Additional Questions - 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

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 $==₹ 134.4$