﻿ Inequality Worksheets | Problems & Solutions Inequality Worksheets

Inequality Worksheets
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1.
Are the two inequalities equivalent? - 7$x$ < 35 and - $x$ < 5 a. Yes b. No

Solution:

- 7x < 35
[First inequality.]

- x < 5
[Divide throughout by 7.]

- x < 5
[Second inequality.]

If two inequalities have the same solution set, then they are called equivalent inequalities.

So, the two inequalities are equal.

2.
Are the two inequalities equivalent? 3$x$ - 5 ≤ 13 and 3$x$ ≥ 18 a. Yes b. No

Solution:

3x - 5 ≤ 13
[First inequality.]

3x ≤ 18
[Add 5 to both sides of the equation.]

3x ≥ 18
[Second inequality.]

If the two inequalities have the same solution set, then they are called equivalent inequalities.

So, the given inequalities are not equal.

3.
Solution for the inequality is given in set-builder notation. Solve and choose the correct one for the inequality given.
7 [2$x$ - ($x$ + 7)] < 3 (2$x$ - 3) a. {$x$: $x$ < - 40} b. {$x$: $x$ < 58} c. {$x$: $x$ < 40} d. {$x$: $x$ < - 58}

Solution:

7 [2x - (x + 7)] < 3(2x - 3)

14x - 7(x + 7) < 3(2x - 3)
[Distributive property.]

14x - 7x - 49 < 6x - 9
[Distributive property.]

7x - 49 < 6x - 9
[Group the like terms.]

x - 49 < - 9
[Subtracting 6x from the two sides of the equation.]

x < 40
[Add 49 to both sides of the equation.]

The solution set is {x: x < 40}

4.
Solution for the inequality is given in set-builder notation. Solve and choose the correct one for the inequality given.
3[2$y$ + (3$y$ - 1)] ≥ 5(2$y$ + 1) a. {$y$: $y$ ≤ 1.6} b. {$y$: $y$ ≥ 2.6} c. {$y$: $y$ ≥ 1.6} d. {$y$: $y$ ≤ - 1.6}

Solution:

3 [2y + (3y - 1)] ≥ 5 (2y + 1)

6y + 3 (3y - 1) ≥ 5 (2y + 1)
[Distributive property.]

6y + 9y - 3 ≥ 10y + 5
[Distributive property.]

15y - 3 ≥ 10y + 5

5y - 3 ≥ 5
[Subtracting 10y from the two sides of the equation.]

5y ≥ 8
[Add 3 to both sides of the equation.]

y8 / 5
[Divide throughout by 5 .]

y ≥ 1.6

The solution set is {y: y ≥ 1.6}

5.
The difference of the measures of any two sides of a triangle is always less than the measure of the third side. In a triangle ABC, BC = 6 and AC = 4 + AB. Then measure of AB will be: a. Less than 1 b. Greater than 1 c. Equal to 1

Solution:

The difference of the measures of any two sides of a triangle is always less than the measure of the third side.

So, BC - AC < AB

6 - (4 + AB) < AB
[Substitute the values.]

2 - AB < AB

2 < 2AB
[Add AB to both sides of the equation.]

1 < AB
[Divide throughout by 2.]

So, AB > 1

6.
Katie can spend at most $788.10 on a television, which includes 6.5% sales tax. Find the maximum price of the television which she can buy. a.$788.10 b. $106.50 c.$704 d. $740 Solution: Let x be the maximum price of the television. So, sales tax = 6.5x100 [Sales Tax = 6.5%.] She can afford a maximum of$788.10.

x + 6.5x100 ≤ 788.10
[Amount she has to spend = Price of the television + sales tax.]

100x + 6.5x100 ≤ 788.10

106.5x100 ≤ 788.10
[Simplify.]

x788.10 ×100106.5

x ≤ $740 So, the maximum price of the television Katie can afford to buy is$740.

7.
Solve for $x$.
5$\mathrm{ax}$ + 8$b$ < 2$\mathrm{ax}$ - 6$b$, $a$ < 0 a. {$x$: $x$ > - $\frac{14b}{3a}$ } b. {$x$: $x$ < - $\frac{14b}{3a}$ } c. {$x$: $x$ < $\frac{14b}{3a}$ } d. {$x$: $x$ > $\frac{14b}{3a}$ }

Solution:

5ax + 8b < 2ax - 6b, a < 0

3ax + 8b < - 6b
[Subtracting 2ax from the two sides of the equation.]

3ax < -14b
[Subtracting 8b from the two sides of the equation.]

Given a < 0, hence it is a negative value. The inequality sign will change when dividing by 3a.

x > - 14b3a

So, the solution set is {x: x > - 14b3a }

8.
Determine whether $a$ - $b$ > 0 when $a$ > $b$. a. No b. Yes

Solution:

a - b > 0 when a > b is true. Because, the addition or subtraction of any real number does not affect the inequality.

a > b

a - b > 0
[Subtracting b from the two sides of the inequality.]

9.
Tim has $25. He wants to buy some pens and each pen costs 73 cents. Find the maximum number of pens that Tim can buy. a. 35 b. 32 c. 33 d. 34 Solution: Let x be the number of pens Tim can buy. Total cost of x pens =$73 / 100 x
[73 cents = $73 / 100.] The amount with Tim is only$25. So, the total price of the pens shall not exceed \$25.

73 / 100 x ≤ 25
[Express as a relation.]

73x ≤ 2500
[Multiply throughout by 100.]

x ≤ 34.24657

So, Tim can buy 34 pens.

10.
Gary has to take four Math tests to complete his seventh grade. On each of the tests he can score a maximum of 25. His scores on three tests are 20, 22 and 16. What should be his score in the fourth test, so that he gets a total score of at least 77 on all four tests? a. atmost 18 b. atmost 19 c. atleast 24 d. atleast 19

Solution:

Let n be the score of Gary in his fourth test.

n + 20 + 22 + 16 ≥ 77

n + 58 ≥ 77
[Simplify.]

n + 58 - 58 ≥ 77 - 58
[Subtracting 58 from the two sides of the equation.]

n ≥ 19
[Simplify.]

Gary should get at least 19 points in his fourth test.