Exercise 3.2 - Chapter 3 Algebra Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.2$
Question 1.
Given that $x>y$. Fill in the blanks with suitable inequality signs.
(i) $y[] x$
(ii) $x+6[] y+6$
(iii) $x^{2}[] x y$
(iv) $-x y[]-y^{2}$
(v) $x-y[] 0$
Answer:
(i) $y[\leq] x$
(ii) $x+6[\geq] y+6$
(iii) $x^{2}[\geq] x y$
(iv) $-x y[\leq]-y^{2}$
(v) $x-y[\geq] 0$

Question $2 .$
Say True or False.
(i) Linear inequation has almost one solution.
Answer:
False
(ii) When $x$ is an integer, the solution set for $x \leq 0$ are $-1,-2, \ldots$
Answer:
False
(iii) An inequation, $-3<x<-1$, where $x$ is an integer, cannot be represented in the number line.
Answer:
True
(iv) $x<-y$ can be rewritten as $-y<x$
Ans:
False

Question $3 .$
Solve the following inequations.
(i) $\mathrm{x} \leq 7$, where $\mathrm{x}$ is a natural number.
(ii) $x-6<1$, where $x$ is a natural number.
(iii) $2 \mathrm{a}+3 \leq 13$, where $a$ is a whole number.
(iv) $6 x-7 \geq 35$, where $x$ is an integer.
(v) $4 x-9>-33$, where $x$ is a negative integer.
Solution:
(i) $\mathrm{x} \leq 7$, where $\mathrm{x}$ is a natural number.
Since the solution belongs to the set of natural numbers, that are less than or equal to 7 , we take the values of $x$ as $1,2,3,4,5,6$ and 7 .
(ii) $x-6<1$, where $x$ is a natural number.
$x-6<1$ Adding 6 on the both the sides $x-6+6<1+6$
$x<7$
Since the solutions belongs to the set of natural numbers that are less than 7 , we take the values of $x$ as $1,2,3,4,5$ and 6
(iii) $2 a+3 \leq 13$, where a is a whole number.
$2 \mathrm{a}+3 \leq 13$
Subtracting 3 from both the sides $2 a+3-3 \leq 13-3$
$2 \mathrm{a} \leq 10$
Dividing both the side by 2. $\frac{2 a}{2} \leq \frac{10}{2}$
$a \leq 5$
Since the solutions belongs to the set of whole numbers that are less than or equal to 5 we take the values of a as $0,1,2,3,4$ and 5
(iv) $6 x-7 \geq 35$, where $x$ is an integer.
$6 x-7 \geq 35$ Adding 7 on both the sides
$6 x-7+7 \geq 35+7$
$6 x \geq 42$
Dividing both the sides by 6 we get $\frac{6 x}{6} \geq \frac{42}{6}$ $x \geq 7$
Since the solution belongs to the set of integers that are greater than or equal to 7 , we take the values of $x$ as 7,8 , 9,10 ...
(v) $4 x-9>-33$, where $x$ is a negative integer.
$4 x-9>-33+9$ Adding 9 both the sides
$4 x-9+9>-33+9$
$4 x>-24$
Dividing both the sides by 4

$\begin{aligned}
&\frac{4 x}{4}>\frac{-24}{4} \\
&x>-6
\end{aligned}$
Since the solution belongs to a negative integer that are greater than $-6$, we take values of u as $-5,-4,-3,-2$ and $-1$
 

Question $4 .$
Solve the following inequations and represent the solution on the number line:
(i) $\mathrm{k}>-5, \mathrm{k}$ is an integer.
(ii) $-7 \leq y, y$ is a negative integer.
(iii) $-4 \leq x \leq 8, x$ is a natural number.
(iv) $3 \mathrm{~m}-5 \leq 2 \mathrm{~m}+1$, $\mathrm{m}$ is an integer.
Solution:
(i) $\mathrm{k}>-5, \mathrm{k}$ is an integer. Since the solution belongs to the set of integers, the solution is $-4,-3$, shown below.

$-6,-5-4-3-2-1 \quad 0 \quad 1 \quad 2 \quad 3 \quad 4$

(ii) $-7 \leq y, y$ is a negative integer. $-7 \leq y$
Since the solution set belongs to the set of negative integers, the solution is $-7,-6,-5,-4,-3,-2,-1$.

(iii) $-4 \leq x \leq 8, x$ is a natural number.
$-4 \leq x \leq 8$
Since the solution belongs to the set of natural numbers, the solution is $1,2,3,4,5,6,7$ and 8 .

Its graph on number line is shown below

(iv) $3 m-5 \leq 2 m+1$, $m$ is an integer.
$3 \mathrm{~m}-5 \leq 2 \mathrm{~m}+1$
Subtracting 1 on both the sides
$3 m-5-1 \leq 2 m+1+1$
$3 m-6 \leq 2 m$
Subtracting $2 m$ on both the sides $3 m-6-2 m \leq 2 m-2 m$
$m-6 \leq 0$
Adding 6 on both the sides $m-6+6 \leq 0+6$
$\mathrm{m} \leq 6$
Since the solution belongs to the set of integers, the solution is $6,5,4,3,2,1,0,-1, \ldots$
Its graph on number line is shown below

Question $5 .$
An artist can spend any amount between ₹ 80 to ₹ 200 on brushes. If cost of each brush is ₹ 5 and there are 6 brushes in each packet, then how many packets of brush can the artist buy?
Solution:
Given the artist can spend any amount between ₹ 80 to ₹ 200
Let the number of packets of brush he can buy be $x$
Given cost of 1 brush $=₹ 5$
Cost of 1 packet brush ( 6 brushes) $=₹ 5 \times 6=₹ 30$
$\therefore$ Cost of $\mathrm{x}$ packets of brushes $=30 \mathrm{x}$
$\therefore$ The inequation becomes $80 \leq 30 \mathrm{x} \leq 200$
Dividing throughout by 30 we get $\frac{80}{30} \leq \frac{30 x}{30} \leq \frac{200}{30}$
$\frac{8}{3} \leq \mathrm{x} \leq \frac{20}{3}$
$2 \frac{2}{3} \leq x \leq 6 \frac{2}{3}$
brush packets cannot get in fractions.
$\therefore$ The artist can buy $3<\mathrm{x}<6$ packets of brushes,
or $x=3,4,5$ and 6 packets of brushes.
 

Objective Type Questions
Question 1.
The solutions set of the inequation $3 \leq \mathrm{p} \leq 6$ are (where $\mathrm{p}$ is a natural number)
(i) 4,5 and 6
(ii) 3,4 and 5
(iii) 4 and 5
(iv) $3,4,5$ and 6
Answer:
(iv) $3,4,5$ and 6
 

Question $2 .$
The solution of the inequation $5 x+5 \leq 15$ are (where $x$ is a natural number)
(i) 1 and 2
(ii) 0,1 and 2
(iii) $2,1,0,-1,-2$
(iv) $1,2,3$..
Answer:
(i) 1 and 2

Question $3 .$
The cost of one pen is ₹ 8 and it is available in a sealed pack of 10 pens. If Swetha has only ₹ 500 , how many packs of pens can she buy at the maximum?
(i) 10
(ii) 5
(iii) 6
(iv) 8
Answer:
(iii) 6
Hint:
Price of 1 pen $=₹ 8$
Price of 1 pack $=10 \times 8=80$
Number of packs Swetha can buy $=x$
$80 \mathrm{x} \leq 500$
$8 \mathrm{x} \leq 50$
$x \leq \frac{50}{8}=6.25$
$x$ is a natural number $x=1,2,3,4,5,6$
 

Question $4 .$
The inequation that is represented on the number line as shown below is

(i) $-4<\mathrm{x}<0$
(ii) $-4 \leq \mathrm{x} \leq 0$
(iii) $-4<x \leq 0$
(iv) $-4 \leq x<0$
(v) $-4 \leq x \leq 2$
Answer:
(v) $-4 \leq x \leq 2$