In This Chapter
^ Making sense of problems involving comparisons of ages
^ Adding and subtracting the years to solve problems
^ Dealing with complications of real puzzlers in age problems
Ne of my favorite Funky Winkerbean cartoons starts out with something like, "When Henry was twice as old as Hank was two years after half of
Hank’s age. . . ." And the punch line: "How many apples is an orange worth?"
Some people think this is hilarious. Others don’t find a bit of humor in these
Very involved, contrived, convoluted problems.
In this chapter, I acquaint you with techniques and procedures to use when tackling word problems involving the ages of one, two, or more people. When broken down into their component parts, age problems seem much less intimidating. Here you see how to deal with aging — well, ages of people after a certain amount of time. I hope you’ll see the humor in the cartoon, too.
Doing Age Comparisons
Age comparisons are done almost as soon as a person can talk. Children take advantage of being older than one another to get first in line or get to stay up later or get whatever the prize. Doing age comparisons makes for interesting math problems, too. The usual rules apply: Always let the variable represent a number, and use the math words for clues about adding, subtracting, and so on.
Warming up to age
Before getting into some serious algebra word problems, here are a few warm-ups to get you in the mood.
The Problem: A 40-year-old man married a 25-year-old woman. The woman died at the age of 40, and her husband was so saddened that he wept for years after that. He died 5 years after he stopped weeping, on his 80th birthday. How many years was he a widower?
First, just do some mental calculations. The couple was married when he was 40 and she was 25; he was 15 years older than she was. If she died when she was 40, then he was still 15 years older than she was, so he was 55. It doesn’t matter how many years he wept and that for 5 years he had stopped weeping. The only figuring you need to do is to take the age when he died, 80, and subtract the age he was when his wife died, 55. Subtracting 80 – 55 = 25, so he was a widower for 25 years.

The next problem involves more counting and logic than algebra to figure out the ages.
The Problem: A man has three children. The oldest child is three times as old as the youngest. The second child is six years older than the youngest and six years younger than the oldest. How old are the children?
The second or middle child’s age is halfway between two numbers that are six years away in either direction. So the ages could be 1, 7, and 13; 2, 8, and 14; 3, 9, and 15; and so on. Go back to the beginning of the problem. The oldest child is three times as old as the youngest, so the age of the oldest has to be a multiple of 3. The multiple has to be larger than 13, so the oldest is 15, 18, 21, or some other multiple of three. Also, the oldest is 12 years older than the youngest. So what number can you multiply by 3 and have it turn out to be 12 larger than the original number? It’s the number 18, that you want. If the children are 6, 12, and 18, then the oldest is three times as old as the youngest.
Okay, so you can’t stand not using algebra and an equation to do this problem. I give. Here goes: Let the middle child’s age be X. Then the oldest child’s age is X + 6 and the youngest child’s age is X - 6. If the oldest child’s age is equal to three times the youngest child’s age, you write that as X + 6 = 3(x – 6). Distributing the 3 on the right, the equation becomes X + 6 = 3x – 18. Subtract X From each side and add 18 to each side, and you get 24 = 2x. Dividing by 2, 12 = X. The middle child is X Years old, or 12, the oldest is X + 6 years older, or 18, and the youngest is X - 6 years old, or 6.
Brothers and sisters, I have some
If each child in a certain family has at least four smallest number of children that the family brothers and three sisters, then what is the can have?
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Making age an issue
Traditional age word problems use algebraic expressions to write comparisons such as Twice the age of Or Four years older than And then solve for one or more person’s age. You’ll see lots of parentheses in the equations that are first written so that the meaning is clearly defined and the people’s ages are clearly identified. Age problems can get pretty wordy and confusing, so you want to be sure that you’ve written something to parallel the wording.
The Problem: To be a member of a certain club, you have to meet its stringent age requirement. Three more than twice your age must be equal to 21 more than your age. What is the age requirement (in simpler terms)?
^VLA/V First, decide if the club is worth it. That decided, now tackle the problem. Let X Represent your age. Write Three more than twice your age As 3 + 2x. Then 21 more than your age Becomes 21 + X. Set the two expressions equal to one another, and the equation to solve is 3 + 2x = 21 + x. Subtracting X And 3 from each side, you get X = 18.
You double your pleasure and double your fun when you introduce twins into the mix. The ages of twins is the same (I won’t get into the special case of twins born on two different days even though minutes apart), so you have to be sure to include the age of the twins twice into the problem.
The Problem: The mother of a pair of twins is 30 years older than they are. If the sum of the ages of all three of them (the mother and the twins) is 12 greater than the mother’s age, then how old is the mother?
^VLA/V Let the age of the twins be represented with X. Then their mother’s age is represented with X + 30. Adding the ages of the three people together, you get the expression X + X + (x + 30). Because this sum is 12 greater than the age of the mother, you set this sum equal to (x + 30) + 12. The final equation and its solution is as follows.
X + x + (x + 30) = (X + 30) + 12 3x + 30 = X + 42
2X= 12
X= 6
The age of the twins is 6, so their mother is 6 + 30 = 36 years. The sum of all three people is 6 + 6 + 36 = 48, and 48 is 12 greater than the mother’s age of 36.
The Problem: Libin is three times as old as Larry. If the sum of their ages is 16 more than twice Larry’s age plus 4, then how old is Libin?
Even though the problem asks for Libin’s age, it’ll be easier to let the variable, X, Represent Larry’s age and then answer the question after solving the equation. Letting X Represent Larry’s age, then 3x represents Libin’s age. The sum of their ages is X + 3x. You write 16 more than twice Larry’s age plus 4 With numbers and a variable in exactly that same order: 16 + 2x + 4. Now write the entire equation with the two expressions separated by an equal sign and solve.
X + 3x = 16 + 2x + 4 4x = 20 + 2x
2X= 20
X= 10
So Larry is 10 years old, making Libin 30 years old.
Going Back and Forth into the Future and the Past
Age problems get even more exciting when you get out your crystal ball and make predictions about the future or look back in a photo album to reminisce about the past. The equations used to solve age problems about the future and past get more complicated because the same amount of time has to be added to or subtracted from each variable expression.
Looking to the future
How old will you be in seven years? Some of us grimace at the thought of another significant birthday, but it’s better than the alternative. Add seven to your age, and you know how many candles go on your birthday cake.
VLAiV

The Problem: Jake is six years older than Jack. In 7 years, the sum of their ages will be 52. How old are Jake and Jack now?

Let Jack’s age be represented with X And Jake’s age represented with X + 6. (An alternative would be to let Jake’s age be X And Jack’s be X - 6; most people prefer to work with addition instead of subtraction.) Now deal with In 7years. Add 7 to Both Ages, so Jack’s age will be X + 7 and Jake’s age will be X + 6 + 7 or X + 13 in seven years. The sum of their ages in seven years will be 52, so the equation and its solution are as follows:
(X + 7) + (X + 13) = 52 2x + 20 = 52
2X= 32
X= 16
So Jack is 16, and Jake, who is 6 years older, is 22. In seven years, Jack will be 16 + 7 = 23 and Jake will be 22 + 7 = 29. The sum 23 + 29 = 52.
Some age problems also use comparisons where one person is twice as old or four times as old in a number of years.
The Problem: Molly is 30 years older than Margaret. In 18 years, Molly will be twice as old as Margaret. How old will Molly be when she’s twice Margaret’s age?
Let Margaret’s age be represented by X And Molly’s age be represented by X + 30. In 18 years, Margaret will be X + 18 and Molly will be (x + 30) + 18 = X + 48. At that time, 18 years from now, Molly will be twice as old as Margaret, so you write that Molly’s age, X + 48 = 2(x + 18), twice Margaret’s age. Solving the equation, you multiply through by 2 on the right to get X + 48 = 2X + 36. Subtract X From each side and subtract 36 from each side to get 12 = X. So Margaret is 12 years old now, and Molly, who is 30 years older than Margaret is 12 + 30 = 42 years old. In 18 years, when Molly is twice Margaret’s age, Molly will be 42 + 18 = 60 and Margaret will be 12 + 18 = 30.
Catching up in age
In an old Abbott and Costello movie, there’s a discussion about the ages of a husband and wife. When they marry, the husband is four times as old as his bride. Eight years later he’s three times as old as she is. And 24 years after that he’s only
Twice her age. Abbott asks Costello how long it’ll take for her to catch up with him in age. Of course, that’s impossible, but what were the ages of the couple when they married? (Try to keep it legal in most states.)
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The Problem: Stan is four times as old as Steve. In ten years, Stan will be twice as old as Steve. How old are Stan and Steve?
£,?LAiV Let Steve’s age be represented by X And Stan’s age by 4x. In ten years, Steve will be X + 10 and Stan will be 4x + 10. Because Stan’s age in ten years will be twice Steve’s, you write that 4x + 10 = 2(x + 10). Multiplying through by 2 on the right, 4x + 10 = 2x + 20. Subtract 2x from each side and 10 from each side to get 2x = 10, or X = 5. So Steve is 5 and Stan is four times that, or 20. In ten years, Steve will be 15 and Stan will be 30.
Going back in time
Looking back on things, you recall those good times with close friends and what you did when you were all younger — by two years or ten years or more. Just as with adding years in word problems, when you have so many years ago in a problem, you subtract the same number of years from each age involved in the situation.
The Problem: Today, Ernest’s age is four years more than twice Christine’s age. Three years ago, the sum of their ages was 31. How old is Christine right now?
Let X Represent Christine’s age right now. That means that Ernest’s age is represented by 4 + 2x. To adjust for Three years ago, You subtract 3 from each age, so Christine was X - 3 and Ernest was 4 + 2x – 3 = 2x + 1 years old. The sum of those two ages, three years ago, was 31. The equation and solution are:
(X – 3) + (2x + 1) = 31 3x – 2 = 31 3X = 33 X= 11
Christine is 11 years old right now, so Ernest is 4 + 2(11) = 26 years old. Three years ago, Christine was 8 and Ernest was 23. The sum 8 + 23 = 31.
With multiple births comes multiples of the same age. Each person involved has his age adjusted in the same way.

The Problem: Three years ago, the sum of the ages of the quintuplets and their older sister was 155. If Sis is five years older than her siblings, then how old are the quints today?
.jVLAiV First, Quintuplets Refers to five siblings born at the same time. Designate the variable X As the age of the quints today. Their sister’s age is then X + 5 years. Three years ago, the quints were X - 3 years old, and their sister was X + 5 – 3 or X + 2 years old. The sum of the ages of these six children is written by multiplying the quints’ age by five and adding the sister’s age: 5(x – 3) + (x + 2), and that sum is set equal to 155 for the equation.
5 (X - 3) + (X + 2) = 155
5x – 15 + X + 2 = 155
6x – 13 = 155
6x = 168
168 oa X = —Q – = 28
The quints are each 28 today. Their sister is five years older, so she’s 33 years old. Three years ago, the quints were 25 and Sis was 30. Multiplying 25 x five you get 125. Add 30, and the sum of the six ages is 155.
The Problem: Amanda’s age is two years more than twice Bill’s age. Nine years ago, her age was three years more than three times Bill’s age. How old will they be in five years?
^VLA* Let Bill’s age be represented with X So that Amanda’s age can be represented with 2 + 2x. Nine years ago, Bill was X - 9 years old and Amanda was 2 + 2x – 9 or 2x – 7 years old. Amanda’s age nine years ago was equal to Three years more than Bill’s age At that time. The equation to use is 2x – 7 = 3 + 3(x – 9). Notice that both ages in the equation are those representing nine years ago.
2x – 7 = 3 + 3 (X - 9) 2x – 7 = 3 + 3x – 27
2X-7=3X-24
17 = X

Bill is 17 years old right now, and Amanda is 2 + 2(17) or 36 years old. In five years, Bill will be 22 and Amanda will be 41.
Facing Some Challenges of Age
My Funky Winkerbean story, at the beginning of this chapter, is an example of how some age problems can seem awfully complicated and difficult. The problems in this section may seem impossible at first, but, by taking them apart and putting them back together again as logical equations, you see how those challenges of age aren’t all that bad after all.
The Problem: Mike and Ike are good friends and like to challenge new acquaintances with the following riddle, saying that they’ll pick up the tab at the restaurant if the riddle can be solved. Here’s what they pose to the unsuspecting: Mike is twice as old as Ike was when Mike was as old as Ike is now. Right now, Mike is 36 years old. How old is Ike right now?
The most common plan of action is to let Ike’s age be represented with X And then write some expressions and equations involving that variable. In this case, because you know Mike’s age already, it makes more sense to figure out how many years ago the riddle is referring to — solve for X. Here’s what you know:
Mike is twice as old as Ike was. . . So, since Mike is 36, Ike was 18. Mike was as old. . . Letting X Be the number of years ago, Mike was 36 – X. Ike is now. . . The number of years since Ike was 18 is 18 + X.
Does that look like sleight-of-hand? Look at Table 15-1, showing possible ages of Mike and Ike if X Is the number of years since Ike was 18.
|
Table 15-1 |
Mike and Ike’s Ages |
|
|
X Years |
Mike’s Age (36 – x) |
Ike’s Age (18 + x) |
|
1 |
36 – 1 = 35 |
18 + 1 = 19 |
|
2 |
36 – 2 = 34 |
18 + 2 = 20 |
|
3 |
36 – 3 = 33 |
18 + 3 = 21 |
|
8 |
36 – 8 = 28 |
18 + 8 = 26 |
|
9 |
36 – 9 = 27 |
18 + 9 = 27 |
Nine years ago, Mike was 27, which is the same age that Ike is now. When Mike was 27, Ike was 18 (or half Mike’s age now). If I had just written the equation 36 – X = 18 + X And solved it to get 18 = 2x or X = 9, you might not have believed me. For the naturally skeptical, Table 15-1 helps.
The next problem uses more number theory than algebra. Haul out your prime factorizations to do this problem.
The Prime factorization Of a number is the unique set of all the prime numbers whose product is that number. For example, the prime factorization of the number 105 = 3 X 5 X 7. The prime factorization of 100 = 22 X 52.
The Problem: Mike and Ike are back at it again. They point to a family with three children enjoying their dinner at a nearby table. They offer to buy dinner for that entire family and the person they’re making the bet with if this poor, unsuspecting person can figure out the riddle (otherwise, this person pays). The riddle: At the nearby table, the product of the ages of the three children is 72, and the sum of their ages is today’s date. How old are the children? The bet-taker thinks for a while and then says, "But that isn’t enough information!" Mike takes pity and offers him a clue, "Okay, I’ll tell you that the oldest child doesn’t like pizza." Can you determine the ages of the children?
Haul out another list or chart. The prime factorization of the number 72 is 23 X 32. Use the factors to figure out all the different numbers that divide 72 evenly — and include the number 1 (which isn’t a prime number, but is a factor of 72). All the possible ages — numbers that divide 72 evenly are: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. Using these numbers, you create a list of the possible ages of the three children from combinations that multiply to give you 72. For example, 72 and 1 and 1 multiply to give you 72. The numbers aren’t very probable for the ages of three children, but I’m sure it’s happened. Another combination is 36 and 2 and 1. Making this list is fine, but a table is more helpful, because you want the Sum Of the ages, too. Table 15-2 lists all of the possible ages and their sums. Notice that the ages multiplied together are always 72.
|
|
Ages of Three Children and the Sum of Their Ages |
|
|
Child 2 Child 3 Sum of Ages |
|
72 |
1 1 74 |
|
36 |
2 1 39 |
|
24 |
3 1 28 |
|
18 |
4 1 23 |
(continued)
|
Table 15-2 (continued) |
|||
|
Child 1 |
Child 2 |
Child 3 |
Sum of Ages |
|
18 |
2 |
2 |
22 |
|
12 |
6 |
1 |
19 |
|
12 |
3 |
2 |
17 |
|
9 8 1 18 |
|||
|
9 |
4 |
|
15 |
|
8 |
3 |
3 |
14 |
|
6 |
6 |
2 |
14 |
|
6 |
4 |
3 |
13 |
You see all the possibilities for the ages. So, if the bet-taker had worked out all the possible ages and sums and compared the sum with today’s date, he could have answered the question — unless today is the 14th. That’s why he needed more information. By giving the hint that the oldest child doesn’t like pizza, the answer to the riddle is clear. The children must be 8, 3, and 3 years old, because the other combination with a sum of 14 is 6, 6, and 2 where there are Two Older children.
One more age problem to consider mixes in a little number sense with ages now and in the past.
The Problem: Ben is a teenager. Bob is one-third as old as Ben was when Bob was half as old as Ben is now. How old are Ben and Bob?
Identical applications
Two men apply for the same job. They are iden – hiring asks, "Are you two twins?" They respond, tical in appearance. They have the same mother quite honestly, "No, we aren’t." How is this and father and birth date. The person doing the possible.
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^VLA* First, you can narrow down Ben’s age to being either 14, 16, or 18. This conclusion comes from the fact that he’s a teenager and Bob is half his age; Ben’s age has to be evenly divisible by two. Of course, this assumes that the problem only deals with whole numbers, and that’s a good assumption. You can just play around with the three guesses to see whatever numbers work, or you can be a bit more systematic. Let X Represent the number of years ago in the problem:
Bob’s age = ^(Ben’s age — X) Bob is one-third as old as Ben was X Years ago.
3Ben’sage — Bob’sage = i X 33
Ben’s age —3 Bob’s age =x, Simplifying the equation by distributing the fraction and then multiplying each term by 3
Ben’s age — X = ^ Ben’s age, when Bob was half as old as Ben is now. 2 Ben’s age = X, simplifying the previous equation in terms of Ben’s age.
Ben’s age —3 Bob’s age =x 2
A Ben’s age = x, so
Ben’s age — 3 Bob’s age = ^ Ben’s age, letting the two equations ending in X Equal one another and simplifying
2 Ben’s age = 3 Bob’s age Ben’s age = 6 Bob’s age
So, if Ben is six times as old as Bob and Ben is a teenager, then Ben must be 18 years old and Bob must be 3.
Chapter 16
Table 15-2
Child 1
2






