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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) x^{2} [ ] xy

(iv) -xy [ ] – y^{2}

(v) x – y [ ] 0

Answer:

(i) y [<] x

(ii) x+ 6 [>] y + 6

(iii) x^{2} [>] xy

(iv) -xy [<] – y^{2}

(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. \(\frac { 2a }{ 2 } \) *<* \(\frac { 10 }{ 2 } \)

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

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 < \(\frac { 10 }{ 5 } \) = 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

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 < \(\frac { 50 }{ 8 } \) = 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

Answer:

(v) -4 < x < 2