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

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

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
Answer:
(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.
Answer:
False

(ii) When x is an integer, the solution set for x < 0 are -1, -2,..
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) 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. 2a/2 < 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 6x/6 > 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
4x/4 > −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.

(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

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
Answer:
(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..
Answer:
(i) 1 and 2
Hint: 5x + 5 < 15
5x < 15 – 5 = 10
x < 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
Answer:
(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 < 50/8 = 6.25
x is a natural number x = 1, 2, 3, 4, 5, 6