ML AGGARWAL CLASS 8 CHAPTER 8 Simple and Compound Interest Exercise 8.3

Exercise 8.3

Question 1

Sol :

(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

Compound 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 annum

Compound 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 quarterly means 3 months 

R= $\frac{4}{4}$=1% per 3 months

$n=\frac{9 \mathrm{months}}{3 \text { months }}=3$

$\begin{aligned} \text { Amount }(A)^{\circ} &=P\left(1+\frac{R}{100}\right)^{n} \\ &=100000\left(1+\frac{1}{100}\right)^{3} \\ A &=103030.1 \\ C &=A-P \\ &=103030.1-100000 \\ C &=3030.1 \end{aligned}$

Hence , amount Rs 103030.1 and compound interest Rs 3030.1

Question 2

Sol :

Given 

principal (p) = Rs 4800

rate of interest (R) = 5% p.a , Time = 2 years

simple interest (S.I) = $\frac{P T R}{100}$

$\begin{aligned} &=\frac{4800 \times 2 \times 5}{100} \\ \text { S.I } &=480 \end{aligned}$

Compounded interest $(c \cdot I)=A-P$

$\begin{aligned} \text { Amount }(A) &=P\left(1+\frac{R}{100}\right)^{n} \\ &=4800\left(1+\frac{5}{100}\right)^{2} \\ A &=5292 \end{aligned}$

$\begin{aligned} \text { Compound interest } C . I &=A-P \\ &=5292-4800 \\ C . I &=492 \end{aligned}$

$\begin{aligned} \text { Difference between C. I .E  S.I } &=492-480 \\ &=12 \end{aligned}$

Question 3

Sol :

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 $\left(R_{3}\right)=61$

$\begin{aligned} \text { Amount }(A)=& P\left(1+\frac{R_{1}}{100}\right)\left(1+\frac{R_{2}}{100}\right)\left(1+\frac{R_{3}}{100}\right) \\=& 3125\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{6}{100}\right) \\ A &=73617.25 \\ \text { Compound Interest }(C . I) &=A-P \\ &=3617.25-3125 \\ C . I &=492.25 \end{aligned}$

Question 4

Sol :

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

Sol :

Principal (P)=218000

Rate of interest (R)=8 % 

Time =2 year

Simple interest $SI=\frac{P T R}{100}$

$=\frac{18000 \times 2 \times 8}{100}$

S.I=2880

Compound Interest=A-P

Amount $A=P\left(1+\frac{R}{100}\right)^{n}$

$=18000\left(1+\frac{8}{100}\right)^{2}$

A =Rs 20995.2

 $\begin{aligned} \text { Compound interest }(C \cdot I) &=20995.2-18000 \\ &=2995.2 \end{aligned}$

Difference between C . I . and   S . I 

= 2995.2 - 2880 

∴ The extra amount anil has to  pay in Rs 115.2 

Question 6

Sol :

Given 

Principal (P)=75000

Rate of interest =12 % p.a

(i) Compounded annually

Time $=1 \frac{1}{2}$ year

n=1 ; x=1 / 2

$\begin{aligned} \text { Amount }(A) &=P\left(1+\frac{R}{100}\right)^{n}\left(1+\frac{x \cdot R}{100}\right) \\ &=75000\left(1+\frac{12}{100}\right)\left(1+\frac{\frac{1}{2} \times 12}{100}\right) \\ A_{1} &=₹ 89040 \end{aligned}$

(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 

Sol :

Given 

Principal (P)=10000

rate of interest (R)=7 % p.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 

$\begin{aligned} A_{3} &=P\left(1+\frac{R}{100}\right)^{3} \\ &=10000 \times\left(1+\frac{7}{100}\right)^{3} \\ A_{3} &=12250.43 \end{aligned}$

Interest for 3rd year = $A_{3}-A_{2}$

=12250.43-11449 

=Rs 801.49

Hence , aryma receives interest for 3rd year is Rs 801.49

Question 8

Sol :

Given 

Amount (A)=9261

Time$=3 year : Compounded annually

Rate of interest (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

Sol :

Given 

Amount $(A)=2140608$

Time $=1 \frac{1}{2}$ years

Compounded half-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 principal amount is Rs 125000

Question 10

Sol :

Given 

principal (p) = Rs 2000

Total amount (A)= Rs 2315 . 25 

Time = 3 years : n = 3 (Compounded annually)

$A=P\left(1+\frac{R}{100}\right)^{n}$

$2315.25=2000\left(1+\frac{R}{100}\right)^{3}$

$1+\frac{R}{100}=\frac{21}{20}$

$\frac{R}{100}=\frac{1}{20}$

R=5% Per annum.

Hence , rate interest in 5 % per annum

Question 11

Sol :

Given 

Principal (P)=₹ 40000

Amount (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 % Per annum.

Hence , rate of interest 10 % per annum

Question 12

Sol :

Given 

Amount (A)=217576

Principal (P)=₹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 both sides

$\begin{array}{c}\log _{e}(1.1248)=n \cdot \log _{e}(1.04) \\ n=3\end{array}$

Hence , the required time is 3 years

Question 13

Sol :

Principal =216000

Amount =218522

Rate of interest (R)=10 % p.a

Compounded Semi. annually

R=5 % per half - 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 both 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.