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Mathematics [5 marks sum]

Compound Interest · ICSE STD 10 (ICSE - English Medium). 17 questions.

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17 questions · timed · auto-graded

Question 15 Marks
The population of a city is $24,000.$ In the next $3$ years it will be $27,783.$ Find the rate of growth of the population.
Answer
$ V _{ n }=27,783 ; V _0=24,000 ; r=? ; t =3 \text { years } $
$V _{ n }= V _0\left(1+\frac{ r }{100}\right)^{ n } $
$ 27783=24000\left(1+\frac{ r }{100}\right)^3 $
$\frac{27783}{24000}=\left(1+\frac{ r }{100}\right)^3$
$ \frac{21^3}{20^3}=\left(1+\frac{ r }{100}\right)^3 $
$\left(1+\frac{ r }{100}\right)=\frac{21}{20} $
$\frac{ r }{100}=\frac{21}{20}-1 $
$\frac{ r }{100}=\frac{1}{20} $
$ r=\frac{1}{20} \times 100$
$ r=5 \%$
The rate of growth of population is $5 \%$.
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Question 25 Marks
In a factory the production of scooters rose to $46,305$ from $40,000$ in $3$ years. Find the annual rate of growth of the production of scooters.
Answer
$V _{ n }=46,305 ; V _0=40,000 ; r=? ; t =3 \text { years } $
$V _{ n }= V _0 \times\left(1+\frac{ r }{100}\right)^{ n }$
$ 46305=40000\left(1+\frac{ r }{100}\right)^3 $
$ \frac{46305}{40000}=\left(1+\frac{ r }{100}\right)^3 $
$ \frac{21^3}{20^3}=\left(1+\frac{ r }{100}\right)^3 $
$ \left(1+\frac{ r }{100}\right)=\frac{21}{20} $
$ \frac{ r }{100}=\frac{21}{20}-1$
$ \frac{ r }{100}=\frac{1}{20} $
$r =\frac{1}{20} \times 100 $
$ r ={5 \%}$
The annual rate of growth of scooters is $5 \%$
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Question 35 Marks
A building worth $Rs.1, 33,100$ is constructed on a plot of land worth $Rs.72,900.$ After how many years will the values of both be same, if the land appreciates at $10\%$ p.a. and the building depreciates at $10\%$ p.a.?
Answer
For the building:
$ V_n=V_0\left(1-\frac{r}{100}\right)^n$
$V_n=\operatorname{Rs} 133100\left(1-\frac{10}{100}\right)^n$
$V_n=\operatorname{Rs} 133,100 \times(0.9)^n $
For the plot:
$ V_n=V_0\left(1+\frac{r}{100}\right)^n$
$V_n=\operatorname{Rs} 72900\left(1+\frac{10}{100}\right)^n$
$V_n=\operatorname{Rs} 72900 \times(1.1)^n $
Since, value becomes same:
$ 1,33,100 \times(0.9)^{ n }=72,900 \times(1.1)^{ n } \frac{\lim}{x \rightarrow \infty}$
$\frac{(1.1)^{ n }}{(0.9)^{ n }}=\frac{133100}{72900}$
$\frac{(1.1)^{ n }}{(0.9)^{ n }}=\frac{1331}{729}=\frac{11^3}{9^3}$
$t =3 $
Hence, after $3$ years value of both will be same.
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Question 45 Marks
Find the difference between the compound interest compounded yearly and half-yearly for the following:
Rs 20,000 for $1 \frac{1}{2}$ years at $16 \%$ p.a.
Answer
$P=\operatorname{Rs} 20,000 ; t=1 \frac{1}{2}$ years
When compounded yearly: $r=16 \%$ p.a.
$ A = P \left(1+\frac{ r }{100}\right)^{ n } $
$A=R s 20000\left(1+\frac{16}{100}\right)\left(1+\frac{16}{100}\right)^{\frac{1}{2}}$
$=$ Rs $20000 \times 1.16 \times\left(1+\frac{1}{2} \times \frac{16}{100}\right)$
$=$ Rs $20,000 \times 1.16 \times 1.08$
$=$ Rs 25,056
C.I. $= A - P$
$=\operatorname{Rs}(25,056-20,000)$
$=$ Rs 5,056
When compounded half-yearly:
$A = P \left(1+\frac{ r }{100}\right)^{ n }$
$A=\operatorname{Rs} 20000\left(1+\frac{8}{100}\right)^3$
$=$ Rs $20,000 \times 1.08 \times 1.08 \times 1.08$ $=$ Rs $25,194.24$
C.I. $= A - P$
$=\operatorname{Rs}(25,194.24-20,000)$
$=$ Rs 5,194.24
Hence the difference in the interest=Rs $(5,194.24$ - $ 5,056)=\operatorname{Rs} 138.24 $
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Question 55 Marks
Find the difference betlween the compound interest compounded yearly and half-yearly for the following:
$Rs.15,000$ for $1 \frac{1}{2}$ years at $12 \%$ p.a.
Answer
$P=\operatorname{Rs} 15,000 ; t=1 \frac{1}{2}$ years
When compounded yearly : $r=12 \%$ p.a.
$A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 15000\left(1+\frac{12}{100}\right)\left(1+\frac{12}{100}\right)^{\frac{1}{2}}$
$=\text { Rs } 15000 \times 1.12 \times\left(1+\frac{1}{2} \times \frac{12}{100}\right)$
$=\text { Rs } 15,000 \times 1.12 \times 1.06$
$=\text { Rs } 17,808$
$\text { C.I. }=A-P$
$=\text { Rs }(17,808-15,000)$
$=\text { Rs } 2808 $
When compounded half-yearly :
$ A=P\left(1+\frac{r}{100}\right)^n$
$A=\operatorname{Rs} 15000\left(1+\frac{6}{100}\right)^3$
$=\operatorname{Rs} 15,000 \times 1.06 \times 1.06 \times 1.06$
$=\operatorname{Rs} 17,865.24$
$\text { C.I. }=A-P$
$=\text { Rs }(17,865.24-15,000)$
$=\text { Rs } 2,865.24 $
Hence the difference in the interest=Rs $(2,865.24-2,808)$
$ =\operatorname{Rs} 57.24 $
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Question 65 Marks
A sum of money placed at compound interest compounded annually amounts to $Rs.26,460$ in $2$ years and to $Rs.29,172.15$ in $4$ years. Calculate the rate of interest and the sum.
Answer
$P=x ; r=? ; t=2$ and $4$ years; $A=Rs.26,460$ ($2$ years) and $Rs.29, 172.15$ ($4$ years)
$ A = P \left(1+\frac{ r }{100}\right)^{ n }$
$26460= x \left(1+\frac{ r }{100}\right)^2 .......(i)$
$29172.15=x\left(1+\frac{r}{100}\right)^4 .......(ii)$
$\therefore \frac{ x \left(1+\frac{ r }{100}\right)^4}{ x \left(1+\frac{ r }{100}\right)^2}=\frac{29172.15}{26460}$
$\Rightarrow\left(1+\frac{ r }{100}\right)^2=\frac{194481}{176400}$
$\Rightarrow\left(1+\frac{ r }{100}\right)^2=\left(\frac{441}{420}\right)^2$
$\Rightarrow 1+\frac{ r }{100}=\frac{441}{420}$
$\Rightarrow \frac{ r }{100}=\frac{441}{420}-1$
$\Rightarrow \frac{ r }{100}=\frac{441-420}{420}$
$r=\frac{21}{420} \times 100$
$r=5 \%$
$ \text { Using (i) }$
$x\left(1+\frac{r}{100}\right)^2=\text { Rs } 26460$
$x\left(1+\frac{5}{100}\right)^2=\text { Rs } 26460$
$x\left(\frac{105}{100}\right)^2=\text { Rs } 26460$
$1.1025 x=\text { Rs } 26,460$
$x=\text { Rs } 24,000 $
The sum $=$ Rs $24,000$ and rate of interest $=5 \%$
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Question 75 Marks
A sum of money placed at compound interest compounded annually amounts to Rs 31,360 in 2 years and to Rs 35,123.20 in 3 years. Calculate the rate of interest and the sum.
Answer
$ P=x ; r=? ; t=2 \text { and } 3 \text { years; } A=\operatorname{Rs} 31,360(2 $
years) and Rs 35, 123.20 ( 3 years)
$ A = P \left(1+\frac{ r }{100}\right)^{ n }$
$31360=x\left(1+\frac{ r }{100}\right)^2 .... (i) $
$ 35123.20= x \left(1+\frac{ r }{100}\right)^3 .... (ii) $
$ \therefore \frac{ x \left(1+\frac{ r }{100}\right)^3}{x\left(1+\frac{ r }{100}\right)^2}=\frac{35123.20}{31360} $
$\Rightarrow\left(1+\frac{r}{100}\right)=\frac{35123.20}{31360}$
$ \Rightarrow \frac{ r }{100}=\frac{35123.20}{31360}-1 $
$ \Rightarrow \frac{ r }{100}=\frac{35123.20-31360}{31360} $
$ r=\frac{3763.20}{31360} \times 100$
$r=12 \% $
Using (i)
$ x\left(1+\frac{r}{100}\right)^2=\operatorname{Rs} 31,360$
$x\left(1+\frac{12}{100}\right)^2=\operatorname{Rs} 31,360$
$x\left(\frac{112}{100}\right)^2=\operatorname{Rs} 31,360$
$1.2544 X=\operatorname{Rs} 31,360$
$x=\text { Rs } 25,000 $
The sum $=$ Rs 25,000 and rate of interest $=12 \%$.
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Question 85 Marks
A certain sum of money invested at compound interest compounded annually amounted to $Rs.26,450$ in $2$ years and to $Rs. 30,417.50$ in $3$ years. Calculate the rate of interest and the sum invested.
Answer
Here, $r=? P=x$ (say)
$T=2$ years and $3$ years
$A=R s 26,450$ in $2$ vears and $Rs.30,417.50$ in $3$ years.
$ A=P\left(1+\frac{r}{100}\right)^n $
$ 26450=x\left(1+\frac{r}{100}\right)^2 ......(i) $
$ 30417.50=x\left(1+\frac{r}{100}\right)^3 ......(ii) $
Dividing (ii) by (i)
$ \frac{x\left(1+\frac{r}{100}\right)^3}{x\left(1+\frac{r}{100}\right)^2}=\frac{30417.50}{26450} $
$\Rightarrow 1+\frac{r}{100}=\frac{30417.50}{26450} $
$ \Rightarrow \frac{ r }{100}=\frac{30417.50}{26450}-1 $
$ \Rightarrow \frac{ r }{100}=\frac{30417.50-26450}{26450} $
$ \Rightarrow \frac{ r }{100}=\frac{3967.50}{26450}$
$ \Rightarrow r =\frac{3967.50}{26450} \times 100$
$\Rightarrow r =15 \%$
$x\left(1+\frac{r}{100}\right)^2=R s 26450$
$ x\left(1+\frac{15}{100}\right)^2=\text { Rs } 26450 $
$x \times \frac{23}{20} \times \frac{23}{20}=\text { Rs } 26450 $
$x \times \frac{529}{400}=\text { Rs } 26450 $
$ x=\text { Rs } \frac{26450 \times 400}{529} $
$ x=\text { Rs } 20,000 $
$\text { Hence, rate of interest }=15 \% \text { and sum invested } $
$=\text { Rs } 20,000$
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Question 95 Marks
A certain sum of money invested at compound interest compounded annually amounted to Rs 5,082 after 2 years and to Rs 5,590.20 after 3 years. Calculate the rate of interest and the sum invested.
Answer
Here, $r=? P=x$ (say)
$T=2$ years and 3 years
$A=R s 5,082$ in 2 years and Rs 5,590.20 in 3 years.
$ A = P \left(1+\frac{ r }{100}\right)^{ n }$
$5082= x \left(1+\frac{ r }{100}\right)^2 \ldots ......(i)$
$5590.20= x \left(1+\frac{ r }{100}\right)^3  ........(ii) $
Dividing (ii) by (i)
$ \frac{x\left(1+\frac{r}{100}\right)^3}{x\left(1+\frac{r}{100}\right)^2}=\frac{5590.20}{5082}$
$\Rightarrow 1+\frac{r}{100}=\frac{5590.20}{5082}$
$\Rightarrow \frac{r}{100}=\frac{5590.20}{5082}-1$
$\Rightarrow \frac{r}{100}=\frac{5590.20-5082}{5082}$
$\Rightarrow \frac{r}{100}=\frac{508.20}{5082}$
$\Rightarrow r=\frac{508.20}{5082} \times 100$
$\Rightarrow r=10 \% $
$ \text { using (i) }$
$x\left(1+\frac{r}{100}\right)^2=\operatorname{Rs} 5082$
$x\left(1+\frac{10}{100}\right)^2=\operatorname{Rs} 5082$
$x \times \frac{11}{10} \times \frac{11}{10}=\operatorname{Rs~} 5082$
$x \times \frac{121}{100}=\operatorname{Rs} 5082$
$x=\operatorname{Rs} \frac{5082 \times 100}{121}$
$x=\text { Rs } 4200 $
Hence, rate of interest $=10^{\circ} 10$ and sum invested $=$ Rs $4,200$.
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Question 105 Marks
What sum of money will amount to Rs 13,675.20 in 3 years at compound interest, if the rates of interest are 10%, 11% and 12% for the successive years?
Answer
For the third year
Here $P=x ; A=R s 13,675.20 ; t=1$ year; $r=12 \%$ p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n }$
$\Rightarrow Rs 13675.20= x \left(1+\frac{12}{100}\right)^1$
$\Rightarrow \operatorname{Rs} 13675.20= x \left(\frac{112}{100}\right)$
$\Rightarrow x=\operatorname{Rs} \frac{13675.20 \times 100}{112}$
$\Rightarrow x =\operatorname{Rs} 12,210 $
The sum of money will be Rs 12,210 at the end of the second year or beginnino;; of the third year.
For the second year
Here $P=x ; A=R s 12,210 ; t=1$ year ; $r=11 \%$ p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $ $\Rightarrow \operatorname{Rs} 12210= x \left(1+\frac{11}{100}\right)^1$
$\Rightarrow \operatorname{Rs} 12210= x \left(\frac{111}{100}\right)$
$\Rightarrow x= Rs \frac{12210 \times 100}{111}$
$\Rightarrow x=\operatorname{Rs} 11,000$
The sum of money will be Rs 11,000 at the end of the first year or beginning of the second year.
For the first year
Here $P=x ; A=R s 11,000 ; t=1$ year; $r=10 \%$
p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $
$ \Rightarrow \text { Rs } 11000= x \left(1+\frac{10}{100}\right)^1$
$\Rightarrow \text { Rs } 11000= x \left(\frac{11}{100}\right) $
$ \Rightarrow x=\operatorname{Rs} \frac{11000 \times 10}{11} $
$ \Rightarrow x=\operatorname{Rs} 10,000 $
The sum of money will be Rs 10,000 at the beginning of the first year.
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Question 115 Marks
What sum of money will amount to Rs 3,326.40 in 3 years at compound interest, if the rates of interest are 8%, 10% and 12% for the successive years?
Answer
For the third year
Here $P=x ; A=R s ~ 3,326.40 ; t=1$ year ; $r=12 \%$ p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n }$
$\Rightarrow Rs 3326.40= x \left(1+\frac{12}{100}\right)^1$
$\Rightarrow Rs 3326.40= x \left(\frac{112}{100}\right)$
$\Rightarrow x =\operatorname{Rs} \frac{3326.40 \times 100}{112}$
$\Rightarrow x \operatorname{Rs} 2970 $
The sum of money will be Rs 2, 970 at the end of the second year or beginning of the third year.
For the second year
Here $P=x ; A=R s 2,970 ; t=1$ year ; $r=10 \%$ p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n } $
$\Rightarrow$ Rs $2970= x \left(1+\frac{10}{100}\right)^1$
$\Rightarrow$ Rs $2970= x \left(\frac{11}{10}\right)$
$\Rightarrow x = Rs \frac{2970 \times 10}{11}$
$\Rightarrow x=\operatorname{Rs} 2700$
The sum of money will be Rs 2,700 at the end of the first year or beginning of the second year.
For the first year
Here $P=x ; A=R s 2,700 ; t=1$ year ; $r=8 \%$ p.a.
$ \therefore A = P \left(1+\frac{ r }{100}\right)^{ n }$
$\Rightarrow Rs 2700= x \left(1+\frac{8}{100}\right)^1$
$\Rightarrow Rs 2700= x \left(\frac{108}{100}\right)$
$\Rightarrow x = Rs \frac{2700 \times 100}{108}$
$\Rightarrow x =\operatorname{Rs} 2500 $
The sum of money will be Rs 2,500 at the beginning of the first year.
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Question 125 Marks
QUESTION Archana borrowed Rs $18,000$ from Ritu at $12 \%$ p.a. compound interest. If at the end of the $1^{\text {st }}, 2^{\text {nd }}$, and $3^{\text {rd }}$ years, Archana returned $Rs 5,250 , Rs 5,875$ and $Rs 6,875$ respectively, find the amount Archana has to pay Ritu at the end of the $4^{\text {th }}$ year to clear her debt.
Answer
$ P=\text { Rs. } 18,000, R=12 \% \text { p.a. } $
Interest for first year
$ =\frac{\operatorname{Rs} 18000 \times 12 \times 1}{100}$
$=\text { Rs 2,160 }$
Amount due after 1st year
$ =\text { Rs. } 18,000+\text { Rs. } 2,160=\text { Rs 20,160 } $
Amount paid after 1st year $=$ Rs. 5,250
Balance amount $=$ Rs. 20, $160-$ Rs. 5,250 = Rs. 14,910
Interest for seoond year
$ =\frac{\operatorname{Rs} 14910 \times 12 \times 1}{100}$
$=\text { Rs } 1789.20 $
Amount due after 2 nd year
$ =\text { Rs. } 14,910+\text { Rs. } 1,789.20$
$=\text { Rs } 16,699.20 $
Amount paid after 2nd year = Rs. 5,875
Balance amount $=$ Rs. $16,699.20-$ Rs. $5,875=$ Rs. $ 10,824.20 $
Interest for third year
$ =\frac{\operatorname{Rs} 10824.20 \times 12 \times 1}{100}$
$=\text { Rs } 1,298.904 $
Amount due after 3rd year
$ =\text { Rs. } 10,824.20 \text { + Rs. } 1,298.904=\text { Rs 12, } 123.10 $
Amount paid after 3rd year $=$ Rs. 6,875
Balance amount = Rs. 12, 123.10- Rs. 6,875 = Rs. 5,248.104
Interest for fourth year
$ =\frac{\operatorname{Rs} 5248.104 \times 12 \times 1}{100}$
$=\text { Rs } 629.7725 $
Amount due after 4th year =Rs. 5,248.104 +Rs. $ 629.7725=\text { Rs } 5877.876$
$= Rs.5877.87$
Archana needs to pay $Rs.5877.87$ to Ritu at the end of $4$ th year to clear
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Question 135 Marks
Rajan borrowed Rs 90,000 at 15% p.a. compound interest. If he repays Rs 35,000 at the end of each year, find the amount of loan outstanding at the beginning of the fourth year.
Answer
$ P=\text { Rs. } 90,000, R=15 \% \text { p.a. } $
Interest for first year
$ =\frac{\operatorname{Rs} 90000 \times 15 \times 1}{100}$
$=\operatorname{Rs} 13500 $
Amount due after 1st year
$ =\text { Rs. } 90,000+\text { Rs. } 13,500=\text { Rs } 103,500 $
Amount paid after 1 st year $=$ Rs. 35,000
Balance amount= Rs. 103,500 - Rs. $35,000=$ Rs. $ 68,500 $
Interest for second year
$ =\frac{\operatorname{Rs} 68500 \times 15 \times 1}{100}$
$=\operatorname{Rs} 10,275 $
Amount due after 2 nd year
$ =\text { Rs. } 68,500+\text { Rs. } 10,275=\text { Rs } 78,775$
$\text { Amount paid after 2nd year =Rs. } 35,000$
$\text { Balance amount= Rs. } 78,775-\text { Rs. } 35,000=\text { Rs. } 43,775$
$\text { Interest for third year }$
$=\frac{\text { Rs } 43775 \times 15 \times 1}{100}$
$=\text { Rs } 6,566.25 $
Amount due after 3rd year
$ =\text { Rs. } 43,775+\text { Rs. 6,566.25 = Rs 50,341.25 }$
$\text { Amount paid after 3rd year = Rs.35, 000 }$
$\text { Balance amount = Rs. 50,341.25 - Rs.35, } 000$
$=\text { Rs. 15,341.25 } $
Loan outstanding at the beginning of the fourth year$ =\operatorname{Rs} 15,341.25 $
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Question 145 Marks
Mohan borrowed Rs 25,000 at 10% p.a. compound interest. If he pays back Rs 7,500 every year, find the amount of loan outstanding at the beginning of the fourth year.
Answer
$ P=\text { Rs. } 25,000, R=10 \% \text { p.a. } $
Interest for first year
$ =\frac{\operatorname{Rs} 25000 \times 10 \times 1}{100}$
$=\text { Rs } 2,500 $
Amount due after 1 st year
$ =\text { Rs. } 25,000+\text { Rs. } 2,500$
$=\text { Rs } 27,500 $
Amount paid after 1 st year= Rs. 7,500
Balance am
ount $=$ Rs. $27,500-$ Rs. $7,500=$ Rs. 20,000
Interest for seoond year
$ =\frac{\operatorname{Rs} 20000 \times 10 \times 1}{100}$
$=\operatorname{Rs} 2000 $
Amount due after 2 nd year
$ =\text { Rs. } 20,000+\text { Rs. } 2,000=\text { Rs } 22,000 $
Amount paid after 2 nd year= Rs. 7,500
Balance amount= Rs. 22,000 - Rs. 7,500 = Rs. 14,500
Interest for third year
$ =\frac{\operatorname{Rs} 14500 \times 10 \times 1}{100} $
Amount due after 3rd year
$ =\text { Rs. } 14,500+\text { Rs. } 1,450=\text { Rs } 15,950 $
Amount paid after 3rd year $=$ Rs. 7,500
Balance amount = Rs. 15, $950-$ Rs. $7,500=$ Rs. 8,450
Loan outstanding at the beginning of the fourth year $=$ Rs 8,450 .
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Question 155 Marks
Calculate the arnount and the cornpound interest for the following:
Rs 20,000 for 3 years at $7 \frac{1}{2} \%$ for the first year, $8 \%$ for the second year and $10 \%$ for the third year.
Answer
$ P=\operatorname{Rs} 20000 ; $
(i) Interest for he first year
$ T=1 \text { year, } R =7 \frac{1}{2} \% \text { for first year }=\frac{15}{2} \%$
$=\frac{\operatorname{Rs} 20000 \times \frac{15}{2} \times 1}{100}$
$=\frac{\operatorname{Rs} 20000 \times 15 \times 1}{2 \times 100}$
$=\text { Rs } 1500 $
(ii) Principal for the second year
$=$ Amount after one year
$ =\text { Rs } 20000+\text { Rs } 1500$
$=\text { Rs } 21500 $
(iii) Interest for the second year
$T=1$ year, $R=18 \%$ for second year
$ =\frac{\operatorname{Rs} 21500 \times 8 \times 1}{100}$
$=\operatorname{Rs} 1720 $
(iv) Principal for the third year
$=$ Amount after second year
$ =\text { Rs } 21500+\text { Rs } 1720$
$=\text { Rs } 23220 $
(v) Interest for the third year
$ t=1 \text { year, } R=10 \% \text { for second year }$
$=\frac{\operatorname{Rs} 23220 \times 10 \times 1}{100}$
$\text { = Rs } 2322$
Therefore Amount at the end of 3rd year
$ =\text { Rs } 23220+\text { Rs } 2322$
$=\text { Rs } 25542$
$\text { Amount }=\text { Rs } 25542$
$\text { C.I. }=\text { A }- \text { P }$
$=\text { Rs }(25542-20000)$
$\text { C.I. }=\text { Rs } 5542 $
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Question 165 Marks
Calculate the amount and the compound interest for the following:
Rs. 16,000 at $15 \%$ p.a. in $2 \frac{2}{3}$ years
Answer
Here, $P=$ Rs. 16,$000 ; r=15 \%$ p.a. $; t=2 \frac{2}{3}$ years
For the first year: $t=1$ year
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 16000 \times 15 \times 1}{100}$
$\text { S.I. }=\text { Rs21400 } $
$ A=P+S . I .$
$=\operatorname{Rs}(16,000+2,400)=\operatorname{Rs} 18,400=\text { new principal } $
For the second year: $t=1$ year; $P=R s 18,400$
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 18400 \times 15 \times 1}{100}$
$\text { S.I. }=\text { Rs2, } 760 $
$ A=P+S . I .$
$A=\operatorname{Rs}(18,400+2,760)=\operatorname{Rs} 21,160=\text { new principal } $
For the third year: $t=2 / 3$ year; $P=R s 21,160$
$ \text { S.I. }=\frac{P \times r \times t}{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 21160 \times 15 \times 2}{100 \times 3}$
$\text { S.I. }=\text { Rs2116 } $
$ A=P+S . I .$
$A=\operatorname{Rs}(21,160+2116)=\operatorname{Rs} 23,276 $
C.I. $=$ Interest in first year + interest in second year + interest in third year
$ \text { C.I. }=\operatorname{Rs}(2,400+2,760+2116)=\operatorname{Rs} 7,276 $
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Question 175 Marks
Calculate the amount and the compound interest for the following:
Rs. 7500 at 12% p.a. in 3 years.
Answer
Here, $P=$ Rs. $7500 ; r=12 \%$ p.a.; $t=3$ years
For the first year : $t=1$ year
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 7500 \times 12 \times 1}{100}$
$\text { S.I. }=\text { Rs } 900 $
$ A=P+S . I .$
$=\operatorname{Rs}(7500+900)=\operatorname{Rs} 8400=\text { new principal } $
For the seoond year: $t=1$ year; $P=R s 8,400$
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs} 8400 \times 12 \times 1}{100}$
$\text { S.I. }=\operatorname{Rs} 1,008 $
$ A=P+S . I .$
$A=\operatorname{Rs}(8,400+1,008)=\operatorname{Rs} 9,408=\text { new principal } $
For the third year: $t=1$ year; $P=R s 9,408$
$ \text { S.I. }=\frac{ P \times r \times t }{100}$
$\text { S.I. }=\frac{\operatorname{Rs~} 9408 \times 12 \times 1}{100}$
$\text { S.I. }=\text { Rs1, 128.96 } $
$A=P+S . I .$
$A=R s(9,408+1,128.96)=\operatorname{Rs} 10,536.96$
$\text { C.I. = Interest in first year + interest in second year + }$
$\text { interest in third year }$
$\text { C.I. }=\operatorname{Rs}(900+1,008+1,128.96)=\operatorname{Rs} 3,036.96$
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[5 marks sum] - Mathematics STD 10 Questions - Vidyadip