# Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra Ex 3.2

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## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 3 Algebra 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) x2 [ ] xy
(iv) -xy [ ] – y2
(v) x – y [ ] 0
(i) y [<] x
(ii) x+ 6 [>] y + 6
(iii) x2 [>] xy
(iv) -xy [<] – y2
(v) x – y [>] 0 Question 2.
Say True or False.

(i) Linear inequation has almost one solution.
False

(ii) When x is an integer, the solution set for x < 0 are -1, -2,..
False

(iii) An inequation, -3 < x < -1, where x is an integer, cannot be represented in the number line.
True

(iv) x < -y can be rewritten as – y < x
Ans :
False

Question 3.
Solve the following inequations.
(i) x < 7, where x is a natural number.
(ii) x – 6 < 1, where x is a natural number.
(iii) 2a + 3 < 13, where a is a whole number.
(iv) 6x – 7 > 35, where x is an integer.
(v) 4x – 9 > -33, where x is a negative integer.
Solution:
(i) x < 7, where 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) 2a + 3 < 13, where a is a whole number.
2a + 3 < 13
Subtracting 3 from both the sides 2a + 3 – 3 < 13 – 3
2a < 10
Dividing both the side by 2. $$\frac { 2a }{ 2 }$$ < $$\frac { 10 }{ 2 }$$
a < 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) 6x – 7 > 35, where x is an integer.
6x – 7 > 35 Adding 7 on both the sides
6x – 7 + 7 > 35 + 7
6x > 42
Dividing both the sides by 6 we get $$\frac { 6x }{ 6 }$$ > $$\frac { 42 }{ 6 }$$
x > 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) 4x – 9 > -33, where x is a negative integer.
4x – 9 > – 33 + 9 Adding 9 both the sides
4x – 9 + 9 > -33 + 9
4x > – 24
Dividing both the sides by 4
$$\frac { 4x }{ 4 }$$ > $$\frac { -24 }{ 4 }$$
x > -6
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) k > -5, k is an integer.
(ii) -7 < y, y is a negative integer.
(iii) -4 < x < 8, x is a natural number.
(iv) 3m – 5 < 2m + 1, m is an integer.
Solution:
(i) k > -5, k is an integer.
Since the solution belongs to the set of integers, the solution is -4, -3, -2, -1, 0,… It’s graph on number line is shown below. (ii) -7 < y, y is a negative integer.
-7 < y
Since the solution set belongs to the set of negative integers, the solution is
-7, -6, -5, -4, -3, -2, -1.
Its graph on the number line is shown below (iii) -4 < x < 8, x is a natural number.
-4 < x < 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) 3m – 5 < 2m + 1, m is an integer.
3m – 5 < 2m + 1
Subtracting 1 on both the sides
3m – 5 – 1 < 2m + 1 + 1
3m – 6 < 2m
Subtracting 2m on both the sides 3m- 6 – 2m < 2m -2m
m – 6 < 0
Adding 6 on both the sides m – 6 + 6 < 0 + 6
m < 6
Since the solution belongs to the set of integers, the solution is
6, 5, 4, 3, 2, 1, 0,-1,…
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 × 6 = ₹ 30
∴ Cost of x packets of brushes = 30 x
∴ The inequation becomes 80 < 30x < 200
Dividing throughout by 30 we get $$\frac { 80 }{ 30 }$$ < $$\frac { 30x }{ 30 }$$ < $$\frac { 200 }{ 30 }$$
$$\frac { 8 }{ 3 }$$ < x < $$\frac { 20 }{ 3 }$$ ;
2 $$\frac { 2 }{ 3 }$$ < x < 6 $$\frac { 2 }{ 3 }$$
brush packets cannot get in fractions.
∴ The artist can buy 3 < 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 < p < 6 are (where 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
(iv) 3,4,5 and 6

Question 2.
The solution of the inequation 5x + 5 < 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..
(i) 1 and 2
Hint: 5x + 5 < 15
5x < 15 – 5 = 10
x < $$\frac { 10 }{ 5 }$$ = 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
(iii) 6
Hint:
Price of 1 pen = ₹ 8
Price of 1 pack = 10 × 8 = 80
Number of packs Swetha can buy = x
80x < 500
8x < 50
x < $$\frac { 50 }{ 8 }$$ = 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 < x < 0
(ii) -4 < x < 0
(iii) -4 < x < 0
(iv) -4 < x < 0
(v) -4 < x < 2