# SOLVING WORD PROBLEMS WITH INEQUALITIES

Many of the real-life word problems can be solved algebraically by translating the given information into an inequality and then solving the inequality.

Example 1 :

Alex Car Rental charges a flat fee of \$40.00 per day plus \$0.54 per mile to rent a car. Jack Car Rental charges a flat fee of \$65.00 per day plus \$0.36 per mile to rent a car. If a car is rented for three days, at least how many miles would you have to drive, to the nearest mile, to make the Jack Car Rental company the better option ?

Solution :

The problem asks for how many miles you would have to drive to make Jack Car Rental the better option.

Let x be the number of miles of driving which would make Jack Car Rental the better option.

Alex Car Rental's rental charge for 3 days

>

Jack Car Rental's rental charge for 3 days

40 ⋅ 3 + 0.54x  >  65 ⋅ 3 + 0.36x

Simplify.

120 + 0.54x  >  195 + 0.36x

Subtract 0.36x from each side.

120 + 0.18x  >  195

Subtract 120 from each side.

0.18x  >  75

Divide each side by 0.18.

x  >  416.67

You need to drive 417 miles or more to make Jack Car Rental the better option.

Example 2 :

A 38 inch long wire is cut into two pieces. The longer piece has to be at least 3 inches longer than twice the shorter piece. What is the maximum length of the shorter piece, to the nearest inch ?

Solution :

The problem asks for the maximum length of the shorter piece, to the nearest inch.

Let x be the length of the shorter piece.

Then, the length of the longer piece is (38 - x).

The longer piece is at least 3 inches longer than twice the shorter piece.

Then,

38 - x  ≥  2x + 3

38  ≥  3x + 3

Subtract 3 from each side.

35  ≥  3x

Divide each side by 3.

11.67  ≥  x

The maximum length of the shorter piece, to the nearest inch, is 11.

Example 3 :

Tom wants to rent a truck for two days and pay no more than \$300. Find the maximum distance (in miles) Tom can drive the truck if the truck rental cost \$49 a day plus \$0.40 a mile.

Solution :

Let x be the total number of miles travelled in 2 days and y be the total cost.

y = 0.40x + 2(49)

y = 0.40x + 98

Given : The total cost should be no more than \$300.

≤ 300

0.40x + 98 ≤ 300

0.4x ≤ 202

x ≤ 505

The maximum distance Tom can drive is 505 miles.

Example 4 :

Tim has 140 paperback and hard cover copies in his book shelf. If the hard cover copies do not exceed one sixth the number of paperback copies, find the minimum number of paperback copies in Tim’s book shelf.

Solution :

Let p be the number of paper back copies and h be the number hard cover copies.

Given : Tim has 140 paperback and hard cover copies in his book shelf.

p + h = 140

h = 140 - p

Given : The hard cover copies do not exceed one sixth the number of paperback copies.

()p

ᵖ⁄₆

6h ≤ p

p ≥ 6h

Substitute h = 140 - p.

p ≥ 6(140 - p)

p ≥ 840 - 6p

7p ≥ 840

p ≥ 120

The minimum number of paperback copies in Tim’s book shelf is 120.

Example 5 :

The number of students in a geometry class is four fifths the number of students in a Spanish class. The total number of students in both classes does not exceed 54. What is the greatest possible number of students in the Spanish class?

Solution :

Let g be the number of students in geometry class and s be the number of students in Spanish class.

Given : The number of students in geometry class is four fifths the number of students in Spanish class.

g = ()s

5g = 4s

Given : The total number of students in both classes does not exceed 54.

g + s ≤ 54

Multiply both sides by 5.

5(g + s) ≤ 5(54)

5g + 5s ≤ 270

Substitute 5g = 4s.

4s + 5s ≤ 270

9s ≤ 270

s ≤ 30

The greatest possible number of students in the Spanish class is 30.

Example 6 :

At a sporting goods store, Jay paid \$172 for a pair of shoes and a pair of pants. The pants cost less than two thirds of what the shoes cost. What is the minimum price of the shoes to the nearest dollar?

Solution :

Let s be the price of shoes and p be the price of pants.

Given : Jay paid \$172 for a pair of shoes and a pair of pants.

s + p = 172

p = 172 - s

Given : The pants cost less than two thirds of what the shoes cost.

p < ()s

3p < 2s

Substitute p = 172 - s.

3(172 - s) < 2s

516 - 3s < 2s

516 < 5s

103.2 < s

s > 103.2

The minimum price of the shoes to the nearest dollar is \$104.

Example 7 :

Ty earns \$14 an hour working on weekdays and \$21 an hour working on weekends. If he wants to make at least \$600 by working a total of 36 hours in a week, to the nearest hour, at least how many hours does he need to work on the weekends?

Solution :

x ---> number of hours Ty works on weekdays

y ---> number of hours Ty works on weekends

Given : Ty works a total of 36 hours in a week.

x + y = 36

x = 36 - y

Given : Ty wants to make at least \$600.

14x + 21y ≥ 600

Substitute x = 36 - y.

14(36 - y) + 21y ≥ 600

504 - 14y + 21y ≥ 600

504 + 7y ≥ 600

7y ≥ 96

y ≥ ⁹⁶⁄₇ ≈ 13.74

Ty needs to work at least 14 hours on the weekends.

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