Question 13 Marks
The present population of a town is $1,15200.$ If it increases at the rate of $6 \frac{2}{3} \%$ per annum , find Its population after $2$ years
Answer$ V_n=7 \cdot V_o=115200 ; r=6 \frac{2}{3} \%=20 / 3 \% ; t=2 \text { years }$
$V_n=V_0 \times\left(1+\frac{r}{100}\right) $
$ V_n=115200\left(1+\frac{20}{100 \times 3}\right)^2 $
$ V_n 1,15,200 \times 1.06667 \times 1.06667 $
$ V_n=1,31,072 $
The population $2$ years later $=1,31,072$
View full question & answer→Question 23 Marks
The cost of a T.V. was quoted $Rs.17,000$ at the beginning of the year $1999.$ In the beginning of $2000$ the price was hiked by $5\%.$ Because of decrease in demand the cost was reduced by $4\%$ in the beginning of $2001.$ What is the cost of the T.V. in $2001?$
Answer$V _{ n }=? ; V _0= Rs 17,000 ; t =2$ years ( 1 for increment and 1 for decrement); $r=5 \%$ tor increase and $4 \%$ tor decrease.
$V_n=V_0\left(1+\frac{r}{100}\right)^n\left(1-\frac{r}{100}\right)^n$
$ V_n=\operatorname{Rs} 17000\left(1+\frac{5}{100}\right)\left(1-\frac{4}{100}\right) $
$ V_n=\operatorname{Rs~} 17,000 \times 1.05 \times 0.96$
$ V_n=\operatorname{Rs~} 17136$
The cost of the T.V. in $2001$ is Rs $17,136$.
View full question & answer→Question 33 Marks
What sum of money will amount to $Rs.8,073$ in $2$ years at compound interest if the rates of interest for the successive years are $15\%$ and $17\%?$
AnswerHere $P=? ; t=2$ years; $r=15 \%$ and $17 \%$ successively;
$A=\text { Rs } 8,073$
$A=P\left(1+\frac{r}{100}\right)^n $
$ \text { Rs } 8073= P \left(1+\frac{15}{100}\right)\left(1+\frac{17}{100}\right) $
Rs $8,073=P \times 1.15 \times 1.17$
Rs $8,073=1.3455 P$
$ P=R s \frac{8073}{13455} $
$ P=\operatorname{Rs} 6,000 $
Hence, the sum of money is $Rs.6,000$
View full question & answer→Question 43 Marks
Calculate the amount and the compouncl interest of the following:
$Rs.9,125$ for $2$ years if tl1e rates of interest are $12\%$ and $14\%$ for the successive years.
Answer$P=\operatorname{Rs} 9,125 ; t=2$ years $; r=12 \%$ and $14 \%$
successively.
$A=P\left(1+\frac{r}{100}\right)^n$
$ A=\text { Rs } 9125\left(1+\frac{12}{100}\right)\left(1+\frac{14}{100}\right)$
$=\text { Rs } 9,125 \times 1.12 \times 1.14 $
$=\text { Rs } 11,650.80$
$ \text { C.I. }=A-P$
$=\text { Rs }(11,650.80-9,125)$
$=\text { Rs 2,525.80 } $
Hence, Amount $= Rs.11,650.80$ and C.I. $=Rs.2,525.80$
View full question & answer→Question 53 Marks
Calculate the amount and the compound interest for the following, when cornpounded half-yearly:
$Rs.25,000$ for $1 \frac{1}{2}$ years at $12 \%$
Answer$P=\operatorname{Rs} 25,000 ; t=1 \frac{1}{2}$ years $; r=12 \%$ p.a. $=6 \%$ half-early.
$A=P\left(1+\frac{r}{100}\right)^n $
$A=\operatorname{Rs} 25000\left(1+\frac{6}{100}\right)^2\left(1+\frac{12}{100}\right)^{\frac{1}{2}} $
$ =\text { Rs } 25,000 \times 1.06 \times 1.06 \times\left(1+\frac{1}{2} \times \frac{12}{100}\right)$
$=\text { Rs } 25,000 \times 1.06 \times 1.06 \times 1.06 $
$=\text { Rs29,775.40 } \\ \text { C.I. }=A-P $
$=\text { Rs }(29,775.40-25,000) $
$ =\text { Rs } 4,775.40$
Hence, Amount $= Rs.29,775.40$ and C.I. $=Rs.4,775.40$
View full question & answer→Question 63 Marks
Calculate the amount and the compound interest for the following, when cornpounded half-yearly:
$Rs.6,000$ for $1 \frac{1}{2}$ years at $10 \%$ p.a.
Answer$Rs.6,000$ for $1 \frac{1}{2}$ years at $10 \%$ p.a.
$P=Rs.6,000 ; t=1 \frac{1}{2}$ years $; r=10 \%$ p.a. $=5 \%$
half-yearly.
$A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 6000\left(1+\frac{5}{100}\right)^2\left(1+\frac{10}{100}\right)^{\frac{1}{2}}$
$=\text { Rs } 6000 \times 1.05 \times 1.05 \times\left(1+\frac{1}{2} \times \frac{10}{100}\right)$
$=\text { Rs } 61000 \times 1.05 \times 1.05 \times 1.05$
$=\text { Rs } 6945.75$
$\text { C.I. }=A-P$
$=\text { Rs }(6,945.75-6,000)$
$=\text { Rs } 945.75$
Hence, Amount $= Rs.6,945.75$ and C.I. $=Rs.945.75$
View full question & answer→Question 73 Marks
Calculate the amount and cornpound interest for the following, when cornpounded annually:
$Rs.7,500$ for $2 \frac{1}{2}$ years ; $r=16 \%$ p.a.
Answer$P=\operatorname{Rs} 7,500 ; t=2 \frac{1}{2} \text { years } ; r=16 \% \text { p.a. } $
$A=P\left(1+\frac{r}{100}\right)^n$
$ A=\operatorname{Rs} 7500\left(1+\frac{16}{100}\right)^2\left(1+\frac{16}{100}\right)^{\frac{1}{2}} $
$=\operatorname{Rs} 7500 \times 1.16 \times 1.16 \times\left(1+\frac{1}{2} \times \frac{16}{100}\right) $
$=\text { Rs } 7,500 \times 1.16 \times 1.16 \times 1.08 $
$ =\text { Rs } 10,899.36 \\ \text { C.I. }=A-P$
$ =\text { Rs }(10,899.36-7,500) $
$=\text { Rs } 3,399.36 $
Hence, Amount $= Rs.10,899.36$ and C.I. $=Rs.3,399.36$
View full question & answer→Question 83 Marks
Calculate the amount and cornpound interest for the following, when cornpounded annually:
$Rs.8,000$ for $1 \frac{1}{2}$ years at $12 \%$ p.a.
Answer$P=\operatorname{Rs} 8,000 ; t=1 \frac{1}{2} \text { years } ; r=12 \% \text { p.a. }$
$ A=P\left(1+\frac{r}{100}\right)^{ n } $
$ A=\operatorname{Rs} 8000\left(1+\frac{12}{100}\right)\left(1+\frac{12}{100}\right)^{\frac{1}{2}}$
$=\text { Rs } 8000 \times 1.12 \times\left(1+\frac{1}{2} \times \frac{12}{100}\right) $
$=\text { Rs } 8,000 \times 1.12 \times 1.06$
$=\text { Rs } 9,497.60$
$ \text { C.I. }=\text { A }-P$
$=\operatorname{Rs}(9,497.60-8,000) $
$=\operatorname{Rs~} 1,497.60$
Hence, Amount $= Rs.9,497.60$ and C.I. $=Rs.1,497.60$
View full question & answer→Question 93 Marks
Calculate the amount and cornpound interest for the following, when cornpounded annually:
$Rs.16,000$ for $3$ years at $7 \frac{1}{2} \%$ p.a.
Answer$Rs.16,000$ for $3$ years at $7 \frac{1}{2}$ p.a.
$P=\operatorname{Rs} 16,000 ; t=3$ years $; r=7 \frac{1}{2}$ p.a.
$A=P\left(1+\frac{r}{100}\right)^n $
$A=\text { Rs } 16000\left(1+\frac{15}{2 \times 100}\right)^3$
$=\text { Rs } 16,000 \times 1.075 \times 1.075 \times 1.075 $
$=\text { Rs } 19,876.75$
$ \text { C.I. }=A-P $
$ =\text { Rs }(19,876.75-16,000) $
$ =\text { Rs 3,876. } 75 $
Hence, Amount $= Rs.19,876.75$ and C.I. $= Rs.3,876.75$
View full question & answer→Question 103 Marks
Calculate the amount and cornpound interest for the following, when cornpounded annually:
Rs 25,000 for 3 years at 8 % p.a.
AnswerRs 25,000 for 3 years at $8 \%$ p.a.
$P=R s 25,000 ; t=3$ years; $r=8 \%$ p.a.
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ A=\operatorname{Rs} 25000\left(1+\frac{8}{100}\right)^3 $
$ =\text { Rs } 25,000 \times 1.08 \times 1.08 \times 1.08 $
$ =\text { Rs } 31,492.80 $
C.I $= A - P$
$ =\operatorname{Rs}(31,492.80-25,000) $
$ =\text { Rs 6,492.80 } $
Hence, Amount $=$ Rs 31,492.80 and C.I. $=$ Rs 6,492.80
View full question & answer→Question 113 Marks
Calculate the Amount and Cornpound Interest for the Following, when Cornpounded Annually:
$Rs.12,000$ for $3$ years at $15\%$ p.a.
Answer$Rs.12,000$ for $3$ years at $15 \%$ p.a.
$ P=\text { Rs } 12,000 ; t=3 \text { years; } r=15 \% \text { p.a. } $
$A=P\left(1+\frac{r}{100}\right) $
$ A=\text { Rs } 12000\left(1+\frac{15}{100}\right)$
$=\text { Rs } 12000 \times 1.15 \times 1.15 \times 1.15 $
$ =\text { Rs } 18250.50$
$\text { C.I. }=A-P$
$ =\text { Rs }(18,250.50-12,000) $
$ =\text { Rs } 6,250.50 $
Hence. Amount $= Rs.18.250.50$ and C.I. $= Rs.6.250 .50$
View full question & answer→Question 123 Marks
Anand borrows $Rs.20,000$ at $9 \%$ p.a. simple interest for $3$ years. He immediately gave it to Prakash at $8 \frac{1}{2} \%$ p.a. compound interest compounded annually. Find Anand's loss or gain.
AnswerHere, $P=Rs.20,000 ; t=3$ years
For simple interest: $r=9 \%$
$\text { S.I. }=\frac{ P \times r \times t }{100} $
$ \text { S.I. }=\operatorname{Rs} \frac{20000 \times 9 \times 3}{100} $
$ \text { S.I. }=\operatorname{Rs} 5400$
For compound interest : $r=8 \frac{1}{2} \%$
$A=P\left(1+\frac{r}{100}\right)^n $
$ A=\text { Rs } 20000\left(1+\frac{17}{2 \times 100}\right)^3 $
$ A=\text { Rs } 20000 \times \frac{217}{200} \times \frac{217}{200} \times \frac{217}{200} $
$A=\text { Rs } 25545.70 \\ \text { C.I. = A - P } $
$ \text { C.I. = Rs (25,545.70 - 20,000) } $
$\text { C. I. = Rs 5,545.70 }$
The difference in the compound interest and the simple interest $= Rs.(5,545.705-400)= Rs.145.70$ Anand gained $Rs.145.70$
View full question & answer→Question 133 Marks
The simple interest and the compound interest on a certain sum of money for $2$ years at the same rate of interest are $Rs.8,000$ and $Rs.8,640$ respectively. Calculate the rate of interest and the sum.
AnswerThe extra interest earned = C.I. - S.I. $=$ Rs $(8,640-$ $8,000)= Rs.640.$
The interest for the first year $=S$. I. for $2$ years $/ 2=$ Rs $\frac{8000}{2}= Rs.4000$
Therefore, the rate of interest $=\frac{640}{4000} \times 100=16 \%$
Now,
$\text { S.I. }=\frac{ P \times r \times t }{100}$
$ \Rightarrow Rs.8000=\frac{ P \times 16 \times 2}{100} $
$ \Rightarrow P=R s \frac{8000 \times 100}{32}$
$\Rightarrow P=\operatorname{Rs} 25000$
The rate of interest was $16 \%$ and the original sum was $Rs.25,000 .$
View full question & answer→Question 143 Marks
The simple interest on a certain sum in $2$ years is $Rs.1,300,$ whereas the compound interest on the same sum at the same rate and for the same time is $Rs.1,365.$ Find the rate per cent and the sum.
AnswerThe extra interest earned = C. I. - S. I. $= Rs ( 1,365 - 1,300)=\text { Rs } 65 \text {. } $
The interest for the first year = S. I. for $2$ years $/ 2=$ Rs $ \frac{1300}{2}=\text { Rs } 650 $
Therefore, the rate of interest $=\frac{65}{650} \times 100=10 \%$
Now,
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\Rightarrow \text { Rs } 1300=\frac{P \times 10 \times 2}{100}$
$\Rightarrow P=\operatorname{Rs} 1300 \times 5$
$\Rightarrow P=\operatorname{Rs} 6500 $
The rate of interest was $10 \%$ and the original sum was Rs 6,500.
View full question & answer→Question 153 Marks
The cost of a machine dep reciated by $Rs.2592$ during the third year and by $Rs.2332.80$ during the fourth year. Calculate :
The cost at the end of the fourth year.
Answer$\text { Here, } P=\text { Rs } 32,000 ; r=10 \% ; t=4 \text { years } $
$A=P\left(1+\frac{r}{100}\right)^n $
$ A=Rs.32000\left(1-\frac{10}{100}\right)^4 $
$ A=\text { Rs } 32,000 \times 0.9 \times 0.9 \times 0.9 \times 0.9$
$ A=\text { Rs } 20,995.20 $
$ \text { Cost at the end of the fourth year = Rs } 20,995.20$
View full question & answer→Question 163 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 rate of depreciation.
AnswerDifference in the depreciation $=\operatorname{Rs}(2,592-2,332.80)$ $=$ Rs 259.20
Rate of depreciation $=\frac{259.20}{2592} \times 100=10 \%$
Rate of depreciation $=10 \%$
View full question & answer→Question 173 Marks
The cost of a scooter depreciated by $Rs.5100$ during the second year and by $Rs.4,335$ during the third year. Calculate
AnswerThe oost of the soooter at the end of the third year. Here, $P=\operatorname{Rs} 40,000 ; r=15 \% ; t=3$ years
$ A=P\left(1+\frac{r}{100}\right)^n$
$ A=\operatorname{Rs} 40000\left(1-\frac{15}{100}\right) $
$ A=\operatorname{Rs} 40,000 \times 0.85 \times 0.85 \times 0.85 $
$ A=\operatorname{Rs} 24,565 $
Cost of the soooter at the end of third year $=$ Rs $ 24,565 $
View full question & answer→Question 183 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 rate of depreciatlon
AnswerThe rate of depreciation. Difference in the depreciation $=\operatorname{Rs}(5,100-4,335)=$ Rs 765
Rate of depreciation $=\frac{765}{5100} \times 100=15 \%$
Rate of depreciation $=15 \%$
View full question & answer→Question 193 Marks
What sum of money will amount to $Rs.16,637.50$ in $3$ years at $10\%$ p.a. compound interest?
AnswerHere $P=x ; A=R s 16,637.50 ; t=3$ years $; r=10 \%$ p.a
$\therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \Rightarrow Rs 16637.50= x \left(1+\frac{10}{100}\right)^3 $
$ \Rightarrow Rs 16637.50= x \left(\frac{11}{10}\right)^3 $
$ \Rightarrow \operatorname{Rs} 16637.50= x \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$
$ \Rightarrow \operatorname{Rs} 16637.50= x \times \frac{1331}{1000} $
$ \Rightarrow x =\operatorname{Rs} \frac{16637.50 \times 1000}{1331} $
$\Rightarrow x =\operatorname{Rs} 12500 $
The sum of money will be $Rs.12,500 .$
View full question & answer→Question 203 Marks
What sum of money will amount to $Rs.9,447.84$ in $3$ years at $8\%$ p.a. compound interest?
AnswerHere $P=x ; A=Rs.9,447.84 ; t=3$ years; $r=8 \%$ p.a
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \Rightarrow Rs.9447.84= x \left(1+\frac{8}{100}\right)^3 $
$ \Rightarrow \operatorname{Rs} 9447.84= x \left(\frac{108}{100}\right)^3 $
$\Rightarrow \operatorname{Rs} 9447.84= x \times \frac{27}{25} \times \frac{27}{25} \times \frac{27}{25}$
$\Rightarrow \operatorname{Rs} 9447.84= x \times \frac{19683}{15625} $
$ \Rightarrow x =\operatorname{Rs} \frac{9447.84 \times 15625}{19683}$
$\Rightarrow x =\operatorname{Rs} 7500$
The sum of money will be $Rs.7,500.$
View full question & answer→Question 213 Marks
The value of a machine depreciates by $10\%, 12\%$ and $15\%$ in the first $3$ years. Express the total depreciation of the machine as a single per cent during the three years.
AnswerLet value of machine be Rs $x$.
$V_0=R s x ; n=3 ; r=10 \%$ for first year, $12 \%$
for $2$ nd year and $15 \%$ for $3$rd year
$\therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n }$
$ \Rightarrow V _{ t }= Rs \times\left(1-\frac{10}{100}\right)\left(1-\frac{12}{100}\right)\left(1-\frac{15}{100}\right) $
$ \Rightarrow V _{ t }= Rs \times \frac{9}{10} \times \frac{22}{25} \times \frac{17}{20}$
$ \Rightarrow V _{ t }= Rs \times \frac{3366}{5000} $
$\Rightarrow V _{ t }=\text { Rs. } 0.6732 x$
Depreciation in the value of car
$ =\operatorname{Rs}(x-0.6732 x)=\operatorname{Rs} 0.3268 x $
Percentage change in depreciation
$ =\frac{0.3268 x }{ x } \times 100$
$=32.68 \%$
Percentage change $=32.68 \%$
View full question & answer→Question 223 Marks
The value of a car depreciated by $10\%$ in the first $2$ years and by $8\%$ in the third year. Express the total depreciation of the car as a single per cent during the three years.
AnswerLet value of car be Rs $x$.
$V_0=R s\ x ; n=3 ; r=10 \%$ for first $2$ years and $8 \%$ for $3$rd year.
$ \therefore V _{ t }= V _0 \times\left(1-\frac{ r }{100}\right)^{ n } $
$ \Rightarrow V _{ t }= Rs\ x \times\left(1-\frac{10}{100}\right)^2\left(1-\frac{8}{100}\right) $
$ \Rightarrow V _{ t }= Rs\ x \times \frac{9}{10} \times \frac{9}{10} \times \frac{23}{25} $
$ \Rightarrow V _{ t }= Rs\ x \times \frac{1863}{2500} $
$ \Rightarrow V _{ t }=\operatorname{Rs} 0.7452 x $
Depreciation in the value of car
$ =\operatorname{Rs}(x-0.7452 x)=\operatorname{Rs} 0.2548 x $
Percentage change in depreciation
$=\frac{0.2548 x }{ x } \times 100$
$ =25.48 \%$
Percentage change $=25.48 \%$
View full question & answer→Question 233 Marks
Prerna borrowed $Rs.16000$ from a friend at $15\%$ p.a. compound interest. Find the amount , to the nearest rupees, that she needs to return at the end of $2.4$ years to clear the debt.
Answer$C _1=\frac{16000 \times 15 \times 1}{100}=2400 $
$P _1=18400 $
$ C _2=\frac{18400 \times 15 \times 1}{100}=2760$
$ P _2=21160 $
$ C _3=\frac{21160 \times 15 \times 1}{400}=7935 $
$ P _3=29095$
View full question & answer→Question 243 Marks
Natasha gave $Rs.60,000$ to Nimish for $3$ years at $15\%,$ p.a. compound interest.
Calculate to the nearest rupee :
The amount Natasha receives at the end of $3$ years.
Answer$ C _1=\frac{60000 \times 15 \times 1}{100}=9000 $
$ P _1=69000$
$ C _2=\frac{69000 \times 15 \times 1}{100}=10350 $
$ P _2=79350 $
$ C _3=\frac{79350 \times 1 \times 15}{100}=1190.25 $
$ P _3=91252.5$
View full question & answer→Question 253 Marks
Harijyot deposited $Rs.27500$ in a deposite scheme paying $12\%$ p.a. compound interest . If the duration of the deposite is $3$ years , calculate :
The amount received by him at the end of three years.
Answer$ C _1=\frac{27500 \times 12 \times 1}{100}=3300 $
$ P _1=27500+3300=30800 $
$ C _2=\frac{30800 \times 12 \times 1}{100}=3696 $
$ P _2=34496 $
$ C _3=\frac{34496 \times 12 \times 1}{100}=4139.52$
$ P _3=38636$
View full question & answer→Question 263 Marks
Ameesha loaned $Rs. 24,000$ to a friend for $2 \frac{1}{2}$ at $10 \%$ p.a. compound interest.
Calculate the interest earned by Ameesha.
Answer$C _1=\frac{ P \times R \times T }{100}=\frac{24000 \times 1 \times 10}{100}=2400$
$ P _1=24000+2400=26400$
$C _2=\frac{26400 \times 1 \times 10}{100}=2640 $
$ P _2=26400+2640=29040 $
$C _3=\frac{29040 \times 1 \times 10}{100}=2904$
$P _4=29040+2904=31944$
View full question & answer→Question 273 Marks
Alisha invested $Rs.75000$ for $4$ years at $8\%$ p.a. compound interest ,
Find the amount at the end of third year.
Answer$ C _3=\frac{ P \times R \times T }{100}=\frac{87480 \times 1 \times 8}{100}=6998.4 $
$ P _3=87480+6998.4=94478.4$
View full question & answer→Question 283 Marks
Alisha invested $Rs.75000$ for $4$ years at $8\%$ p.a. compound interest ,
Find the amount at the end of the second year.
Answer$ C _1=\frac{ P \times R \times T }{100}=\frac{75000 \times 1 \times 8}{100}=6000 $
$ P _1=75000+6000=81000 $
$ C _2=\frac{ P \times R \times T }{100}=\frac{81000 \times 1 \times 8}{100}=6480 $
$ P _2=81000+6480=87480$
View full question & answer→Question 293 Marks
A sum of $Rs. 65000$ is invested for $3$ years at $8\%$ p.a. compound interest.
Find the sum due at the end of the second year.
Answer$C _2=\frac{ P \times R \times T }{100}=\text { Rs } 5616$
$P _2= Rs.75816$
View full question & answer→Question 303 Marks
A sum of $Rs. 65000$ is invested for $3$ years at $8\%$ p.a. compound interest.
Find the sum due at the end of the first year.
Answer$C _1=\frac{ P \times R \times T }{100} $
$=\frac{65000 \times 8 \times 1}{100}=\text { Rs. } 5200 $
$P_1=5200+65000 $
$ =\text { Rs } 70200$
View full question & answer→Question 313 Marks
Pradeep gave $Rs.16000$ to a friend for $1.5$ years at $15\%$ p.a. compounded semi-annually. Find the interest earned by him at the end of $1.5$ years.
Answer$\text { Amount }= P \left(1+\frac{ r }{200}\right)^{2 t } $
$ \text { Amount }=16000\left(1+\frac{15}{200}\right)^3=\text { Rs. } 19876.75 $
$C =19876.75-16000=3876.75$
View full question & answer→Question 323 Marks
Shekhar had a fixed deposit of Rs 24000 for 3 years . If he received interest at 10% p.a compounded annually, find the amount received by him at the time of maturity.
AnswerAmount $= P \left(1+\frac{ r }{100}\right)^{ t }$
Amount $=24000\left(1+\frac{10}{100}\right)^3=31944$
Therefore, Shekhar received Rs . 31944 at the time of maturity.
View full question & answer→Question 333 Marks
Calculate the amount and the compound interest for the following:
$Rs. 40,000$ at $5 \frac{1}{4} \%$ p.a. in $1 \frac{1}{3}$ years
AnswerHere, $P=$ Rs. $40,000 ; r=5 \frac{1}{4} \%$ p.a. $=\frac{21}{4} \% ; t=1 \frac{1}{3}$ years
For the first year: $t=1$ year
$ \text { S.I. }=\frac{ P \times r \times t }{100} $
S.I. $\frac{\operatorname{Rs~} 40000 \times 21 \times 1}{100 \times 4}$
$ \text { S.I. }=\text { Rs2, } 100$
$A=P+S . I . $
$=\operatorname{Rs}(40,000+2,100)=\operatorname{Rs} .42,100=$ new principal
For the second year: $t=1 / 3$ year; $P=R s 42,100$
$ \text { S.I. }=\frac{ P \times r \times t }{100} $
$ \text { S.I. } \frac{\text { Rs } 42100 \times 21 \times 1}{100 \times 4 \times 3}$
$\text { S.I. }=\text { Rs736. } 75$
$A=P+S . I .$
$A=\operatorname{Rs}(42,100+736.75)=\operatorname{Rs}.42,836.75 $
C.I. $=$ Interest in first year +interest in second year
$ \text { C.I. }=\operatorname{Rs}(2,100+736.75)=\operatorname{Rs}.2,836.75 $
View full question & answer→