Inequalities MCQ -

Last Updated on: 26th December 2024, 12:29 pm

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Inequalities MCQ

1. Inequalities are statements which shows an ___________ relationship between any two or more given quantities.

(a) direct

(b) unequal

(d) indirect.

Inequalities are statements which shows an inequal relationship between any two or more given quantities. So, (b) is the correct option.

2. An inequation of the from $\displaystyle \mathbf{ax}\text{ }+\mathbf{b}\ge \mathbf{0}$ is known as inequation in one _________________ .

(a) Constant

(b) Real number

(d) Dependent.

In the inequation $\displaystyle \mathbf{ax}\text{ }+\mathbf{b}\ge \mathbf{0}$ , b & a are constant number, x is the only variable. So, (c) is the correct option.

3. If x > y and z > 0 then

(a) x – z > y – z

(b) $\displaystyle \frac{x}{y}<\frac{y}{x}$

(d) none of the above

x > y and z > 0, So, z is a positive number. So, (x – z) > (y – z). So, (a) is the correct option.

4. If a >0 and b <0, it follows that:

(a) $\displaystyle ~\frac{1}{a}>\frac{1}{b}$

(b) $\displaystyle \frac{1}{a}<\frac{1}{b}$

(d) a = b

a>0 means a is a positive number, b<0 means b is negative number. So, $\displaystyle \frac{1}{a}>\frac{1}{b}$ . So, (a) is the correct option.

5. The linear relationship between two variables in an inequality:

(a) $\displaystyle {ax+\text{ }by\le c}$

(b) $\displaystyle {ax\text{ }by\le c}$

(d) $\displaystyle {ax+\text{ }by+\text{ }c\le 0}$

The linear relationship between two variables x & y, in an inequality is ax + by <= c. So, (a) is the correct option.

6. Solution of 3(x – 2) > 4x – 3 is

(a) x > 6

(b) x < -3

(d) x > 3

3(x – 2) > 4x – 3, Or, 3x – 6 > 4x – 3, Or, – 6 + 3 > 4x – 3x, Or, – 3 > x, Or, x < – 3. So, (b) is the correct option.

7. If x+ $\displaystyle \frac{1}{4}>\frac{7}{4}$ , then

(a) $\displaystyle x>-\frac{3}{2}$

(b) $\displaystyle x>\frac{3}{2}$

(d) None of these

$\displaystyle x+~\frac{1}{4}>\frac{7}{4},\text{ }so,\text{ }x>~\frac{7}{4}~-\frac{1}{4}.\text{ }Or\text{ }x>\frac{6}{4},\text{ }or\text{ }x>\frac{3}{2}.$
So, (b) is the correct option.

8. For the inequation $\displaystyle \frac{{3x-4}}{4}\le \frac{5}{{12}}$

(a) {x: $\displaystyle {\frac{{19}}{{18}}\le }$ x $\displaystyle {\le \frac{{29}}{{18}}}$ }

(b) {x: $\displaystyle {\frac{{7}}{{9}}\le }$ x $\displaystyle {\le \frac{{17}}{{9}}}$ }

(d) None of the these

So, (b) is the correct option.

9. Assuming, x, y and b are real numbers and x < y, b < 0, then:
(a) x/b < y/b
(b) x/b ≤ y/b
(c) x/b >y/b
(d) x/b ≥ y/b
x < y
Divide both sides of the inequality by “b”
x/b < y/b {since b < 0}
So, option (a) is correct

10. If – 8 ≤ 5x – 3 < 7, then
(a) – $\displaystyle \frac{1}{2}$ ≤ x ≤ 2
(b) 1 ≤ x < 2
(c) –1 ≤ x < 2
(d) –1 < x ≤ 2
– 8 ≤ 5x –3 < 7
Adding 3, we get
– 8 + 3 ≤ 5x – 3 + 3 < 7 + 3
Or, –5 ≤ 5x < 10
Dividing by 5, we get
–1 ≤ x < 2
So, option (c) is correct

11. If x is a whole number and 10x≤50 then find solution set of x.
a) {0,1,2,3,4,5}
b) {1,2,3,4,5}
c) {1,2,3,4}
d) {0,1,2,3,4}

10x≤50
Dividing by 10 on both sides, x ≤ ( $\displaystyle \frac{{50}}{{10}}$ ) => x≤5
Since x is a whole number so x = 0,1,2,3,4,5
So, option (a) is correct

12. If x-1>-x+7 then which is true?
a) x>4
b) x<4 c) x>2
d) x<2
Ans: a
x-1>-x+7
or, 2x>8, or x>4.
So, option (a) is correct

13. Ravi got 20 and 25 marks in first two tests. How much he must get in third test to secure qualifying avrage of 30 marks to pass
a) 40
b) 35
c) 55
d) 45

Let the minum marks to get in third test be x.
So, his average marks would be (20+25+x)/3
So, (20+25+x)/3 ≥30
or, 45+x≥90, or x≥90-45, or x≥45.
So, to qualify in test , Ravi must get Minimum 45 marks in 3rd test.
So, option (d) is correct

14. The longest side of a triangle is 2 times the shortest side and the third side is 4 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
a) 7
b) 9
c) 11
d) 13

Let shortest side be x. So, longest side = 2x, third side = 2x – 4.
Perimeter of triangle ≥ 61 cm (at least 61 cm)
or, (x)+(2x)+(2x-4) ≥ 61, or 5x≥65 = x≥13.
Minimum length of the shortest side is 13 cm.
So, option (d) is correct

15. The solution set of 14 + 2a > 4 where a is a negative integer is
a) { – 4, -3, – 2, – 1 }.
b) { – 5, -4, – 2, – 1 }.
c) { – 6, -4, – 2, -1 }.
d) None of the above
14 + 2a > 4, or, 14 + 2a – 14 > 4 –14 [subtract 14 from both sides]
Or, 2a > – 10 [simplifying the expression]. So, a > – 5 [Divide both sides by 2]
As ‘a’ is a negative integer, the solution set is { – 4, -3, – 2, – 1 }. [ because -4 is greater than -5, satisfying the condition, a belongs to the set of negative numbers for -5 onwards to -1)
So, option (a) is correct

16. The solution set of inequation $\displaystyle \frac{{(2a-1)}}{4}$ $\displaystyle \le$ 5, a $\displaystyle \in$ W (W is set of positive whole numbers, including zero)
a. { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11 }.
b. { 2, 3, 4, 5, 6, 7, 8, 9, 10,11,12 }.
c. { 3, 4, 5, 6, 7, 8, 9, 10,11,12,13 }.
d. { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }.
$\displaystyle \frac{{(2a-1)}}{4}\le 5$
Or, 2a – 1 $\displaystyle \le$ 20 [multiply both sides by 4]
Or, 2a – 1 + 1 $\displaystyle \le$ 20 + 1 [add 1 to both sides to eliminate constant values from left side]
Or, 2a $\displaystyle \le$ 21, or a $\displaystyle \le$ 10.5
As a $\displaystyle \in$ W, the solution set is { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }, whole number of value less than 10.5
So, option (d) is correct

17. Solve the inequality 4 – 2x $\displaystyle \ge$ x – 12 ,x $\displaystyle \in$ N (Natural Numbers)
a) {0, 1, 2, 3, 4, 5 }
b) { 1, 2, 3, 4, 5 }
c) { 2, 3, 4, 5,6 }
d) None of the above
4 – 2x $\displaystyle \ge$ x – 12
Or, 4 – 2x – 4 $\displaystyle \ge$ x – 12 – 4 [ Subtract 4 from both sides]
Or, : – 2x $\displaystyle \ge$ x – 16
Or, -3x $\displaystyle \ge$ – 16 [Subtract x from both sides]
Or, x< $\displaystyle \frac{{16}}{3}$ [Divide by – 3 and reverse the symbol]
As x $\displaystyle \in$ N, the solution set is { 1, 2, 3, 4, 5 } [ because 5 is less than $\displaystyle \frac{{16}}{3}$ , satisfying the condition, number elements 6 …onwards does not include in the set, as it is greater than $\displaystyle \frac{{16}}{3}$ ). Zero is not included in set of Natural numbers N.
So, option (b) is correct

18. The shaded region represents

Mathematical Inequalities graphical representation-dvidya.com.

(a) 2x + 3y > 6
(b) 2x + 3y $\displaystyle \ge$ 6
(c) 2x + 3y < 6
(d) 2x + 3y $\displaystyle \le$ 6
Shaded area touches line AB. Putting x = 0 & y = 0, we test, 2x + 3y > 6, We find, 2.0 + 3.0 > 6 . 0>6 is false. This not satisfied. We test 2x + 3y < 6, 0 $\displaystyle \le$ 6 is Satisfied. Shaded area is origin side and it touches line AB. So, 2x + 3y $\displaystyle \le$ 6 is the required inequation.

So, option (d) is correct

19. The shaded region represents

Mathematical Inequalities graphical representation-dvidya.com.

(a) 3x + 4y $\displaystyle \ge$ 12
(b) 3x – 4y > 12
(c) 3x + 4y $\displaystyle \le$ 12
(d) 3x + 4y < 12

Shaded area touches equation line. So, we test two equation with sign either $\displaystyle \ge$ or $\displaystyle \le$
For inequation 3x + 4y $\displaystyle \ge$ 12, Putting x = 0, y = 0, 0 + 0 $\displaystyle \ge$ 12 not satisfied
So, shaded area will be non origin side.
For inequation 3x + 4y $\displaystyle \le$ 12, 0 + 0 $\displaystyle \le$ 12 Satisfied. Shaded area will be origin side. So, required inequation is 3x + 4y $\displaystyle \ge$ 12
So, option (a) is correct

20. The shaded region represents

Mathematical Inequalities graphical representation-dvidya.com.

(a) 3y – 5x < 30
(b) 3y – 5x $\displaystyle \le$ 30
(c) 3y – 5x > 30
(d) 3y – 5x $\displaystyle \ge$ 30

Shaded area don’t touches equation line.
So, we test the equation with either > or < sign.
Option (a) 3y – 5x < 30
0 < 30
Satisfied Shaded area origin sided.
Option (c) 3y – 5x > 30
0 > 0
Not satisfied
So, shaded area non-origin side.
So, picture is for
equation 3y – 5x < 30
So, option (a) is correct

21. The graph of y <5 is

(a)

(b)

(c)

(d)

Replacing the inequality sing we get
equation y = 5
we draw the line for y = 5 which is parallel to x axis.
Now putting x = 0 & y = 0 in equation y < 5 we get 0 < 5 which
is true. Hence solution set for y < 5 is origin side.
If the equation was y $\displaystyle \le$ 5 option (a) will be answer.
As y < 5, option (c) will be answer.
So, option (c) is correct

22. The graph of x $\displaystyle \le$ 2y < 3 will be

(a)

(b)

(c)

(d) None of these

Equation x + 2y < 3
Putting x = 0, y = 0
0 + 2.0 $\displaystyle \le$ 3
Or, 0 $\displaystyle \le$ 3 Satisfied
Shaded area origin sided.
So, True option is (c)
So, option (c) is correct

23. The graph of 3x – 4y $\displaystyle \ge$ 0 is

(a)

(b)

(c)

(d)

Equation 3x – 4y $\displaystyle \ge$ 0
Putting x = 0, y = 0
3.0 – 4.0 $\displaystyle \ge$ 0
0 $\displaystyle \ge$ 0 Satisfied
Option (a) is right answer.
So, option (a) is correct

Inequalities MCQ

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