Test: Linear Inequalities One Variable - JEE MCQ

Let x be the smaller of the two consecutive even positive integers, then the other even integer is x + 2. 
 Given x < 10  and x + (x + 2) > 11.  
⇒ x < 10, and 2x + 2 > 11.  
⇒ x < 10, 2x > 9 
⇒ x < 10, x > 9/2 
⇒  x > 9/2   and    x<10 
∴ the required parity even integers is (6, 8)

Quadratic inequalities can be of the following forms:
ax² + bx + c > 0
ax² + bx + c ≥ 0
ax² + bx + c < 0
ax² + bx + c ≤ 0

Let marks required be x 
= (45 + 62 + x)/3 > 60
45+62+ x > 60*3
107 + x > 180 
x > 180 - 107
x > 73

So, 73 is the minimum score required in the third attempt to qualify.

12(x) >30

=> x>30/12 

=> x>2.5

=>x=3 

The given inequality is 5x– 3 < 7
=> 5x – 3 + 3 < 7 + 3                             [3 is added both sides]
=> 5x < 10
=> x < 10/5
=> x < 2
When x is a real number, the solutions of the given inequality are given by x < 2, i.e., all real numbers x which are less than 2.

-5x + 2 < 7x - 4
2 + 4 < 7x + 5x
6 < 12x
x > 1/2

We have 5x−3<3x+1
⇒ 5x − 3 + 3 < 3x + 1 + 3
⇒ 5x < 3x + 4
⇒ 5x − 3 × < 3x + 4 − 3x
⇒ 2x < 4 ⇒ x < 2
 When x is an integer the solutions of the given inequality are {.............,−4,−3,−2,−1,0,1}

To solve the inequality 3(a - 6) < 4 + a, we need to simplify and isolate the variable a on one side of the inequality. Here are the steps:
1. Distribute the 3 on the left side of the inequality: 3a - 18 < 4 + a
2. Subtract a from both sides of the inequality: 3a - a - 18 < 4
3. Combine like terms on the left side of the inequality: 2a - 18 < 4
4. Add 18 to both sides of the inequality: 2a < 22
5. Divide both sides of the inequality by 2: a < 11
Therefore, the solution to the inequality 3(a - 6) < 4 + a is a < 11. This means that a is less than 11. Therefore, the correct answer to the question is D: a < 11
 

3x - 7> x + 3
2 x > 10
x > 5
so x- coordinate should be > 5

5x + 6 < 2x - 3
5x - 2x < - 3- 6
⇒ 3x < -9
x < -3

As we move to right side on the number line the value increases. i.e. To the right of the point (-3,0)

4x + 3 < 5x + 7
subtract 4 both sides,
4x + 3 - 3 < 5x + 7 - 3
⇒ 4x < 5x + 4
subtract ' 5x ' both sides ,
[ equal number may be subtracted from both sides of an inequality without affecting the sign of inequality]
4x - 5x < 5x + 4 - 5
-x < 4
now, multiple with (-1) then, sign of inequality change .
-x.(-1) > 4(-1)
x > -4
hence, x€ ( -4 , ∞)

Two less than 5 times a number is greater than the third multiple of the number.
5x - 2 >3x
5x - 3x > 2
2x > 2
x > 1

To solve the inequality 6x - 7 > 5, we need to isolate x on one side of the inequality sign. Here are the steps:
1. Add 7 to both sides of the inequality:
6x - 7 + 7 > 5 + 7
6x > 12
2. Divide both sides of the inequality by 6:
6x/6 > 12/6
x > 2
Therefore, the solution to the inequality 6x - 7 > 5 is x > 2. This means that x is greater than 2. Therefore, the correct answer to the question is D: x > 2