SOLVING TWO STEP INEQUALITY WORD PROBLEMS

Problem 1 :

Lee is y years old. Toby is 8 years younger than Lee. The sum of their ages is less than 41.

(a) Write down in terms of y, an inequality to show this information.

(b) Work out the oldest age that Lee can be. Give your answer as a whole number of years

Solution :

Age of Lee = y

Age of Toby = y - 8

Sum of their ages is less than 41

a) y + (y - 8) < 41

2y - 8 < 41

b)  Solving for y,

2y < 41 + 8

2y < 49

y < 49/2

y < 24.5

So, Lee will be 24 years old maximum.

Problem 2 :

Annie, Beth and Carly go shopping. Annie spend m pounds. Beth spend twice as much as Annie. Carly spend 5 pounds more than Annie. The total amount of money spent, in pounds, is more than £60.

(a) Write down, in terms of m, an inequality to show this information. Each girl spends an whole number of pounds.

(b) Work out the least each girl could have spent.

Solution :

Amount that Annie spends = m

Amount that Beth spends = 2m

Carly spends = m + 5

Total amount spent = more than 60

m + 2m + m + 5 > 60

4m + 5 > 60

4m > 60 - 5

4m > 55

m > 55/4

m > 13.75

Annie can spent more than $13.75

Beth will spend more than 2(13.75) ==> $27.5

Carly will spend more than 13.75 + 5 ==> $18.75

Problem 3 :

Lauren goes shopping and has £50 to spend. She bought a T-shirt and 3 pairs of leggings. The T-shirt cost £23. Each pair of leggings cost £x

(a) Form an inequality in terms of x.

(b) Solve the inequality to find the possible price of the leggings.

Solution :

Cost of T-shirt = 23

Cost of pair of leggings = £x

Amount they can spend = £50

23 + 3x ≤ 50

Solving for x,

3x ≤ 50 - 23

3x ≤ 27

x ≤ 27/3

x ≤ 9

So, the maximum price of each pair of legging will be $9.

Problem 4 :

Farmer Taylor is placing a fence around his field. He has 300 meters of fencing but this is not enough.

(a) Form an inequality in terms of x.

(b) Solve the inequality to find the possible width of the field.

Solution :

He has to more than 300 meters.

Perimeter of the field > 300

2x + 5 + x + 2x + 5 + x > 300

6x + 10 > 300

6x > 300 - 10

6x > 290

x > 290/6

x > 48.33

The measure of width will be more than 48.33 m.

Problem 5 :

The perimeter of the regular pentagon is larger than the perimeter of the equilateral triangle.

(a) Form an inequality in terms of x

(b) Solve the inequality to find the possible range of values of x.

Solution :

Pentagon will have 5 sides, triangle will have 3 sides.

5(x + 2) > 3(x + 6)

5x + 10 > 3x + 18

5x - 3x > 18 - 10

2x > 8

x > 8/2

x > 4

So, the value of x will be greater than 4.

Problem 6 :

Which statement can be modeled by x + 3 ≤12?

A)   Sam has 3 bottles of water. Together, Sam and Dave have at most 12 bottles of water.

B)   Jennie sold 3 cookbooks. To earn a prize, Jennie must sell at least 12 cookbooks.

C)   Peter has 3 baseball hats. Peter and his brothers have fewer than 12 baseball hats.

D)   Kathy swam 3 laps in the pool this week. She must swim more than 12 laps.

Solution :

x + 3 ≤12

The maximum value is 12.

So, option A is correct.

Problem 7 :

Julia has $80. She wants to purchase a nail set for $16 and earrings. She spends the rest of the money on earrings. Each pair of earrings costs $8. Write an inequality for the number of pairs of earrings she can purchase and solve.

Solution :

Amount Julia has = $80

She can spend maximum $80 or less than 80.

Inequality sign to be used = ≤

Cost of nail set = $16

Number of pair of earrings = x

Cost of each pair of earring = $8

16 + 8x ≤ 80

8x ≤ 80 - 16

8x ≤ 64

x ≤ 64/8

x ≤ 8

So, the number of pairs of earrings can be purchased is 8.

Problem 8 :

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

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