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.
0 ≤ 210-35x < 140
Subtract 210 everywhere.
0-210) ≤ 210-35x < 140
-210 ≤ -35x < 70
Divide by -35 everywhere.
6 ≤ -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
Additional charge = $3
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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