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 = 38 28 - 0.75(38) < 7.75 28 - 28.5 < 7.75 It is not possible. |
When x = 37 28 - 0.75(37) < 7.75 28 - 27.75 < 7.75 It 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
x ≤ 30/0.60
x ≤ 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.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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