Exercise 8.1
Q1 | Ex-8.1 | Class 8 | ML AGGARWAL | chapter 8 | Simple And Compound Interest | myhelper
Question 1
Find the simple interest on ₹4000 at 7.5% p.a. for 3 years 3 months. Also, find the amount.
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 $(I)=\frac{P\times R \times T}{100}$
$=\frac{4000 \times \frac{15}{2}\times \frac{13}{4}}{100}$
$=\frac{4000\times 15\times 13}{2\times 4\times 100}$
=5×15×13
=975
Amount = Principal + Interest
= 4000 + 975
Amount = 4975
Q2 | Ex-8.1 | Class 8 | ML AGGARWAL | chapter 8 | Simple And Compound Interest | myhelper
Question 2
What sum of money will yield ₹170.10 as simple interest in 2 years 3 months at 6% per annum?
Sol :
Given
Given
Simple Interest (I) = 170 .10
Time (T) = 2 year and 3 month
$=2 \cdot \frac{3}{12}=2 \cdot \frac{1}{4}=\frac{9}{4}$ 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
Q3 | Ex-8.1 | Class 8 | ML AGGARWAL | chapter 8 | Simple And Compound Interest | myhelper
Question 3
Find the rate of interest when ₹800 fetches ₹130 as a simple interest in 2 years 6 months.
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%
Q4 | Ex-8.1 | Class 8 | ML AGGARWAL | chapter 8 | Simple And Compound Interest | myhelper
Question 4
Find the time when simple interest on ₹3.3 lakhs at 6.5% per annum is ₹75075.
Sol :
Given
principal (p) = 3.3 lakhsP= 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
Q5 | Ex-8.1 | Class 8 | ML AGGARWAL | chapter 8 | Simple And Compound Interest | myhelper
Question 5
Find the sum of money when
(i) simple interest at $7\frac{1}{4}$% p.a. for $2\frac{1}{2}$ years is ₹2356.25
(ii) the final amount is ₹ 11300 at 4% p.a. for 3 years 3 months.
Sol :
(i) simple interest (I) = 2356.25
$\begin{aligned}(T) &=2 . \frac{1}{2} \text { yeers } \\ T &=\frac{5}{2} \text { years } \\(R) &=7 \cdot \frac{1}{4} \% \\ R &=\frac{29}{4}\\ P=& \frac{I \times 100}{T \times R} \\=& \frac{2356.25 \times 100}{\frac{5}{2} \times \frac{29}{4}} \\ P &=13,000\end{aligned}$
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 (P)+Simple interest (I)
$11300=P+\frac{P \times R \times T}{100}$
$11300=P\left(1+\frac{T R}{100}\right)$
$11300=P\left(1+\frac{\frac{13}{4} \times 4}{100}\right)$
$11300=P\left(\frac{113}{100}\right)$
$P=11300 \times \frac{100}{113}$
P=10,000
Hence , the required principal is 10 ,000
Q6 | Ex-8.1 | Class 8 | ML AGGARWAL | chapter 8 | Simple And Compound Interest | myhelper
Question 6
How long will it take a certain sum of money to triple itself at $13\frac{1}{3}$% per annum simple interest?
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.
Q7 | Ex-8.1 | Class 8 | ML AGGARWAL | chapter 8 | Simple And Compound Interest | myhelper
Question 7
At a certain rate of simple interest ₹4050 amounts to ₹4576.50 in 2 years. At the same rate of simple interest, how much would ₹1 lakh amount to in 3 years?
Sol :
Given
Final Amount =4576.501
$\begin{aligned} \text { Prinupal }+\text { Simple Interest }&=4576.50 \\ 4050+I_{1} &=4576.50 \\ I_{1} &=526.50\end{aligned}$
Time =2 year
$I_{1}=\frac{P_{1} T_{1} R}{100}$
$526.50=\frac{4050 \times 2 \times R}{100}$
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
Q8 | Ex-8.1 | Class 8 | ML AGGARWAL | chapter 8 | Simple And Compound Interest | myhelper
Question 8
What sum of money invested at 7.5% p.a. simple interest for 2 years produces twice as much interest as ₹9600 in 3 years 6 months at 10% p.a. simple interest?
Sol :
Let the money invested to be Rs p
Let the money invested to be Rs p
Given
Principal $P_{1}=P$
Rate $R_{1}=7.5$
Time $T_{1}=2$ years
Let Interest $=I_{1}$
Principal $P_{2}=₹ 9600$
Rate $R_{2}=10$
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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