ONE VARAIBLE INEQUATLITY WORD PROBLEMS

Problem 1 :

Tamara has a cell phone that charges $0.07 per minute plus a monthly fee of $19.00. She budgets $29.50 per month for total cell phone expenses without taxes. What is the maximum number of minutes Tamara could use her phone each month in order to stay within her budget?

Solution : 

Monthly fee = $19

Cost per minute = $0.07

Her budget = $29.50

Let x be the number of minutes.

19 + 0.07x ≤ 29.50

0.07x ≤ 29.50 - 19

0.07x ≤ 10.50

Dividing by 0.07 on both sides, we get 

x ≤ 10.50 / 0.07

x ≤ 150

So, the maximum number of minutes is 150.

Problem 2 :

An online music club has a one-time registration fee of $13.95 and charges $0.49 to buy each song. If Emma has $50 to join the club and buy songs, what is the maximum number of songs she can buy?

Solution :

One time registration fee = $13.95

Charge for each song = $0.49

Let x be the number of songs.

Maximum amount spent = 50

13.95 + 0.49 x ≤ 50

0.49x ≤ 50 - 13.95

0.49 x ≤ 36.05

x ≤ 36.05/0.49

x ≤ 73.5

So, the maximum number of songs is 73.

Problem 3 :

Peter begins his kindergarten year able to spell 10 words. He is going to learn to spell 2 new words every day.

Write an inequality that can be used to determine how many days, d, it takes Peter to be able to spell at least 75 words. Use this inequality to determine the minimum number of whole days it will take for him to be able to spell at least 75 words.

Solution :

10 + 2d ≥ 75

Solving for d,

2d  ≥ 75 - 10

2d ≥ 65

d  ≥ 65/2

d ≥ 32.5

So, the maximum number of days is 32.

Problem 4 :

Chelsea has $45 to spend at the fair. She spends $20 on admission and $15 on snacks. She wants to play a game that costs $0.65 per game. Write an inequality to find the maximum number of times, x, Chelsea can play the game. Using this inequality, determine the maximum number of times she can play the game.

Solution :

x be the number of times she can play the game.

(20 + 15) + 0.65 x ≤ 45

35 + 0.65 x ≤ 45

0.65x ≤ 45 - 35

0.65x ≤10

x ≤ 10/0.65

x ≤ 15.3

So, 15 times she can play the games.

Problem 5 :

Melissa wants to spend no more than $300 on school clothes. She spends $75 on a coat and then wants to buy some sweaters that are on special for $10 each. Solve the inequality 75 + 10s ≤ 300 to find the greatest number of sweaters s she can buy.

Solution :

75 + 10s ≤ 300

Subtracting 75 on both sides.

10s ≤ 300 - 75

10s ≤ 225

s ≤ 225/10

s ≤ 22.5

So, the maximum number of sweaters is 22.

Problem 6 :

A small airplane can carry less than 1,050 pounds of luggage and mail. The mail for the day weighs 490 pounds. If each passenger brings 70 pounds of luggage, what is the greatest possible number of passengers that can be taken?

Solution :

Let x be the number of passengers

490 + 70x ≤ 1050

Solving for x.

70x ≤ 1050 - 490

70x ≤ 560

x ≤ 560/70

x ≤ 8

Problem 7 :

You rent a car and are offered 2 payment options. You can pay $25 a day plus 15 cents a mile or you can pay $10 a day plus 40 cents a mile. For what amount of daily miles will Option A be the cheaper plan? Write and solve an inequality to determine the answer.

Solution :

Option A < Option B

25 + 0.15m < 10 + 0.40 m

25 - 10 + 0.15m < 0.40m

15 < 0.40m - 0.15m

15 < 0.25m

15/0.25 < m

60 < m

Option A will be cheaper for less than 60 miles.

Problem 8 :

Keith and Michelle went out to dinner. The total amount they had to spend on the cost of the meal, including the tip, was to $53.70. If the combined tip came out to $9.60, and each friend spent an equal amount, how much could each friend pay, not including the tip?

Solution :

Number of friends come out = x

2x + 9.60  ≥ 53.70

2x  ≥ 53.70 - 9.60

2x ≥ 44.1

x ≥ 44.1/2

x  ≥ 22.05

Each friend pays 22.05