The entire article deals with the Problems on Law of Inequality. All the concepts related to the Law of Inequality are given in previous articles. Now, you can see different practice questions, solved problems on law of inequality n this article. Students who are lagging in Inequality topics can read our articles and practice well for the exam. Various problems and different methods will let you understand the concept in deep. So, without waiting a minute, begin your practice by referring to our 10th grade math concepts.

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## Solving Law of Inequality Questions

Check out all the below problems. The solution and explanation are hidden below every question. First, practice every question on your own and then go through the given explanation and answers to cross-verify your answers.

Question 1. Mark the statement true or false. Justify your answer.

(i) If 2x + 12 > 30 then 2x – 12 > 6

(ii) If 8m > – 48 then – 2m > 12.

**Solution:**

(i) Given equation is 2x + 12 > 30 where x is the variable.

2x + 12 > 30

Now, subtract 24 on both sides of the above equation.

2x + 12 – 24 > 30 – 24

2x – 12 > 6

Therefore, the given statement if 2x + 12 > 30 then 2x – 12 > 6 is true.

(ii) Given equation is 8m > – 48 where m is the variable.

8m > – 48

Now, divide the above equation with -4 into both sides. The inequality reverses on multiplying or dividing both sides by -1.

8m/(-4) > – 48/-4

-2m < 12

Therefore, the given statement if 8m > – 48 then – 2m > 12 is false.

Question 2. If 6k + 8 < 32 and k ∈ N then find k.

**Solution:**

(i) Given inequation is 6k + 8 < 32 and k ∈ N.

6k + 8 < 32

Now, subtract 8 on both sides of the above equation.

6k + 8 – 8 < 32 – 8

6k < 24

Now, divide the above equation with 6 into both sides.

6k/6 < 24/6

k < 4

The value of k is 4 or less than 4 or k < 4.

But given that k ∈ N, k = 1, 2, 3.

Question 3. If (x – 1) (6 – x) > 0 and x ∈ N then find x.

**Solution:**

(i) Given inequation is (x – 1) (6 – x) > 0 and x ∈ N.

We know that xy > 0 then x > 0, y > 0 or x < 0, y < 0

(x – 1) (6 – x) > 0

x – 1 > 0 and 6 – x > 0 ………………….. (i)

or, x – 1 < 0 and 6 – x < 0………………….. (ii)

Now, solve the equation

Now, subtract 8 on both sides of the above equation.

6k + 8 – 8 < 32 – 8

6k < 24

Now, divide the above equation with 6 into both sides.

6k/6 < 24/6

k < 4

The value of k is 4 or less than 4 or k < 4.

But given that k ∈ N, k = 1, 2, 3.

Question 4. Solve the below linear inequality 6m − 10 ≤ 6 − 2m with addition.

**Solution:**

Given inequation is 6m − 10 ≤ 6 − 2m where m is a variable.

6m − 10 ≤ 6 − 2m

Now, add 10 on both sides of the above equation.

6m − 10 + 10 ≤ 6 − 2m + 10

6m ≤ 16 − 2m

Move 2m to the left side of the above equation.

6m + 2m ≤ 16

8m ≤ 16

Now, divide the above equation with 8 into both sides.

8m/8 ≤ 16/8

m ≤ 2.

The value of m is 2 or less than 2.

Therefore, the final answer is m ≤ 2.

Question 5. Calculate the range of values of x, which satisfies the inequality: 2x − 8 < 4x + 10.

**Solution:**

Given inequation is 2x − 8 < 4x + 10 where x is a variable.

2x − 8 < 4x + 10

Now, add 8 on both sides of the above equation.

2x − 8 + 8 < 4x + 10 + 8

2x < 4x + 18

Now, subtract 4x on both sides of the above equation.

2x – 4x < 4x – 4x + 18

-2x < 18

Now, divide the above equation with -2 into both sides. The inequality reverses on multiplying or dividing both sides by -1.

-2x/-2 < 18/-2

x > -9

The value of x is greater than -9.

Therefore, the final answer is x > -9.

Question 6. Solve the below linear inequalities with subtraction.

(i) Solve 2y + 16 > 10.

(ii) Solve 10y + 20 > 6y + 48.

**Solution:**

(i) Given inequation is 2y + 16 > 10 where y is a variable.

2y + 16 > 10

Now, subtract 16 on both sides of the above equation.

2y + 16 – 16 > 10 – 16

2y > -6

Now, divide the above equation with 2 into both sides.

2y/2 > -6/2

y > -3

The value of y is greater than -3.

Therefore, the final answer is y > -3.

(ii) Given inequation is 10y + 20 > 6y + 48 where y is a variable.

10y + 20 > 6y + 48

Now, subtract 20 on both sides of the above equation.

10y + 20 – 20 > 6y + 48 – 20

10y > 6y + 28

Now, subtract 2y on both sides of the above equation.

10y – 6y > 6y – 6y + 28

4y > 28

Now, divide the above equation with 4 into both sides.

4y/4 > 28/4

y > 7

The value of y is greater than 7.

Therefore, the final answer is y > 7.

Question 7. A student scored 40 marks in the first test and 25 marks in the second test of the terminal examination. How many minimum marks should the student score in the third test get a mean of at least 42 marks?

**Solution:**

Given that a student scored 40 marks in the first test and 25 marks in the second test of the terminal examination.

Let us take the marks scored in the third test be x marks.

(30 + 25 + x)/3 ≥ 42

(55 + x)/3 ≥ 42

Now, multiply 3 on both sides of the above equation.

(55 + x) ≥ 126

Now, subtract 55 on both sides of the above equation.

55 – 55 + x ≥ 126 – 55

x ≥ 71

Therefore, the student must score 71 marks to maintain a mean of at least 42 marks.

Question 8. Ram requires at least 1000 to hold his birthday party. If already he has saved 300 and 7 months are left to this date. What is the minimum amount he must save monthly?

**Solution:**

Given that Ram requires at least 1000 to hold his birthday party.

He already saved 300 and 7 months are left to this date.

Let the minimum amount saved monthly = x

300 + 7x ≥ 1000

Now, subtract 300 on both sides of the above equation.

300 – 300 + 7x ≥ 1000 – 300

7x ≥ 700

Now, divide the above equation with 7 into both sides.

7x/7 ≥ 700/7

x ≥ 100

Therefore, Ram should save $100 or more money.

Question 9. Find two consecutive odd numbers which are greater than 20 and have the sum of less than 60.

**Solution:**

Let the smaller odd number = x.

Therefore, the next number will be x + 2

x > 20 ………. greater than 20

x + (x + 2) < 60 ……sum is less 60

Solve the inequation x + (x + 2) < 60.

2x + 2 < 60

Now, subtract 2 on both sides of the equation.

2x + 2 – 2 < 60 – 2

2x < 58

Now, divide the above equation with 2 into both sides.

2x/2 < 58/2

x < 29.

By combining the two expressions.

20 < x < 29.

Therefore, the consecutive odd numbers are 21 and 23, 23 and 25, 25 and 27, 27 and 29.

Question 10. Draw the Number Line for the given Inequalities.

(i) x ≥ 2.

(ii) –3 < x < 3

(iii) –2 ≤ x ≤ 3

**Solution:**

(i) Given that x ≥ 2.

The x value is either 2 or greater than 2.

Firstly, take the number line and place the point at 2. Then, draw a line that must present a right side of the 2.

The number line for the given inequality is

(ii) Given that –3 < x < 3.

The x value is in between -3 and 3.

Firstly, take the number line and place the points at -3 and 3. Place -3 on the left side of the number line and place the 3 on the right side of the number line. Then, draw a line between -3 and 3.

The number line for the given inequality is

(iii) Given that –2 ≤ x ≤ 3.

The x value is in between -2 and 3.

Firstly, take the number line and place the points at -2 and 3. Place -2 on the left side of the number line and place the 3 on the right side of the number line. Then, draw a line between including -2 and 3.

The number line for the given inequality is

Question 11. Choose the correct answer for the below linear inequality,

3x > 11

(i) x > 11/3

(ii) x < 11/3

(iii) x < 11

(iv) x > 11

**Solution:**

Given inequation is 3x > 11

Now, divide 3 into both sides of the above equation.

3x/3 > 11/3

x > 11/3.

Therefore, the correct answer is (i) x > 11/3.

Question 12. Choose the correct answer for the below linear inequality,

2x + 10 > 26

(i) x > 18

(ii) x < 18

(iii) x < 8

(iv) x > 8

**Solution:**

Given inequation is 2x + 10 > 26

Now, subtract 10 from both sides of the above equation.

2x + 10 – 10 > 26 – 10

2x > 16

Now, divide 2 into both sides of the above equation.

2x/2 > 16/2

x > 8.

Therefore, the correct answer is (iv) x > 8.