Average Ages and Age Problems

Average Age Moves with Time

Ages behave like any other numbers, so everything you know about averages still works. The one new thing is time: every year, every person gets one year older.

For a fixed group of people:

• After n years, each age goes up by n, so the average goes up by n and the total goes up by (number of people) × n.
• n years ago, the average was n less and the total was (number of people) × n less.

For a family of 5 with average age 26 today:

A typical question. "Three years ago the average age of a family of 6 was 21 years. What is the sum of their ages now?" Each of the 6 is 3 years older, so the average is 24 now and the sum is 6 × 24 = 144. Adding just 3 to the old total (6 × 21 + 3 = 129) is the trap; the total grows by 6 × 3 = 18.

From one time to another. "The average age of 4 friends will be 30 years after 5 years. What was their average age 2 years ago?" Move the average, not the total: 30 − 5 − 2 = 23 years. Between "2 years ago" and "after 5 years" there are 7 years, so you can also go 30 − 7 = 23 in one step.

One person from two totals. "The average age of A and B today is 28 years. Three years from now, the average age of A, B and C will be 33 years. How old is C?" Three years from now the three ages add up to 3 × 33 = 99, so today they add up to 99 − 9 = 90. A and B make 56 today, so C is 34.

When the average does not move. If a question says the average age is the same as it was some years ago, the group cannot be the same people. Someone has left or died, someone younger has replaced an older member, or a baby has been born. The parts below take each case in turn.

Months. Age questions often use months. 6 months = 0.5 year, 4 months = $\frac{1}{3}$ year and 3 months = 0.25 year. Change months into years before you multiply.

A Baby Born, a Member Who Dies or Leaves, a Bride Who Joins

Here the group itself changes, so work with totals at one fixed time, usually today.

A baby is born. "Four years ago the average age of a family of 5 was 22 years. A baby has been born since then, and today the average age of the family is still 22 years. How old is the baby?"

• The 5 old members: 5 × 22 = 110 four years ago, so 110 + 5 × 4 = 130 today.
• The family of 6 today: 6 × 22 = 132.
• Baby = 132 − 130 = 2 years.

Quick check: the baby was born less than 4 years ago, so its age must be below 4. If your answer is bigger, one of your totals is at the wrong time.

A member dies. "Six years ago the average age of a family of 6 was 31 years. The grandmother has died since then, and the average age of the other 5 today is 30 years. How old was the grandmother six years ago?"

• The other 5 today: 5 × 30 = 150, so six years ago they made 150 − 30 = 120.
• The family of 6 six years ago: 6 × 31 = 186.
• Grandmother six years ago = 186 − 120 = 66 years. If she died two years ago, she was 70 when she died.

A member leaves. "The average age of a family of 5 is 30 years. When the daughter leaves for college, the average age of the other 4 becomes 33 years." Daughter = 5 × 30 − 4 × 33 = 150 − 132 = 18 years. The average went up because a younger member left.

A bride joins. "The average age of a family of 4 is 38 years. After the son marries, the average age of the family of 5 is 36 years." Bride = 5 × 36 − 4 × 38 = 180 − 152 = 28 years.

Quick formulas. If n people have average a and one person joins so that the average changes by c (take c as minus if it falls):

$$\text{newcomer} = a + (n + 1)c$$

For the bride: 38 + 5 × (−2) = 28. If one person leaves a group of n and the average changes by c:

$$\text{person who left} = a - (n - 1)c$$

For the daughter: 30 − 4 × 3 = 18. The same "change × count" idea runs through the section Change in Average: Joining, Leaving, Replacement and Errors.

[SSC Tip] Bring every total to the same moment before you add or subtract. Mixing a total from four years ago with today's total is the usual slip in these questions.

A Teacher Joins or Leaves, and New Students Arrive

A class with its teacher is an age question SSC sets often.

Teacher included. "The average age of 32 students is 13 years. When the teacher's age is included, the average becomes 14 years." Teacher = 33 × 14 − 32 × 13 = 462 − 416 = 46 years.

Shortcut: the teacher brings the new average for himself plus 1 extra year for each of the 32 students: 14 + 32 × 1 = 46. Or start from the old average: 13 + 33 × 1 = 46.

The teacher's age given, the old average asked. "The average age of 8 trainees goes up by 3 years when their 47-year-old instructor is included. What was the trainees' average age?" The instructor = old average + 9 × 3, so the old average is 47 − 27 = 20 years.

Teacher excluded. "The average age of 24 students and their teacher is 16 years. Without the teacher, the average is 15 years." Teacher = 25 × 16 − 24 × 15 = 400 − 360 = 40 years.

Teacher replaced. "The average age of 40 students and their teacher is 16 years. The teacher, aged 58, retires and a new teacher joins. The average falls by 6 months." There are 41 people and the average falls by 0.5 year, so the total drops by 41 × 0.5 = 20.5 years. The only change is the teacher, so the new teacher is 58 − 20.5 = 37.5 years, that is, 37 years 6 months.

New students join. "The average age of 40 students is 15 years. When 8 new students join, the average falls by 3 months." The new average is 14.75 for 48 students, a total of 48 × 14.75 = 708. The old total is 40 × 15 = 600, so the newcomers add up to 108 and their average is 13.5 years.

Quick way: newcomers' average = new average − (old count × fall) ÷ number of newcomers = 14.75 − 40 × 0.25 ÷ 8 = 14.75 − 1.25 = 13.5.

Students leave. "The average age of 36 students is 13 years. Six students whose average age is 16 leave." The rest have 36 × 13 − 6 × 16 = 468 − 96 = 372 years for 30 students, an average of 12.4 years.

[SSC Tip] Count the teacher. "32 students and a teacher" is 33 people. Most wrong options come from multiplying by 32 instead of 33.

Replacement over Time: Committees and Teams

The same average after some years. "The average age of a 9-member committee is the same today as it was 4 years ago, because one old member has been replaced by a younger one. How much younger is the new member than the member he replaced?"

In 4 years, the 9 members' total should rise by 9 × 4 = 36 years. It did not rise at all, so the swap took away exactly 36 years. The new member is 36 years younger than the old member (comparing their ages today).

A smaller rise than time allows. "The average age of an 8-member board is 2 years more than it was 5 years ago, although one member has been replaced." Time should add 8 × 5 = 40 years, but the total rose by only 8 × 2 = 16. The swap took away 24 years, so the new member is 24 years younger than the one replaced.

Replacing two members. "The average age of 12 players goes up by 1.5 years when two players aged 22 and 25 are replaced by two new players." The total rises by 12 × 1.5 = 18 years, so the new pair adds up to 22 + 25 + 18 = 65, an average of 32.5 years. Plain replacement questions like this one are covered in detail in the Change in Average section.

The captain and the wicketkeeper. "The captain of an 11-member cricket team is 31 years old and the wicketkeeper is 3 years older. If these two are left out, the average age of the other 9 players is 1 year less than the average age of the whole team. Find the team's average age."

Let the team average be A. The whole team's total is 11A, and the other 9 have 9(A − 1):

$$11A - 31 - 34 = 9(A - 1)$$

So $11A - 65 = 9A - 9$, which gives $2A = 56$ and A = 28 years. Check: the team total is 308, the other 9 have 308 − 65 = 243, and 243 ÷ 9 = 27, one less than 28.

Quick way: if the two left out were exactly at the team average, they would have 2A between them and the other 9 would also average A. The other 9 are 1 year below A each, so the two are 9 years above 2A: their 65 years equal $2A + 9$, so $2A = 65 - 9 = 56$.

In short: "same average as n years ago" means the new member is (members × n) years younger, and "the average rose by only r" means (members × (n − r)) years younger.

Husband, Wife and Children: Averages at Different Times

These questions give averages of different sets of people at different times. Bring every total to today, then subtract.

Marriage, then a child. "At the time of their marriage, the average age of a husband and wife was 24 years. Six years later they have a 3-year-old child. What is the average age of the family now?"

• The couple's total at marriage was 2 × 24 = 48.
• Six years later each of them is 6 years older, so their total is 48 + 12 = 60.
• With the child the family total is 63 for 3 people, an average of 21 years.

Adding only 6 to the couple's total is the usual mistake. It gives 57 ÷ 3 = 19, a typical wrong option.

Three people, two times. "Two years ago, the average age of a man, his wife and their daughter was 25 years. Four years ago, the average age of the wife and daughter was 19 years. Find the man's present age."

• All three today: 3 × (25 + 2) = 81.
• Wife and daughter today: 2 × (19 + 4) = 46.
• Man = 81 − 46 = 35 years.

Parents and children. "The average age of a couple and their two children is 25 years. The couple's average age is 39 years, and the elder child is 4 years older than the younger." The family total is 100 and the couple's is 78, so the children add up to 22. Two children 4 years apart who add up to 22 are 9 and 13.

Before and after a child's birth. "Five years ago the average age of a husband and wife was 30 years. Today the average age of the husband, the wife and their child is 24 years. How old is the child?" The couple today: 2 × 35 = 70. The family today: 3 × 24 = 72. The child is 2 years old.

Grandparents in the picture. The same steps work for bigger families. Bring each average to today, turn it into a total, and subtract the people you already know. If the average age of a grandfather, a father and a son is 46 today, and the father and son together average 32, the grandfather is 3 × 46 − 2 × 32 = 138 − 64 = 74 years.

Ages from Sums, Differences and 'Times as Old' Equations

Some age questions have no averages at all; they give sums, differences, products or "times as old" statements. Use one letter for the younger person, write the other ages from it, and remember that the difference between two people's ages never changes.

Sum and difference. "A is 8 years older than B, and their ages add up to 52." Then $2B + 8 = 52$, so B = 22 and A = 30. Quick way: the elder is half of the sum plus the difference, so (52 + 8) ÷ 2 = 30.

"Times as old" at another time. "A man is 24 years older than his son. Two years ago, he was three times as old as his son. Find their present ages." Take the son as S, so the man is $S + 24$. Two years ago they were $S - 2$ and $S + 22$:

$$S + 22 = 3(S - 2)$$

So $S + 22 = 3S - 6$, which gives $2S = 28$. The son is 14 and the man is 38. Check: two years ago they were 12 and 36, and 36 is three times 12.

Two conditions. "A father's age is 3 years more than three times his son's age. Three years from now, the father's age will be 10 years more than twice the son's age." Write $F = 3S + 3$. In three years, $3S + 6 = 2(S + 3) + 10$, so the son is 10 and the father 33.

Product of ages. "Two sisters' ages differ by 3 years and multiply to 270." Try numbers near $\sqrt{270}$, which is a little over 16: 15 × 18 = 270, so the sisters are 15 and 18. Testing the options is often faster than solving the quadratic.

[Tier 2] Products at two times. If two ages are a and b today, then n years ago their product was

$$(a - n)(b - n) = ab - n(a + b) + n^2$$

So the product today and the product n years ago together give the sum $a + b$.

Digits reversed. "A father's age and his son's age have the same two digits in reverse order. Their ages add up to 77 and differ by 45." Write them as $10a + b$ and $10b + a$. The sum is $11(a + b) = 77$, so $a + b = 7$; the difference is $9(a - b) = 45$, so $a - b = 5$. Then a = 6 and b = 1: the father is 61 and the son 16.

So the sum of two reversed-digit ages is always a multiple of 11 and the difference a multiple of 9. Use that to strike out options quickly.

"Times as old" questions written as ratios (A : B = 3 : 1 after some years) are quicker with the units method of Chapter 4.

[SSC Tip] In "n years ago he was k times as old" questions, change both ages by n before multiplying. Writing $S + 24 = 3(S - 2)$, which moves only the son back in time, is the common slip.

Youngest, Eldest and Siblings Born at Intervals

Siblings born at equal gaps. If children are born a fixed number of years apart, their ages go up in equal steps. So the average age is the age of the middle child.

• 5 children born 2 years apart with average age 10: the third child is 10, so the ages are 6, 8, 10, 12 and 14. The youngest is 6 and the eldest 14.
• With an even number of children, the average sits halfway between the two middle children. 4 siblings born 3 years apart with average age 13.5: the middle two are 12 and 15, so the ages are 9, 12, 15 and 18.
• If the sum is given, divide by the count first. 7 children born 2 years apart with ages adding up to 98 have average 14, so the youngest is 14 − 3 × 2 = 8 and the eldest 14 + 3 × 2 = 20.

In general, with n children born g years apart:

$$\text{youngest} = \text{average} - \frac{(n - 1)g}{2}, \quad \text{eldest} = \text{average} + \frac{(n - 1)g}{2}$$

Unequal gaps. "The average age of three sisters is 15 years. The eldest is 5 years older than the middle one, who is 2 years older than the youngest." With the youngest as y, the ages are y, $y + 2$ and $y + 7$, which add up to 45. So $3y + 9 = 45$ and y = 12. The sisters are 12, 14 and 19.

Eldest or youngest from two averages. "The average age of four cousins is 21 years. The average age of the three youngest is 18 years. How old is the eldest?" All four add up to 84 and the three youngest to 54, so the eldest is 30. Leaving out the youngest works the same way.

Ages of boys and girls in a class. This is a weighted average, done with the distance rule from Weighted Average and Combined Groups. "In a class of 40, the boys' average age is 16 years, the girls' is 14.5 years and the class average is 15.1 years." Boys are 0.9 above the class average and girls 0.6 below, so boys : girls = 0.6 : 0.9 = 2 : 3. There are 16 boys and 24 girls. Check: 16 × 16 + 24 × 14.5 = 256 + 348 = 604 = 40 × 15.1.

[Tier 2] Multi-Step Age Problems

Tier 2 likes questions where people join, leave and get replaced one after another, or where a long stretch of time passes with births and deaths in between. Keep a running list of totals.

Joining, then replacing. "The average age of A, B and C is 30 years. When D joins them, the average becomes 28 years. E, who is 2 years younger than D, then replaces A, and the average age of B, C, D and E becomes 26 years. Find A's age."

• A, B and C add up to 90, and A, B, C and D add up to 112, so D = 22 and E = 20.
• B, C, D and E add up to 104, so B and C together are 104 − 22 − 20 = 62.
• A = 90 − 62 = 28 years.

Births and deaths over a long gap. "Ten years ago, the average age of a family of 5 was 28 years. Since then the grandmother, who was 62 at that time, has died, and a baby has been born who is now 4 years old. Find the present average age of the family."

• If all 5 were still alive, their total today would be 140 + 5 × 10 = 190.
• Remove the grandmother at the age she would be today, 62 + 10 = 72: 190 − 72 = 118.
• Add the baby: 118 + 4 = 122 for 5 people, an average of 24.4 years.

When she died does not matter. Taking away the age she would have today gives the same result as tracking every single year.

Time and a replacement together. "Three years ago, the average age of a 5-member committee was 40 years. Since then, a member who is 55 today has been replaced by a member who is 35 today. Find the present average age." Without the change, the committee's total today would be 200 + 15 = 215. The swap takes away 55 and adds 35: 215 − 55 + 35 = 195, an average of 39 years.

Merging two groups. Bringing two families or two classes together is a weighted average: families of 4 (average 30 years) and 6 (average 25 years) together average (120 + 150) ÷ 10 = 27 years.

Solved Examples

Question 1

The average age of a family of 5 was 24 years four years ago. What will be the sum of their ages six years from now?

Answer. From four years ago to six years from now is 10 years, so every member, and so the average, goes up by 10. The average will be 34, and the sum will be 5 × 34 = 170 years. Adding just 10 to the old total of 120 would give 130, which is the trap.

Question 2

Five years ago, the average age of a family of 4 was 23 years. A baby was born after that, and the present average age of the family of 5 is still 23 years. How old is the baby?

Answer. Five years ago the 4 members added up to 4 × 23 = 92. Each is now 5 years older, so today they add up to 92 + 20 = 112. The family of 5 today adds up to 5 × 23 = 115. So the baby is 115 − 112 = 3 years old, which makes sense because the baby was born less than 5 years ago.

Question 3

The present average age of a family of 6 is 35 years. Seven years ago the family had 7 members, including the grandfather, and their average age was 34 years. The grandfather died 3 years ago. How old was he when he died?

Answer. Today the 6 add up to 6 × 35 = 210, so seven years ago they added up to 210 − 42 = 168. Seven years ago all 7 added up to 7 × 34 = 238, so the grandfather was 238 − 168 = 70 then. He died 3 years ago, which is 4 years after that, so he was 74 years old when he died.

Question 4

The average age of a family of 6 is 29 years. When the eldest son moves to another city for a job, the average age of the other 5 becomes 30.6 years. How old is the son?

Answer. All 6 add up to 6 × 29 = 174 and the other 5 add up to 5 × 30.6 = 153. So the son is 174 − 153 = 21 years old. It makes sense that the average went up, because he is younger than the family average.

Question 5

The average age of a family of 4 is 41 years. When the son gets married, the average age of the family becomes 38 years. How old is the bride?

Answer. Before the wedding the total is 4 × 41 = 164; after it, 5 × 38 = 190. So the bride is 190 − 164 = 26 years old. Quick way: the average fell by 3 for 5 people, so the bride is 41 − 5 × 3 = 26.

Question 6

The average age of 35 students in a class is 13 years. When the teacher's age is included, the average goes up by 1 year. Find the teacher's age.

Answer. Now there are 36 people with average 14, a total of 36 × 14 = 504. The students alone add up to 35 × 13 = 455. So the teacher is 504 − 455 = 49 years old. Quick way: 13 + 36 × 1 = 49.

Question 7

The average age of 29 students and their class teacher is 15 years. The class teacher, who is 50, is transferred and a new teacher joins. The average age then falls by 6 months. How old is the new teacher?

Answer. There are 30 people, and the average falls by half a year, so the total falls by 30 × 0.5 = 15 years. The only change is the teacher, so the new teacher is 15 years younger than the old one: 50 − 15 = 35 years.

Question 8

The average age of 45 students is 16 years. When 9 new students join, the average age falls by 4 months. What is the average age of the new students?

Answer. 4 months is $\frac{1}{3}$ of a year, so the new average is $15\frac{2}{3}$ years for 54 students. That is a total of $54 \times \frac{47}{3} = 846$ years. The old students add up to 45 × 16 = 720, so the 9 newcomers add up to 846 − 720 = 126, and their average is 126 ÷ 9 = 14 years.

Question 9

The average age of the 7 members of a society's committee is the same as it was 5 years ago, because an old member has been replaced by a new one. How much younger is the new member than the member who was replaced?

Answer. In 5 years the 7 members should have added 7 × 5 = 35 years to their total. The total did not change at all, so the swap took away 35 years. The new member is 35 years younger than the old member would be today.

Question 10

The average age of a 10-member board today is 2 years more than it was 6 years ago, because one member has been replaced. By how many years is the new member younger than the member replaced?

Answer. Six years should have added 10 × 6 = 60 years to the total, but the total actually went up by only 10 × 2 = 20. The missing 40 years went out with the swap, so the new member is 40 years younger.

Question 11

The captain of an 11-member cricket team is 28 years old and the wicketkeeper is 5 years older. If the ages of these two are left out, the average age of the remaining players is 1 year less than the average age of the whole team. What is the average age of the team?

Answer. The captain and wicketkeeper are 28 and 33, which is 61 years together. If the team average is A, then $11A - 61 = 9(A - 1)$, so $2A = 61 - 9 = 52$ and A = 26 years. Check: the team total is 286, the other 9 have 286 − 61 = 225, and 225 ÷ 9 = 25, one less than 26.

Question 12

At the time of their marriage, the average age of a husband and wife was 27 years. Six years later, they have a son who is 3 years old. What is the present average age of the family?

Answer. At the wedding the couple added up to 2 × 27 = 54. Six years on, both are 6 years older, so they add up to 54 + 12 = 66. With the 3-year-old son, the family total is 69 for 3 people, so the average is 23 years.

Question 13

Four years ago, the average age of a husband, his wife and their son was 26 years. Six years ago, the average age of the wife and the son was 17 years. What is the husband's present age?

Answer. I bring both totals to today. All three today: 3 × (26 + 4) = 90. Wife and son today: 2 × (17 + 6) = 46. So the husband is 90 − 46 = 44 years old.

Question 14

The average age of a man, his wife and their two children is 24 years. The average age of the man and his wife is 40 years, and the elder child is 6 years older than the younger. How old is the younger child?

Answer. The family total is 4 × 24 = 96 and the parents' total is 2 × 40 = 80, so the children add up to 16. Two children 6 years apart who add up to 16 are 5 and 11. The younger child is 5 years old.

Question 15

A father is 28 years older than his son. Four years ago, the father was five times as old as the son. What are their present ages?

Answer. Say the son is S, so the father is $S + 28$. Four years ago they were $S - 4$ and $S + 24$, so $S + 24 = 5(S - 4)$. That gives $S + 24 = 5S - 20$, so $4S = 44$ and the son is 11. The father is 39. So the ages are 11 and 39 years. Check: four years ago they were 7 and 35, and 35 is five times 7.

Question 16

The product of the present ages of A and B is 240. Four years ago, the product of their ages was 128. What is the sum of their present ages?

Answer. If the ages are a and b, then four years ago the product was $(a - 4)(b - 4) = ab - 4(a + b) + 16$. So $128 = 240 - 4(a + b) + 16$, which gives $4(a + b) = 128$. The sum is 32 years. The ages are in fact 12 and 20: 12 × 20 = 240 and 8 × 16 = 128.

Question 17

The present ages of a father and his son are two-digit numbers with the same digits in reverse order. Three years from now, the father will be three times as old as the son. What is the father's present age?

Answer. Write the father as $10a + b$ and the son as $10b + a$. In three years: $10a + b + 3 = 3(10b + a + 3)$, which simplifies to $7a = 29b + 6$. Only b = 1 works (it gives a = 5; for b = 2 or more, a would be bigger than 9). So the father is 51 and the son 15. Check: in three years they will be 54 and 18, and 54 is three times 18.

Question 18

The ages of 5 children born at intervals of 3 years each add up to 65 years. How old is the youngest child?

Answer. Equal gaps mean the average is the middle child's age: 65 ÷ 5 = 13. The youngest is two gaps below the middle, so 13 − 2 × 3 = 7 years. The ages are 7, 10, 13, 16 and 19, which add up to 65.

Question 19

The average age of three brothers is 20 years. The eldest is 6 years older than the middle brother, and the middle brother is 3 years older than the youngest. How old is the eldest?

Answer. Call the youngest y. Then the middle one is $y + 3$ and the eldest is $y + 9$. Their total is 60, so $3y + 12 = 60$ and y = 16. The brothers are 16, 19 and 25, so the eldest is 25 years old.

Question 20

In a class of 39 students, the average age is 14 years. The average age of the boys is 14.5 years and that of the girls is 13.2 years. How many boys are in the class?

Answer. The boys are 0.5 above the class average and the girls are 0.8 below it. The sizes go the other way round, so boys : girls = 0.8 : 0.5 = 8 : 5. That is 13 parts for 39 students, so 1 part is 3 students, and there are 24 boys (and 15 girls). Check: 24 × 14.5 + 15 × 13.2 = 348 + 198 = 546 = 39 × 14.

Question 21

[Tier 2] The average age of A, B and C is 32 years. When D joins them, the average age becomes 30 years. E, who is 3 years older than D, then replaces A, and the average age of B, C, D and E becomes 29 years. What is A's age?

Answer. A, B and C add up to 3 × 32 = 96, and with D the four add up to 4 × 30 = 120, so D is 24 and E is 27. B, C, D and E add up to 4 × 29 = 116, so B and C together are 116 − 24 − 27 = 65. That makes A = 96 − 65 = 31 years.

Question 22

[Tier 2] Eight years ago, a family of 4 had an average age of 45 years. Since then the grandfather, who was 64 at that time, has died, and the couple in the family have had two children, who are now 5 and 2 years old. What is the present average age of the family?

Answer. If all 4 were still alive, they would add up to 4 × 45 + 4 × 8 = 180 + 32 = 212 today. The grandfather would be 64 + 8 = 72 today, so I take that out: 212 − 72 = 140. Adding the children gives 140 + 5 + 2 = 147. The family now has 5 members, so the average is 147 ÷ 5 = 29.4 years.