SOLVING WORD PORBLEMS INVOLVING AGES

Problem 1 :

Jan has 6 more dimes than quarters. If the value of the dimes is the same as the value of the quarters, how many dimes and quarters does Jan have?

Solution :

Let x be the number of quarters.

number of dimes will be x + 6

value of dimes = 0.10(x + 6)

value of quarters = x (0.25)

0.25 (x) = 0.10(x + 6)

0.25x = 0.10x + 0.6

0.25x - 0.10x = 0.6

0.15x = 0.6

x = 0.6/0.15

x = 4 (number of quarters)

Number of dimes = x + 6 ==> 10

Problem 2 :

I have 7 more nickels than dimes. If the value of the nickels and dimes is the same, how many of each coin do I have?

Solution :

Let x be the number of dimes

x + 7 be the number of nickels

1 dime = 0.10 and 1 nickel = 0.05

x (0.10) = 0.05 (x + 7)

0.10x = 0.05x + 0.35

0.10x - 0.05x = 0.35

0.05 x = 0.35

x = 0.35/0.05

x = 7 (number of dimes)

Number of nickels = x + 7 ==> 14

Problem 3 :

Alex broke open his piggy bank. He had 30 more dimes than nickels and 8 more quarters than nickels. The value of the quarters is exactly the same as the value of the sum of the dimes and nickels. How many of each coin does Alex have? 

Solution :

Let x be the number of nickels 

Number of dimes = x + 30

Number of quarters = x + 8

Value of quarters = sum of value of dimes and nickels

0.25 (x + 8) = 0.05 x + (x + 30) 0.10

0.25x + 2 = 0.05x + 0.10x + 3

0.25x - 0.15x = 3 - 2

0.10 x = 1

x = 1/0.10

x = 10 (number of nickels)

Number of dimes = x + 30 ==> 40

Number of quarters = x + 8 ==> 18

Problem 4 :

Mr. Price deposited $170 in his bank. The number of $5 bills was 3 times the number of $10 bills, and the number of $1 bills was 30 more than the number of $5 bills. How many bills of each type did he deposit?

Solution :

Let x be the number of $10 bill

Number of $5 bill = 3 x

Number of $1 = 30 + 3x

Value of all bills = 170

10x + 5(3x) + 1(30 + 3x) = 170

10x + 15x + 3x = 170 - 30

28x = 140

x = 140/28

x = 5

Number of $5 bill = 3x ==> 15

Number of $1 bill = 30 + 3x = 45

Problem 5 :

Marie has $5.05 in quarters and dimes. The number of quarters exceeds twice the number of dimes by 1. Find the number she has of each kind.

Solution :

Let x be the number number of dimes

Number of quarters = 2x + 1

Total value = 5.05

0.10x + 0.25(2x + 1) = 5.05

0.10x + 0.5x + 0.25 = 5.05

0.6x = 5.05 - 0.25

0.6x = 4.8

x = 4.8/0.6

x = 8 (number of dimes)

Number of quarters = 2x + 1 ==> 17

Problem 6 :

A purse contains $4.70 in nickels and quarters. There are 30 coins in all. How many of each kind are there?

Solution :

Number of nickels be x, then number quarters be 30 - x

Value = 4.70

0.05x + 0.25(30 - x) = 4.70

0.05x + 7.5 - 0.25x = 4.70

-0.2x = 4.70 - 7.5

0.20x = 2.8

x = 2.8/0.20

x = 14 (number of nickels)

Number of quarters = 30 - x ==> 16

Problem 7 :

Brian wants to earn at least $20 this week to go to the fair. His father said he will pay him $9 for mowing the lawn and $2.75 an hour to weed the flower bed. Brian has decided to do both chores. What is the minimum number of hours he will need to weed to earn at least $20?

Solution :

He will get $9 for moving the lawn

Let x be the number of hours he is working to weed the flower bed.

9 + 2.75x ≥ 20

2.75x ≥ 20 - 9

2.75x ≥ 11

x  ≥ 11/2.75

x ≥ 4

So, minimum number of hours he has to work is 4.

Problem 8 :

An employee earns $2 for every magazine sold and a salary of $10 a week. How many magazines will the employee need to sell in order to earn at least $40 in one week?

Solution :

Let x be the number of magazines he sold.

Weekly salary = 10

10 + 2x ≥ 40

2x ≥ 40 - 10

2x ≥ 30

x ≥ 15

He has to sell minimum 15 magazines.

Problem 9 :

Juan spent $2.50 on apples and oranges. He bought 5 apples at $0.36 each. What is the most he spent on the oranges?

Solution :

Cost of each apple = 0.36 and number of apples he bought = 5

Cost of 5 apples = 5(0.36) ==> 1.8

Cost of apples and oranges = 2.50

Cost spent only for orange = 2.50 - 1.8

= 0.7

Amount spent for oranges = 0.7

Problem 10 :

The cashier in a movie box office sold 200 more adult admission tickets at $11.00 each than children’s admission tickets at $8.00 each. What is the minimum number of each type of ticket that the cashier had to sell for the total receipts to be at least $5000

Solution :

Let x be the number of children tickets sold.

x + 200 will be the number of adult tickets.

8 x + 11(x + 200) ≥ 5000

8x + 11x + 2200 ≥ 5000

19x ≥ 5000 - 2200

19x ≥ 3400

x ≥ 147.3

x = 148, x + 200 ==> 348 adult tickets.

So, minimum number of children ticket sold is 148 and 348 adult tickets.

Problem 11 :

You must have an average score of at least 80 to get a B- on your report card. You have scores of 61, 70, 99 and 70. What is the minimum score you must get on the last test to get a B-on your report card? 

Solution :

Average score must be at least 80

Average = sum of all quantities/total number of quantities

Let x be the 5th quantity.

(61 + 70 + 99 + 70 + x)/5 ≥ 80

300 + x ≥ 80(5)

300 + x ≥ 400

x ≥ 100

The minimum score is 100.