# WORD PROBLEMS ON LINEAR EQUATIONS AND INEQUALITIES

Problem 1 :

Polly has \$210 in her checking account. After writing a check for tickets to a concert, she has less than \$140 in her account but she is not overdrawn. If each ticket cost \$35, how many tickets could she have bought?

Solution :

Let x be the number of ticket he buys.

Since he is not overdrawn, after he spent amount for concert, he will have the balance more than 0.

210-35x < 140

Subtract 210 everywhere.

0-210)  210-35x < 140

-210  -35x < 70

Divide by -35 everywhere.

-x < 2

Divide by -1, we get

2 < x ≤ 6

The possible number of tickets are 3, 4, 5 and 6.

Problem 2 :

Find all positive integers that are solutions of the inequality 4n+7 < 27.

Solution :

Step 1:

Subtract 7 from both sides.

4n+7-7 < 27-7

4n < 20

Step 2:

Divide both sides by 4.

4n/4 < 20/4

n < 5

n = 1, 2, 3, 4

Problem 3 :

peter had 156 cents in coins. After he bought 3 packs of gum he had no more than 9 cents left. What is the minimum cost of gum?

Solution :

No more than, so ≤.

156-3g ≤ 9

Subtract 9 from both sides.

156-3g-9 ≤ 9-9

147-3g ≤ 0

-3g ≤ -147

Divide by -3 on both sides.

-3g/-3 ≤ -147/-3

g ≤ 49

minimum cost of gum is 49 cents.

Problem 4 :

In an algebra class, 3 students are working on a special project and the remaining students are working in groups of five. If there are 18 students in class, how many groups of five are there?

Solution :

3+5g = 18

5g = 18-3

5g = 15

g = 15/5

g = 3

3 groups of five are there.

Problem 5 :

Andy paid a reservation fee of \$8 plus \$12 a night to board her cat while she was on vacation. If Andy paid \$80 to board her cat, how many nights was Andy on vacation?

Solution :

8+12n = 80

12n =80-8

12n = 72

n = 72/12

n = 6

6 nights was Andy on vacation.

Problem 6 :

At a parking garage, parking costs \$5 for the first hour and \$3 for each additional hour or part of an hour. Mr. Kanesha paid \$44 for parking on Monday. For how many hours did Mr. Kanesha park his car?

Solution :

Parking cost for the first hour = \$5

5+3h = 44

3h = 44-5

3h = 39

h = 39/3

h = 13

13 hours Mr. Kanesha park his car.

Problem 7 :

Kim wants to buy an azalea plant for \$19 and some delphinium plants for \$5 each. She wants to spend less than \$49 for the plants. At most how many delphinium plants can she buy?

Solution :

19+5d < 49

Subtract 19 from both sides.

19+5d-19 < 49-19

5d < 30

d < 30/5

d < 6

At most 5 delphinium plants can she buy.

Problem 8 :

To prepare for a tennis match and have enough time for schoolwork. Priscilla can practice no more than 14 hours. If she practices the same length of time Monday through Friday, and then spends 4 hours on Saturday, what is the most she can practice on Wednesday?

Solution :

Let number of hours be h. Accordingly the question, it is no more than 14 hour. So,  ≤ 14h

5d+4h ≤ 14h

Subtract 4 from both sides.

5d+4h-4 ≤ 14h-4

5d ≤ 10

d ≤ 10/5

d ≤ 2

Only 2 hours she can practice on Wednesday.

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