The sum of the ages of A and B (in years) is 22. The product of their ages two years back was 77. Which one of the following is the value of the difference of their current ages?
- (a)2
- (b)3
- (c)4
- (d)5
Correct — C, 4. Work in the past, where the product is given. Two years ago the two ages summed to 22 − 4 = 18 and multiplied to 77. Numbers with sum 18 and product 77 are the roots of t² − 18t + 77 = 0, which factorises as (t − 11)(t − 7), giving 11 and 7. Adding the two years back makes the present ages 13 and 9, whose sum is 22 as required. The difference is 13 − 9 = 4.
- (a)2 — A difference of 2 with a sum of 22 gives present ages 12 and 10, so two years ago they were 10 and 8 with a product of 80, not 77.
- (b)3 — A difference of 3 gives ages of 12.5 and 9.5, and two years ago 10.5 × 7.5 = 78.75. The product does not match.
- (d)5 — A difference of 5 gives 13.5 and 8.5, so two years ago the product was 11.5 × 6.5 = 74.75.
Any pair of numbers with a known sum s and product p are the roots of t² − st + p = 0. Ages shift the frame: subtracting the same number of years from both ages reduces their sum by twice that number and leaves their difference untouched, which is why the difference asked for today is the same as the difference two years ago.
The trap is to use the present sum of 22 with the past product of 77, which belong to different dates. Move both to the same date first. The difference is the one quantity that survives the shift, so it can be read off from either frame — 11 − 7 in the past or 13 − 9 today. Checking backwards is quick: 13 and 9 sum to 22, and 11 × 7 = 77.
- Two years ago the ages summed to 18 and multiplied to 77.
- Sum 18 with product 77 gives the roots of t² − 18t + 77 = 0, namely 11 and 7.
- Present ages are 13 and 9, summing to the given 22.
- The difference between two ages never changes with the passage of time.
- For numbers with sum s and difference d, the values are (s + d)/2 and (s − d)/2.
The sum changes with time; the difference does not, which is why the answer can be read at either date.
- Combining today's sum with the past product without shifting one of them.
- Subtracting only two years from the total instead of four.
- Solving for the ages and then reporting a sum or a product instead of the difference.
An ages item with two time frames, testing whether both facts are brought to the same date before any equation is written.
Two years ago, the age of A was three times the age of B. If B is currently 9 years old, then after how many years, the age of A will be double of the age of B?
- (a) 2 years
- (b) 3 years
- (c) 4 years
- (d) 5 years
Answer(d) 5 years
The same two-year shift on an earlier CAPF paper, this time forward as well as back. Both items are solved by choosing one date, writing every fact at that date, and only then forming the equation.
- practice — not a real PYQ
The sum of two numbers is 15 and their product is 56. Their difference is
- (a)1
- (b)2
- (c)3
- (d)4
Answer(a) 1 — the numbers are 8 and 7, roots of t² − 15t + 56 = 0.
- practice — not a real PYQ
The present ages of two people sum to 40. Five years ago their product was 216. What is the elder's present age?
- (a)21
- (b)22
- (c)23
- (d)24
Answer(c) 23 — five years ago they summed to 30 with product 216, giving 18 and 12, so today they are 23 and 17.