Question 15 Marks
Calculate the compound interest for the second year on $Rs. 15000$ invested for $5$ years at $6\%$ per annum.
AnswerPrincipal $(P)= Rs. 15000$
Rate $(R)=6 \% p.a.$
Period $(n)=5$ years
Interest for the first year $=\frac{P R T}{100}$
$=\frac{15000 \times 6 \times 1}{100}$
$ = Rs. 900$
$\therefore$ Amount for the first year $= Rs. 15000+900$
$= Rs. 15900$
Principal for the second year$= Rs. 15900$
Interest for the second year $=\frac{15900 \times 6 \times 1}{100}$
$=159 \times 6$
$= Rs. 954$
View full question & answer→Question 25 Marks
Mohan borrowed $Rs. 16,000$ for $3$ years at $5\%$ per annum compound interest. Calculate the amount that Mohan will pay at the end of $3$ years.
AnswerFor $1^{st}$ year
Principal $(P) =Rs. 16,000$ , Rate $(R)=5 \%$, Time $(T)=1$ year
$\therefore$ Interest $=\frac{16000 \times 5 \times 1}{100}=160 \times 5= Rs. 800$
$\therefore$ Amount at the end of $1^{st}$ year $= Rs. (16,000+800)= Rs. 16,800$
For $2 ^{nd}$ year
$P= Rs. 16,800, R=5 \%, T=1$ year
$\therefore$ Interest $=\frac{16,800 \times 5 \times 1}{100}=168 \times 5=Rs. 840$
$\therefore$ Amount at the end of $2^{nd}$ year $= Rs. (16,800+840)= Rs. 17640$
For $3^{rd}$ year
$P=17640, R=5 \%, T=1$ year
$\therefore$ Interest $=\frac{17640 \times 5 \times 1}{100}=\frac{1764}{2}=Rs. 882$
$\therefore$ Amount at the end of $3^{rd}$ year $= Rs. (17640+882)= Rs. 18522$
Hence reqd. amount $= Rs. 18522$
View full question & answer→Question 35 Marks
Calculate the compound interest on $Rs. 15,000$ in $3$ years; if the rates of interest for successive years be $6\%, 8\%$, and $10\%$ respectively.
AnswerFor $1 ^{st}$ year
Principal $(P)= Rs. 15,000$ , Rate $(R)=6 \%$, Time $(T)=1$ year
$\therefore$ Interest $=\frac{15,000 \times 6 \times 1}{100}=150 \times 6= Rs. 900$
$\therefore$ Amount at the end of $1 ^{nd}$ year
$= Rs. 15,000+ Rs. 900$
$= Rs. 15900$
For $2 ^{nd}$ year
$P= Rs. 15900, R=8 \%, T=1$ year
$\therefore$ Interest $=\frac{15,900 \times 8 \times 1}{100}=159 \times 8= Rs. 1272$
$\therefore$ Amount at the end $2^{nd}$ year
$= Rs. (15900+1272)$
$= Rs. 17172$
For $3^{rd}$ year
$P= Rs.17172, R=10 \%, T=1$ year
$ \therefore$ Interest $=\frac{17172 \times 10 \times 1}{100}=\text { Rs. } 1717.20$
$\therefore$ Amount at the end of $3^{rd}$ year
$=Rs.(17172+1717.20)$
$ =Rs. 18889.20$
$\therefore$ Compound interest $=18889.20-15,000$
$= Rs. 3889.20$
View full question & answer→Question 45 Marks
Calculate the amount and the compound interest on $Rs. 10,000$ in $3$ years at $8\%$ per annum.
AnswerFor $1^{st}$ year
Principal $(P)= Rs. 10,000$ , Rate $(R)=8 \%$
Time $(T)=1$ year
$\therefore$ Interest $=\frac{10,000 \times 8 \times 1}{100}$
$=100 \times 8$
$= Rs. 800$
For $2^{nd}$ year
$P=Rs. 10,000+Rs. 800= Rs. 10,800$
Rate $(R)=8 \%$, Time $(T)=1$ year
$\therefore$ Interest $=\frac{10,800 \times 8 \times 1}{100}$
$=108 \times 8$
$=Rs. 864$
For $3^{rd}$ year
$\therefore P= Rs.10,800+Rs. 864=Rs. 11664,$
$R=8 \%, T=1$ year
$\therefore$ Interest $=\frac{11664 \times 8 \times 1}{100}=\frac{11664 \times 2}{25}$
$=Rs. 933.12$
$\therefore$ Amount $= Rs. 11664+933.12= Rs. 12597.12$
Hence required amount $= Rs. 12597.12$
$\therefore$ Compound interest $= Rs. 12597.12-10000$
$=Rs. 2597.12$
View full question & answer→Question 55 Marks
Calculate the amount and the compound interest on $Rs. 12,000$ in $2$ years and at $10\%$ per year.
AnswerFor $I\ st$ year
Principal $(P)= Rs. 12,000$
Rate $(\mathrm{R})=10 \%$
Time $(T)=1$ year
$I=$ Interest $=\frac{12,000 \times 10 \times 1}{100}$
$=120 \times 10$
$= Rs. 1200$
Amount $= P +\mathrm{I}= Rs. 12,000+ Rs. 1200= Rs. 13,200$
For $II^{nd}$ year
$P= Rs. 13,200, R=10 \%$, Time $(T)=1$ year
$\therefore$ Interest $=\frac{13,200 \times 10 \times 1}{100}=132 \times 10$
$= Rs. 1320$
$\therefore$ Amount in $2$ years $= Rs. (13,200)+(1320)$
$= Rs. 14520$
Compound interest in $2$ years $= Rs. 1200 + Rs. 1320= Rs. 2520$
$[$or directly $= Rs. 14520 - Rs. 12000 = Rs.2520]$
View full question & answer→Question 65 Marks
A man borrowed $Rs. 20,000$ for $2$ years at $8\%$ per year compound interest. Calculate : $(i)$ the interest of the first year.$(ii)$ the interest of the second year.$(iii)$ the final amount at the end of the second year.$(iv)$ the compound interest of two years.
AnswerHere Principal $(P)= Rs. 20,000$ , Time $=1$ year
Rate $=8 \%$
$(i) \therefore$ The interest of the first year $=\frac{20,000 \times 8 \times 1}{100}$
$= Rs. 1600$
$(ii) \therefore$ Amount after one year
i.e. Principal for second year $= Rs. 20,000+ Rs. 1,600= Rs. 21,600$
$\therefore$ Interest for second year $=\frac{21,600 \times 8 \times 1}{100}$
$=216 \times 8$
$= Rs. 1728$
$(iii)$ Final amount at the end of second year
$= Rs. (21,600+1728)= Rs. 23,328$
$(iv)$ Interest of two years $= Rs. 23,328-R s .20,000= Rs. 3,328$
View full question & answer→Question 75 Marks
A sum of $Rs. 8,000$ is invested for $2$ years at $10\%$ per annum compound interest. Calculate : $(i)$ interest for the first year.$(ii)$ principal for the second year. $(iii)$ interest for the second year. $(iv)$ the final amount at the end of the second year $(v)$ compound interest earned in $2$ years.
Answer$(i)$ Here Principal $(P)=Rs. 8,000$
Rate of interest $=10 \%$
Interest for the first year $=\frac{8,000 \times 10 \times 1}{100}$
$= Rs. 800$
$(ii) \therefore$ Amount $= Rs. 8,000+R s .800=R s .8,800$
Thus Principal for the second year $= Rs. 8,800$
$(iii)$ Interest for the second year
$=\frac{8,800 \times 10 \times 1}{100}$
$ =Rs. 880$
$(iv)$ Amount at the end of second year $= Rs. 8,800+ Rs. 880= Rs. 9,680$
$(v)$ Hence compound interest earned in $2$ years
$= Rs. 9,680-Rs. 8,000= Rs.1680 X X$
View full question & answer→Question 85 Marks
A certain sum of money invested for $5$ years at $8\%$ p.a. simple interest earns an interest of $₹ 12,000$. Find: $(i)$ the sum of money.$(ii)$ the compound interest earned by this money in two years and at $10\%$ p.a. compound interest.
AnswerRate $(R) = 8\%$ p.a.
Period $(T) = 5$ years
Interest $(I) = ₹.12000$
$(i) \therefore$ Sum $=\frac{1 \times 100}{R \times T}$
$=\text {₹}$$\frac{12000 \times 100}{8 \times 5}$
$= ₹.30000$
$(ii)$ Rate $(R) = 10\%$ p.a.
Time $(T) = 2$ years
Principal $(P) = ₹.30000$
Interest for the first year $=\frac{P R T}{100}$
$=₹.\frac{30000 \times 10 \times 1}{100}$
$= ₹.3000$
$\therefore$ Amount after one year $= ₹.30000 + 3000$
$= ₹.33000$
Principal for the second year $= ₹.33000$
Interest for the second year =$\frac{33000 \times 10 \times 1}{100}$
$= ₹.3300$
$\therefore$ Compound Interest for two years
$= ₹.3000 + 3300$
$= ₹.6300$
View full question & answer→Question 95 Marks
Gautam takes a loan of $₹ .16,000$ for $2$ years at $15\%$ p.a. compound interest. He repays $₹. 9,000$ at the end of the first year. How much must he pay at the end of the second year to clear the debt?
AnswerLoan taken $(P) = ₹ 16000$
Rate $(R) = 15\%$ p.a.
Time $(T) = 2$ years
$\therefore$ Interest for the first year
$=\frac{ PRT }{100}=\frac{16000 \times 15 \times 1}{100}$
$= ₹.2400$
Amount after one year $= ₹.16000 + 2400$
$= ₹.18400$
At the end of one year amount paid back $= ₹.9000$
Balance amount $= ₹.18400 − 9000$
$= ₹.9400$
Interest for the second year $= \frac{9400 \times 15 \times 1}{100}$
$= ₹.1410$
Amount after second year $= ₹.9400 + 1410$
$= ₹.10810$
View full question & answer→Question 105 Marks
Peter borrows $₹ 12,000$ for $2$ years at $10\%$ p.a. compound interest. He repays $₹ .8,000$ at the end of the first year. Find:$(i)\ $the amount at the end of the first year, before making the repayment.$\ (ii)\ $the amount at the end of the first year, after making the repayment.$\ (iii)\ $the principal for the second year.$\ (iv)\ $the amount to be paid at the end of the second year, to clear the account.
AnswerSum borrowed $= ₹. 12000$
Rate$ (R) = 10\%$ p.a. compound annually
Time $(T) = 2 $years
Interest for the first year =$\frac{ PRT }{100}$
$=\frac{12000 \times 10 \times 1}{100}$
$= ₹ .1200$
$(i) $Amount $= ₹. 12,000 + 1,200 = ₹ .13,200$
Amount paid $= ₹ .8,000$
$(ii)$ balance amount $= ₹ .13,200 − 8,000 = ₹ .5,200$
$(iii) \therefore$ Principal for the second year $= ₹.5,200$
$(iv)$ Interest for the second year $=\frac{5200 \times 10 \times 1}{100}$
$= ₹ 520$
$\therefore$ Amount $= ₹ 5200 + 520 = ₹ 5720$
View full question & answer→Question 115 Marks
Mr. Sharma lends $₹.24,000$ at $13\%$ p.a. simple interest and an equal sum at $12\%$ p.a. compound interest. Find the total interest earned by Mr. Sharma in $2$ years.
AnswerSum lends $(P)=₹ 24000$
Rate $(R)=13 \%$ p.a.
Time $(T)=2$ years
In case of simple interest,
Simple interest for $2$ years $=\frac{\text { PRT }}{100}$
$=₹. \frac{24000 \times 13 \times 2}{100}$
$=₹. 6240$
In the case of compound interest,
Interest for the first year $=\frac{24000 \times 12 \times 1}{100}$
$= ₹.2880$
Amount after the first year,
$=₹. 24000+2880$
$= ₹ .26880$
Interest for the second year $=₹ \frac{26880 \times 12 \times 1}{100}$
$=₹. \frac{322560}{100}$
$=₹. 3225.60$
$\therefore \text { C.I}$. for $2$ years$ =₹. 2880+3225.60$
$ =₹ .6105.60$
Total interest $=₹ .6240+6105.60$
$=₹ .12345.60$
View full question & answer→Question 125 Marks
Rohit borrowed $₹. 40,000$ for $2$ years at $10\%$ per annum $\text{C.I}$. and Manish borrowed the same sum for the same time at $10.5\%$ per annum simple interest. Which of these two gets less interest and by how much?
AnswerSum borrowed $(P) = ₹.40000$
Rate $(R) = 10\%$ p.a. compounded annually
Time $(T) = 2$ years
$\therefore$ Interest for the first year =$\frac{ PRT }{100}$
$=\text {₹.}$$\frac{40000 \times 10 \times 1}{100}$
$= ₹.4000$
Amount after one year $= ₹.40000 + 4000$
$= ₹.44000$
Principal for the second year $= ₹.44000$
$\therefore$ Interest for the second year
$=\frac{44000 \times 10 \times 1}{100}$
$= ₹.4400$
$\therefore$ Compound Interest for $2$ years $= ₹.4000 + 4400$
$= ₹.8400$
In the second case,
Principal $(P) = ₹.40000$
Rate $(R) = 10.5\%$ p.a.
Time $(T) = 2$ years
$\therefore$ Simple Interest $ = \frac{ PRT }{100}=\frac{40000 \times 10.5 \times 2}{100}$
$=\text {₹.}\frac{40000 \times 105 \times 2}{100 \times 10}$
$= ₹.8400$
In both the cases, interest is same.
View full question & answer→Question 135 Marks
Calculate the difference between the compound interest and the simple interest on $₹. 8,000$ in three years and at $10\%$ per annum.
AnswerPrincipal $(P) = ₹.8000$
Rate $(R) = 10\%$ p.a.
Period $(T) = 3$ years
$\therefore \text{S.I}$. for $3$ years $=\frac{\text { PRT }}{100}=\frac{8000 \times 10 \times 3}{100}$
$= ₹.2400$
Now, $\text{S.I}$. for $1^{st}$ year $=₹.\frac{8000 \times 10 \times 1}{100}$
$= 80 \times 10 \times 1$
$= ₹.800$
Amount for the first year $= P + \text{S.I}$.
$= ₹.8000 + ₹.800$
$= ₹.8800$
Principal for the second year $= ₹.8800$
Interest for the second year =$\frac{8800 \times 10 \times 1}{100}$
$= ₹.880$
$\therefore$ Amount after second year $= ₹.8800 + ₹.880$
$= ₹.9680$
Principal for the third year $= ₹.9680$
Interest for the third year
$=₹.\frac{9680 \times 10 \times 1}{100}$
$= ₹.968$
$\therefore \text{C.I}$. for $3$ year $= ₹.800 + ₹.880 + ₹.968$
$= ₹.2648$
$\therefore$ Differeence between $\text{C.I}$. and $\text{S.I}$. for $3$ year
$= ₹.2648 − ₹.2400$
$= ₹.248$
View full question & answer→Question 145 Marks
Calculate the difference between the compound interest and the simple interest on $₹.7,500$ in two years and at $8\%$ per annum.
AnswerPrincipal $(P) = ₹.7500$
Rate $(R) = 8\%$ p.a.
Period $(T) = 2$ year
$\therefore$ Simple interest $=\frac{ PRT }{100}=\frac{7500 \times 8 \times 2}{100}$
$= ₹.1200$
Interest for the first year =$\frac{7500 \times 8 \times 1}{100}$
$= ₹.600$
$\therefore$ Amount at the end of first year $= P + \text{S.I}.$
$= ₹.7500 + ₹.600$
$= ₹.8100$
Principal for the second year $= ₹.8100$
$\therefore$ Interest for the second year =$\frac{8100 \times 8 \times 1}{100}$
$= ₹.648$
$\therefore$ Total $\text{C.I}$. for $2$ years $= ₹.600 + ₹.648$
$= ₹.1248$
$\therefore$ Difference between $\text{C.I}$. and $\text{S.I}$. for $2$ years
$= ₹.1248 − ₹.1200$
$= ₹.48$
View full question & answer→Question 155 Marks
A person invests $Rs. 5,000$ for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to $Rs. 5,600$. Calculate : $(i)$ the rate of interest per year. $(ii)$ the amount at the end of the second year.
AnswerPrincipal $(P)= Rs. 5000$
Period $(T)=2$ years
Amount at the end of one year $=Rs. 5600$
$\therefore$ Interest for the first year $=\mathrm{A}-\mathrm{P}$
$= Rs. 5600-5000$
$=Rs. 600$
$(i) \therefore$ Rate of interest $=\frac{\text { S.I. } \times 100}{\mathrm{P} \times \mathrm{T}}$
$=\frac{600 \times 100}{5000 \times 1}$
$ =12 \% $ p.a.
$(ii)$ Principal for the second year $= Rs. 5600$
Interest for the second year $=\frac{5600 \times 12 \times 1}{100}$
$ = ₹ 672$
$\therefore$ Amount at the end of the second year
$=\mathrm{P}+\mathrm{S} . \mathrm{I} \text {. }$
$=5600+672$
$= ₹6272$
View full question & answer→Question 165 Marks
A man invests $Rs. 9600$ at $10\%$ per annum compound interest for $3$ years. Calculate : $(i)$ the interest for the first year. $(ii)$ the amount at the end of the first year. $(iii)$ the interest for the second year. $(iv)$ the interest for the third year. the interest for the first year.
AnswerRate $(R)=10 \%$ p.a.
Period $(n)=3$ years
$(i) \therefore$ Interest for the first year $=\frac{\mathrm{PRT}}{100}$
$=\frac{9600 \times 10 \times 1}{100}$
$ =Rs. 960$
$(ii)$ Amount at the end of first year
$=P+\text{S.I} .$
$ =Rs. 9600+960$
$ =Rs.10560$
$(iii)$ Principal for the second year $= Rs. 10560$
Interest for the second year $=\frac{10560 \times 10 \times 1}{100}$
$=Rs. 1056$
$\therefore$ Amount after second year $= Rs. 10560+1056= Rs. 11616$
$(iv)$ Principal for the third year $= Rs. 11616$
Interest for the third year $=\frac{11616 \times 10 \times 1}{100}$
$= 116.16 \times 10$
$= Rs.1161.60$
View full question & answer→Question 175 Marks
If the interest on $Rs.2400$ is more than the interest on $Rs.2000$ by $Rs.60$ in $3$ years at the same rate percent ; find the rate.
AnswerIn first case:
$P=Rs. 2400$
$\mathrm{R}=\mathrm{x} \% ($Assume$)$
$\mathrm{T}=3$ years
Interest $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$=Rs. \frac{2400 \times \mathrm{x} \times 3}{100}$
$=Rs. 72 x$
In second case :
$P=Rs. 2000$
$\mathrm{R}=\mathrm{x} \%....($Rate same as in first case$)$
$\mathrm{T}=3$ years
Interest $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =Rs. \frac{2000 \times \mathrm{x} \times 3}{100}$
$=Rs. 60 \mathrm{x}$
According to the statement,
$72 x=60 x+60$
$\Rightarrow 72 \mathrm{x}-60 \mathrm{x}=60$
$\Rightarrow 12 x=60$
$\Rightarrow \mathrm{x}=\frac{60}{12}$
$\Rightarrow \mathrm{x}=5$
$\therefore$ Rate $=5 \%$
View full question & answer→Question 185 Marks
On what date will $₹.1950$ lent on $5^{th}$ January $2011$ amount to $₹.2125.50$ at $5$ percent per annum simple interest?
Answer$P=Rs .1950$
$A=Rs. 2125.50$
$R=5 \% \text { P.a. }$
$I=\text{A-P}$
$= Rs. 2125.50 - Rs, 1950$
$= Rs. 175.50$
$T=\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{R}}$
$ =\frac{100 \times 175.50}{1950 \times 5}$
$ =\frac{17550}{9750}=\frac{1755}{975}=\frac{117}{65}$
$=\frac{9}{5}$ years $=1 \frac{4}{5}$ years
$=1$ years $292$ days
$\because \frac{4}{5}$ years
$=\frac{4}{5} \times 365$ days $=292$ days
Jan. $+$ Feb. $+$ March $+$ April $+$ May $+$ June $+$ July $+$ Aug. $+$ Sept. $+$ Oct.
$ (31-5)+29+31+30+31+30+31+31+30+23$
$ =292 $ days
$\therefore$ Required date $=23^{rd}$ October $2012$
View full question & answer→Question 195 Marks
At what rate percent of simple interest will the interest on $Rs.3750$ be one$-$fifth of itself in $4$ years? To what will it amount in $15$ years?
Answer$P= Rs. 3750$
$ I=Rs. 3750 \times \frac{1}{5}= Rs. 750$
$ T=4$ years
$ R=\frac{100 \times I}{P \times T}$
$ =\frac{100 \times 750}{3750 \times 4}$
$ =5 \%$
Again, $P= Rs. 3750$
The interest of $4$ years $= Rs. 750$
The interest of $1$ year $= Rs.\frac{750}{4}$
Interest of $15$ years $= Rs. \frac{750}{4} \times 15$
$= Rs. \frac{750 \times 15}{4}$
$= Rs. \frac{5625}{2}$
$= Rs. 2812.50$
Amount in $15$ years will be $= Rs. 3750+ Rs. 2812.50$
$= Rs.6562.50$
$\therefore$ Rate $= 5\%$
The amount in $15$ years will be $= Rs.6562.50$
View full question & answer→Question 205 Marks
What sum of money lent at $6.5\%$ per annum will produce the same interest in $4$ years as $Rs.7500$ produce in $6$ years at $5\%$ per annum?
AnswerIn first case :
$P= Rs. 7500$
$R=5 \%$
$T=6$ years
Interest $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$= Rs. \frac{7500 \times 5 \times 6}{100}$
$= Rs. 75 \times 5 \times 6$
$= Rs. 2250$
In the second case:
According to the statement, interest $= Rs. 2250$
$R=6.5 \%$ P.A.
$\mathrm{T}=4$ years
$P=\frac{100 \times \mathrm{I}}{\mathrm{R} \times \mathrm{T}}$
$=Rs. \frac{100 \times 2250}{6.5 \times 4}$
$= Rs. \frac{225000}{26}$
$= Rs. \frac{112500}{13}$
$=Rs. 8653.85$
Required principal $= Rs. 8653.85$
View full question & answer→Question 215 Marks
Rupees $4000$ amount to $Rs.5000$ in $8$ years ; in what time will $Rs.2100$ amount to $Rs.2800$ at the same rate?
AnswerIn first case :
$A=Rs .5000$
$ P=Rs .4000$
$I=A-P$
$=Rs. 5000 - Rs. 4000$
$= Rs. 1000$
$\mathrm{T}=8$ years
$R=\frac{100 \times I}{P \times R}$
$=\frac{100 \times 1000}{4000 \times 8}$
$=\frac{25}{8} \%$
In the second Case :
$A=R s .2800$
$ P=Rs. 2100$
$ I=Rs. 2800-Rs. 2100=Rs. 700$
$ R=\frac{25}{8} \%$
$ T=\frac{100 \times I}{P \times R}=\frac{100 \times 700}{2100 \times \frac{25}{8}}$
$ =\frac{100 \times 700 \times 8}{2100 \times 25}=\frac{32}{3}$ years
$ =10 \frac{2}{3}$ years
$ =10 \frac{2}{3} \times 12$ months $=10 \frac{24}{3}$ months
$ =10$ years $8$ months
View full question & answer→Question 225 Marks
What sum of money borrowed on $24\ th$ May will amount to $Rs.10210.20$ on $17\ th$ October of the same year at $5$ percent per annum simple interest.
Answer$A=\text { Rs. } 10210.20$
$ R=5 \% \text { P.A. }$
$ T=\text { May + June + July + August + Sept. + Oct. }$
$ =7+30+31+31+30+17$
$ =\frac{146}{365}$ days $=\frac{2}{5}$ year
We know that:
$\mathrm{P}+\mathrm{I}=\mathrm{A}$
$ \Rightarrow \mathrm{P}+\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}=\mathrm{A}$
$ \Rightarrow \mathrm{P}\left(1+\frac{\mathrm{R} \times \mathrm{T}}{100}\right)=\mathrm{A}$
$ \Rightarrow \mathrm{P}\left(1+\frac{5 \times \frac{2}{5}}{100}\right)=\text { Rs. } 10210.20$
$ \Rightarrow \mathrm{P}\left(1+\frac{2}{100}\right)=\text { Rs. } 10210.20$
$ \Rightarrow \mathrm{P} \times \frac{102}{100}=\text { Rs. } 10210.20$
$ \Rightarrow \mathrm{P}=\text { Rs. } 10210.20 \times \frac{100}{102}$
$ \Rightarrow \mathrm{P}=\text { Rs. } \frac{1021020}{102}$
$\Rightarrow P=\text { Rs. } 10010$
$\therefore$ Money to be borrowed $= Rs. 10010$
View full question & answer→Question 235 Marks
John lent $Rs. 2550$ to Mohan at $7.5$ percent per annum. If Mohan discharges the debt after $8$ months by giving an old black and white television and $Rs. 1422.50\ ;$ find the price of the television.
Answer$P=Rs. 2550$
$ R=7.5 \%$
$ T=8 $ months $=\frac{8}{12} $ years
$ =\frac{2}{3} $ years
$ \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =Rs. 2550 \times 7.5 \times \frac{2}{3} \times \frac{1}{100}$
$ =Rs. \frac{2550 \times 7.5 \times 2}{3 \times 100}$
$ =Rs. \frac{2500 \times 5}{100}$
$ =Rs. \frac{12750}{100}$
$ =Rs. 127.50$
Amount $=\mathrm{P}+\mathrm{I}$
$=Rs. 2550 + Rs. 127.50$
$=Rs. 2677.50$
Mohan paid in cash $= Rs. 1422.50$
Price of the television
$=$ Amount $-$ Paid in cash
$= Rs. 2677.50 - Rs. 1422.50$
$= Rs. 1255$
View full question & answer→Question 245 Marks
Mohan lends $Rs. 4800$ to John for $4 \frac{1}{2}$ years and $Rs. 2500$ to Shy am for $6$ years and receives a total sum of $Rs. 2196$ as interest. Find the rate percent per annum, it being the same in both the cases.
AnswerIn the first case :
$P=Rs. 4800$
$ R=x \% ($Suppose$)$
$ T=4 \frac{1}{2} \text { years }=\frac{9}{2}$ years
Interest $=\frac{P \times R \times T}{100}$
$ =Rs. \frac{4800 \times x \times 9}{100 \times 2}$
$ =Rs. 24 \times x \times 9$
$ =Rs. 216 x$
In the second case :
$P=Rs. 2500$
$ R=x \%$
$\mathrm{T}=6$ years
Interest $=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ =Rs. \frac{2500 \times \mathrm{x} \times 6}{100}$
According to statement,
Interest in first case $+$ Interest in second case
$=Rs. 2196$
$\therefore Rs. 216 x+ Rs. 150 x= Rs. 2196$
$\Rightarrow Rs. 366 x= Rs. 2196$
$\Rightarrow x=\frac{2196}{366}$
$\Rightarrow x=6$
$\therefore$ Rate $=6 \%$
View full question & answer→Question 255 Marks
Raj borrows $Rs.8,000;$ out of which $Rs. 4500$ at $5\%$ and the remainder at $6\%.$ Find the total interest paid by him in $4$ years.
AnswerTotal sum borrowed by Raj $= Rs. 8000$
In the First Case $:$
$P=Rs. 4500$
$ R=5 \%$
$ T=4$ years
$ \text { S.I. }=\frac{P \times R \times T}{100}$
$ =Rs. \frac{4500 \times 5 \times 4}{100}$
$ =Rs. 900$
In the Second Case :
$p=Rs. 8000-R s .4500$
$ =Rs. 3500$
$R=6 \%$
$\mathrm{T}=4$ years
$ \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$= Rs. \frac{3500 \times 6 \times 4}{100}$
$=35 \times 6 \times 4$
$=Rs. 840$
Total interest paid by Raj
$=Rs. 900+Rs. 840$
$ =Rs. 1740$
View full question & answer→Question 265 Marks
Find the interest and the amount on : $₹ .5,000$ at $8\%$ per year from $23\ rd$ December $2011$ to $29\ th$ July $2012.$
AnswerPrincipal $(P) = ₹.5000$
Rate $(R) = 8\%$ p.a.
Time $(T) = 23$ December $2011$ to $29$ July $2012$
| Dec. |
Jan. |
Feb. |
March |
April |
May |
June |
July |
| $8$ |
$31$ |
$29$ |
$31$ |
$30$ |
$31$ |
$30$ |
$29$ |
Total of $219$ days $=\frac{219}{365}$ years
$\text { S.I }=\frac{P \times R \times T}{100}$
$\therefore$ Interest $=\frac{\text { PRT }}{100}=\frac{5000 \times 8 \times 219}{100 \times 365}$
$= 10 \times 8 \times 3$
$= ₹.240$
$\therefore$ Amount $= P + I$
$= ₹.5000 + 240$
$= ₹.5240$
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