# Algebra Word Problem

Topics: Miles per hour, Inch, Equals sign Pages: 32 (7266 words) Published: March 4, 2012
ALGEBRA WORD PROBLEMS:
I. AGE PROBLEMS
1. Age Problems (Involving A Single Person)
Five years ago, John’s age was half of the age he will be in 8 years. How old is he now? Solution:
Step 1: Let x be John’s age now. Look at the question and put the relevant expressions above it.

Step 2: Write out the equation.

Isolate variable x

Answer: John is now 18 years old.
2. Age Problems (Involving More Than One Person)
John is twice as old as his friend Peter. Peter is 5 years older than Alice. In 5 years, John will be three times as old as Alice. How old is Peter now? Solution:
Step 1: Set up a table.
| age now | age in 5 yrs |
John|  |  |
Peter|  |  |
Alice|  |  |
Step 2: Fill in the table with information given in the question. John is twice as old as his friend Peter. Peter is 5 years older than Alice. In 5 years, John will be three times as old as Alice. How old is Peter now? Let x be Peter’s age now. Add 5 to get the ages in 5 yrs.  | age now | age in 5 yrs |

John| 2x | 2x + 5|
Peter| x | x + 5|
Alice| x – 5 | x – 5 + 5|
Write the new relationship in an equation using the ages in 5 yrs. In 5 years, John will be three times as old as Alice.
2x + 5 = 3(x – 5 + 5)
2x + 5 = 3x
Isolate variable x
x = 5
Answer: Peter is now 5 years old.
3. Age Problems
John’s father is 5 times older than John and John is twice as old as his sister Alice. In two years time, the sum of their ages will be 58. How old is John now? Solution:
Step 1: Set up a table.
| age now | age in 2 yrs |
John’s father|  |  |
John|  |  |
Alice|  |  |
Step 2: Fill in the table with information given in the question. John’s father is 5 times older than John and John is twice as old as his sister Alice. In two years time, the sum of their ages will be 58. How old is John now? Let x be John’s age now. Add 2 to get the ages in 2 yrs.

| age now | age in 2 yrs |
John’s father| 5x | 5x + 2|
John| x | x + 2|
Alice| | |
Write the new relationship in an equation using the ages in 2 yrs. In two years time, the sum of their ages will be 58.

Answer: John is now 8 years old.
4. Age Problems
Nona is one-third as old as her mother. Five years ago, she was only one-fifth of the age of her mother. How old is Nona now? "

It is easier to set this up with two unknowns say x and y and have 2 equations.

The 1st equation would be:  x = 1/3(y) or x = y/3   since Nona is currently 1/3 as old as her mother.  Now 5 years ago she was only 1/5th the mother’s age so the other equation would be.

x -5 = 1/5(y - 5)   multiply by 5..

5x - 25 = y - 5 so

y = 5x - 20 plug this back into the 1st equation and you get..

x = (5x - 20)/3 multiply by 3

3x = 5x - 20 so 2x = 20 and x = 10

So Nona is 10 years old currently and the mom is 30.   To check this go 5 years back. Nona would be 5 and the mother would be 25   5/25 = 1/5ths of her mother’s age. 5.Age Problems
This is the answer to the age algebra word problem that asked, "Bob is one third the age of his father. In 12 years he will be half the age of his father. How old is each now?"

Ok Bob being 1/3 the age of his father is the same thing as saying the father is 3 times as old as him.  It makes it easier to work with.  So let Bob be x and his father be 3x. And in 12 years bob will be half as old as his father.  So we have..

x + 12 = (3x + 12) / 2

2x + 24 = 3x + 12

x = 12

So Bob is 12 years old and his father is 3(12) or 36 years old now.  In 12 years Bob will be 12 + 12 or 24 years old and his father will be 36 + 12 or 48 years old which means Bob will be half as old so it checks out.

II. COIN PROBLEMS
1. Coin Problems
Jane bought a pencil and received change for \$3 in 20 coins, all nickels and quarters. How many of each kind are given? Solution:
Step 1: Set up a table with quantity and value.
| quantity | value | total |
nickels|  |  |  |
quarters|  |  |  |
together|...