Problems on Ages

The simplest age questions give you two facts about two people: usually the sum of their ages and the difference between them. You already know how to split a number when its sum and difference are given.

Elder = Sum + Difference2, and Younger = Sum − Difference2.

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For example, Radhika and her brother together are 32 and she is 6 years older. Radhika is 32 + 62 = 19 and her brother is 32 − 62 = 13. Check it: 19 + 13 = 32 and 19 − 13 = 6.

The only extra step in this type is time. If the sum is given for "4 years ago", first bring it to the present. Two people each gained 4 years, so the present sum is the old sum + 2 × 4. The difference needs no adjustment at all, since it is the same at every moment.

Questions sometimes ask "after how many years will the sum be S?". Each year adds 2 to the sum of two people, so the number of years is S − present sum2.

When one age is a multiple of another, think in units. "The father is 4 times as old as the son" means: son = 1 unit, father = 4 units. Now the sum is 5 units and the difference is 3 units. Whatever the question gives you, sum or difference, divide by the right number of units.

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Take Radhika's grandmother, who is 7 times as old as Radhika, and say the grandmother is 54 years older. The difference is 7 − 1 = 6 units, so 1 unit = 546 = 9. Radhika is 9 and the grandmother 63.

The care needed in this type is about when the multiple holds. If the sentence says "6 years from now the mother will be 3 times as old as her son", the multiple applies to the ages 6 years from now. Make the units at that moment, find them, and only then move back to the present.

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A multiple that is true today is not true at another time. If a father is 4 times his son today, then n years earlier the same n is taken off both, and 4x − n is no longer 4 times x − n. So never carry the "4 times" to another moment.

When two ages are in the ratio a : b, write them as ax and bx. The x is the size of one unit, and finding it is the whole job. The sum is (a + b) units and the difference is (b − a) units, so a single extra fact gives you x.

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For instance, Radhika and her aunt are in the ratio 3 : 8 and the aunt is 25 years older. The difference is 8 − 3 = 5 units, so 1 unit is 5 years: Radhika is 15 and the aunt 40.

Some questions give a product instead: ages 3 : 4 with product 300. Then 3x × 4x = 12x² = 300, so x² = 25 and x = 5, giving 15 and 20. Others give the average: an average of 21 for two people means a sum of 42.

A fractional statement is also a ratio. "13 of A's age equals 14 of B's age" means A3 = B4, so A : B = 3 : 4. Cross the fractions this way and the question belongs to this type.

This is the type the exam likes most. The question gives the ratio now and the ratio some years later (or earlier), and asks for the ages. Write the present ages as ax and bx, add the years to both, and set up the new ratio:

ax + nbx + n = cd, which gives d(ax + n) = c(bx + n).

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For example, Radhika and her cousin are 4 : 7 now, and in 9 years they will be 7 : 10. Then 10(4x + 9) = 7(7x + 9), so 40x + 90 = 49x + 63, which gives 9x = 27 and x = 3. They are 12 and 21; in 9 years 21 and 30, which is 7 : 10.

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There is a faster way that comes from the foundation: the difference stays the same. Make the differences of the two ratios equal. In 4 : 7 the gap is 3 units; in 7 : 10 the gap is also 3 units. So the units are the same size in both ratios, and the change from 4 to 7 (3 units) is the 9 years: 1 unit = 3. If the gaps are not equal, scale the ratios first so that they are. With 2 : 5 and 3 : 4, the gaps are 3 and 1; write 3 : 4 as 9 : 12 so both gaps are 3 units. Now 2 → 9 is 7 units.

Now both facts are multiples, and each holds at a different moment: "4 years ago the mother was 7 times as old as Radhika; 6 years from now she will be 3 times as old." Take the present ages as R and M and write each sentence at its own moment:

M − 4 = 7(R − 4) and M + 6 = 3(R + 6).

Subtracting, 10 = 3R + 18 − 7R + 28, so 4R = 36 and R = 9, which makes M = 39. Check: 4 years ago 35 and 5, which is 7 times; 6 years later 45 and 15, which is 3 times.

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The quicker way again uses the fixed gap. At the first moment the gap is (7 − 1) = 6 units of Radhika's age then; at the second moment it is (3 − 1) = 2 units of her age then. Make the gaps equal: 7 : 1 stays, 3 : 1 becomes 9 : 3. Now Radhika's age moves from 1 unit to 3 units, a jump of 2 units, while the clock moves 4 + 6 = 10 years. So 1 unit = 5 and Radhika was 5 four years ago, which makes her 9 now.

Whenever you see two "times as old" statements at two moments, this unit picture gives the answer in two lines. Keep the equation method ready for checking.

With three or more people, the trick is to choose one person as x and write everyone else in terms of x. Pick the person the others are compared with. If A is compared with B and C is compared with B, then B is the natural x.

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Here is Radhika's family. Her father is twice her age, and her mother is 5 years younger than her father. The three ages add up to 85. Take Radhika as x: father 2x, mother 2x − 5. So 5x − 5 = 85, x = 18. Radhika is 18, her father 36 and her mother 31.

When the facts are ratios, join them into one ratio first. A : B = 2 : 3 and B : C = 4 : 5 share B; make B equal by writing 8 : 12 and 12 : 15, so A : B : C = 8 : 12 : 15. Now it is the ratio type with three terms.

Time adjustments work the same as before: for three people the sum changes by 3n in n years. If an average age of the three is given, turn it into a sum first.

This type is average in the form of ages. Turn every average into a total: total age = average × number of people. The person who joined or left is the difference of two totals.

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Say a class of 24 children has an average age of 11, and when the teacher is counted the average becomes 12. The totals are 24 × 11 = 264 and 25 × 12 = 300, so the teacher is 300 − 264 = 36.

The deviation view is quicker. Each of the 24 children is 1 below the new average 12, so together they are 24 short. The teacher has to fill that gap, so the teacher is 12 + 24 = 36. If the average falls instead, the newcomer is below the new average by the total excess of the old members.

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When someone leaves, run the same idea backwards: the person who left = old total − new total. If several people join with an average of their own, find their total and add it.

Ages have one more point to note: time. If the same group is looked at n years later, every member is n years older and the average also rises by n. So "the average age of 6 members was 30 three years ago" means the same 6 now average 33.

When two groups are put together, never average the two averages directly unless the groups are the same size. Add the totals and divide by the total number:

Combined average = n₁A₁ + n₂A₂n₁ + n₂.

For example, 12 boys average 13 years and 8 girls average 18 years. The totals are 156 and 144, so the class average is 30020 = 15.

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Picture it as a seesaw. The combined average is the balance point between 13 and 18. It sits 2 from the boys and 3 from the girls, and the counts are 12 : 8 = 3 : 2, the distances turned around. So counts are in the inverse ratio of the distances from the combined average. This lets you find a missing count or a missing average in one line, without any equation.

If the groups are given as percentages, treat the percentages as counts out of 100. Then start from the first group's average and move towards the second by the second group's share of the gap. With 75% of a club at 32 and 25% at 40, the average is 32 + 25% of 8 = 34.

Missing-member questions also live here: total of the family minus the known ages gives the one left out.

"The father was 30 when his son was born" gives the gap between them: father − son = 30, today and always. Most birth questions are based on this one line.

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Radhika was born in 2004 and her father in 1975, so he is 29 years older. In the year Radhika turned 10, her father was 39. With years of birth, age in a year = that year − year of birth, and the gap between two people is the difference of their birth years.

When two children and two parents are involved, fix one moment and compare everyone at that moment. Say the mother was 27 at Radhika's birth and the father was 33 at the birth of her brother, born 2 years after her. At the brother's birth Radhika is 2, so the mother is 29 and the father 33: the parents differ by 4.

A death works like someone leaving a group. Find what the total of the whole family would be today if no one had died (the old total plus count × years), then subtract today's actual total.

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Digit puzzles use the two-digit number 10a + b. Its reverse is 10b + a. The difference of the two is 9(a − b) and the sum is 11(a + b). So if a person's age after 9 years has reversed digits, the digits differ by exactly 1.