Exercise 3.3 - Chapter 3 - Algebra - Term 1 - 7th Maths Guide Samacheer Kalvi Solutions

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Ex $3.3$
Question $1 .$
Fill in the blanks.
(i) An expressions equated to another expression is called
(ii) If $a=5$, the value of $2 a+5$ is
(iii) The sum of twice and four times of the variable $x$ is
Solution:
(i) an equation
(ii) 15
(iii) $6 \mathrm{x}$

Question 2:
Say True or False
(i) Every algebraic expression is an equation.
(ii) The expression $7 x+1$ cannot be reduced without knowing the value of $x$.
(iii) To add two like terms, its coefficients can be added.
Solution:
(i) False
(ii) True
(iii) True

Question $3 .$
Solve (i) $x+5=8$
(ii) $\mathrm{p}-3=1$
(iii) $2 \mathrm{x}=30$
(iv) $\frac{m}{6}=5$
(v) $7 x+10=80$
Solution:
(i) Given $x+5=8$; Subtracting 5 on both the sides

$\begin{aligned}
&x+5-5=8-5 \\
&x=3
\end{aligned}$
(ii) Given $\mathrm{p}-3=7$; Adding 3 on both the sides,
$\begin{aligned}
&\mathrm{p}-3+3=7+3 \\
&\mathrm{p}=10
\end{aligned}$
(iii) Given $2 \mathrm{x}=30$; Dividing both the sides by 2 ,
$\begin{aligned}
&\frac{2 x}{2}=\frac{30}{2} \\
&x=15
\end{aligned}$
(iv) Given $\frac{m}{6}=5 ;$ Multiplying both the sides by 6 ,
$\begin{aligned}
&\frac{m}{6} \times 6=5 \times 6 \\
&m=30
\end{aligned}$
(v) Given $7 x+10=80$; Subtracting 10 from both the sides,
$\begin{aligned}
&7 x+10-10=80-10 \\
&7 x=70
\end{aligned}$
Dividing both sides by 7 ,
$\begin{aligned}
&\frac{7 x}{7}=\frac{70}{7} \\
&x=10
\end{aligned}$

Question $4 .$
What should be added to $3 x+6 y$ to get $5 x+8 y$ ?
Solution:
To get the expression we should subtract $3 x+6 y$ from $5 x+8 y$
$\begin{aligned}
&5 x+8 y-(3 x+6 y)=5 x+8 y+(-3 x-6 y) \\
&=5 x+8 y-3 x-6 y=(5-3) x+(8-6) y \\
&=2 x+2 y
\end{aligned}$
So $2 x+2 y$ should be added.

Question $5 .$
Nine added to thrice a whole number gives 45 . Find the number

Solution:
Let the whole number required be $x$.
Thrice the whole number $=3 \mathrm{x}$
Nine added to it $=3 x+9$
Given $3 x+9=45$
$3 x+9-9=45-9$ [Subtracting 9 on both sides]
$3 x=36$
$\begin{aligned}
&\frac{3 x}{3}=\frac{36}{3} \\
&x=12
\end{aligned}$
$x=12$
$\therefore$ The required whole number is 12
 

Question $6 .$
Find the two consecutive odd numbers whose sum is 200
Solution:
Let the two consecutive odd numbers be $x$ and $x+2$

$\therefore$ Their sum $=200$
$x+(x+2)=200$
$x+x+2=200$
$2 x+2=200$
$2 \mathrm{x}+2-2=200-2[\because$ Subtracting 2 from both sides $]$
$2 \mathrm{x}=198$
$\frac{2 x}{2}=\frac{198}{2}$ [Dividing both sides by 2 ]
$x=99$ The numbers will be 99 and $99+2$.
$\therefore$ The numbers will be 99 and 101 .

Question $7 .$
The taxi charges in a city comprise of a fixed charge of $₹ 100$ for $5 \mathrm{kms}$ and $₹ 16$ per $\mathrm{km}$ for ever additional $\mathrm{km}$. If the amount paid at the end of the trip was $₹ 740$, find the distance traveled.
Solution:
Let the distance travelled by taxi be ' $x$ ' $k m$
For the first $5 \mathrm{~km}$ the charge $=₹ 100$
For additional $\mathrm{kms}$ the charge $=₹ 16(\mathrm{x}-5)$
$\therefore$ For $\mathrm{x} \mathrm{kms}$ the charge $=100+16(\mathrm{x}-5)$
Amount paid $=₹ 740$
$\begin{aligned}
&\therefore 100+16(\mathrm{x}-5)=740 \\
&100+16(\mathrm{x}-5)-100=740-100 \\
&16(\mathrm{x}-5)=640 \\
&\frac{16(x-5)}{16}=\frac{640}{16} \\
&\mathrm{x}-5=40 \\
&\mathrm{x}-5+5=45+5 \\
&\mathrm{x}=45 \\
&\mathrm{x}=45 \mathrm{~km} \\
&\therefore \text { Total distance travelled }=45 \mathrm{~km}
\end{aligned}$

Objective Type Questions
Question 8 .
The generalization of the number pattern $3,6,9,12, \ldots \ldots \ldots \ldots$ is
(i) $\mathrm{n}$
(ii) $2 \mathrm{n}$
(iii) $3 \mathrm{n}$
(iv) $4 \mathrm{n}$
Solution:
(iii) $3 \mathrm{n}$

Question $9 .$
The solution of $3 x+5=x+9$ is $t$
(i) 2
(ii) 3
(iii) 5
(iv) 4
Solution:
(i) 2
Hint: $3 x+5=x+9 \Rightarrow 3 x-x=9-5 \Rightarrow 2 x=4 \Rightarrow x=2$

Question $10 .$
The equation $\mathrm{y}+1=0$ is true only when $\mathrm{y}$ is
(i) 0
(ii) $-1$
(iii) 1
(iv) $-2$
Solution:
(ii) $-1$