Exersice 8.1
Question 1
Principal (P)=4000
Rate of Interest $(R)=7.5 \%$
Time $(T)=$ 3years and 3 months
$\begin{aligned} 1 \operatorname{month}&=\frac{1}{12} \text { year } \\ 3 \text { months } =\frac{3}{12} \text { year } &=\frac{1}{4} \text { year } \\ \therefore \text { Time } &=3 \cdot \frac{1}{4} \text { years }=\frac{13}{4} \text { years } \end{aligned}$
Simple interest $(I)=\frac{P T R}{100}$
Simple interest (I) $=975$
Amount = prinicipal + interest
= 4000 + 975
Amount = 4975
Question 2
Given
simple interest (I) = 170 .10
Time (T) = 2 year and 3 month
(T) $2 \cdot \frac{3}{12}=2 \cdot \frac{1}{4}=\frac{9}{4}$ years
Rate (R)= 6%
$\begin{aligned} P &=\frac{I \times 100}{R \times T} \\ &=\frac{170.10 \times 100}{6 \times \frac{9}{4}} . \end{aligned}$
Principal == p = 1260
Question 3
Given
principal = 800
simple interest (I) = 130
Time = 2 year and 6 months
Time = $2 \cdot \frac{6}{12}=2 \cdot \frac{1}{2}=\frac{5}{2}$ years
$\begin{aligned} R &=\frac{I \times 100}{P \times T} \\ &=\frac{130 \times 100}{800 \times \frac{5}{2}} \\ R &=6.5 \% \end{aligned}$
hence , the required rate of interest = 6.5%
Question 4
Given
principla (p) = 3.3 lakhs
p = 330000
rate (R) 6.5 %
$\begin{aligned} \text { Simple Interest } &=75075 /-\\ \text { Time }(T) &=\frac{I \times 100}{P \times R} \\ &=\frac{75075 \times 100}{330000 \times 6.5} \\ T &=3.5 \text { years } \end{aligned}$
Hence , the required time = 3 years and 6 months
Question 5
(i) simple interest (I) = 2356.25
Time $\begin{aligned}(T) &=2 . \frac{1}{2} \text { yeers } \\ T &=\frac{5}{2} \text { years } \\(R) &=7 \cdot \frac{1}{4} \% \\ R &=\frac{29}{4}-1 . \\ P=& \frac{I \times 100}{T \times R} \\=& \frac{2356.25 \times 100}{\frac{5}{2} \times \frac{29}{4}} \\ P &=13,000\end{aligned}$
Hence , the required principal is Rs 13000
(ii) Given
rate (R) = 4 %
$\begin{aligned} \text { Time }(T) &=3 \text { years and } 3 \text { month } s \\ &=3 \cdot \frac{3}{12}=3 \cdot \frac{1}{4}=\frac{13}{4} \\ T &=\frac{13}{4} \text { years } \end{aligned}$
Final Amount $=113001-$
Fimel Amount $=$ Principal $(D)+$ Simple interest $(I)$
$11300=P+\frac{P T R}{100}$
$11300=P\left(1+\frac{T R}{100}\right)$
$11300=P\left(1+\frac{\frac{13}{4} \times 4}{100}\right)$
$11300=P\left(\frac{113}{100}\right)$
$P=11300 \times \frac{100}{113}$
$P=10,000$
Hence , the required principal is 10 ,000
Question 6
Given
Interest rate (R) = $13 \frac{1}{3}=\frac{40}{3}$ %
Final amount = 3 $\times $principal (p)
$\begin{aligned} P+I &=3 P \\ I &=3 p-P \\ I &=2 P \\ \frac{P T R}{100} &=2 P \quad\left(\because I=\frac{P T R}{100}\right) \\ T &=\frac{200}{R} \\ T &=\frac{200}{\frac{40}{3}} \end{aligned}$
Times = T = 15 years
Hence , the required time to triple itself for given
Interest rate is 15 years.
Question 7
Given
principal $\left(P_{1}\right)=4050$
Final Amount $=4576.501$ -
$\begin{aligned} \text { Prinupal }+\text { Simple Interest }=4576.50 \\ 4050+I_{1} &=4576.50 \\ I_{1} &=526.50\\ \text { Time }=& 2 \text { years. } \end{aligned}$
$I_{1}=\frac{P_{1} T_{1} R}{100}$
$526.50=\frac{4050 \times 2 \times R}{100}$
$R=6.5 \%$ Per Anum
Now we have to calculate simple interest for 1 lakh for 3 years at a rate of 65 % per anum
$\begin{aligned} I_{2} &=\frac{P_{2} T_{2} R}{100} \\ I_{2} &=\frac{1,001000 \times 3 \times 6.5}{100} \\ I_{2} &=19500\end{aligned}$
$\begin{aligned} \text { Total Amount } &=P_{2}+I_{2} \\ &=1,00,000+19,500\\ &=1,19,500\end{aligned}$
∴ Hence , the 1 lakh will amount Rs 1,19,500 for 3 years
Question 8
Let the money invested to be Rs p
Given
Prinupel $P_{1}=P$
Rate $R_{1}=7.5-1$
Time $T_{1}=2$ years
Let Interest $=I_{1}$
Principal $P_{2}=₹ 9600$
Rate $R_{2}=10-1 .$
Time $=3$ years and 6 months.
. Time $T_{z}=3 \cdot \frac{1}{2}=\frac{7}{2}$ years
let Simple Interest $=I_{2}$
$\therefore \quad I_{1}=2 \times I_{2}$
$\frac{P_{1} T_{1} R_{1}}{100}=2 \times \frac{P_{2} T_{2} R_{2}}{100}$
$\frac{P \times 2 \times 7.5}{100}=\frac{2 \times 9600 \times \frac{7}{2} \times 10}{100}$
P = Rs 4,4800
= Rs 44,800
Hence the required sum of money is Rs 44,800
Exersice 8.2
Question 1
Given
principal p = 6000
Rate R = 10 %
Time T = 2 Year
Interest for first year = $\frac{P T R}{100}$
$\frac{6000 \times 4 \times 10}{100}$
$I_{1}$ = 6000
Amowt at the end of the Girst year $=6000+600$
Principal for second year $=266001-$
Interest for Second year $=\frac{P T R}{100}$
$=\frac{6600 \times 1 \times 10}{100}$
$I_{2}=660$
$\begin{aligned} \text { Amount at the end of second yean } &=6600+660 \\ &=7260 /-\end{aligned}$
∴ Compound interest for 2 year = final amount - (original) principa
= 7260 - 6000
= 1260
Question 2
Sol-
$\begin{aligned} \text { Principal for first } \operatorname{year}\left(P_{1}\right)=& ₹ 1875 /-\\ \text { Rate }(R)=4 \% & \text { Time }(T)=2 \text { years. } \\ \text { Interest for first year }\left(I_{1}\right)=\frac{P_{1} \times T_{1} \times R}{100} \\=& \frac{1875 \times 1 \times 4}{100} \\ I_{1}=& ₹ 75\end{aligned}$
Amount at the end of first year $=1875+75$
=1950
Interest for Second your $I_{2}=\frac{1950 \times 1 \times 4}{100}$
$I_{2}=78 /-$
$\therefore$ Compound Interest after 2 year $=I_{1}+I_{2}$
$\begin{array}{l}=75+78 \\=153\end{array}$
Hence, Salma has to pay 2153 as compound Intrest to the Mahila samit
Question 3
Principal for first year $\left(R_{1}\right)=12000$
$Rate=10 \%, \text { Time }=3 \text { yars }$
$\begin{aligned} \text { interest for first year }\left(I_{1}\right) &=\frac{12000 \times 1 \times 10}{100} \\ I_{1} &=₹ 1200\end{aligned}$
Total amout at the end of first year $12000+1200$
$=13200$
principal for second year = Rs 13200
interest for second year $I_{2}=$ $\frac{13200 \times 1 \times 10}{100}$
=₹ 1320
Total amount at the end of the second year
is 13200 + 1320 = Rs 14520
princial amount for third year = Rs 14520
Interest for third year $I_{3}=\frac{14520 \times 1 \times 10}{100}$
$I_{3}=₹ 1452$
$\begin{aligned} \text { Total amount at the end of third year } &=14520+1452 \\ &=₹ 15972 \end{aligned}$
$\begin{aligned} \text { Compound Interest for } 3 \text { years } &=\text { Final amount }-\text { Principal }\left(P_{1}\right) \\ &=15972-12000 \\ &=23972 \end{aligned}$
Hence , At the end of 3 year jacob will get total
Amount as Rs 15972 and compound interest Rs 3972
Question 4
Given
Prineipal $\left(P_{1}\right)=46875$
Rate $(R)=4 \%$, Time $=3$ years.
principal amount for first year $\left(p_{1}\right)$ = Rs 46875
Rate (R) = 4 % time = 3 years
(i) Interest for first year $\left(I_{1}\right)$
$I_{1}=1875$
$\begin{aligned} \text { Total amount at the end of Girst year } &=46875+1875 \\ &=48750 \end{aligned}$
Principal amount for second $\operatorname{year}\left(R_{2}\right)=$ 2 48750
Interest for Second year $\left(I_{2}\right)=\frac{48750 \times 4 \times 1}{100}$
$I_{2}$= Rs 21950
(ii) Total amount at the end of second year
= 48750 + 1950
= Rs 50700
principal for third year $\left(P_{3}\right)$= Rs 50700
(iii) $\begin{aligned} \text { Interest for third year }\left(I_{3}\right) &=\frac{50700 \times 1 \times 4}{100} \\ &=2028 \end{aligned}$
Hence ,
i. The interest for first year $=21875$
ii. The Amourt standing to his credit at the end of
second year is Rs 50700
iii. The interest for third year =Rs 2028
Question 5
Given
Principal $=260001-$, Time $=3$ years
Rate $=10$
PrinCIpal amout for first $\operatorname{year}\left(P_{1}\right)=26000$
Interest for first year $\left(I_{1}\right)=\frac{6000 \times 1 \times 10}{100}$
Total amount at the end of first year $6000+600$
$=6600$
Principal for second year $\left(P_{2}\right)=6600$
(i) Interest for second year $\left(I_{2}\right)=\frac{6600 \times 1 \times 10}{100}$
$I_{2}=₹ 660$
Total amount at the end of second year $=6600+660$
$=7260$
$\begin{array}{ll}\text { Principal for third year } & \left(P_{3}\right)=27260 \\ \text { Interest for Third year } & \left(-I_{3}\right)=\frac{7260 \times 1 \times 10}{100} \\ & I_{3}=726 /-\end{array}$
(ii) Total amount at the end of third year = 7260 + 726
= Rs 7986
hence
(i) Compound interest for second year = Rs 660
(ii) Total amount at the end of the third year = Rs 7986
Question 6
principal amount for first year $\left(P_{1}\right)$=Rs 5000
Rate of interest for first year $\left(R_{1}\right)=6 \%$
$\begin{aligned} \text { Interest for first year } &=\frac{5000 \times 1 \times 6}{100} \\\left(I_{1}\right) &=300 \end{aligned}$
$\begin{aligned} \text { Total amout at the end of first year } &=5000+300 \\ &=5300\\ \text { Principal amourt for second year }\left(P_{2}\right)=& 5300 \\ \text { Rate of Interest for Second year} &=8 \% . \\ \text { Compomd Interest for Second year } &=\frac{5300 \times 1 \times 8}{100} \end{aligned}$
$\left(I_{2}\right)=424$
$\begin{aligned} \text { Total amount at the endof seconed year } &=5300+424 \\ &= 5724 \end{aligned}$
$\begin{aligned} \text { Compound interest for } 2 \text { years } &=\text { final amount }-\text { Principal }\left(P_{1}\right) \\ &=5724-5000 \\ &=₹ 724 \end{aligned}$
Hence total amount after 2 year = Rs 5724
Compound interest for 2 year = Rs 724
Question 7
Given principal = Rs 20000
Time = 2 year
Rate of interest = 8%
Simple Interest $(I)=\frac{P T R}{100}$
$=\frac{20000 \times 2 \times 8}{100}$
Simple Interest (I.) $=33200$
Let compound interest as 'C'
Interest for first year = $\frac{20000 \times 1 \times 8}{100}$
$I_{1}=1600$
Total amount at the end of first year = 20000+ 1600
= 21600
Compound interest for 2 year = $I_{1}+I_{2}$
$=1600+1728$
$C=3328$
∴ Difference between the compound interest and
simple interest is (c - I ) = 3328 - 3200
= Rs 128
Exersice 8.3
Question 1
(i) Given
principal = Rs 15000
Time = 2 year ; n = 2
Rate of interest = 10 %
Amount
$\begin{aligned}(A) &=P\left(1+\frac{R}{100}\right)^{n} \\ &=15000\left(1+\frac{10}{100}\right)^{2} \end{aligned}$
Total Amount $(A)=18150$
Compomd interest $C=A-P$
$=18150-15000$
Compound interest $=3150$
Hence , amount is Rs 18150 and compound interest Rs 2496
(ii) Given
principal (p) = Rs 156250
Time $(n)=1 \frac{1}{2}=3 / 2$
Rate of Interest $(R)=8%$ per anum
Compomd half yearly so
So
$R=\frac{81}{2}=4 \%$ (HalF YEARLY)
n =$\frac{3 / 2}{1 / {2}}=3$
A = $P\left(1+\frac{R}{100}\right)^{n}$
$=156250\left(1+\frac{4}{100}\right)^{3}$
A=Rs 175760
$\begin{aligned} C \cdot I &=A-P \\ &=175760-156250 \\ C . I &=19510 \end{aligned}$
Hence , Total amount Rs 175760 and compound interest is Rs 19510
(iii) Given
P = 100000, Time = 9 months , R = 4 % P.A
Compounded for quaterely means 3 months
R= $\frac{4}{4}$=1% per 3 months
$n=\frac{9 \mathrm{mon} \mathrm{th} \mathrm{s}}{3 \text { monits }}=3 .$
$\begin{aligned} \text { Amount }(A)^{\circ} &=P\left(1+\frac{R}{100}\right)^{n} \\ &=100000\left(1+\frac{1}{100}\right)^{3} \\ A &=103030.1 \\ C &=A-P \\ &=103030.1-100000 \\ C &=3030.1 \end{aligned}$
Hence , amount Rs 103030.1 and compound interest Rs 3030.1
Question 2
Given
principal (p) = Rs 4800
rate of interest (R) = 5% p.a , Time = 2 years
simple interest (S.I) = $\frac{P T R}{100}$
$\begin{aligned} &=\frac{4800 \times 2 \times 5}{100} \\ \text { S.I } &=480 \end{aligned}$
Compomded interest $(c \cdot I)=A-P$
$\begin{aligned} \text { Amout }(A) &=P\left(1+\frac{R}{100}\right)^{n} \\ &=4800\left(1+\frac{5}{100}\right)^{2} \\ A &=5292 \end{aligned}$
$\begin{aligned} \text { Compound interest } C . I &=A-P \\ &=5292-4800 \\ C . I &=492 \end{aligned}$
$\begin{aligned} \text { Difference between C. I .E S.I } &=492-480 \\ &=12 \end{aligned}$
Question 3
Principal $(P)=23125$
Rate interest for first year $\left(R_{1}\right)=4%$
Rate of interest for second year $\left.R_{2}\right)=51$
Rate of interest for Third year $\left(R_{3}\right)=61$
$\begin{aligned} \text { Amout }(A)=& P\left(1+\frac{R_{1}}{100}\right)\left(1+\frac{R_{2}}{100}\right)\left(1+\frac{R_{3}}{100}\right) \\=& 3125\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{6}{100}\right) \\ A &=73617.25 \\ \text { Compound Interest }(C . I) &=A-P \\ &=3617.25-3125 \\ C . I &=492.25 \end{aligned}$
Question 4
Principal =26400
Rate of interest (R)=15% p.a
Time =2 years and 4 months
Amount $(A)=P\left(1+\frac{R}{100}\right)^{n}\left(1+\frac{x \cdot R}{100}\right)$
$n=2 ; x=\frac{4}{12}$ when time infraction
$A=26400\left(1+\frac{15}{100}\right)^{n}\left(\begin{array}{c} \\ \left(1+\frac{\frac{4}{12} \times 15}{100}\right)\end{array}\right.$
A = Rs 36659.7
Hence , kamala has to pay Rs 36659.7 to clear the law
Question 5
Principal $(P)=218000$
Rateof interest $(R)=8 \% .0 .0$
Time $=2$ year
Simple interest $(s \cdot I)=\frac{P T R}{100}$
$=\frac{18000 \times 2 \times 8}{100}$
$S.I=2880$
Compound Iuterest $=A-P$
Amomt $A=P\left(1+\frac{R}{100}\right)^{n}$
$=18000\left(1+\frac{8}{100}\right)^{2}$
A = Rs 20995.2
$\begin{aligned} \text { Compound interest }(C \cdot I) &=20995.2-18000 \\ &=2995.2 \end{aligned}$
Difference between C . I . E S . I = 2995.2 - 2880
∴ The extra amount anil has to pay in Rs 115.2
Question 6
Given
Prinupal (P)=75000
Rateof interess $=12 \% \mathrm{p} . a$
(i) Compounded annually
Time $=1 \frac{1}{2}$ year
$n=1 ; x=1 / 2$
$\begin{aligned} \text { Amout }(A) &=P\left(1+\frac{R}{100}\right)^{n}\left(1+\frac{x \cdot R}{100}\right) \\ &=75000\left(1+\frac{12}{100}\right)\left(1+\frac{\frac{1}{2} \times 12}{100}\right) \\ A_{1} &=₹ 89040 \end{aligned}$
(ii) Compounded half - yearly
$n=\frac{3 / 2}{1 / 2}=3$
$A=P\left(1+\frac{R}{100}\right)^{n}$
$=75000\left(1+\frac{12}{100}\right)^{3}$
$A_{2}=105369.6$
Hence , mukesh has to pay more when compounded
half yearly than that of when compounded yearly
Question 7
Given
Principal $(P)=10000$
rate of interest $(R)=7 \%$. $a$
(i) Amount received by aryman at end of 2 years
$\begin{aligned} A &=P\left(1+\frac{R}{100}\right)^{n} \\ &=10000\left(1+\frac{7}{100}\right)^{2} \\ A &=11449 \end{aligned}$
(ii) interest for 3rd year
Amount received at the end of 3rd year
$\begin{aligned} A_{3} &=P\left(1+\frac{R}{100}\right)^{3} \\ &=10000 \times\left(1+\frac{7}{100}\right)^{3} \\ A_{3} &=12250.43 \end{aligned}$
Interestfor 3rd year = $A_{3}-A_{2}$
=12250.43-11449
=Rs 801.49
Hence , aryma receives interest for 3 rd year is Rs 801.49
Question 8
Given
Amomt $(A)=9261$
Time $=3$ year : Compouned anually
Rate ofinterest $(R)=5 %$. p.a
n= 3
$A=P\left(1+\frac{R}{100}\right)^{n}$
$9261=P\left(1+\frac{5}{100}\right)^{3}$
P=Rs 8000
Hence , the sum of money is Rs 8000
Question 9
Given
Amount $(A)=2140608$
Time $=1 \frac{1}{2}$ years
Compomded halfs - yearly
n=3
Rate of interest = 8% p.a
R = 4 % per half year
$A=P\left(1+\frac{R}{100}\right)^{n}$
$140608=P\left(1+\frac{4}{100}\right)^{3}$
$P=\frac{140608}{1.124}$
P=Rs 125000
Hence the prinicipal amount is Rs 125000
Question 10
Given
prinicipal (p) = Rs 2000
Total amount (A)= Rs 2315 . 25
Time = 3 years : n = 3 (Compounded anually)
$A=P\left(1+\frac{R}{100}\right)^{n}$
$2315.25=2000\left(1+\frac{R}{100}\right)^{3}$
$1+\frac{R}{100}=\frac{21}{20}$
$\frac{R}{100}=\frac{1}{20}$
R=5% Per anum.
Hence , rate interest in 5 % per anum
Question 11
Given
Princpal $(P)=₹ 40000$
Amout $(A)=₹ 46305$
Time $=1 \frac{1}{2}$ year
Compounded hall - yearly
$n=3$
$A=P\left(1+\frac{R}{100}\right)^{n}$
$46305=40000\left(1+\frac{R}{100}\right)^{3}$
$1+\frac{R}{100}=\frac{21}{20}$
$R=5 \%$ per half - year
$R=10 \%$ Pev Anum.
Hence , rate of interest 10 % per anum
Question 12
Given
Amouit $(A)=217576$
Principal $(P)=₹ \quad 15625$
Rate of interest $=4 \times P . a$
$A=P\left(1+\frac{R}{100}\right)^{n}$
$17576=15625\left(1+\frac{4}{100}\right)^{n .}$
$1.1248=(1.04)^{n}$
Apply 'log' on boths side
$\begin{array}{c}\log _{e}(1.1248)=n \cdot \log _{e}(1.04) \\ n=3\end{array}$
Hence , the required time is 3 years
Question 13
Principal $=216000$
Amount $=218522$
Rate of interest $(R)=10 \%$ p.a
Compounded Semi. anually
$R=5 \%$ per halt - year
$A=P\left(1+\frac{R}{100}\right)^{n}$
$18522=16000\left(1+\frac{5}{100}\right)^{n}$
$(1 \cdot 05)^{\eta}=1.1576$
Apply 'log' on boit sides
$n \cdot \log _{e}(1.05)=\log _{e}(1.1576)$
n=3
$\therefore$ Time $=3 \times \frac{1}{2}=1 \frac{1}{2}$ fran.
ஃ The required time =$1 \frac{1}{2}$ years.
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