In Text Questions Try These (Text Book Page No.29, 30,32,33,34,41,44) - Chapter 2 - Percentage and Simple Interest - Term 3 - 7th Maths Guide Samacheer Kalvi Solutions

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Text Questions : Chapter 2  Percentage and Simple Interest Term 3 Class 7th std Maths Guide Samacheer Kalvi Solutions
Exercise 2.1
Try These (Text book Page No. 28)
Question $1 .$
Find the percentage of children whose scores fall in different categories given in table below.

Solution:

Try These (Text book Page No. 29)
Question $1 .$
There are 50 students in class VII of a school. The number of students involved in these activities are :
Scout: 7
Red Ribbon Club: 6
Junior Red Cross : 9
Green Force : 3
Sports: 14
Cultural activity : 11
Find the percentage of students who involved in various activities.

Solution:

Try These (Text book Page No. 30)
Question 1.
Convert the fractions as percentage.
(i) $\frac{1}{20}$
(ii) $\frac{13}{25}$
(iv) $\frac{18}{5}$
(v) $\frac{27}{10}$
(vi) $\frac{72}{90}$
Solution:
(i) $\frac{1}{2 U}$
$\begin{aligned}
&\text { (i) } \frac{1}{20} \\
&=\frac{1}{20} \times \frac{100}{100} \\
&=\frac{1}{20} \times 100 \% \\
&=5 \%
\end{aligned}$
(ii) $\frac{13}{25}$ $=\frac{13}{25} \times \frac{100}{100}$
$=\frac{13}{25} \times \frac{100}{100}$
$=\frac{13}{25} \times \frac{100}{100}$ $=\frac{13}{25} \times 100 \%$ $=52 \%$ (iii) $\frac{45}{50}$
$\%=$
$\begin{aligned}
&\text { (iii) } \frac{45}{50} \\
&=\frac{45}{50} \times \frac{100}{100} \\
&=\frac{45}{50} \times 100 \% \\
&=90 \%
\end{aligned}$

(iv) $\frac{18}{5}$
$\begin{aligned}
&=\frac{18}{5} \times \frac{100}{100} \\
&=\frac{18}{50} \times 100 \% \\
&=360 \%
\end{aligned}$
(iv) $\frac{27}{10}$
$\begin{aligned}
&=\frac{27}{10} \times \frac{100}{100} \\
&=\frac{27}{10} \times 100 \% \\
&=270 \%
\end{aligned}$
(iv) $\frac{27}{10}$
$\begin{aligned}
&=\frac{27}{10} \times \frac{100}{100} \\
&=\frac{27}{10} \times 100 \% \\
&=270 \%
\end{aligned}$
(vi) $\frac{72}{90}$
$\begin{aligned}
&=\frac{72}{90} \times \frac{100}{100} \\
&=\frac{72}{90} \times 100 \% \\
&=80 \%
\end{aligned}$

Question $2 .$
Convert the following percentage as fractions.
(i) $50 \%$
(ii) $75 \%$
(iii) $250 \%$
(iv) $30 \frac{1}{5} \%$
(v) $\frac{7}{20} \%$
(vi) $90 \%$
Solution:

(i) $50 \%$
$\begin{aligned}
&=\frac{50}{100} \\
&=\frac{5}{10} \\
&=\frac{1}{2}
\end{aligned}$
(i1) $750 \%$
$\begin{aligned}
&=\frac{75}{100} \\
&=\frac{3}{4}
\end{aligned}$
(iii) $250 \%$
$\begin{aligned}
&=\frac{250}{100} \\
&=\frac{25}{10} \\
&=\frac{5}{2}
\end{aligned}$
$\begin{aligned}
&\text { (iv) } 30 \frac{1}{5} \% \\
&=\frac{30 \frac{1}{5}}{100}=\frac{\left(\frac{151}{5}\right)}{100} \\
&=\frac{151}{500}
\end{aligned}$
(v) $\frac{7}{20} \%$
$\begin{aligned}
&=\frac{\frac{7}{20}}{100}=\frac{7}{20 \times 100} \\
&=\frac{7}{2000}
\end{aligned}$
(vi) $90 \%=\frac{90}{100}=\frac{9}{10}$

Think (Text book Page No. 32)
Question 1.
What is the difference between $0.01$ and $1 \%$.
Solution: $0.01=\frac{1}{100}=1 \%$
$0.01=\frac{1}{100}=1 \%$
$0.01$ and $1 \%$ are the same.
 

Question $2 .$
In a readymade shop there will be a board showing upto $50 \%$ off. Most of the people will realize that everything is half of its original price, Is that true?
Solution:
No. Only some of them are half of its original price.

Exercise $2.2$
Try These (Text book Page No. 33)
Question 1.
Convert these decimals to percentage.
(i) $0.25$
(ii) $0.07$
(iii) $0.8$
(iv) $0.375$
(v) $3.75$
Solution:
(i) $0.25$
$=\frac{25}{100}=25 \%$
(ii) $0.07$
$=\frac{7}{100}=7 \%$
(iii) $0.8$
$=\frac{80}{100}=80 \%$
(iv) $0.375$
$=\frac{375}{1000}$
$=\frac{375}{10} \times \frac{1}{100}$
$=37.5 \%$
(v) $3.75$
$=\frac{375}{100}=375 \%$

Try These (Text buok Page No. 34)
Question 1.
Write these perecntage as decimals
(i) $3 \%$
(ii) $25 \%$
(iii) $80 \%$
(iv) $67 \%$
(v) $17.5 \%$
(vi) $135 \%$
(vii) $0.5 \%$
Solution:
(i) $3 \%$
$=\frac{3}{100}=0.03$
(ii) $25 \%$
$=\frac{25}{100}=0.25$
(iii) $80 \%$
$=\frac{80}{100}=0.8$

$\begin{aligned}
&\text { (iv) } 67 \% \\
&=\frac{67}{100}=0.67
\end{aligned}$
(v) $17.5 \%$
$=\frac{17.5}{100}=0.175$
$\begin{aligned}
&\text { (vi) } 135 \% \\
&=\frac{135}{100}=1.35 \\
&\text { (vii) } 0.5 \% \\
&=\frac{0.5}{100}=0.005
\end{aligned}$

Exercise $2.3$
Try These (Text book Page No. 38)
Question 1.
Level of water in a tank is increased from 35 litres to 50 litres in 2 minutes, what is the Percentage of increase?
Solution:
Level of water in the tank originally $=35$ litres.
Increase in the water level = amount of change $=50-35=15$ litres

$\begin{aligned}
\text { Percentage of Increase } &=\frac{\text { Amount of change }}{\text { Original amount }} \times 100 \\
&=\frac{3}{35} \times 100=\frac{300}{7}=42.86 \%
\end{aligned}$

Exercise $2.4$
Try These (Text book Page No. 41)

Question $1 .$
Arjun borrowed a sum of $₹ 5,000$ from a bank at $5 \%$ per annum. Find the interest and amount to be paid at the end of three year.
Solution:
Here Principal $(\mathrm{P})=₹ 5,000$
Rate of interest $(r)=5 \%$ Per annum
Time $(n)=3$ years
Simple Interest I = $\frac{p m r}{100}$
$=\frac{5000 \times 3 \times 5}{100}$
Amount to be paid $\mathrm{A}=\mathrm{P}+\mathrm{I}=₹ 5,000+₹ 750=₹ 5,750$
$\mathrm{I}=₹ 750 ; \mathrm{A}=₹ 5,750$
 

Question $2 .$
Shanti borrowed $₹ 6,000 /$ - from a Bank for 7 years at $12 \%$ per annum. What amount will clear off her debt?
Solution:
Here principal $(P)=₹ 6,000$
Rate of Interest $(r)=12 \%$ Per annum
Time $(n)=4$ Years
Simple Interest (I) $=\frac{p n r}{100}=$
$=\frac{6000 \times 7 \times 12}{100}$ $I=₹ 5,040$
Amount to be paid $\mathrm{A}=\mathrm{P}+\mathrm{I}=6,000+5,040=₹ 11,040$
 

Think (Text book Page No. 43)
Question $1 .$
In simple interest, a sum of money doubles itself in 10 years. In how many years it will get triple itself.
Solution:
Let the Principal be P and Rate of interest be $\mathrm{r} \%$ per annum.
Here the number of years $n=10$ years
Given in 10 years $P$ becomes $2 \mathrm{P}$.
$\mathrm{A}=\mathrm{P}+\mathrm{I}$
After 2 years $A=2 P$
i.e. $2 \mathrm{P}=\mathrm{P}+1$
$2 \mathrm{P}-\mathrm{P}=\mathrm{I}$

$\begin{aligned}
\mathrm{P} &=\mathrm{I} \\
\mathrm{P} &=\frac{\mathrm{P} \times n \times r}{100} \\
\mathrm{P} &=\frac{\mathrm{P} \times 10 \times r}{100} \\
r &=\frac{\mathrm{P} \times 100}{\mathrm{P} \times 10} \\
r &=10 \%
\end{aligned}$
Now if the amount becomes triple then $\mathrm{A}=\mathrm{P}+\mathrm{I}=3 \mathrm{P}$
$3 P=P+I$
$3 \mathrm{P}-\mathrm{P}=\mathrm{I}$
$2 P=1$
$2 \mathrm{P}=\frac{\mathrm{P} \times n \times 10}{100}$
$\begin{aligned} \frac{2 \mathrm{P} \times 100}{\mathrm{P} \times 10} &=n \\ n &=20 \end{aligned}$
$n=20$ years
$\therefore$ After 20 years the amount get tripled.