Question 14 Marks
Rekha borrowed $Rs. 40,000$ for $3$ years at $10\%$ per annum compound interest. Calculate the interest paid by her for the second year.
AnswerFor $1st$ year
Principal $= Rs. 40,000$ , Rate $=10 \%$, Time $=1$ year
$\therefore$ Interest $=\frac{40,000 \times 10 \times 1}{100}$
$=400 \times 10$
$Rs. 4000$
$\therefore$ Amount at the end of $1 st$ year $= Rs. (40,000+4000)=Rs. 44,000$ For $2nd$ year
$P= Rs.44,000, R=10 \%, T=1$ year
$\therefore$ Interest $= Rs.\frac{44,000 \times 10 \times 1}{100}$
$=440 \times 10$
$=Rs. 4400$
Thus interest earned in the second year $= Rs. 4400$
View full question & answer→Question 24 Marks
Calculate the compound interest on $Rs. 5,000$ in $2$ years; if the rates of interest for successive years be $10\%$ and $12\%$ respectively.
AnswerFor $1st$ year
Principal $(P)=Rs. 5,000$ , Rate $(R)=10 \%$, Time $(T)=1$ year
$\therefore$ Interest $=\frac{5,000 \times 10 \times 1}{100}=50 \times 10= Rs. 500$
$\therefore$ Amount at the end of $1 st$ year $= Rs.(5000+500)= Rs. 5500$
For $2nd$ year
$P= Rs. 5500$, Rate $=12 \%, T=1$ year
$\therefore$ Interest $=\frac{5500 \times 12 \times 1}{100}=55 \times 12= Rs. 660$
$\therefore$ Amount at the end of $2nd$ year $=Rs.5500+ Rs.660=Rs. 6160$
Hence compound interest $= Rs. 6160 - Rs. 5000= Rs. 1160$
View full question & answer→Question 34 Marks
Find the amount and the compound interest on $₹ 10,000$ in $3$ years, if the rates of interest for the successive years are $10\%, 15\%,$ and $20\%$ respectively.
AnswerPrincipal $(P) = ₹10,000$
Time $(t) = 3$ years
Rate $(R_1) = 10\%$
Rate $(R_2) = 15\%$
Rate $(R_3) = 20\%$
Amount $=P \times\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\left(1+\frac{R_3}{100}\right)$
$=\text{₹} 10,000 \times\left(1+\frac{10}{100}\right)\left(1+\frac{15}{100}\right)\left(1+\frac{20}{100}\right)$
$=\text{₹}10,000 \times \frac{11}{10} \times \frac{23}{20} \times \frac{6}{5}$
$= ₹15,180$
$C.I$. = Amount − Principal
$= ₹15,180 − ₹10,000$
$= ₹5180$
View full question & answer→Question 44 Marks
Find the amount and the compound interest on $₹ 4,000$ in $2$ years, if the rate of interest for the first year is $10\%$ and for the second year is $15\%.$
AnswerPrincipal $(P) = ₹4,00$
Time $(t) = 2$ years
Rate $(R_1) = 10\%$ and rate $(R_2) = 15\%$
Amount $=P \times\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)$
$=\text{₹} 4,000\left(1+\frac{10}{100}\right)\left(1+\frac{15}{100}\right)$
$=\text{₹} 4,000 \times \frac{11}{10} \times \frac{23}{20}$
$= ₹5060$
$C.I. =$ Amount $−$ Principal
$= ₹5060 − ₹4000$
$= ₹1060$
View full question & answer→Question 54 Marks
Find the amount and the compound interest on $₹ 32,000$ for $1$ year at $20\%$ per annum compounded half$-$yearly.
AnswerPrincipal $(P) = ₹32,000$
Time $(t) = 1$ year
Rate $(r) = 20\%$
Amount $=$Principal $\times\left(1+\frac{r}{2 \times 100}\right)^{ n \times 2}$
$=\text{₹} 32,000 \times\left(1+\frac{20}{200}\right)^{1 \times 2}$
$=\text{₹} 32,000 \times\left(\frac{11}{10}\right)^2$
$=\text{₹} 32,000 \times \frac{11}{10} \times \frac{11}{10}$
$=\text{₹} 38,720$
$C.I. =$ Amount $−$ Principal $= ₹38,720 − ₹32,000 = ₹6,720$
View full question & answer→Question 64 Marks
Find the amount and the compound interest on $₹ 20,000$ for $1 \frac{1}{2}$ years at $10 \%$ per annum compounded half$-$yearly.
AnswerPrincipal $(P) = ₹20,000$
Time $(t)=1 \frac{1}{2}$ years $=\frac{3}{2}$ years
Rate $(r) = 10\%$
Amount $=P \times\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}$
$=\text{₹} 20,000 \times\left(1+\frac{10}{200}\right)^{\frac{3}{2} \times 2}$
$=\text{₹} 20,000 \times\left(\frac{210}{200}\right)^3$
$=\text{₹} 20,000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$
$= ₹23,152.50$
$C.I. =$ Amount $−$ Principal
$= ₹23,152.50 − ₹20,000$
$= ₹3,152.50$
View full question & answer→Question 74 Marks
Find the amount and the compound interest on ₹ $16,000$ for $3$ years at $5\%$ per annum compounded annually.
AnswerPrincipal $(P) = ₹16,000$
Time $(t) = 3$ years
Rate $(r) = 5\%$
Amount $=$ Principal $\times\left(1+\frac{R}{100}\right)^n$
$=16000 \times\left(1+\frac{5}{100}\right)^3$
$=16000 \times\left(\frac{21}{20}\right)^3$
$=16000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20}$
$= ₹18,522$
$C.I.=$ Amount $−$ Principal
$= ₹18,522 − ₹16,000$
$= ₹2522$
View full question & answer→Question 84 Marks
Find the amount and the compound interest on $₹ 24,000$ for $2$ years at $10\%$ per annum compounded yearly.
Answer$P = ₹24,000$
$n = 2$ years
$R = 10\%$ P.a.
$A =P\left(1+\frac{R}{100}\right)^n$
$A =24000\left(1+\frac{10}{100}\right)^2$
$=24000\left(\frac{10+1}{10}\right)^2$
$=24000\left(\frac{11}{10}\right)^2$
$=24000 \times \frac{11}{10} \times \frac{11}{10}$
$=240 \times 121$
$A = 29040$
Amount $= ₹29040$
$C.I = A - P$
$= 29040 - 24000$
$= ₹5040$
View full question & answer→Question 94 Marks
Find the amount and the $C.I$. on $₹ 8,000$ in $1 \frac{1}{2}$ years at $20 \%$ per year compounded halfyearly.
AnswerPrincipal $(P) = ₹8000$
Rate $= 20\%$
Time $=1 \frac{1}{2}$ years $=\frac{3}{2}$ years
Amount $=$ Principal $\times\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}$
$=\text {₹} 8000$$\times\left(1+\frac{20}{200}\right)^{\frac{3}{2} \times 2}$
$=\text {₹} 8000$$\left(\frac{220}{200}\right)^3$
$=\text {₹} 8000$$\times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$
$= ₹10648$
$C.I. =$ Amount $−$ Principal
$= ₹10648 − ₹8000$
$= ₹2648$
View full question & answer→Question 104 Marks
Find the amount and the $C.I$. on $₹ 12,000$ at $10\%$ per annum compounded half$-$yearly.
AnswerPrincipal $(P) = ₹12,000$
Rate $(r) = 10\%$
Time $(t) = 1 $years
Amount $=P \times\left(1+\frac{r}{2 \times 100}\right)^{n \times 2}$
$=\text {₹} 12000$$\times\left(1+\frac{10}{200}\right)^2$
$=\text {₹} 12000$$\times\left(\frac{210}{200}\right)^2$
$=\text {₹} 12000$$\times \frac{21}{20} \times \frac{21}{20}$
$= ₹13,230$
$C.I. =$ Amount $−$ Principal
$= ₹13230 − ₹12000$
$= ₹1230$
View full question & answer→Question 114 Marks
A certain sum amounts to $Rs.3825$ in $4$ years and to $Rs.4050$ in $6$ years. Find the rate percent and the sum.
AnswerIn $6$ years sum amounts to $= Rs. 4050$
In $4$ years sum amounts to $=Rs. 3825$
$\therefore$ Interest of $2$ years $= Rs.4050- Rs.3825= Rs. 225$
The interest of $4$ years $= Rs.\frac{225}{2} \times 4$
$= Rs. 450 ... (\because Rs. 225$ is interest for $2$ years $)$
Now $P=A-I$
$=Rs.3825-Rs. 450$
$= Rs. 3375$
$I = Rs. 450$
$\mathrm{T}=4$ years
$\mathrm{R}=\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{T}}$
$=\frac{100 \times 450}{3375 \times 4}$
$=\frac{45000}{13500} \%=\frac{450}{135} \%$
$=\frac{10}{3} \%=3 \frac{1}{3} \%$
$\therefore \mathrm{R}=3 \frac{1}{3} \%$
$P = Rs.3375$
View full question & answer→Question 124 Marks
If $₹3,750$ amount to $₹ 4,620$ in $3$ years at simple interest. Find:$(i)$ the rate of interest $(ii)$ the amount of $Rs.7,500$ in $5 \frac{1}{2}$ years at the same rate of interest
Answer$(i) \mathrm{A}= Rs. 4620$
$P=Rs.3750$
$I=A-P=R s .4620-R s .3750= Rs. 870$
$\mathrm{T}=3$ years
$\mathrm{R}=\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{T}}=\frac{100 \times 870}{3750 \times 3}=\frac{100 \times 290}{3750}=\frac{4 \times 29}{15}$
$ =\frac{116}{15}=7 \frac{11}{15} \%$
$(ii) P= Rs. 7500$
$\mathrm{R}=\frac{116}{15} \%$
$ \mathrm{~T}=5 \frac{1}{2} \text { years }=\frac{11}{2} \text { years }$
Interest $=\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}$
$=Rs. \frac{7500 \times 11 \times 116}{2 \times 15 \times 100}=\frac{250 \times 116 \times 11}{100}$
$ =10 \times 29 \times 11=290 \times 11=\text { Rs. } 3190$
Amount $=Rs.7500+3190= Rs. 10,690$
View full question & answer→Question 134 Marks
Divide $Rs. 15,600$ into two parts such that the interest on one at $5$ percent for $5$ years may be equal to that on the other at $4 \frac{1}{2}$ percent for $6$ years.
AnswerLet one part $= Rs. x$
$\therefore$ Second part $= Rs. (15,600-\mathrm{x})$
By the given condition
$=\frac{x \times 5 \times 5}{100}=\frac{(15,600-x) \times \frac{9}{2} \times 6}{100}$
$ \Rightarrow 25 x=27 \times 15,600-27 x$
$ \Rightarrow 25 x+27 x=27 \times 15,600$
$ \Rightarrow 52 x=27 \times 15,600$
$ \Rightarrow x=\frac{27 \times 15,600}{52}$
$ =27 \times 300$
$ =8100$
Hence one part $= Rs. 8100$ and second part $Rs.(15,600-8,100)=Rs.7,500$
View full question & answer→Question 144 Marks
Find the principal which will amount to $Rs. 4,000$ in $4$ years at $6.25\%$ Per annum.
Answer$A= Rs. 4000$
$R=6.25 \%$
$\mathrm{T}=4$ years
$A=P\left(1+\frac{R T}{100}\right)$
$4000=P\left(1+\frac{6.25 \times 4}{100}\right)$
$4000=P(1.25)$
$P=\frac{4000}{1.25}$
$P=Rs. 3200$
View full question & answer→Question 154 Marks
On what principal will the simple interest be $Rs. 7,008$ in $6$ years $3$ months at $5\%$ per year?
AnswerLet Principal $= Rs. \mathrm{P}$
Time $(T)=6$ years $3$ months $=6$ year $+\frac{3}{12}$
year $=\frac{75}{12}=\frac{25}{4}$ year $=6 \frac{1}{4}$ years
Rate $(R)=5 \%$
Simple interest $=\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}$
$\Rightarrow 7,008=\frac{\mathrm{P} \times \frac{25}{4} \times 5}{100}$
$\Rightarrow \mathrm{P}=\frac{7008 \times 100 \times 4}{25 \times 5}$
$=\frac{7008 \times 16}{5}=\frac{112128}{5}$
$= Rs. 22425.60$
View full question & answer→Question 164 Marks
Ashok lent out $Rs.7000$ at $6\%$ and $Rs.9500$ at $5\%$. Find his total income from the interest in $3$ years.
AnswerIn $I$ case:
$P=Rs.7000$
$ R=6 \%$
$ T=3$ years
$ \text { S.I. }=\frac{P \times R \times T}{100}$
$ =Rs. \frac{7000 \times 6 \times 3}{100}$
$ = Rs.1260$
In $II$ case :
$P=Rs. 9500$
$ R=5 \%$
$T=3 \text { years }$
$ \text { S.I. }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}$
$ = Rs. \frac{9500 \times 5 \times 3}{100}$
$ =Rs.1425$
Total income from the interest
$= Rs. 1260 + Rs. 1425$
$= Rs.2685$
View full question & answer→Question 174 Marks
In what time will the interest on a certain sum of money at $6 \%$ be $\frac{5}{8}$ of itself?
AnswerLet $P= Rs. 8$
Interest $= Rs.8 \times \frac{5}{8}= Rs. 5$
$R=6 \%$
$\mathrm{T}=\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{R}}$
$=\frac{100 \times 5}{8 \times 6}$
$=\frac{500}{48}=\frac{125}{12}$ years
$=10 \frac{5}{12}$ years
$=10$ years $5$ months
$[\because \frac{5}{12}$ year $=\frac{5}{12} \times 12 $months $=5 $months $]$
$\therefore$ Time $=10$ years $5$ months
View full question & answer→Question 184 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 194 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 204 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 214 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 224 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 234 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 $= Rs. 23,328-Rs .20,000= Rs. 3,328$
View full question & answer→Question 244 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 254 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 $=\frac{1 \times 100}{R \times T}$
$=₹\frac{12000 \times 100}{8 \times 5}$
$= ₹30000$
$(ii)$ 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 264 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 274 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 284 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 294 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 $=\frac{ PRT }{100}$
$=₹.\frac{40000 \times 10 \times 1}{100}$
$= ₹.4000$
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 $= \frac{ PRT }{100}=\frac{40000 \times 10.5 \times 2}{100}$
$\frac{40000 \times 105 \times 2}{100 \times 10}$
$= ₹.8400$
In both the cases, interest is same.
View full question & answer→Question 304 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 $=\text {₹}$$\frac{8000 \times 10 \times 1}{100}$
$= 80 \times 10 \times 1$
$= ₹.800$
Amount for the first year $= P + \text{S.I}.$
$= ₹.8000 + ₹.800$
$= ₹.8800$
Principal for the second year $= ₹.8800$
Interest for the second year =$\frac{8800 \times 10 \times 1}{100}$
$= ₹.880$
$\therefore$ Amount after second year $= ₹.8800 + ₹.880$
$= ₹.9680$
Principal for the third year $= ₹.9680$
Interest for the third year
$=\text {₹}$$\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 314 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 324 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 334 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 344 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 354 Marks
On what date will $₹.1950$ lent on $5\ th$ January $2011$ amount to $₹ 2125.50$ at $5$ percent per annum simple interest?
Answer$P=R s .1950$
$A=\text { Rs. } 2125.50$
$R=5 \% $ P.a.
$I=A-P$
$=Rs. 2125.50 - Rs, 1950$
$=Rs. 175.50$
$T=\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{R}}$
$ =\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 364 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$P=\text { Rs. } 3750$
$ I=\text { Rs. } 3750 \times \frac{1}{5}=\text { 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 374 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 384 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 394 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$A=Rs. 10210.20$
$ R=5 \% $ P.A.
$ T=$ 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)=Rs. 10210.20$
$ \Rightarrow \mathrm{P}\left(1+\frac{2}{100}\right)=Rs.10210.20$
$ \Rightarrow \mathrm{P} \times \frac{102}{100}= Rs.10210.20$
$ \Rightarrow \mathrm{P}= Rs. 10210.20 \times \frac{100}{102}$
$ \Rightarrow \mathrm{P}= Rs. \frac{1021020}{102}$
$\Rightarrow P= Rs.10010$
$\therefore$ Money to be borrowed $= Rs. 10010$
View full question & answer→Question 404 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 414 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}$ 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 424 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 434 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
$\text { S.I }=\frac{P \times R \times T}{100}$
$\therefore \text { Interest }=\frac{\text { PRT }}{100}$
$=\frac{5000 \times 8 \times 219}{100 \times 365}$
$= 10 \times 8 \times 3$
$= ₹240$
$\therefore$ Amount $= P + I$
$= ₹5000 + 240$
$= ₹5240$ View full question & answer→