SOLVING WORD PROBLEMS INVOLVING LINEAR INEQUALITIES

Problem 1 :

The Yellow Taxi Cab Co. charges a $2.75 flat rate and $.65 for each additional mile. Emma has no more than $14 to spend on a ride. How many miles can Emma ride without exceeding her spending limit ?

Solution :

Let m be number of miles driven.

Cost spent for each additional mile = $0.65

Emma doesn't want to spend more than $14.

Rent of taxi = 2.75 + 0.65m

2.75 + 0.65m ≤ 14

Solving it, we get

Subtracting 2.75 on both sides.

0.65m ≤ 14 - 2.75

0.65m ≤ 11.25

m ≤ 11.25/0.65

m ≤ 17.31

So, without exceeding the spending limit she can travel 17 miles.

Problem 2 :

Tom wants to rent a truck for two days and pay no more than $300. How far can he drive the truck if the truck rental cost $49 a day plus $0.40 a mile?

A) 490     B) 505     C) 520       D) 535

Solution :

Maximum amount spent by Tom = $300.

Let m be number of miles driven.

Rent costs per day = 49

Tom wants to rent for 2 days = 2(49) = 98

Rental cost = 98 + 0.40m

98 + 0.40m ≤ 300

Subtracting 49, 

0.40m ≤ 300 - 98

0.40m ≤ 202

m ≤ 202/0.40

m ≤ 505

So, option B is correct.

Problem 3 :

Charlie has $500 in a savings account at the beginning of the summer, but withdraws $25 each week to spend on meals at Duchess (his favorite). He wants to have at least $200 in the account at the end of the summer for the start of school. How many weeks can Charlie withdraw money from the account?

Solution :

The amount that he has initially = $500

Each week he is withdrawing $25.

Let x be the number of weeks she is withdrawing the same amount.

He needs to have at least $200.

500 - 25x ≥ 200

-25x ≥ 200 - 500

-25x ≥ -300

x ≤ 300/25

x ≤ 12

Charlie can withdraw the amount for 12 months.

Problem 4 :

The length of a rectangle is 4 cm longer than the width. The perimeter is no more than 28 cm. What are the maximum possible dimensions for the rectangle?

Solution :

Let x be the width.

length = x + 4

Perimeter is no more than 28

2(x + x + 4) ≤ 28

2x + 4 ≤ 14

2x ≤ 10

x ≤ 5

Maximum value of x is 5.

Width = 5

Length = 5 + 4 => 9

So, the required dimension is 5 cm and 9 cm.

Problem 5 :

There are three exams given in a marking period. Ryan received an 85 and a 91 on the first two exams. What grade must he earn on the last exam in order to get an average of no less than a 90 for the marking period?

Solution :

Marks received in two exams are 85 and 91.

Let x be his score in the last exam. So, the average should be less than 90.

(85 + 91 + x)/3 ≥ 90

176 + x ≥ 270

x ≥ 270 - 176

x ≥ 94

So, at least he has to get 94.

Problem 6 :

The length of a rectangle is 5 cm less than twice its width. If the perimeter of the rectangle is no more than 70 cm, what are the maximum possible dimensions?

Solution :

Let x be the width.

length = 2x - 5

Perimeter of the rectangle ≤ 70 cm

2(x + 2x - 5) ≤ 70

2(3x - 5) ≤ 70

3x - 10 ≤ 35

3x ≤ 45

x ≤ 45/3

x ≤ 15

Maximum vaue for x = 15 (width)

length = 2(15) - 5 ==> 25 cm

So, the required dimension of the rectangle is 15 cm and 25 cm.

Problme 7 :

Find two consecutive integers such that 7 times the smaller is less than 6 times the greater. What are the greatest such integers?

Solution :

Let x and x + 1be two consecutive integers.

x - smaller and x + 1 - larger.

7x < 6(x + 1)

7x < 6x + 6

Subtracting 6x on both sides, we get

7x - 6x < 6

x < 6

So, the required consecutive numbers are 5 and 6.

Problem 8 : 

The sum of two consecutive even integers is at most 400. Find the pair of integers with the greatest sum.

Solution :

Let x be the even number, its consecutive even number be x + 2.

x + x + 2 ≤ 400

2x + 2 ≤ 400

2x ≤ 398

Dividing by 2 on both sides, we get

x ≤ 398/2

x ≤ 199

So, the required consecutive even numbers are 198 and 200.

Problem 9 :

The length of a rectangle is 4 cm more than three times its width. If the perimeter is more than 56 cm, what are the minimum possible dimensions of the rectangle (not allowing for fractional side measures)?

Solution :

Let x be the width of the rectangle.

length = 3x + 4

Perimeter is more than 56 cm.

2(x + 3x + 4) > 56

2(4x + 4) > 56

4x + 4 > 28

4x > 24

x > 6

Since we find minimum possible dimension, let us take x = 7.

length = 3(7) + 4

= 21 + 4

= 25 cm

So, the dimension is 7 cm and 25 cm.

Problem 10 :

The sum of three consecutive odd integers is no less than 51. What is the middle integer?

Solution :

Let x be the odd numbers. Its consecutive odd numbers be x + 2 and x + 4.

x + x + 2 + x + 4 ≥ 51

3x + 6  ≥ 51

3x ≥ 51 - 6

3x ≥ 45

x ≥ 15

x = 15, x + 2 = 17 and x + 4 = 19

Som the required middle number is 17.