Additional Questions - Chapter 2 - Percentage and Simple Interest - Term 3 - 7th Maths Guide Samacheer Kalvi Solutions

You can Download the Additional Questions - Chapter 2 - Percentage and Simple Interest - Term 3 - 7th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Additional Questions and Answers
Exercise 2.1
Question $1 .$
$72 \%$ of 25 students are good at science. How many are not good at science?
Solution:
Number of students who are good at science
$=72 \%$ of $25=\frac{72}{100} \times 25=18$ students
$\therefore$ Number of students who are not good at science
$=25-18=7$ students
 

Question $2 .$
A flower garden has 1000 plants. $5 \%$ of the plants are roses and $1 \%$ are daisy plants. What is the total number of other plants.
Solution:
Total plants $=1000$
Number of rose plants $=5 \%$ of $1000=\frac{5}{100} \times 1000=50$
Number of Daisy plants $=1 \%$ of $1000=\frac{1}{100} \times 1000=10$
Total of rose and daisy $=50+10=60$
Number of other plants $=1000-60=940$
 

Question $3 .$
Find $135 \%$ of $80 ₹$.

Solution:
$135 \%$ of $80=\frac{135}{100} \times 80=₹ 108$
 

Exercise 2.2
Question 1.
Neka bought $72.3 \mathrm{~m}$ of cloth from a role of $100 \mathrm{~m}$. Express the cloth bought in terms of percentage.
Solution:
Total length of the cloth $=100 \mathrm{~m}$
Length of cloth bought $=72.3 \mathrm{~m}$
Percentage of cloth bought $=\frac{72.3}{100}=72.3 \%$
 

Question $2 .$
Convert (i) $88 \%$ (ii) $1.86 \%$ into decimals.
Solution:
(i) $88 \%=\frac{88}{100}=0.88$
(ii) $1.86 \%=\frac{1.86}{100}=0.0186$
 

Question $3 .$
Convert (i) $3.35$ (ii) $0.5$ into percentage.
Solution:
(i) $3.35=\frac{335}{100} \times 100 \%=335 \%$
(ii) $0.5=\frac{5}{10} \times 100 \%=50 \%$
 

Exercise $2.3$
Question 1.
If Gayathri had $₹ 600 \mathrm{left}$ after spending $75 \%$ of her money, how much did she have in the beginning?

Solution:
Suppose Gayathri had ₹ X in the beginning.
Then money Spend $=75 \%$ of $\mathrm{X}=\frac{75}{100} \mathrm{X}=\frac{3 X}{4}$
Money left with her $-\mathrm{X}-\begin{gathered}3 X \\ 4\end{gathered}-4 X-3 X-\frac{X}{4}$
But it is given that money left $=₹ 600$
i.e. $\frac{X}{4}=600$
$X=600 \times 4=2400$
$\therefore$ Gayathri had ₹ 2,400
 

Question $2 .$
Mohan gets 98 marks in her exams. This amounts to $56 \%$ of the total marks, What are the maximum marks?
Solution:
Let the maximum marks be X. $56 \%$ of $X=98$
$\frac{56}{100} \times(X)=98$
$\begin{aligned}
&100 \times(X)=98 \\
&\Rightarrow X=98 \times \frac{100}{56} \\
&X=175
\end{aligned}$
$\therefore$ Maximum marks $=175$

Exercise $2.4$
Question $1 .$
On what sum of money lent out at $9 \%$ per annum for 6 years does the simple interest amount to ₹ 810 ?
Solution:
Given Simple Interest I $=₹ 810$
Let the sum of money (Principal) be P
Rate of interest $r=9 \%$ Per annum.
Time $n=6$ years
$\begin{aligned}
&\mathrm{I}=\frac{p a r}{100} \\
&810=\frac{P \times 6 \times 9}{100} \\
&\mathrm{P}=\frac{810 \times 100}{6 \times 9} \\
&\mathrm{P}-₹ 1500
\end{aligned}$
Sum of money required $=₹ 1500$
 

Question $2 .$
Find the simple interest on $₹ 1120$ for $2 \frac{2}{5}$ years at the rate of $5 \%$ per annum.
Solution:
Simple Interest I $=\frac{p r r}{100}$
Principal $P=₹ 1120$
Time $\mathrm{n}=2 \frac{2}{5}$ years
$=\frac{12}{5}$ years
Rate of Interest $\mathrm{r}=5 \%$

$\begin{aligned}
&\therefore I=1120 \times \frac{2}{5} \times \frac{5}{100} \\
&=\frac{672}{5} \\
&=₹ 134.4
\end{aligned}$
$\therefore \mathrm{I}=1120 \times \frac{12}{5} \times \frac{5}{100}$ $=\frac{6 \pi 2}{5}$ $=₹ 134.4$ Simple interest $==₹ 134.4$