Question 14 Marks
A new car is purchased for $Rs.4, 00,000.$ Its value depreciates at the rate of $10\%$ per annum. What will be its value after $4$ years?
Answer$V_n=? ; V_0=\operatorname{Rs} 4,00,000 ; r=10 \% ; t=4 \text { years } $
$ V_n=V_0\left(1-\frac{r}{100}\right)^n $
$ V_n=\operatorname{Rs} 400000\left(1-\frac{10}{100}\right)^4 $
$V_n=\operatorname{Rs} 4,00,000 \times 0.9 \times 0.9 \times 0.9 \times 0.9$
$V_n=\operatorname{Rs} 2,62,440$
The value of car after four years will be $Rs.2,62,440.$
View full question & answer→Question 24 Marks
Under the electrification programme of villages, the number of villages with electricity rose to $27,040$ from $25,000$ in $2$ years. Find the rate of growth in the number of villages with electricity.
Answer$V _{ n }=27,040 ; V _0=25,000 ; r=? ; t =2 \text { years } $
$ V _{ n }= V _0\left(1+\frac{ r }{100}\right)^{ n }$
$ 27040=25000\left(1+\frac{ r }{100}\right)^2 $
$ \frac{27040}{25000}=\left(1+\frac{ r }{100}\right)^2 $
$ \left(\frac{164.43}{158}\right)^2=\left(1+\frac{ r }{100}\right)^2 $
$ \left(1+\frac{ r }{100}\right)=\frac{164.43}{158}$
$ \frac{ r }{100}=\frac{164}{158}-1$
$ \frac{ r }{100}=\frac{6.43}{158} $
$ r =0.040 \times 100 $
$ r =4 \%$
The rate of growth in number of villages with electricity is $4 \%$.
View full question & answer→Question 34 Marks
A machine was purchased $2$ years ago. Its value decreases by $10\%$ every year. Its present value is $Rs.19,083.60.$ For how much money was the machine purchased?
Answer$V_n=\operatorname{Rs} 19,083.60 ; V_0=? ; r=10 \% ; t=2 \text { years } $
$ V_n=V_0\left(1+\frac{r}{100}\right)^n $
$ \text { Rs } 19,083.60=V_0\left(1-\frac{10}{100}\right)^2 $
$ V_0=\text { Rs } 19083.60 \times \frac{10}{9} \times \frac{10}{9} $
$ V_0=\text { Rs } 23560$
The machine was purchased for $Rs.23,560$ i.e. Rs $(23,560-19083.60)=\operatorname{Rs} 4,476.40$ more than the present value.
View full question & answer→Question 44 Marks
The present population of a town is $1$
, 0 /o per annum $1,15,200$. If it increases at the rate of $6 \frac{2}{3} \%$ per annum find Its population $2$ years ago.
Answer$ V _{ n }=? ; V _0=115200 ; r =6 \frac{2}{3} \%=\frac{20}{3} \% ; t=2 \text { years }$
$V _{ n }= V _0\left(1+\frac{ r }{100}\right)^{ n }$
$V _{ n }=115200\left(1-\frac{20}{100 \times 3}\right)^2$
$V _{ n }=1,15,200 \times 0.933333 \times 0.933333$
$V _{ n }=1,00,352$
$\text { The population } 2 \text { years ago was }=1,00,352$
View full question & answer→Question 54 Marks
The population of a village increases at the rate of $50$ per thousand. Its population after $2$ years will be $22,050.$ Find the present population.
Answer$ \text { Rate of increase }= $
$ r=\frac{50}{1000} \times 100=5 \% $
$V_n=22,050 ; V_0=? ; r=5 \% ; t=2 \text { years } $
$ V_n=V_0\left(1+\frac{r}{100}\right)^t $
$22050=V_0\left(1+\frac{5}{100}\right)^2$
$ 22,050=v \cdot \times 1.05 \times 1.05 $
$V_0=\frac{22050}{11025} $
$ V_0=20000 $
The present population is $20,000 .$
View full question & answer→Question 64 Marks
The population of a city is $1, 25,000.$ If the annual birth rate and death rate are $5.5\%$ and $3.5\%$ respectively, calculate the population of the city after $3$ years.
Answer$ V_n=? ; V_0=1,25,000 ; r=5.5 \% \text { (birth) and } 3.5 \%$
$ (\text { death) } ; t=3 \text { years } $
$ V_n=V_0\left(1+\frac{r}{100}\right)^t$
$V_n=125000\left(1+\frac{5.5}{100}\right)^3\left(1-\frac{3.5}{100}\right)^3$
$ V_n=1,25,000 \times 10.55 \times 10.55 \times 10.55 \times 9.65 \times 9.65 \times 9.65$
$V_n=1,25,000 \times 1174.241 \times 898.6321 $
$ V_n=1,32,651$
The population in $2007$ is $1,32,651$
View full question & answer→Question 74 Marks
The population of a town in the year $2005$ was $4, 25,000.$ Find its population in the year $2007$ if the rate of annual increase is $4\%$ per year.
Answer$V_n=? ; V_0=4,25,000 ; r=4 \% ; t=2 \text { years }$
$V_n=V_0\left(1+\frac{r}{100}\right)^t $
$ V_n=425000\left(1+\frac{4}{100}\right)^2 $
$ V_n=4,25,000 \times 1.04 \times 1.04 $
$ V_n=4,59,680$
The population in $2007$ is $4,59,680$
View full question & answer→Question 84 Marks
A mango tree was planted $3$ years ago. The rate of growth is $20\%$ per annum. If at present, the height of the tree is $1$ m $8$ cm, how high was it when planted?
Answer$V_n=1 m 8 cm =108 cm ; V_0=? ; t=2 \text { years; } r=20 \%$
$ V_n=V_0\left(1-\frac{r}{100}\right)^{ n } $
$108 cm =V_0\left(1+\frac{20}{100}\right)^3 $
$V_0=108 cm \times 0.8333 \times 0.8333 \times 0.8333 $
$ V_0=62.5=m $
$ \text { The height of tree was } 62.5\ cm \text { when planted. }$
View full question & answer→Question 94 Marks
The value of the refrigerator which was purchased $2$ years ago, depreciates at $12\%$ per annum. If its present value is $Rs.9,680.$ For how much was it purchased?
Answer$V_n=\operatorname{Rs} 9,680 ; V_0=? ; r=12 \% ; t=2 \text { years } $
$ V_n=V_0\left(1-\frac{r}{100}\right)^n$
$ \text { Rs } 9680=V_0\left(1-\frac{12}{100}\right)^2 $
$ V_0=\operatorname{Rs} 9680 \times \frac{100}{88} \times \frac{100}{88} $
$ V_0=\operatorname{Rs} 9,680 \times 1.136364 \times 1.136364 $
$ V_0=\operatorname{Rs} 12,500 $
The refrigerator was purchased for $Rs.12,500$
View full question & answer→Question 104 Marks
The value of a property decreases every year at the rate of $5\%.$ If its value at the end of $3$ years be $Rs.44,540,$ what was the original value at the beginning of these $3$ years?
Answer$V _{ n }= Rs.44,540 ; V _0=? ; r =5 \% ; t =3 \text { years }$
$ V _{ n }= V _0\left(1-\frac{ r }{100}\right)^{ n } $
$Rs 44540= V _0\left(1-\frac{5}{100}\right)^3 $
$ V _0=\operatorname{Rs} 44540 \times \frac{100}{95} \times \frac{100}{95} \times \frac{100}{95}$
$V _0=\operatorname{Rs} 44,540 \times 1.052632 \times 1.052632 \times 1.052632$
$ V_0=\operatorname{Rs} 51,949.26 $
$\text { The original value of the property was Rs } 51,949.26$
View full question & answer→Question 114 Marks
What sum of money will amount to $Rs.15,746.40$ at $16\%$ p.a. compounded half-yearly?
Answer$ P=? ; A=\operatorname{Rs} 15,746.40 ; t=\frac{1}{2} \text { years } ; r=16 \%$
$A=P\left(1+\frac{r}{100}\right)^n$
$ \text { Rs } 15746.40=P\left(1+\frac{16}{100}\right)^{\frac{1}{2}} $
$\text { Rs1 } 5,746.40-P\left(1+\frac{1}{2} \times \frac{16}{100}\right) $
$ \text { Rs } 15,746.40=P \times 1.08$
$ P=\text { Rs } \frac{15746.40}{1.08} $
$ P=\text { Rs } 14,580$
Hence, the sum of money will be $Rs.14,580$
View full question & answer→Question 124 Marks
What sum of money will amount to $Rs.18,792$ in $1 \frac{1}{2}$ years at $16 \%$ p.a. compounded yearly?
Answer$P=? ; A=\operatorname{Rs} 18792 ; t=1 \frac{1}{2} \text { years } ; r=16 \% $
$ A=P\left(1+\frac{r}{100}\right)^n $
$ \Rightarrow \operatorname{Rs} 18972=P\left(1+\frac{16}{100}\right)\left(1+\frac{16}{100}\right)^{\frac{1}{2}} $
$\Rightarrow \operatorname{Rs} 18972=P\left(1+\frac{16}{100}\right)\left(1+\frac{1}{2} \times \frac{16}{100}\right)$
$ \Rightarrow \operatorname{Rs} 18,972=P \times 1.16 \times 1.08$
$ \Rightarrow \operatorname{Rs} 18,972=1.2528 P$
$\Rightarrow P=\operatorname{Rs} \frac{18972}{12528} $
$ \Rightarrow P=\operatorname{Rs} 15,143.68 $
Hence, the sum of money will be Rs $15,143.68$
View full question & answer→Question 134 Marks
What sum of money will amount to $Rs.10,256.40$ in $3$ years at compound interest if the rates of interest for the successive years are $10\%, 11\%$ and $12\%?$
AnswerHere $P=? ; t=3$ years $; r=10 \%, 11 \%$ and $12 \%$ successively; $A= Rs.10.256 .40$
$A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \operatorname{Rs~} 10256.40= P $
$ \left(1+\frac{10}{100}\right)\left(1+\frac{11}{100}\right)\left(1+\frac{12}{100}\right) $
$ \operatorname{Rs} 10,256.40= P \times 1.1 \times 1.11 \times 1.12 $
$\operatorname{Rs} 10,256.40=1.36752 P$
$ P =\operatorname{Rs} \frac{10256.40}{1.36752}$
$ P =\operatorname{Rs} 7,500$
Hence, the sum of money is $Rs.7,500 .$
View full question & answer→Question 144 Marks
Find the principal which will amount to Rs 22,344 in 2 years at compound interest if the rates of interest for the successive years are 12% and 14%?
AnswerHere $P=? ; t=2$ years; $r=12 \%$ and $14 \%$
successively; $A=$ Rs 22,344
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
Rs $22344= P \left(1+\frac{12}{100}\right)\left(1+\frac{14}{100}\right)$
Rs $22344= P \times 1.12 \times 1.14$
Rs22,344 $=1.2768 P$
$P=\operatorname{Rs} \frac{22344}{1.2768}$
$P=\operatorname{Rs} 17500$
Hence, the principal is Rs 17, 500.
View full question & answer→Question 154 Marks
Calculate the amount and the compound interest for the following:
$Rs.10,000$ for $3$ years if tl1e rates of interest are $10\%, 11\%$ and $12\%$ for the successive years.
Answer$ P=\operatorname{Rs} 10,000 ; t=3 \text { years; } r=10 \%, 11 \% \text { and } 12 \% $ successively.
$ A=P\left(1+\frac{r}{100}\right)^n $
$ A=\operatorname{Rs} 10000\left(1+\frac{10}{100}\right)\left(1+\frac{11}{100}\right)\left(1+\frac{12}{100}\right) $
$=\text { Rs } 10000 \times 1.1 \times 1.11 \times 1.12 $
$ =\text { Rs } 13,675.20 \\ \text { C.I. }=A-P$
$ =\text { Rs }(13,675.20-10,000)$
$=\text { Rs3,675.20 } $
Hence, Amount $= Rs.13,675.20$ and C.I. $= Rs.3,675.20$
View full question & answer→Question 164 Marks
Calculate the amount and the compound interest for the following:
$Rs.12,500$ for $3$ years if the rates for the successive years are $8\%, 9\%$ and $10\%$ respectively.
Answer$P=\operatorname{Rs} 12,500 ; t=3$ years; $r=8 \%, 9 \%$ and $10 \%$ successively.
$A=P\left(1+\frac{r}{100}\right)^n $
$A=\operatorname{Rs} 12500\left(1+\frac{8}{100}\right)\left(1+\frac{9}{100}\right)\left(1+\frac{10}{100}\right)$
$=\text { Rs } 12,500 \times 1.08 \times 1.09 \times 1.1 $
$=\text { Rs } 16,186.50 $
$ \text { C.I. }=A-P$
$=\text { Rs }(16,186.50-12,500) $
$ =\text { Rs } 3,686.50 $
Hence, Amount $=$ Rs $16,186.50$ and C.I. $=$ Rs $3,686.50$
View full question & answer→Question 174 Marks
Calculate the amount and the compound interest for the following:
$Rs.20,000$ for $2$ years if the rates of interest are $12 \frac{1}{4}$ and $5 \frac{1}{2}$ for the successive years.
Answer$P=\operatorname{Rs} 20,000 ; t=2 \text { years } ; r=12 \frac{1}{4} \% \text { and } 5 \frac{1}{2} \% $
$ \text { successively }=\frac{49}{4} \% \text { and } \frac{11}{2} \% \text { successively. } $
$ A=P\left(1+\frac{r}{100}\right)^n $
$A=R s 20000\left(1+\frac{49}{4 \times 100}\right)\left(1+\frac{11}{2 \times 100}\right) $
$ =\text { Rs } 20,000 \times 1.1225 \times 1.055 $
$ =\text { Rs } 23,684.75 $
$\text { C.I. }=A-P $
$=\text { Rs }(23,684.75-20,000) $
$ =\text { Rs } 3,684.75$
Hence, Amount= Rs $23,684.75$ and C.I. =Rs $3,684.75.$
View full question & answer→Question 184 Marks
The compound interest on a certain sum of money at $10\%$ p.a. for $3$ years is $Rs.4,965.$ What will be the simple interest on the same sum for $3$ years at $11\%$ p.a.?
Answer$P=x ; t=3 \text { years; } r=10 \% A=R s(x+4965)$
$ A=P\left(1+\frac{r}{100}\right)^n$
$ \operatorname{Rs}(x+4,965)=x\left(1+\frac{10}{100}\right)^3$
$ \operatorname{Rs}(x+4,965)=x \times 1.1 \times 1.1 \times 1.1 $
$ \operatorname{Rs}(x+4,965)=1.331 x$
$ 0.331 x=\operatorname{Rs} 4,965 x$
$ =\operatorname{Rs} 15,000$
$I=\frac{P \times r \times t}{100}$
$I=R s \frac{15000 \times 11 \times 3}{100} $
$I=\text { Rs } 4,950 $
Simple interest will be $Rs.4,950$
View full question & answer→Question 194 Marks
The compound interest on a certain sum of money at $5\%$ p.a. for $2$ years is $Rs.512.50.$ What will be the simple interest on the same sum for $3$ years at $6\%$ p.a.?
Answer$P=x ; t=2 \text { years } ; r=5 \% ; A=\operatorname{Rs}(x+512.50) $
$ A=P\left(1+\frac{r}{100}\right)^n$
$ \operatorname{Rs}(x+512.50)=x\left(1+\frac{5}{100}\right)^2$
$ \operatorname{Rs}(x+512.50)=x \times 1.05 \times 1.05$
$ \operatorname{Rs}(x+512.50)=1.1025 x $
$ 0.1025 x=\operatorname{Rs} 512.50 $
$ x=\operatorname{Rs} 5,000 $
$ I=\frac{P \times r \times t}{100} $
$ I=\operatorname{Rs} \frac{5000 \times 6 \times 3}{100} $
$ I=\operatorname{Rs~} 900 $
Simple interest will be $Rs.900$
View full question & answer→Question 204 Marks
A sum of money placed at compound interest compounded annually amounts to $Rs.47,610$ in $2$ years and to $Rs.54,751.50$ in $3$ years. Calculate the rate of interest and the sum.
Answer$P=x ; r=? ; t=2 \text { and } 3 \text { years; } A=R s ~ 47,610 \text { (2 years) }$
$(3 \text { years) }$
$A=P\left(1+\frac{r}{100}\right)^n$
$47610=x\left(1+\frac{r}{100}\right)^2 \ldots \ldots \ldots . .(i)$
$54751.50=x\left(1+\frac{r}{100}\right)^3 \ldots \ldots \ldots \ldots \ldots . . \text { (ii) }$
$\therefore \frac{x\left(1+\frac{r}{100}\right)^3}{ x \left(1+\frac{r}{100}\right)^2}=\frac{54751.50}{47610}$
$\Rightarrow\left(1+\frac{r}{100}\right)=\frac{54751.50}{47610}$
$\Rightarrow \frac{r}{100}=\frac{54751.50}{47610}-1$
$\Rightarrow \frac{r}{100}=\frac{54751.50-47610}{47610}$
$r=\frac{7141.50}{47610} \times 100$
$r=15 \%$
$ \text { Using (i) }$
$x\left(1+\frac{r}{100}\right)^2=\operatorname{Rs} 47610$
$x\left(1+\frac{15}{100}\right)^2=\operatorname{Rs} 47610$
$x\left(\frac{115}{100}\right)^2=\operatorname{Rs} 47610$
$1.3225 x=\text { Rs } 47,610$
$x=\operatorname{Rs} 36,000$
The sum $=Rs.36,000$ and rate of interest $=15 \%$
View full question & answer→Question 214 Marks
On what sum will the difference between compound interest and the simple interest for $3$ years at $12\%$ be $Rs.1,123.20?$
Answer$ P=x ; t=3 \text { years; } r=12 \% \text { For S.I.: }$
$I=\frac{P \times r \times t}{100}$
$=\frac{x \times 12 \times 3}{100}$
$=\frac{9 x}{25} $
For C.I. :
$ \text { C.I. }= P \left(1+\frac{ r }{100}\right)^{ t }- P$
$= x \left(1+\frac{12}{100}\right)^3- x$
$= x \left(1+\frac{3}{25}\right)^3- x$
$=( x \times 1.12 \times 1.12 \times 1.12)- x$
$=1.404928 x - x$
$=0.404928 x $
Given C.1.- S. I. $= Rs.22.50$
$ \Rightarrow 0.404928 x-\frac{9 x}{25}=\operatorname{Rs} 1123.20$
$\Rightarrow 0.404928 x-0.36 x=\text { Rs } 1123.20$
$\Rightarrow 0.044928 x=\text { Rs } 1123.20$
$\Rightarrow x=\text { Rs } 25000 $
Hence, sum $= Rs.25,000$
View full question & answer→Question 224 Marks
On what sum will the difference between compound interest and the simple interest for $2$ years at $7 \frac{1}{2} \%$ be $Rs.22.50 ?$
Answer$P=x ; t=2$ years $; r=7 \frac{1}{2} \%=\frac{15}{2} \%$
For S.I. :
$I=\frac{P \times r \times t}{100} $
$=\frac{x \times \frac{15}{2} \times 2}{100} $
$ =\frac{3 x}{20}$
For C.I. :
$ \text { C.I. }= P \left(1+\frac{ r }{100}\right)^{ t }- P $
$= x \left(1+\frac{\frac{15}{2}}{100}\right)^2- x$
$ = x \left(1+\frac{15}{2 \times 100}\right)^2- x $
$= x \left(1+\frac{3}{40}\right)^2- x$
$=( x \times 1.075 \times 1.075)- x $
$ =1.155625 x - x$
$ =0.155625 x$
$\text { Given C.I. - S.I. }=\text { Rs } 22.50$
$ \Rightarrow 0.155625 x-\frac{3 x}{20}=\operatorname{Rs} 22.50$
$ \Rightarrow 0.155625 x-0.15 x=\operatorname{Rs} 22.50$
$ \Rightarrow 0.005625 x=\operatorname{Rs} 22.50$
$\Rightarrow x-R s.4,000$
Hence, sum $= Rs.4,000$
View full question & answer→Question 234 Marks
In what time will $Rs.50,000$ yield an interest of $Rs.32,151.60$ at $18\%$ per annum interest compounded annually?
Answer$P=\operatorname{Rs} 50,000 ; A=\operatorname{Rs}(50,000+32,151.60)=\operatorname{Rs}$
$82,151.60 ; r=18 \% ; t=? $
$A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \text { Rs } 72,151.60=\text { Rs 50,000 }\left(1+\frac{18}{100}\right)^n$
$ \frac{82151.60}{50000}=\left(1+\frac{18}{100}\right)^n $
$ \frac{8215160}{50000 \times 100}=\left(1+\frac{18}{100}\right)^{ n } $
$ \frac{205379}{125000}=\left(1+\frac{18}{100}\right)^{ n } $
$\frac{(59)^3}{(50)^3}=\left(\frac{59}{50}\right)^{ n }$
$ t =3$
$ T =3 \text { years } $
View full question & answer→Question 244 Marks
In what time will $Rs.8,000$ amount to $Rs.12,167$ at $15\%$ per annum compounded annually?
Answer$P=R s.8000 ; A=Rs.12167 ; r=15 \% ; t=? $
$ \therefore A=P\left(1+\frac{r}{100}\right)^n$
$ \text { Rs } 12,167=\operatorname{Rs}8,000\left(1+\frac{15}{100}\right)^n $
$ \frac{12167}{8000}=\left(1+\frac{15}{100}\right)^n$
$ \frac{(23)^3}{(20)^3}=\left(\frac{23}{20}\right)^3 $
$ t=3 $
$ T=3 \text { years }$
View full question & answer→Question 254 Marks
Calculate the rate per cent at which $Rs.16,000$ will yield $Rs.3,876.75$ as compound interest in $3$ years.
Answer$ P =\operatorname{Rs} 16,000 ; A =\operatorname{Rs}(16,000+3,876.75)= Rs.19,876.75 ; t =3 \text { years; } r =? $
$A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \text { Rs } 19,876.75=\operatorname{Rs} 16,000\left(1+\frac{ r }{100}\right)^3 $
$ \frac{19876.75}{16000}=\left(1+\frac{ r }{100}\right)^3$
$\frac{(27.08)^3}{(25.19)^3}=\left(1+\frac{ r }{100}\right)^3$
$ \frac{2708}{2519}=1+\frac{ r }{100}$
$ \frac{ r }{100}=\frac{2708}{2519}-1=\frac{2708-2519}{2519}=\frac{189}{2519} $
$ r =\frac{18900}{2519}=7.5 \%$
Hence, $r=7.5 \%$
View full question & answer→Question 264 Marks
Calculate the rate percent at which $Rs.15,000$ will yield $Rs.8,413.44$ as compound interest in $3$ years.
Answer$ P =\operatorname{Rs} 15,000 ; A =\operatorname{Rs}(15,000+8,413.44)=\operatorname{Rs}$
$ 23,413.44 ; t =3 \text { years; } r =? $
$ A = P \left(1+\frac{ r }{100}\right)^{ n }$
$ \operatorname{Rs} 23,413.44=\operatorname{Rs} 15,000\left(1+\frac{ r }{100}\right)^3 $
$ \frac{23413.44}{15000}=\left(1+\frac{ r }{100}\right)^3 $
$ \frac{(29)^3}{(25)^3}=\left(1+\frac{ r }{100}\right)^3 $
$\frac{29}{25}=1+\frac{ r }{100}$
$ \frac{ r }{100}=\frac{29}{25}-1=\frac{29-25}{25}=\frac{4}{25} $
$r =\frac{400}{25}=16 \%$
Hence, $r=16 \%$
View full question & answer→Question 274 Marks
Calculate the rate per cent at which $Rs.12,250$ will yield $Rs.3,116.40$ as compound interest in $2$ years.
Answer$P=\operatorname{Rs} 12,250 ; A=\operatorname{Rs}(12,250+3,116.40)=\operatorname{Rs} 15,366.40$
$; t =2 \text { years; } r=?$
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \text { Rs } 15,366.40=\operatorname{Rs} 12250\left(1+\frac{ r }{100}\right)^2 \\ \frac{15366.40}{12250}=\left(1+\frac{ r }{100}\right)^2 $
$ \frac{(196)^2}{(175)^2}=\left(1+\frac{ r }{100}\right)^2 $
$ \frac{196}{175}=1+\frac{ r }{100} $
$\frac{ r }{100}=\frac{196}{175}-1=\frac{196-175}{175}=\frac{21}{175}$
$r =\frac{2100}{175}=12 \%$
Hence, $r=12 \%$
View full question & answer→Question 284 Marks
On what sum of money will the compound interest for $2$ years at $10\%$ p.a. compounded half-yearly amount to $Rs.3,448.10?$
Answer$P = x ; t =2 \text { years }=4 x 6 \text { months } ; r =10 \% $
$ \text { compounded half-yearly }=\frac{10}{2} \%=5 \% ; A=\operatorname{Rs}(x+ 3,448.10) $
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \operatorname{Rs}( x +3,448.10)= x \left(1+\frac{5}{100}\right) $
$ \operatorname{Rs}( x +3,448.10)= x \times 1.05 \times 1.05 \times 1.05 \times 1.05 $
$ \operatorname{Rs}( x +3,448.10)=1.215506 x $
$ 0.215506 x =\operatorname{Rs3}, 448.10 $
$ x=\operatorname{Rs} 16,000.02=\operatorname{Rs} 16,000 $
$\text { On Rs } 16,000 \text { the C.I. for } 2 \text { years at } 10 \% \text { compounded } $
$\text { half-yearly will be Rs 3,448.10 }$
View full question & answer→Question 294 Marks
On what sum of money will the compound interest for $1 \frac{1}{2}$ years at $16 \%$ p.a. compounded half-yearly amount to Rs $649.28 ?$
Answer$P=x ; t=1 \frac{1}{2} \text { years }=3 \times 6 \text { months } ; r=16 \% $
$ \text { compounded half-yearly }=\frac{16}{2} \%=8 \% ; A=R s(x+ 649.28) $
$ A=P\left(1+\frac{r}{100}\right)^n $
$ \operatorname{Rs}(x+649.28)=x\left(1+\frac{8}{100}\right)^3 $
$ \operatorname{Rs}(x+649.28)=x \times 1.08 \times 1.08 \times 1.08 $
$\operatorname{Rs}(x+649.28)=1.259712 x$
$ 0.259712 x=\operatorname{Rs} 649.28 $
$x=\operatorname{Rs} 2,500$
On $Rs.2,500$ the C.I. for $1 \frac{1}{2}$ years at $16 \%$ compounded half-yearly will be $Rs.649.28$
View full question & answer→Question 304 Marks
On what sum of money will the compound interest for $2 \frac{1}{2}$ years at $12 \frac{1}{2} \%$ per annum amount to Rs $82,734.37 ?$
Answer$P=x ; t=2 \frac{1}{2}$ years $; r=12 \frac{1}{2} \%=\frac{25}{2} \% ; A=R s(x+$ 82734.37)
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$\operatorname{Rs}( x +82,734.37)= x$
$ \left(1+\frac{25}{2 \times 100}\right)^2\left(1+\frac{25}{2 \times 100}\right)^{\frac{1}{2}} $
Rs $(x+82,734.37)-x$
$ \left(1+\frac{25}{2 \times 100}\right)^2\left(1+\frac{25}{2 \times 100}\right)^{\frac{1}{2}} $
$\operatorname{Rs}(x+82,734.37)=x \times 1.125 \times 1.125 \times$
$ \left(1+\frac{1}{2} \times \frac{1}{8}\right) $
$\operatorname{Rs}(x+82,734.37)-x \times 1.125 \times 1.125 \times 1.0625$
Rs $(x+82,734.37)-1.344727 x$
$0.344727 x$ - Rs $82,734.37$
$x=\operatorname{Rs} 2,39,999.7=$
Rs 2,40, 000
On Rs $2,40,000$ the C.I. for $2 \frac{1}{2}$ years at $12 \frac{1}{2} \%$ will be Rs $82,734.37$
View full question & answer→Question 314 Marks
On what sum of money will the compound interest for $2 \frac{1}{2}$ years at $12 \%$ per annum amount to Rs $8,241.60 ?$
Answer$P=x ; t=2 \frac{1}{2} \text { years } ; r=12 \% ; A=\operatorname{Rs}(x+8,241.60) $
$A=P\left(1+\frac{r}{100}\right)^n$
$ \operatorname{Rs}(x+8,241.60)=x\left(1+\frac{12}{100}\right)^2\left(1+\frac{12}{100}\right)^{\frac{1}{2}} $
$ \operatorname{Rs}(x+8,241.60)=x \times 1.12 \times 1.12 \times $
$ \left(1+\frac{1}{2} \times \frac{12}{100}\right) $
$\operatorname{Rs}(x+8,241.60)=x x 1.12 \times 1.12 \times 1.06 $
$ \operatorname{Rs}(x+8,241.60)=1.329664 x $
$ 0.329664 x=\operatorname{Rs} 8,241.60 $
$ x=\operatorname{Rs} 25,000 $
On $Rs.25,000$ the C.I. for $2 \frac{1}{2}$. years at $12 \%$ will be $Rs.8241.60 .$
View full question & answer→Question 324 Marks
On what sum of money will the compound interest for $2$ years at $8\%$ per annum amount to $Rs.1399.68?$
Answer$P=x ; t=2 \text { years } ; r=8 \% ; A=R s(x+1399.68) $
$ A=P\left(1+\frac{r}{100}\right)^n $
$ \operatorname{Rs}(x+1399.68)=x\left(1+\frac{8}{100}\right)^2 $
$ \operatorname{Rs}(x+1399.68)=x \times 1.08 \times 1.08$
$ \operatorname{Rs}(x+1399.68)=1.1664 x $
$0.1664 x=\operatorname{Rs} 1399.68 $
$x=\text { Rs } 81411.538$
On Rs $8,411.538$ the C.I. for $2$ years at $8 \%$ will be $Rs.1399.68$
View full question & answer→Question 334 Marks
The simple interest on an amount for $2$ years at $8\%$ is $Rs.320.$ Calculate the compound interest on the same amount at the same rate for $1$ year if the interest is compounded half-yearly.
AnswerHere, $P=? ; t=2$ years; $r=8 \%$ p.a.,
$\text { S.I. }=\text { Rs } 320 $
$P=\operatorname{Rs} \frac{\text { S.I. } \times 100}{r \times t} $
$P=\operatorname{Rs} \frac{320 \times 100}{8 \times 2}$
$ P=\operatorname{Rs} 2,000$
Now, $P=\operatorname{Rs} 2,000 ; t=1$ year
$n =2 t =2 \times 1=2 $
$ r =\frac{1}{2} \times 8 \%=4 \% \text { Per conversion period. } $
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ A=\operatorname{Rs} 2000\left(1+\frac{4}{100}\right)^2 $
$ =\text { Rs } 2,000 \times 1.04 \times 1.04=\text { Rs } 2,163.20 $
$ \text { C.I. }=A-P $
$ =\operatorname{Rs}(2,163.20-2,000) $
$ =\operatorname{Rs} 163.20 $
Hence, compound interest $= Rs.163.20$
View full question & answer→Question 344 Marks
Find the sum invested at $12 \frac{1}{2}$ p.a. compound interest on which the interest for the third year exceeds that of the first year by $Rs.531.25.$
AnswerLet the sum be $P$
Interest for first year:
$ P \left(1+\frac{25}{2 \times 100}\right)- P ........(i) $
Interest for third year:
$ P \left(1+\frac{25}{2 \times 100}\right)^3- P \left(1+\frac{25}{2 \times 100}\right)^2 ........(ii) $
Subtracting $(ii)$ from $(i)$
$P \left(1+\frac{25}{2 \times 100}\right)^3- P \left(1+\frac{25}{2 \times 100}\right)^2- P \left(1+\frac{25}{2 \times 100}\right)$
$ =\text { Rs } 531.25 $
$ \text { Rs 531.25 }=1.423828 P -1.265625 P -1.125 P + P$
$ \text { Rs 531.25 }=0.033203 P$
$P =\text { Rs } 16,000$
Hence the sum is $Rs.16,000$
View full question & answer→Question 354 Marks
Find the sum invested at $8\%$ p.a. compound interest on which the interest for the third year exceeds that of the first year by $Rs. 166.40.$
AnswerLet the sum be $P$
Interest for first year :
$ P \left(1+\frac{8}{100}\right)- P ........(i) $
Interest for third year:
$ P \left(1+\frac{8}{100}\right)^3- P \left(1+\frac{8}{100}\right)^2 ........(ii) $
Subtracting $(ii)$ from $(i)$
$ P \left(1+\frac{8}{100}\right)^3- P \left(1+\frac{8}{100}\right)^2- P \left(1+\frac{8}{100}\right)+ P$
$ =\text { Rs } 166.40$
$ \text { Rs } 166.40=1.259712 P -1.1664 P -1.08 P + P $
$ \text { Rs } 166.40=0.013312 P$
$ P =\text { Rs } 12,500$
Hence the sum is $Rs.12,500$
View full question & answer→Question 364 Marks
Rajan borrowed Rs 32,000 at 12% compound interest for 2 years. At the end of the first year he returned some amount and on paying Rs 17,920 at the end of the second year, he cleared the loan. Calculate the amount Rajan paid at the end of the first year.
Answer$ \text { Interest for first year : }$
$\text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{32000 \times 12 \times 1}{100}$
$\text { S.I. }=3,840 $
Principal amount for second year
$=\operatorname{Rs}(32,000+3,840)=\operatorname{Rs} 35,840$
Rajan paid $=$ Rs $\times$ (say)
Therefore, new principal $=\operatorname{Rs} 35,840-x$
$A=R s.17,920 ; r=12 \% ; n=1$ year
$A = P \left(1+\frac{ r }{100}\right)^{ n }$
$\operatorname{Rs} 17,920=\operatorname{Rs}(35,840-x)\left(1+\frac{12}{100}\right)$
Rs $17,920=\operatorname{Rs}(35,840-x) \times 1.12$
Rs $17,920=$ Rs $40,140.80-$ Rs $1.12 x$
$1.12 x=\operatorname{Rs}(40,140.80-17,920)$
$x =\frac{\operatorname{Rs} 22220.80}{1.12}$
$x=\operatorname{Rs} 19,840$
Therefore, Amount Rajan paid at the end of first year $=\operatorname{Rs} 19,840$
View full question & answer→Question 374 Marks
Ramesh borrowed Rs 12,000 at 15% compound interest for 2 years. At the end of the first year he returned some amount and on paying Rs 9,200 at the end of the second year, he cleared the loan. Calculate the amount of money Ramesh returned at the end of the first year.
AnswerInterest for first year:
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{12000 \times 15 \times 1}{100}$
$\text { S.I. }=1800 $
Principal amount for second year$=\operatorname{Rs}(12,000+ 1800) = Rs.13,800$
Ramesh paid $=$ Rs $x$ (say)
Therefore, new principal $=$ Rs 13,800 - $x$
$A=R s 9200 ; r=15 \% ; n=1$ year
$A = P \left(1+\frac{ r }{100}\right)^{ n }$
Rs $9,200=\operatorname{Rs}(13,800-x)\left(1+\frac{15}{100}\right)$
Rs $9,200=\operatorname{Rs}(13,800-x) \times 1.15$
Rs $9,200=$ Rs 15,870 - Rs $1.15 x$
$1.15 x=\operatorname{Rs}(15870-9,200)$
$x =\operatorname{Rs} \frac{6670}{1.15}$
$x=\operatorname{Rs} 5,800$
Therefore, Amount Ramesh paid at the end of first
$ \text { year }=\text { Rs 5,800 } $
View full question & answer→Question 384 Marks
Find the difference between the compound interest and the simple interest in $3$ years on $Rs.15,000$ at $8\%$ p.a. compounded yearly.
AnswerHere, $P=\operatorname{Rs} 15,000 ; r=8 \% ; t=3$ years
For simple interest :
$ \text { S.I. }=\frac{ P \times r \times t }{100} $
$ \text { S.I. }=\operatorname{Rs} \frac{15000 \times 8 \times 3}{100} $
$ \text { S.I. }=\operatorname{Rs} 3600 $
For compound interest:
$A=P\left(1+\frac{r}{100}\right)^n $
$ A=\text { Rs } 15000\left(1+\frac{8}{100}\right)^3 $
$ A=\text { Rs } 15000 \times \frac{108}{100} \times \frac{108}{100} \times \frac{108}{100}$
$A=\text { Rs } 18895.68$
$\text { C.I. }=A-P$
$ \text { C.I. }=\text { Rs }(18,895.68-15,000)$
$ \text { C.I. }=\text { Rs3, 895.68 } $
The difference in the compound interest and the simple interest $=$ Rs $(3,895.683 .600)= Rs.295.68$
View full question & answer→Question 394 Marks
Find the difference between the compound interest and the simple interest in $2$ years on $Rs.5,000$ at $8\%$ p.a. compounded annually.
AnswerHere $P=$ Rs $5000 \cdot r=8 \% t=2$ years
For simple interest:
$ \text { S.I. }=\frac{ P \times r \times t }{100} $
$\text { S.I. }=\operatorname{Rs} \frac{5000 \times 8 \times 2}{100} $
$\text { S.I. }=\text { Rs } 800 $
For compound interest:
$ A=P\left(1+\frac{r}{100}\right)^n$
$ A=\text { Rs } 5000\left(1+\frac{8}{100}\right)^2$
$ A=\text { Rs } 5000 \times \frac{108}{100} \times \frac{108}{100} $
$ A=\text { Rs 5,832 } $
$ \text { C.I. }=A-P $
$\text { C.I. }=\text { Rs (5832 -5,000) }$
$\text { C.I. }=\text { Rs 832 }$
The difference in the compound interest and the simple interest
$=\operatorname{Rs}(832-800) $
$=\operatorname{Rs} 32 . $
View full question & answer→Question 404 Marks
The interests on two successive years for a sum invested at compound interest compounded annually are $Rs.840$ and $Rs.940.80. $ Calculate the rate of interest and the sum invested.
AnswerFor the seoond year:
$ A=Rs.940.80 ; P=R s ~ 840 ; n=1 ; r=?$
$A=P\left(1+\frac{r}{100}\right)^n$
$ 940.80=840\left(1+\frac{r}{100}\right)^1 $
$ 94080=84000+840 r $
$840 r=10080$
$ r=12$
Hence, rate of interest $=12 \%$
For the first year:
$I=\operatorname{Rs} 840 ; r=12 \% ; n=I ; P=?$
$I=\frac{P \times r \times n}{100} $
$ \text { Rs } 840=\operatorname{Rs} \frac{P \times 12 \times 1}{100} $
$ P=\operatorname{Rs} \frac{84000}{12}$
$P=\operatorname{Rs} 7000$
The sum invested $= Rs.7,000$
View full question & answer→Question 414 Marks
Mohan invested a certain sum at compound interest, compounded annually. If the interests for two successive years were $Rs.600 $ and $Rs.648,$ calculate the rate of interest and the sum invested.
AnswerFor the second year:
$ A=R s 648 ; P=R s 600 ; n=1 ; r=?$
$\therefore A=P\left(1+\frac{r}{100}\right)^n$
$\Rightarrow 648=600\left(1+\frac{r}{100}\right)^1$
$\Rightarrow 648=600+6 r$
$\Rightarrow 6 r=48$
$\Rightarrow r=8 $
Hence, rate of interest $=8 \%$
For the first year:
$ I=R s 600 ; r=8 \% ; n=1 ; P=?$
$I=\frac{P \times r \times n}{100}$
$\text { Rs } 600=\operatorname{Rs} \frac{P \times 8 \times 1}{100}$
$P=\operatorname{Rs} \frac{60000}{8}$
$P=\text { Rs } 7,500 $
The sum invested $= Rs.7,500.$
View full question & answer→Question 424 Marks
The cost of a machine depreciated by Rs 2592 during the third year and by Rs 2332.80 during the fourth year. Calculate :
The original cost.
AnswerDepreciation for seoond year $=$ Rs $2,592+10 \%$ of Rs 2,592
Here, $10 \%$ of Rs $2,592=$ Rs 259.20
Hence, Depreciation for seoond year
$ =\text { Rs 2,592 + Rs } 259.20=\text { Rs 2,851.20 } $
Depreciation for first year $=$ Rs $2,851.20+10 \%$ of Rs 2,851.20
Here, $10 \%$ of Rs $2,851.20=$ Rs 285.12
Hence, Depreciation for first year
$ =\text { Rs 2,851.20 + Rs } 285.12=\text { Rs 3, } 136.32 $
Total depreciation for 4 years
$ =\operatorname{Rs}(3,136.32+2,851.20+2,592+2,332.80)$
$=\operatorname{Rs} 10,912.32$
$A=P-\operatorname{Rs} 10,912.32 ; P=x$
$A=P\left(1+\frac{ r }{100}\right)^{ n }$
$x -10912.32= x \left(1-\frac{10}{100}\right)^4$
$x -10912.32= x \times 1.1 \times 1.1 \times 1.1 \times 1.1$
$x (1-0.6561)=\operatorname{Rs} 10912.32$
$x =\operatorname{Rs} \frac{10912.32}{0.3439}$
$X =\operatorname{RS} 31731.08$
$\Rightarrow x =\operatorname{Rs} 32,000 \text { (approx) } $
Original cost $=$ Rs 32,000
View full question & answer→Question 434 Marks
The cost of a scooter depreciated by Rs 5100 during the second year and by Rs 4,335 during the third year. Calculate
The original cost of the scooter.
AnswerDepreciation for first year $=$ Rs $5,100+15 \%$ of Rs 5,
100
Here, $15 \%$ of Rs $5,100=$ Rs 765
Hence, Depreciation for first year $=$ Rs $5,100+$ Rs 765
$=$ Rs 5,865
Total depreciation for 3 years
$=$ Rs $(5,865+5,100+4335)=$ Rs 15,300
$A=P-R s 15300 ; P=x$
$A=P\left(1+\frac{r}{100}\right)^n$
$x-15300=x\left(1-\frac{15}{100}\right)^3$
$x-15,300=x \times 0.85 \times 0.85 \times 0.85$
$x(1-0.614)=$ Rs 15,300
$x=R s \frac{15300}{0.386}$
$x=$ Rs $39,637.31$
$\Rightarrow x=$ Rs 40,000 (approx)
Oriqinal cost of soooter =Rs 40,000
View full question & answer→Question 444 Marks
Meera borrowed $Rs.35,000$ at $12.5\%$ p.a. simple interest for $3$ years. She immediately gave it to Archana at $12\%$ p.a. compound interest compounded annually. Find Meera's loss or gain at the end of $3$ years.
AnswerHere, $P= Rs.35,000 ;$ t $=3$ years
For simple interest: $r=12.5 \%$
$\text { S.I. }=\frac{ P \times r \times t }{100} $
$\text { S.I. }=\operatorname{Rs} \frac{35000 \times 12.5 \times 3}{100}$
$ \text { S.I. }=\operatorname{Rs} 13,125$
For compound interest : $r=12 \%$
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ A=\operatorname{Rs} 35000 \times \frac{112}{100} \times \frac{112}{100} \times \frac{112}{100} $
$ A=\operatorname{Rs} 49172.48$
$ \text { C.I. }=\text { A }-P $
$ \text { C.I. }=\operatorname{Rs}(49,172.48-35,000) $
C.I. $=$ Rs $14,172.48$
The difference in the compound interest and the simple interest
$=\operatorname{Rs}(14,172.4813,125)=\operatorname{Rs~1,047.48}$
Meera gained Rs $1,047.48$
View full question & answer→Question 454 Marks
Neena's savings increases by Rs 1,000 every year. If she saves Rs 4,000 in the first year and invests it at 15% compound interest, find her total savings at the end of the third year.
Answer$P=$ Rs. 4,$000 ; R=15 \%$ p.a. $; T=3$ years Interest for the 1 st year
$=\operatorname{Rs} \frac{4000 \times 15 \times 1}{100}$
$=$ Rs 600
Principal for the second year
$=$ Amount at the end of one year + her new savings
$=$ Rs. $4,000+$ Rs. $600+$ Rs. $5,000=$ Rs. 9,600
Interest for the seoond year
$=\frac{\operatorname{Rs} 9600 \times 15 \times 1}{100}$
$=$ Rs 1,440
Compound interest for seoond year =Rs. 1,440 Principal for the third year
$=$ Amount at the end of two years + her new savings
=Rs. $9,600+$ Rs. $1,440+$ Rs. 6000 =Rs. 17.040 Interest for the third year
$=\operatorname{Rs} \frac{17040 \times 15 \times 1}{100}$
$=$ Rs 2,556
Sum due at the end of third year $=$ her savings at the
end of third year
$=$ Rs. $17,040+$ Rs. $2,556=$ Rs 19,5963
View full question & answer→Question 464 Marks
A man's savings increases by Rs 50 every year. If he saves Rs 500 in the first year and puts it at 10% compound interest, find his savings at the end of the third year.
Answer$P=$ Rs. $500 ; R=10^{\circ} \%$ p.a.; $T=3$ years Interest for the 1 st year
$ =\operatorname{Rs} \frac{500 \times 10 \times 1}{100}$
$=\operatorname{Rs} 50 $
Principal for the second year
= Amount at the end of one year + his new saving
$=$ Rs. 500 + Rs. 50 + Rs. $550=$ Rs. 1,100
Interest for the second year
$ =\operatorname{Rs} \frac{1100 \times 10 \times 1}{100}$
$=\operatorname{Rs} 110 $
Compound interest for second year =Rs. 110
Principal for the third year $=$ Amount at the end of two years + his new savings
$ =\text { Rs. } 1,100 \text { + Rs. } 110 \text { +Rs. } 600=\text { Rs. } 1,810 $
Interest for the third year
$ =\operatorname{Rs} \frac{1810 \times 10 \times 1}{100}$
$=\operatorname{Rs} 181 $
Sum due at the end of third year = his savings at the end of third year $ =\text { Rs } 1,810+\text { Rs. } 181=\text { Rs } 1,991 $
View full question & answer→Question 474 Marks
Manoj saves $Rs.5,000$ every year and invests it at $12\%$ p.a. compound interest. Calculate his savings at the end of the third year.
Answer$ P=\text { Rs. } 5,000 ; R=12 \% \text { p.a. } ; T=3 \text { years } $
Interest for the $1$ st year
$ =\operatorname{Rs} \frac{5000 \times 12 \times 1}{100}$
$=\operatorname{Rs} 600$
Principal for the second year
$=$ Amount at the end of one year + his new savings
$ =\text { Rs. } 5,000+\text { Rs. } 600+\text { Rs. } 5,000=\text { Rs. } 10,600 $
Interest for the seoond year
$ =\operatorname{Rs} \frac{10600 \times 12 \times 1}{100} $
$ =\text { Rs 1, } 272 $
Compound interest for seoond year $=Rs. 1,272$
Principal for the third year
$=$ Amount at the end of two years $+$ his new savings
$ =\text { Rs. } 10,600 \text { +Rs. } 1,272+\text { Rs. } 5,000=\text { Rs. } 16,872 $
Interest for the third year
$ =\operatorname{Rs} \frac{16872 \times 12 \times 1}{100}$
$=\text { Rs } 2,024.64 $
Sum due at the end of third year = his savings at the end of third year
$ =\text { Rs. } 16,872+\text { Rs. } 2,024.64=\text { Rs } 18,896.64 $
View full question & answer→Question 484 Marks
Ramesh saves $Rs.4,000$ every year and invests it at $10\%$ p.a. compound interest. Calculate his savings at the end of the third year.
Answer$P=$ Rs. 4,$000 ; R=10^{\circ} /$ p.a.; $T=3$ years Interest for the $1$ st year
$ =\frac{\operatorname{Rs} 4000 \times 10 \times 1}{100}=\operatorname{Rs~} 400 $
Principal for the second year
$=$ Amount at the end of one year + his new savings
$ =\text { Rs. } 4,000+\text { Rs. } 400 \text { + Rs. } 4,000=\text { Rs. 8,400 } $
Interest for the seoond year
$ =\frac{\operatorname{Rs} 8400 \times 10 \times 1}{100} $
$ =\text { Rs } 840$
Compound interest for seoond year $=Rs. 840$ Principal for the third year
$=$ Amount at the end of tvvo years $+$ his new savings
$=$ Rs. $8400+ Rs. 840 + Rs. 4000 =Rs. 13,240$ Interest for the third year
$ =\frac{\operatorname{Rs} 13240 \times 10 \times 1}{100} $
$ =\text { Rs } 1,324$
Sum due at the end of third year $=$ his savings at the end of third year
$ =\text { Rs. } 13,240+\text { Rs. } 1,324=\text { Rs } 14,564 $
View full question & answer→Question 494 Marks
What sum of money will amount to $Rs.7,128$ in $2$ years at compound interest, if the rates of interest are $8\%$ and $10\%$ for successive years?
AnswerFor the second year
Here $P=x ; A=R s 7,128 ; t=1$ year ; $r=10 \%$ p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \Rightarrow Rs 7128= x \left(1+\frac{10}{100}\right)^1$
$ \Rightarrow Rs 7128= x \left(\frac{11}{10}\right) $
$ \Rightarrow x =\operatorname{Rs} \frac{7128 \times 10}{11}$
$\Rightarrow x =\operatorname{Rs} 6480 $
The sum of money will be $Rs 6,480$ at the end of the first year or beginning of the second year.
For the first vear
Here $P=x ; A=R s 6,480 ; t=1$ year ; $r=8 \%$ p.a.
$\therefore A = P \left(1+\frac{ r }{100}\right)^{ n }$
$\Rightarrow Rs 6480= x \left(1+\frac{8}{100}\right)^1$
$ \Rightarrow Rs 6480= x \left(\frac{108}{100}\right)$
$ \Rightarrow x = Rs \frac{6480 \times 100}{108} $
$\Rightarrow x =\operatorname{Rs} 61000$
The sum of money will be $Rs.6,000$ at the beginning of the first year.
View full question & answer→Question 504 Marks
The value of a 'Honda' bike depreciated by $16\%$ in the first year and by $13\%$ in the second year. Find the value of the bike if it depreciated by $Rs.7,098$ in the second year.
AnswerLet value of the bike be Rs $x$.
$ V_0=\text { Rs } x ; n=2 ; r=16 \% $
Depreciation in the first year $=$
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$\Rightarrow V _{ t }= Rs \times\left(1-\frac{16}{100}\right)^2$
$\Rightarrow V _{ t }= Rs \times \frac{21}{25}$
$\Rightarrow V _{ t }= Rs.0.84 x $
Depreciation in the second year when $r$ is $13 \%=$
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$\Rightarrow V _{ t }= Rs.0.84 x \times\left(1-\frac{13}{100}\right)$
$\Rightarrow V _{ t }= Rs.0.84 x \times 0.87$
$\Rightarrow V _{ t }= Rs.0.7308 x $
Depreciation in the value of bike in the seoond year
$ =\operatorname{Rs}(0.84 x-0.7308 x)=\operatorname{Rs} 7,098$
$\Rightarrow 0.1092 x=\operatorname{Rs} 7,098$
$\Rightarrow x=\operatorname{Rs} 65,000 $
The original value of the bike was $Rs.65,000 .$
View full question & answer→