Exercise 2.4 - Chapter 2 Percentage and Simple Interest Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 2.4
Question 1.
Find the simple interest on ₹ 35,000 at $9 \%$ per annum for 2 years?
Solution:
$\begin{array}{r}
350 \\
\times \quad 18 \\
\hline 2800 \\
3500 \\
\hline 6300
\end{array}$
Principal $P=₹ 35,000$
Rate of interest $r=9 \%$ Per annum
Time $(n)=2$ years
Simple Interest $\mathrm{I}=\frac{P_{n r}}{100}=\frac{35000 \times 2 \times 9}{100}=₹ 6300$
Simple intrest $I=₹ 6300$
 

Question $2 .$
Aravind borrowed a sum of ₹ 8,000 from Akash at $7 \%$ per annum. Find the interest and amount to be paid at the end of two years.
Solution:
Here Principal $\mathrm{P}=₹ 8,000$
Rate of interest $\mathrm{r}=7 \%$ Per annum
Time $(n)=2$ Years
Simple Interest $(\mathrm{I})=\frac{P_{n r}}{100}=\frac{8000 \times 2 \times 7}{100}$

$\begin{aligned}
&I=₹ 1120 \\
&\text { Amount }=P+I \\
&I=₹ 8000+1120=9120 \\
&\text { Interest to be paid }=₹ 1,120 \\
&\text { Amount to be paid }=₹ 9,120
\end{aligned}$

Question $3 .$
Sheela has paid simple interest on a certain sum for 4 years at $9.5 \%$ per annum is $₹ 21,280$. Find the sum.
Solution:
Let the Principal be ₹ $\mathrm{P}$
Rate of interest $r=9.5 \%$ per annum
Time $(n)=4$ years
Simple Interest I = $\frac{P r r}{100}$
Given I = ₹ 21,280
$\begin{aligned}
\therefore \frac{P n r}{100} &=₹ 21,280 \\
\frac{P \times 4 \times 9.5}{100} &=₹ 21,280 \\
\frac{P \times 4 \times 95}{1000} &=₹ 21,280 \\
P &=\frac{21280 \times 1000}{4 \times 95}=₹ 56,000
\end{aligned}$
$\therefore$ Sum of money Sheela bought $=₹ 56,000$
 

Question $4 .$
Basha borrowed $₹ 8,500$ from a bank at a particular rate of simple interest. After 3 years, he paid $₹$ 11,050 to settle his debt. At what rate of interest he borrowed the money?
Solution:

Let the rate of interest be $\mathrm{r}^{0} \%$ per annum
Here Principal $\mathrm{P}=₹ 8,500$
Time $n=3$ years
Total amount paid $-₹ 11,050$
$\mathrm{A}=\mathrm{P}+1=₹ 11,050$
i.e. $8,500+1=₹ 11,050$
$I=₹ 11,050-₹ 8,500=₹ 2,550$
Also we know that $\mathrm{I}=\frac{\mathrm{P} n r}{100}=₹ 2,550$
$\begin{aligned}
\frac{8,500 \times 3 \times r}{100} &=₹ 2,550 \\
r &=\frac{2550 \times 100}{8500 \times 3} \\
r &=10 \%
\end{aligned}$
$\begin{aligned} \frac{8,500 \times 3 \times r}{100} &=₹ 2,550 \\ r &=\frac{2550 \times 10}{8500} \\ r &=10 \% \\ \text { Rate of interest } r &=10 \% \end{aligned}$
 

Question $5 .$
In What time will ₹ 16,500 amount to $₹ 22,935$ at $13 \%$ per annum?

Rate of interest $r=13 \%$ per annum
Here Amount $\mathrm{A}=₹ 22,935$
Principal $\mathrm{P}=₹ 16,500$
$\mathrm{A}=\mathrm{P}+\mathrm{I}$
$22935=16,500+\mathrm{I}$
$\therefore$ Interest $\mathrm{I}=22935-16,500=₹ 6,435$
Simple Interest I = $\frac{p m r}{100}$

$\mathrm{n}=3 \text { years }$
$\begin{aligned}
&6435=\frac{16500 \times n \times 13}{100} \\
&\mathrm{n}=\frac{6435 \times 100}{16500 \times 13} \\
&\mathrm{n}=3 \mathrm{ycars}
\end{aligned}$
 

Question $6 .$
In what time will ₹ 17800 amount to ₹ 19936 at $6 \%$ per annum?
Solution:
Let the require time be $n$ years
Here Principal $\mathrm{P}=₹ 17,800$
Rate of interest $r=6 \%$ per annum
Amount $\mathrm{A}=₹ 19,936$
$\begin{aligned}
&A=P+I \\
&19936=17800+1 \\
&19936-17800=I \\
&2136=I
\end{aligned}$
Simple Interest $(\mathrm{I})=\frac{p n r}{100}$
$\begin{aligned}
&2136=\frac{17800 \times n \times 6}{100} \\
&\mathrm{n}=\frac{2136 \times 100}{17800 \times 6}
\end{aligned}$

$\mathrm{n}=2 \text { Years }$
Required time $=2$ years
 

Question $7 .$
A sum of ₹ 48,000 was lent out at simple interest and at the end of 2 years and 3 months the total amount was $₹ 55,560$. Find the rate of interest per year.
Solution:
Given Principal $\mathrm{P}=₹ 48,000$
Time $n=2$ years 3 months
$=2+\frac{3}{12}$ years $=2+\frac{1}{4}$ years
$=\frac{8}{4}+\frac{1}{4}$ years $=\frac{9}{4}$ years
Amount $\mathrm{A}=₹ 55,660$
$\mathrm{A}-\mathrm{p}+1$
$\begin{aligned}
&55660=48000+I \\
&I=55660-48000=₹ 7660 \\
&\therefore \text { Interest for } \frac{9}{4} \text { years }=₹ 7660 \\
&\text { Simple intrest }=\frac{p n r}{100}
\end{aligned}$

$\begin{aligned}
&7660=48000 \times \frac{9}{4} \times \frac{r}{100} \\
&r=\frac{7660 \times 4 \times 100}{9 \times 48000}=7.09 \%=7 \%
\end{aligned}$
Rate of interest $=7 \%$ Per annum
 

Question 8.
A principal becomes $₹ 17,000$ at the rate of $12 \%$ in 3 years. Find the principal.
Solution:
Given the Principal becomes ₹ 17,000
Let the principle initially be $\mathrm{P}$
Rate of Interest $\mathrm{r}$ Time $=12 \%$ Per annum
Time $n=3$ years
According to the problem given $\mathrm{I}=17000-\mathrm{P}=\frac{P \times 3 \times 12}{100}$
$\begin{aligned}
&17000=\frac{36}{100} \mathrm{p}+\mathrm{p} \\
&17000=\mathrm{p}\left(\frac{36}{100}+1\right) \\
&17000=\mathrm{p}\left(\frac{136}{100}\right) \\
&\mathrm{p}=\frac{17000 \times 100}{136}=12,500 \\
&\therefore \text { Principal } \mathrm{P}=₹ 12,500
\end{aligned}$

Objective Type Questions
Question $9 .$
The interest for a principle of? 4,500 which gives an amount of? 5,000 at end of certain period is
(i) ₹ 500
(ii) ₹ 200
(iii) $20 \%$
(iv) $15 \%$
Hint: Interest $=$ Amount $-$ Principle $=₹ 5000-₹ 4500=₹ 500$
Answer:
(i) ₹ 500

Question $10 .$
Which among the following is the simple interest for the principle of $₹ 1,000$ for one year at the rate of $10 \%$ interest per annum?
(i) ₹ 200
(ii) ₹ 10
(iii) ₹ 100
(iv) ₹ 1,000
Hint: Intrest $=\frac{p n r}{100}=\frac{1000 \times 1 \times 10}{100}=₹ 100$
Answer:
(iii) ₹ 100
 

Question 11.
Which among the following rate of interest yields an interest of $₹ 200$ for the principle of $₹ 2,000$ for one year.
(i) $10 \%$
(ii) $20 \%$
(iii) $5 \%$
(iv) $15 \%$
Hint: $\mathrm{r}=\frac{I \times 100}{P \times n}=\frac{200 \times 100}{2000 \times 1}=10 \%$
Answer:
(i) $10 \%$