# FINDING RANGE OF VALUES INEQUALITY PROBLEMS

Problem 1 :

Find the range of values of x that satisfies both

3(x + 2) ≤ 30 and 4x + 3 > 21

Solution :

Solving 3(x + 2) ≤ 30

Dividing by 3 on both sides

x + 2 ≤ 30/3

x + 2 ≤ 10

Subtracting 2 on both sides

x ≤ 8

Solving 4x + 3 > 21

4x > 21 - 3

4x > 18

x > 18/4

x > 4.5

Combing the values of x, we get

(4.5, 8]

Problem 2 :

Liam brought \$28 to the arcade and played games that cost 75 cents each. Laney brought \$12 to the arcade and spent \$4.25 on a slice of pizza and a Coke. How many games can Liam play so that he leaves with less money than Laney?

Solution :

Let x be the number of games that Liam played.

Amount spent on games = 0.75x

Amount she has remaining = (28 - 0.75 x)

Amount spent for pizza and coke = \$4.25

Amount that Laney has remaining = 12 - 4.25

The amount that Liam has to be less than Laney

28 - 0.75x < 12 - 4.25

28 - 0.75x < 7.75

-0.75x < 7.75 - 28

- 0.75x < -20.25

Dividing by - on both sides

x > 20.25/0.75

x > 27

 When x = 3828 - 0.75(38) < 7.7528 - 28.5 < 7.75It is not possible. When x = 3728 - 0.75(37) < 7.7528 - 27.75 < 7.75It is possible.

So, Liam can play more than 27 games and maximum 37 games.

Problem 3 :

Christina goes to the market with \$50. She buys some papayas for \$20 and spends the rest of the money on bananas. Each banana cost \$0.60. Write an inequality for the number of bananas she can purchase and solve.

Solution :

Amount Christina has = \$50

Cost of papayas = \$20

Cost of each banana = \$0.60

Number of bananas she is purchasing = x

20 + 0.60x  50

Solving for x,

0.60x  50 - 20

0.60x  30

30/0.60

50

So, he can buy 50 or lesser than 50 bananas.

Problem 4 :

Billy goes to the store. He has \$90. He wants to purchase a leather jacket for \$45, a hat for \$10, and the rest on jeans. Each pair of jeans costs \$35. Write an inequality for the number of jeans he can purchase and solve.

Solution :

Amount he has = \$90

Cost of leather jacket = \$45

Cost of hat = \$10

Let x be the number of pair of jeans.

Cost of pair of jeans = \$35

45 + 10 + 35x 90

Solving for x,

55 + 35x 90

35x 90 - 55

35x ≤ 35

x 35/35

x 1

So, he can buy maximum one pair of jeans.

Problem 5 :

Adrian works in New York City and makes \$42 per hour. She works in an office and must get her suit dry cleaned every day for \$75. If she wants to end up with more than \$260 a day, at least how many hours must she work?

Solution :

Let x be the number of hours she is working.

75 + 42x > 260

Solving for x,

42x > 260 - 75

42x > 185

x > 185/42

x > 4.40

She has to work more than 4 hours to earn more than 260.

Problem 6 :

Erin has \$50. She wants to purchase a cell phone (\$20) and spend the rest on music CDs. Each music CD costs \$8. Write and solve an inequality for the number of music CDs she can purchase.

Solution :

Amount Erin has = \$50

Cost of cell phone = \$20

Cost of each CD = \$8

Let x be the number of CD's

20 + 8x < 50

Subtracting 20

8x < 50 - 20

8x < 30

x < 30/8

x < 3.75

It is not possible to purchase 3.75 CDs. So, she can buy 3 or less number of CDs. Then the range is [0, 3].

Problem 7 :

Jason is saving up to buy a digital camera that costs \$490. So far, he saved \$175. He would like to buy the camera 3 weeks from now. Write an inequality to represent at least how much he must save every week to have enough money to purchase the camera. Solve

Solution :

Let x be the amount to be saved on each week.

175 + 3x ≥ 490

Solving for x,

3x ≥ 490 - 175

3x ≥ 315

x ≥ 315/3

x ≥ 105

So, the least amount to be saved for each week is 105.

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