ML AGGARWAL CLASS 8 CHAPTER 8 Simple and Compound Interest Exercise 8.1

 Exercise 8.1

Question 1

Sol :

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 = principal + interest 

= 4000 + 975 


Amount = 4975 




Question 2

Sol :
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

Sol :
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

Sol :

Given
principal (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

Sol :

(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 $=113001-$

Final 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

Sol :

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

Sol :

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 annum

Now we have to calculate simple interest for 1 lakh for 3 years at a rate of 65 % per annum 

$\begin{aligned} I_{2} &=\frac{P_{2} T_{2} R}{100} \\ I_{2} &=\frac{1,001000 \times 3 \times 6.5}{100} \\ I_{2} &=19500\end{aligned}$

$\begin{aligned} \text { Total Amount } &=P_{2}+I_{2} \\ &=1,00,000+19,500\\ &=1,19,500\end{aligned}$

∴ Hence , the 1 lakh will amount Rs 1,19,500 for 3 years

Question 8

Sol :
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

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