left(1+\frac{r}{100}\right)^t-P$
$\left.=P\left[\left(1+\frac{ r }{100}\right)\right]^{ t }-1\right]$
$\Rightarrow 40.80= P \left[\left(1+\frac{ r }{100}\right)^2-1\right]$
$\Rightarrow 40.80= P \left(1+\frac{ r ^2}{10000}+\frac{2 r }{100}-1\right)$
$\Rightarrow 40.80= P \left(\frac{ r ^2}{10000}+\frac{2 r }{100}\right)$
$\Rightarrow 40.80= Pr \left(\frac{ r }{10000}+\frac{2}{100}\right)$
$\Rightarrow 40.80=2000\left(\frac{ r +200}{10000}\right)$
$\Rightarrow 40.80=\frac{ r +200}{5}$
$\Rightarrow r =40.80 \times 5-200$
$=204-200$
$=4$
Hence, $r=4 \%$
Now, $Pr =2000$
$ \Rightarrow P=\frac{2000}{ r }$
$=\frac{2000}{4}$
$=500 . $
Thus, sum is $Rs. 500$ and rate of interest is $4 \%$.
View full question & answer→Question 75 Marks
Find the amount and the compound interest on $Rs.17500$ for $3$ years, if the rates for successive years is $4\%, 5\%$ and $6\%$ respectively, the interest is payable annually.
AnswerHere $P_1= Rs. 17500$ and $r=4 \%$
So, Amount after $1$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=17500\left(1+\frac{4}{100}\right)$
$=17500 \times \frac{104}{100}$
$=18200$
Thus, $P_2= Rs. 18200$ and $r=5 \%$
Amount after $2$ year
$= P \left(1+\frac{ r }{100}\right)$
$=18200\left(1+\frac{5}{100}\right)$
$=18200 \times \frac{105}{100}$
$=19110$
Thus, $P_3= Rs. 19110$ and $r=6 \%$
Amount after $3$ year
$= P \left(1+\frac{ r }{100}\right)$
$=19110\left(1+\frac{6}{100}\right)$
$=19110 \times \frac{106}{100}$
$=20256.60$
Hence, Amount $= Rs. 20256.60$
Also, $C.I.$
$= A \cdot P$
$ =\text { Rs. } 20256.60-\text { Rs. } 17500$
$=\text { Rs. } 27256.60 .$
View full question & answer→Question 85 Marks
A man invests $Rs.24000$ for two years at compound interest, If his money amounts to $Rs.27600$ after one year, find the amount at the end of second year.
AnswerSince, Amount after $1$ year
$=P\left(1+\frac{r}{100}\right) $
$\Rightarrow 27600=24000\left(1+\frac{r}{100}\right) $
$\Rightarrow 1+\frac{r}{100}=\frac{27600}{24000}=\frac{23}{20} $
$\Rightarrow \frac{r}{100}=\frac{23}{20}-1=\frac{3}{20} $
$\Rightarrow r=\frac{100 \times 3}{20}=15$
Amount after $2$ year
$= P \left(1+\frac{ r }{100}\right) $
$=27600\left(1+\frac{15}{100}\right) $
$=27600 \times \frac{115}{100} $
$=31740 .$
View full question & answer→Question 95 Marks
Find the amount and compound interest on $Rs.7500$ for $1 \frac{1}{2}$ years at $8 \%$, payable semi$-$annually.
AnswerHere $P_1= Rs. 7500$ and rate of interest for half year $(r)=4 \%$
So, Amount after $\frac{1}{2}$ year
$= P \left(1+\frac{ r }{100}\right)$
$=7500\left(1+\frac{4}{100}\right)$
$=7500 \times \frac{104}{100}=7800 $
Thus, $P_2= Rs. 7800$ and $r=4 \%$
Amount after $1$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=7800\left(1+\frac{4}{100}\right)$
$=7800 \times \frac{104}{100}=8112$
Thus, $P_3= Rs. 8112$ and $r=4 \%$
Amount after $1 \frac{1}{2}$ year
$ = P \left(1+\frac{ r }{100}\right)^2$
$=8112\left(1+\frac{4}{100}\right)$
$=8112 \times \frac{104}{100}=8436.48 $
Hence, Amount $= Rs. 8436.48$
Also, $C.l.$
$ =A-P$
$=R s .8436 .48-R s .7500$
$=R s .936 .48 .$
View full question & answer→Question 105 Marks
Compute the compound interest for the third year on $Rs. 5000$ invested for $5$ years at $10\%$ per annum, the interest being payable annually.
AnswerFor $1^{st}$ year: $P=Rs .5000, R=10 \%$ and $T=1$ year
$\therefore$ Interest $= Rs. \frac{5000 \times 10 \times 1}{100}$
$=\operatorname{Rs.} 500$
And, amount
$=R s .5000+R s .500$
$= Rs. 5500$
For $2^{nd}$ year: $P=Rs .5500, R=10 \%$ and $T=1$ year
$\therefore$ Interest $= Rs.\frac{5500 \times 10 \times 1}{100}$
$=\operatorname{Rs.} 550$
And, amount
$= Rs. 5500+R s .550$
$= Rs. 6050$
For $3^{rd}$ year: $P=R s .6050, R=10 \%$ and $T=1$ year
$\therefore$ Interest $= Rs. \frac{6050 \times 10 \times 1}{100}$
$= Rs. 605$
$\therefore$ Compound interest for $3^{rd}$ year is $Rs. 605 .$
View full question & answer→Question 115 Marks
Calculate the amount and compound interest on $Rs.20000$ for $3$ years at $10\%$ per annum, interest being payable annually.
AnswerHere $P_1=R s, 20000$ and $r=10 \%$
So, Amount after $1$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=20000\left(1+\frac{10}{100}\right)$
$=20000 \times \frac{110}{100}=22000 $
Thus, $P_2= Rs. 22000$ and $r=10 \%$
Amount after $2$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=22000\left(1+\frac{10}{100}\right)$
$=22000 \times \frac{110}{100}=24200 $
Thus, $P_3= Rs. 24200$ and $r=10 \%$
Amount after $3$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=24200\left(1+\frac{10}{100}\right)$
$=24200 \times \frac{110}{100}=26620 $
Hence, Amount $= Rs. 26620$
Also, $C.I.$
$= A - P$
$=\text { Rs. } 26620 \text { - Rs. } 20000$
$=R s, 6620$.
View full question & answer→Question 125 Marks
Samidha borrowed $Rs.7500$ from Shreya at $30\%$ per annum compounded interest. After $2$ years, Samidha gave $Rs.10000$ and a juicer to Shreya to clear the debt. Find the cost of the juicer.
AnswerFor $1^{\text {st }}$ year: $P = Rs. 7500, R =30 \%$ and $T=1$ year
Interest$ =\text { Rs. } \frac{7500 \times 30 \times 1}{100} $
$=\text { Rs. } 2550$
Amount
$=\text { Rs. } 7500+\text { Rs. } 2250 $
$=\text { Rs. } 9750$
For $2^{\text {nd }}$ year: $P=R s .9750 ; R=30 \%$ and $T=1$ year
Interest$=\text { Rs. } \frac{9750 \times 30 \times 1}{100} $
$=\text { Rs. } 2925$
Amount
$=\text { Rs. } 9750+\text { Rs. } 2925 $
$=\text { Rs. } 12675$
Thus, total amount to be paid by Samidha $= Rs. 12675$
But, Samidha gave $Rs. 10000 +$ juicer to Shreya.
$\Rightarrow \text { Rs. } 10000+$ Cost of juicer $=\text { Rs. } 12675 $
$\Rightarrow$ Cost of juicer
$=\text { Rs. }(12675-10000) $
$=\text { Rs. } 2675 .$
View full question & answer→Question 135 Marks
Priyanka lends $Rs.15,500$ at $10\%$ for the first year, at $15\%$ for the second year and at $20\%$ for the third year. If the rates of interest are compounded yearly, find the difference between the compound interest of the second year and the third year.
AnswerFor $1^{\text {st }}$ year: $P=R s .15500, R=10 \%$ and $T=1$ year
Interest $=\text { Rs. } \frac{15500 \times 10 \times 1}{100} $
$=\text { Rs. } 1550$
Amount
$=\text { Rs. } 15500+R s .1550 $
$=\text { Rs. } 17050$
For $2^{\text {nd }}$ year: $P=R s .17050, R=15 \%$ and $T=1$ year
Interest$=\text { Rs. } \frac{17050 \times 15 \times 1}{100} $
$=\text { Rs. } 2557.50$
Amount
$=\text { Rs. } 17050+\text { Rs. } 2557.50 $
$=\text { Rs. } 19607.50$
For $3^{\text {rd }}$ year: $P=R s .19607 .50 ; R=20 \%$ and $T=1$ year
$=\text { Rs } \frac{19607.50 \times 20 \times 1}{100} $
$=\text { Rs. } 3921.50$
Amount
$=\text { Rs. } 19607.50+\text { Rs. } 3921.50 $
$=\text { Rs. } 23529$
Difference between the $C.I.$ of the $2^{\text {nd }}$ year and the $3^{\text {rd }}$ year
$=\text { Rs. }(3921.50-2557.50) $
$=\text { Rs. } 1364 .$
View full question & answer→Question 145 Marks
Ankita bought a gold ring worth $Rs.x$. The value of the ring increased at $10\%$ per year compounded annually, on which the appreciation for the first year plus the appreciation for the second year amounts to $Rs.6300$. Find the value of the ring.
AnswerLet the value of ring $\left(P_1\right)=R s .100$.
Appreciation $(C.I)$ for the $1^{\text {st }}$ year
$=\text { Rs. } \frac{100 \times 10 \times 1}{100}$
$=\text { Rs. } 10$
$\therefore$ Value of the ring at the end of $1^{\text {st }}$ year $\left(A_1\right)$
$=\text { Rs. } 100+\text { Rs. } 10$
$=\text { Rs. } 110$
$\therefore$ Value of the ring at the begging of $2^{\text {nd }}$ year $\left(P_2\right)=R s .110$
Appreciation $(C.I)$ for the $2^{\text {nd }}$ year
$=\text { Rs. } \frac{110 \times 10 \times 1}{100}$
$=\text { Rs. } 11$
Sum of the appreciation $(C.I)$ for the $1^{\text {st }}$ year and appreciation $(C.I)$ of $2^{\text {nd }}$ year
$=\operatorname{Rs} \cdot(10+11)$
$=R s .21$
Thus, when sum of appreciation is $Rs. 21$ , then value of the ring $\left(P_1\right)=R s .100$
And, when sum of appreciation is $Rs. 6300$ , then value of the ring
$=\text { Rs. } \frac{100 \times 6300}{21}$
$=\text { Rs. } 30000$
So, the value of the ring is $R s .30000$.
View full question & answer→Question 155 Marks
A man borrows $Rs.20000$ at $10\%$ per annum compound interest payable annually. If he repays $Rs.5000$ at the end of the first year and $Rs.10000$ at the end of the second year; how much should he pay at the end of the third year in order to clear the account? Find the answer correct to the nearest rupee.
AnswerFor $1^{\text {st }}$ half year: $P = Rs \cdot 2000, R =10 \%$ and $T =1$ year
Interest $=\text { Rs. } \frac{20000 \times 10 \times 1}{100}$
$=\text { Rs. } 2000$
Amount
$=\text { Rs. } 20000+R s .2000$
$=\text { Rs. } 22000$
Money paid at the end of $1^{\text {st }}$ half year $=R s .5000$
Balance money for $2^{\text {nd }}$ half year
$=\text { Rs. } 22000-R s .5000$
$=\text { Rs. } 17000$
For $2^{\text {nd }}$ half year: $P=R s .17000 ; R=10 \%$ and $T=1$ year
Interest$ =\text { Rs. } \frac{17000 \times 10 \times 1}{100}$
$=\text { Rs. } 1700$
Amount
$=\text { Rs. } 17000+\text { Rs. } 1700$
$=\text { Rs. } 18700$
Money paid at the end of $2^{\text {nd }}$ half year $=$ Rs. 10000
Balance money for $3^{\text {rd }}$ half year
$=\text { Rs. } 18700-\text { Rs. } 10000$
$=\text { Rs. } 8700$
For $3^{\text {rd }}$ half year: $P=R s .8700 ; R=10 \%$ and $T=1$ year
$\text { Interest }=\text { Rs. } \frac{8700 \times 10 \times 1}{100}$
$=\text { Rs. } 870$
Amount
$=\text { Rs. } 8700+R s .870$
$=R s .9570$
A man should pay $Rs. 9570$ at the end of $3^{\text {rd }}$ year to clear the account.
View full question & answer→Question 165 Marks
Find the amount and the compound interest payable annually on:$Rs.17500$ for $3$ years at $8\%, 10\%$ and $12\%$ for the successive years.
AnswerFor $1^{st}$ year: $P=R s .17500, R=8 \%$ and $T=1$ year
$\therefore$ Interest $= Rs. \frac{17500 \times 8 \times 1}{100}$
$= Rs. 1400$
And, amount
$=\text { Rs. } 17500+\text { Rs. } 1400$
$=\text { Rs. } 18900$
For $2^{nd}$ year: $P=R s .18900, R=10 \%$ and $T=1$ year
$\therefore$ Interest $=\text { Rs. } \frac{18900}{1}$
$=\text { Rs. } 1890$
And, amount
$=\text { Rs. } 18900+\text { Rs. } 1890$
$=\text { Rs. } 20790$
For $3^{rd}$ year: $P=R s .20790, R=12 \%$ and $T=1$ year
$\therefore \text { Interest }=\text { Rs. } \frac{20790 \times 12 \times 1}{100}$
$=\text { Rs. } 2494.80$
And, amount
$=\text { Rs. } 20790+R s .2494 .80$
$=\text { Rs. } 23284.80$
$\therefore$ Required amount $= Rs. 23284.80$
And, Compound Interest
$=A \cdot P$
$=\text { Rs. } 23284.80-R s .17,500$
$=R s .5784 .80$
View full question & answer→Question 175 Marks
A man borrows $Rs.6500$ at $10\%$ per annum compound interest payable half$-$yearly. He repays $Rs.2000$ at the end of every six months. Calculate the amount outstanding at the end of the third payment. Give your answer to the nearest rupee.
AnswerFor $1^{\text {st }}$ half year: $P=R s, 6500, R=10 \%$ and $T=\frac{1}{2}$ year
Interest $=\text { Rs. } \frac{6500 \times 10 \times 1}{100 \times 2}$
$=\text { Rs. } 325$
Amount
$=\text { Rs. } 6500+\text { Rs. } 325$
$=\text { Rs. } 6825$
Money paid at the end of $1^{st}$ half year $= Rs. 2000$
Balance money for $2^{\text {nd }}$ half year
$=\text { Rs. } 6825-\text { Rs. } 2000$
$=\text { Rs. } 4825$
For $2^{\text {nd }}$ half year : $P=$ Rs. $4825 ; R=10 \%$ and $T=\frac{1}{2}$ year
$\text { Interest }=\text { Rs. } \frac{4825 \times 10 \times 1}{100 \times 2}$
$=\text { RS. } 241.25$
Amount
$=\text { Rs. } 4825+\text { Rs. } 241.25$
$=\text { Rs. } 5066.25$
Money paid at the end of $2^{\text {nd }}$ half year $=Rs. 2000$
Balance money for $3^{\text {rd }}$ half year
$=\text { Rs. } 5066.25-\text { Rs. } 2000$
$=\text { Rs. } 3066.25$
For $3^{\text {rd }}$ half year: $P=R s .3066 .25 ; R=10 \%$ and $T=\frac{1}{2}$ year
$\text { Interest }=\text { Rs. } \frac{3066.25 \times 10 \times 1}{100 \times 2}$
$=\text { Rs. } 153.3125$
Amount
$=\text { Rs. } 3066.25+\text { Rs. } 153.3125$
$=\text { Rs. } 3219.5625$
Money paid at the end of $3^{\text {rd }}$ half year $= Rs. 2000$
Amount outstanding at the end of $3^{\text {rd }}$ payment
$=\text { Rs. } 3219.5625-\text { Rs. } 2000$
$=\text { Rs. } 1219.5625$
$=\text { Rs. } 1220($nearest rupee$)$
View full question & answer→Question 185 Marks
Find the amount and the compound interest payable annually on:$Rs.16000$ for $2$ years at $15\%$ and $12\%$ for the successive years.
AnswerFor $1^{st}$ year: $P=R s .16000, R=15 \%$ and $T=1$ year
$\therefore$ Interest $=\text { Rs. } \frac{16000 \times 15 \times 1}{100} $
$=\text { Rs. } 2400$
And, amount
$=\text { Rs. } 16000+\text { Rs. } 2400 $
$=\text { Rs. } 18400$
For $2^{nd}$ year: $P=18400, R=12 \%$ and $T=1$ year
$\therefore$ Interest $=\text { Rs. } \frac{18400 \times 12 \times 1}{100} $
$=\text { Rs } 2208$
And, amount
$=\text { Rs. } 18400+\text { Rs. } 2208 $
$=\text { Rs. } 20608$
$\therefore$ Required amount $= Rs. 20608$
And, Compound Interest
$=A \cdot P $
$=R s .20608-R s .16000 $
$=\text { Rs. } 4608 .$
View full question & answer→Question 195 Marks
The value of a mobile depreciated by $5\%$ per year during the first two years and $10\%$ per year during the third year. Express the total depreciation of the value of the mobile in percent during the three years.
AnswerLet the value of mobile in the begging be $Rs. 100 .$
For $1^{\text {st }}$ year ${ }_x$ depreciation $=5 \%$ of $Rs. 100$
$=\frac{5}{100} \times 100$
$=\text { Rs. } 5$
Value of machine for second year
$=\text { Rs. } 100-\text { Rs. } 5$
$=\text { Rs. } 95$
For $2^{\text {nd }}$ year, depreciation $=5 \%$ of $Rs. 95$
$=\frac{5}{100} \times 95$
$=\text { Rs.4.75 }$
Value of machine for third year
$=\text { Rs. } 95-R s .4 .75$
$=\text { Rs. } 90.25$
For $3^{\text {rd }}$ year, depreciation $=10 \%$ of $Rs. 90.25$
$=\frac{10}{100} \times 90.25$
$=\text { Rs. } 9.025$
Value of machine at the end of third year
$=R s .90 .25-R s .9 .025$
$=R s .81 .225$
Net depreciation
$=\text { Rs. } 100-\text { Rs. } 81.225$
$=\text { Rs. } 18.775$ or $18.775 \%$
View full question & answer→Question 205 Marks
Sunil borrows $Rs. 50,000$ at $10 \% S.I.$ for $1 \frac{1}{2}$ years. He immediately invests the entire amount for $1 \frac{1}{2}$ years at $10 \%$ compounded annually. What is his gain at the end of the stipulated time, when he repays his loan?
AnswerTo calculate the $S.I.$ paid by Sunil :
$P=\text { Rs. } 50000, R=10 \%$ and $T=1 \frac{1}{2} \text { years }=\frac{3}{2} \text { years }$
$\therefore \text { S.I. }=\text { Rs. } \frac{50000 \times 10 \times 3}{100 \times 2}$
$=\text { Rs. } 7500 $
To calculate the $C.I.$ earned by Sunil :
For $1^{st}$ year: $P=R s .50000, R=10 \%$ and $T=1$ year
$\therefore$ Interest$=\text { RS. } \frac{50000 \times 10 \times 1}{100}$
$=\text { Rs. } 5000$
And, amount
$=\text { Rs. } 50000+\text { Rs. } 5000$
$=\text { Rs. } 55000$
For next half year: $P=R s .55000, R=10 \%$ and $T=\frac{1}{2}$ year
$\therefore$ Interest $=\text { Rs. } \frac{55000 \times 10 \times 1}{100 \times 2}$
$=\text { Rs. } 2750$
And, amount
$=\text { Rs. } 55000+\text { Rs. } 2750$
$=\text { Rs. } 7750 $
$\therefore$ Total $C.I.$ earned
$ =\text { Rs. } 55750+\text { Rs. } 2750$
$=\text { Rs. } 57750 $
$\Rightarrow$ Sunil's gain in $1 \frac{1}{2}$ year
$= C.I$. earned $- S.I.$ paid
$=R s .250$.
View full question & answer→Question 215 Marks
A man borrows $Rs.62500$ at $8\% p.a.$, simple interest for $2$ years. He immediately lends the money out at $CI$ at the same rate and for same time. What is his gain at the end of $2$ years?
AnswerCase $I:$
Simple interest$=\frac{62500 \times 8 \times 2}{100}$
$=10000$
Amount
$=\text { Rs. } 62500+\text { Rs. } 10000 $
$=\text { Rs. } 72500$
Case $II:$
Here $P_1= Rs. 6250$0 and $r=8 \%$
So, Amount after $1$ year
$= P \left(1+\frac{ r }{100}\right) $
$=62500\left(1+\frac{8}{100}\right) $
$=62500 \times \frac{108}{100} $
$=67500$
Thus, $P_2= Rs. 67500$ and $r=8 \%$
Amount after $2$ year
$= P \left(1+\frac{ r }{100}\right) $
$=67500\left(1+\frac{8}{100}\right) $
$=67500 \times \frac{108}{100} $
$=72900$
Hence, Amount $= Rs. 72900$
Thus, gain in amount
$=\text { Rs. } 72900-R s .72500 $
$=\text { Rs. } 400 .$
View full question & answer→Question 225 Marks
Simple interest on a sum of money for $2$ years at $4\%$ is $Rs. 450$. Find the compound interest at the same rate for $1$ year if the interest is reckoned half$-$yearly.
AnswerSince Simple interest
$=\frac{P \times r \times t}{100}$
$\Rightarrow 450=\frac{P \times 4 \times 2}{100}$
$\Rightarrow P=\frac{45000^{10}}{8}$
$=5625$
Now for $C.I, P=R s .5625, r=4, t=1$ year
Here $P_1=Rs. 5625$ and rate of interest for half$-$yearly $=2 \%$
So, Amount after $\frac{1}{2}$ year
$= P \left(1+\frac{ r }{100}\right)^2$
$=5625\left(1+\frac{2}{100}\right)$
$=5625 \times \frac{102}{100}$
$=5737.50$
Thus, $P_2= Rs. 5737.50$ and $r=2 \%$
Amount after $1$ year
$= P \left(1+\frac{ r }{100}\right)$
$=5737.50\left(1+\frac{2}{100}\right)$
$=5737.50 \times \frac{102}{100}$
$=5852.25$
Hence, Amount $=Rs.5852.25$
Also, $C.I.$
$=A-P$
$=R s .5852 .25-R s .5625$
$=R s .227 .25 .$
View full question & answer→Question 235 Marks
Find the difference between simple and compound interest on $Rs. 5000$ invested for $3$ years at $6\%\ \ p.a.$, interest payable yearly.
AnswerCase $1:$
Here $P_1=R s .5000$ and $r=6 \%$
So, Amount after $1$ year
$=P\left(1+\frac{r}{100}\right)$
$=5000\left(1+\frac{6}{100}\right)$
$=5000 \times \frac{106}{100}$
$=5300$
Amount after $2$ year
$=P\left(1+\frac{r}{100}\right)$
$=5300\left(1+\frac{6}{100}\right)$
$=5300 \times \frac{106}{100}$
$=5618$
Thus, $P_3=\text { Rs. } 5618$ and $ r=6 $
Amount after $3$ year
$=P\left(1+\frac{r}{100}\right)$
$=5618\left(1+\frac{6}{100}\right)$
$=5618 \times \frac{106}{100}$
$=5955.08$
Hence, Amount $=\text { Rs. } 5955.08$
Also, $\text {C.I. }$
$=A \cdot P$
$=\text { Rs. } 5955.08-R s .5000$
$=\text { Rs. } 955.08$
Case$ II:$
Simple interest $=\frac{5000 \times 6 \times 3}{100}$
$=900$
Difference between $C.I.$ and $S.I.$
$=\text { Rs. } 955.08 \text { - Rs. } 900$
$=\text { Rs. } 55.08 \text {. }$
View full question & answer→Question 245 Marks
The simple interest on a certain sum of money at $4\%\ \ p.a.$ for $2$ years is $Rs.1500.$ What will be the compound interest on the same sum for the same time?
AnswerSince, Simple interest
$=\frac{ P \times r \times t }{100}$
$\Rightarrow 1500=\frac{ P \times 4 \times 2}{100}$
$\Rightarrow P =\frac{150000^{-1}}{8}$
$=18750$
Now for $C.I, P= Rs. 18750, r=4 \%, t=2$ year
Here $P_1= Rs. 18750$ and $r=4 \%$
So, Amount after $1$ year
$= P \left(1+\frac{ r }{100}\right)$
$=18750\left(1+\frac{4}{100}\right)$
$=18750 \times \frac{104}{100}$
$=19500$
Thus, $P_2= Rs. 19500$ and $r=4 \%$
Amount after $2$ year
$= P \left(1+\frac{ r }{100}\right)$
$=19500\left(1+\frac{4}{100}\right)$
$=19500 \times \frac{104}{100}$
$=20280$
Hence, Amount $= Rs. 20280$
Also, $Cl$.
$= A - P$
$= Rs.20280-R s .18750$
$= Rs. 1530.$
View full question & answer→Question 255 Marks
Find the amount and the compound interest payable annually on the following: $Rs. 24000$ for $1 \frac{1}{2}$ years at $7 \frac{1}{2} \%$ per annum.$[$Hint $: 1$ year $146$ days $=1 \frac{146}{365}$ year $=1 \frac{2}{5}$ year$]$
Answer$Rs. 24000$ for $1 \frac{1}{2}$ years at $7 \frac{1}{2} \%$ per annum.
Here $P= Rs. 24000, t=1 \frac{1}{2}$ years, $r=7 \frac{1}{2} \%=\frac{15}{2} \%$
Now, Amount after $1$ year
$= P \left(1+\frac{ r }{100}\right)$
$=24000\left(1+\frac{15}{2 \times 100}\right)$
$=24000\left(1+\frac{3}{40}\right) $
$=24000\left(\frac{43}{40}\right) $
$=25800$
Thus, principle for the next $6$ months $= Rs. 25800$
Interest for the next $6$ months
$=\frac{25800 \times 15 \times 6}{200 \times 12} $
$=967.50$
Therefore, amount after $1 \frac{1}{2}$ years
$=\text { Rs. } 25800+\text { Rs. } 967.50$
$=\text { Rs. } 26767.50$
And $Cl$
$=A-P$
$=R s .26767 .50-R s .24000$
$=R s .2767 .50 .$
View full question & answer→Question 265 Marks
Find the difference between the compound interest and simple interest on $Rs. 20,000$ at $12\%$ per annum for $3$ years, the compound interest being payable annually.
AnswerCase $1:$
Here $P_1= Rs. 20000$ and $r=12 \%$
So, Amount after $1$ year
$=P\left(1+\frac{r}{100}\right) $
$=20000\left(1+\frac{12}{100}\right) $
$=20000 \times \frac{112}{100} $
$=22400$
Thus, $P_2= Rs. 22400$ and $r=12 \%$
Amount after $2$ year
$=P\left(1+\frac{r}{10}\right) $
$=22400\left(1+\frac{12}{100}\right) $
$=22400 \times \frac{112}{100} $
$=25088$
Thus, $P_3= Rs. 25088$ and $r=12 \%$
Amount after $3$ year
$= P \left(1+\frac{ r }{100}\right) $
$=25088\left(1+\frac{12}{100}\right) $
$=25088 \times \frac{112}{100} $
$=28098.56$
Hence, Amount $= Rs. 28098.56$
Also, $C.I.$
$= A - P$
$=\text { Rs. } 28098.56-R s .20000 $
$=\text { Rs. } 8098.56$
Case $II :$
Simple interest $=\frac{20000 \times 12 \times 3}{100} $
$=7200$
Difference bertween $C.I.$ and $S.I.$
$=R s .8098 .56-R s .7200 $
$=R s .898 .56 .$
View full question & answer→Question 275 Marks
Find the amount and the compound interest payable annually on the following :$Rs.10000$ for $2 \frac{1}{2}$ years at $6\%$ per annum.
Answer$ \text { Rs. } 10000$ for $ 2 \frac{1}{2}$ years at $6 \%$ per annum.
Here $P_1=\text { Rs. } 10000$ and $r=6 \%$
So, Amount after $1$ year
$=P\left(1+\frac{r}{100}\right)$
$=10000\left(1+\frac{6}{100}\right)$
$=10000 \times \frac{106}{100}=10600$
Thus, $P_2=\text { Rs. } 10600$ and $r=6 \%$
So, Amount after $2$ year
$=P\left(1+\frac{ r }{100}\right)$
$=10600\left(1+\frac{6}{100}\right)$
$=10600 \times \frac{106}{100}=11236$
Thus, principle for the next $6$ months $=\text { Rs. } 11236$
Interest for the next $6$ months
$=\frac{11236 \times 6 \times 6}{100 \times 12}$
$=337.08 $
Thus, principle for the next $6$ months $= Rs. 11236$
Interest for the next $6$ months
$ =\frac{11236 \times 6 \times 6}{100 \times 12}$
$=337.08 $
Therefore, amount after $1 \frac{1}{2}$ years
$ =\text { Rs. } 11236+\text { Rs. } 337.08$
$=\text { Rs. } 11573.08$
And $Cl $
$=A \cdot P$
$=\text { Rs. } 11573.08-R s .10000$
$=\text { Rs. } 1573.08 . $
View full question & answer→Question 285 Marks
A man lends $Rs.15000$ at $10.5\%$ per annum $C.I$., interest reckoned yearly, and another man lends the same sum at $10\%$ per annum, interest being reckoned half$-$yearly. Who is the gainer at the end of one year and by how much?
AnswerCase $I:$
Here $P= Rs. 15000$ and $r=10.5 \%$
So, Amount after $1$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=15000\left(1+\frac{10.5}{100}\right)$
$=15000 \times \frac{110.5}{100}$
$=16575 $
Case $II:$
Here $P_1= Rs. 15000$ and rate of intees for half year $(r)=5 \%$
So, Amount after $\frac{1}{2}$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=15000\left(1+\frac{5}{100}\right)$
$=15000 \times \frac{105}{100}$
$=15750 $
Thus, $P_2= Rs. 15750$ and $r=5 \%$
Amount after $1$ year
$= P \left(1+\frac{ r }{100}\right)$
$ =15750\left(1+\frac{5}{100}\right)$
$=15750 \times \frac{105}{100}$
$=16537.50 $
Hence the first man gains by $Rs. 16575 - Rs. 16537.50$
$=R s .37 .50 \text {. }$
View full question & answer→Question 295 Marks
Find the amount and the compound interest payable annually on the following :$Rs.32000$ for $2$ years at $7 \frac{1}{2} \%$ per annum.
Answer$Rs. 32000$ for $2$ years at $7 \frac{1}{2} \%$ per annum.
Here $P_1= Rs. 32000$ and $r=7 \frac{1}{2} \%=\frac{15}{2} \%$
So, Amount after $1$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=32000\left(1+\frac{15}{2 \times 100}\right)$
$=32000\left(1+\frac{3}{40}\right)$
$=32000\left(\frac{43}{40}\right)$
$=34400 $
Thus, $P_2= Rs. 34400$ and $r=\frac{15}{2} \%$
So, Amount after $2$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=34400\left(1+\frac{15}{2 \times 100}\right)$
$=34400\left(1+\frac{3}{40}\right)$
$=34400\left(\frac{43}{40}\right)$
$=36980 $
Hence, Amount $= Rs. 36980$
Also , $Cl$
$= A \cdot P$
$ =\text { Rs. } 36980-\text { Rs. } 32000$
$=\text { Rs. } 4980 . $
View full question & answer→Question 305 Marks
Calculate the amount and compound interest to the nearest rupee on $Rs.42000$ for $2$ years at $8\%$ per annum, interest being payable half$-$yearly.
AnswerHere $P_1= Rs. 4200$ and rate of interest for half year $=4 \%, t=4$ half years
So, Amount after $\frac{1}{2}$ year
$ = P \left(1+\frac{ r }{100}\right)^2$
$=4200\left(1+\frac{4}{100}\right)$
$=42000 \times \frac{104}{100}$
$=43680 $
Thus, $P_2= Rs.43680$
Amount after $1$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=43680\left(1+\frac{4}{100}\right)$
$=43680 \times \frac{104}{100}$
$=\text { Rs. } 45427.20 $
Thus, $P_3= Rs. 45427.20$
Amount after $1 \frac{1}{2}$ year
$ = P \left(1+\frac{ r }{100}\right)^2$
$=45427.20\left(1+\frac{4}{100}\right)$
$=45427.20 \times \frac{104}{100}$
$=47244.29 $
Thus, $P_4 = Rs.47244,29$
Amount after $2$ year
$ = P \left(1+\frac{ r }{100}\right)$
$=47244.29\left(1+\frac{4}{100}\right)$
$=47244.29 \times \frac{104}{100}$
$=49134.06 $
Hence, Amount $= Rs. 49134.06$
Also, $C.I.$
$= A - P$
$ =\text { Rs } 49134.06-R s .42000$
$=R s .7134 .06 . $
View full question & answer→Question 315 Marks
Find the amount and the compound interest payable annually on the following :$Rs.25000$ for $1 \frac{1}{2}$years at $10\%$ per annum.
Answer$Rs. 25000$ for $1 \frac{1}{2}$ years at $10 \%$ per annum.
Here $P=R s .25000, t=1 \frac{1}{2}$ years, $r=10 \%$
Now, Amount after $1$ year
$ =p\left(1+\frac{r}{100}\right)$
$=25000\left(1+\frac{10}{100}\right)$
$=25000\left(1+\frac{1}{100}\right)$
$=25000\left(\frac{11}{100}\right)$
$=27500 $
Thus, principle for the next $6$ months $= Rs. 27500$
Interest for the next $6$ months
$ =\frac{27500 \times 6 \times 10}{100 \times 12}$
$=1375 $
Therefore, amount after $1 \frac{1}{2}$ years
$ =R s: 27500+R s .1375$
$=R s .3875$
And $Cl$
$=A \cdot P$
$=R s .28875-R s .25000$
$=R s .3875 . $
View full question & answer→Question 325 Marks
A man borrows $₹ 4000$ at $14 \%\ \ p.a.$, compound interest, being payable half$-$yearly. Find the amount he has to pay at the end of $1 \frac{1}{2}$ years.
AnswerFor $1^{st}$ half$-$year: $P=R s .4000, R=14 \%$ and $T=\frac{1}{2}$ year
Interest$=\operatorname{Rs} . \frac{4000 \times 14 \times 1}{100 \times 12}$
$=\text { Rs. } 280$
And, amount
$ =R s .4000+R s .280$
$=R s .4280 $
For $2^{nd}$ half$-$year: $P=R s .4280, R=14 \%$ and $T=\frac{1}{2}$ year
Interest $=\text { Rs. } \frac{4280 \times 14 \times 1}{100 \times 2}$
$=\text { Rs. } 299.60 $
And, amount
$ =\text { Rs. } 4280+R s .299 .60$
$=\text { Rs. } 4579.60 $
For $3^{rd}$ half$-$year: $P=R s .4579 .60, R=14 \%$ and $T=\frac{1}{2}$ year
Interest$=\text { Rs. } \frac{4579.60 \times 14 \times 1}{100 \times 2}$
$=\text { Rs. } 320.572 $
And, amount
$ =\text { Rs. } 4579.60+\text { Rs. } 320.572$
$=\text { Rs. } 4900.172 $
Thus, the amount to be paid at the end of $1 \frac{1}{2}$ years is $Rs. 4900.172 .$
View full question & answer→