EPFO APFC 2023 General Studies · Question 76 of 120

Correct — A, (a) 30. Take the present ages as 5x and 3x, which is what the ratio 5 : 3 allows. Fifteen years ago they were 5x minus 15 and 3x minus 15, and the ratio then was 2 : 1, so 5x minus 15 equals twice 3x minus 15, that is 6x minus 30. Solving gives x equal to 15, so Mr. Kumar is 75 today and his son is 45. Checking backwards, fifteen years ago they were 60 and 30, which is 2 : 1 as required. The question asks for the father's age when the son was born, and that is simply the difference between their ages, 75 minus 45, which is 30. The difference is the key to the whole problem, because it is the one quantity in an age problem that never changes: both people grow older at the same rate, so the gap between them is fixed for life, and the father's age at the son's birth is exactly that gap. Recognising this converts the problem into a single equation in one unknown. Call the difference d. Fifteen years ago the ratio was 2 : 1, which means the father was twice the son, so the difference then equalled the son's age then; the son was therefore d and the father 2d at that time. Today, fifteen years later, the son is d plus 15 and the father is 2d plus 15, and their ratio is 5 : 3, so three times 2d plus 15 equals five times d plus 15. That gives 6d plus 45 equal to 5d plus 75, so d equals 30. This second route also gives a fast way of testing any proposed answer: from a candidate difference, reconstruct both ages and check the present ratio.

• (b)35 — Test it against the conditions and it fails. If the age difference were 35, then fifteen years ago, when the ratio was 2 : 1 and therefore the difference equalled the son's age, the son would have been 35 and the father 70. Today they would be 50 and 85, and 85 to 50 is 17 : 10, not 5 : 3. The reconstruction takes about fifteen seconds and it is worth doing on any age problem where the algebra has felt uncertain, because the conditions are simple enough to verify directly. Note the direction of the error as well: as the difference grows, the present ratio grows with it, so the ratios produced by the four options rise steadily above 5 : 3, and only the smallest of them can be right. Seeing that monotone relationship is often faster than solving, and it turns a four-way choice into a single check.
• (c)40 — A difference of 40 would mean that fifteen years ago the son was 40 and the father 80, satisfying the 2 : 1 condition, and that today they are 55 and 95. But 95 to 55 is 19 : 11, which is about 1·73, whereas 5 : 3 is about 1·67, so the present ratio is wrong. This option, like the previous one, is consistent with the second condition in the stem and inconsistent with the first, which is how a well-built age problem is constructed: each wrong option satisfies part of the data so that a candidate who checks only one condition is caught. The discipline is to verify a proposed answer against every condition in the stem, not against the one that was used to generate it, and in a two-condition age problem that means reconstructing both the past and the present ages.
• (d)45 — Forty-five is the son's present age, and it is the trap for a candidate whose algebra is entirely correct but who reports the wrong quantity. Solving the equations gives x equal to 15 and the two present ages as 75 and 45, and the number 45 is sitting there at the end of the working, waiting to be written down. The question, though, asks for the father's age when the son was born, which is the difference between the two ages, 75 minus 45. This is the single most common way of losing a solved age problem, and the remedy is mechanical: after finding the ages, read the final sentence of the stem again and identify which of the numbers now in front of you it is asking for, or better, write down what the answer will represent before starting the algebra.

Age problems are linear equations dressed in words, and they are governed by one structural fact: the difference between two people's ages is constant over time, while the ratio of their ages changes. Every question of this type exploits that asymmetry. A ratio at one date and a ratio at another date give two equations; introducing the ratio as a multiple of a single unknown, so that ages of 5 : 3 become 5x and 3x, keeps the algebra to one variable. The quantities most often asked for are the present ages, the age at some past or future date, the number of years until a stated ratio holds, or — as here — the age of the older person at the birth of the younger, which is precisely the constant difference. A useful reformulation is that a ratio of the form n : 1 means the difference equals the smaller age multiplied by n minus 1; in the special case of 2 : 1 the difference equals the younger person's age at that moment, which is why fifteen years ago the son's age and the age gap were the same number in this problem. Recognising these small identities lets many age problems be solved in one line, and it also provides an independent check on an answer obtained by longer algebra.

The quantitative block of this paper reserves a place for one word problem of this kind in most groups, and the numbers are always chosen so that the solution is a whole number of years and the arithmetic is light. What is being tested is the translation from words to equations and, just as much, the reading of the final question. The option set here is built around that second skill: one option is the son's present age, which appears in the working of every correct solution, and the others are differences that satisfy the past condition while failing the present one. That is a general pattern in this paper and a reliable warning. If an option equals a number that appeared in your working, treat that as a reason for suspicion rather than for confidence, and reread the stem before marking it. The other habit worth building for these items is verification by reconstruction. Age problems are unusual in that checking an answer is nearly as fast as finding it: take the proposed value, rebuild both people's ages at both dates, and confirm both ratios. A candidate who does that will never lose one of these questions to an algebraic slip.

• The difference between two people's ages is constant over time, so the age of a parent when a child was born is exactly the difference between their present ages.
• Writing a ratio of 5 : 3 as 5x and 3x reduces the problem to one unknown; here 5x minus 15 equals twice 3x minus 15 gives x equal to 15, so the present ages are 75 and 45 and the difference is 30.
• A ratio of 2 : 1 at any date means the age difference equals the younger person's age at that date, which is why the son was 30 and the father 60 fifteen years ago.
• The answer can be found directly in the difference: if d is the gap, the pair were d and 2d fifteen years ago and are d plus 15 and 2d plus 15 today, and setting that in the ratio 5 : 3 gives d equal to 30.
• Reconstructing from each option shows why the others fail: a gap of 35 gives present ages 50 and 85, a gap of 40 gives 55 and 95, and a gap of 45 gives 60 and 105, whose ratios are 17 : 10, 19 : 11 and 7 : 4 rather than 5 : 3.

• Reporting the son's present age of 45, or the father's present age of 75, when the question asks for the father's age at the son's birth, which is the difference of 30
• Forgetting that the age difference is constant and setting up an equation in which it changes between the two dates
• Subtracting fifteen years from only one of the two ages when moving the situation back in time
• Checking a proposed answer against only one of the two ratios given; each wrong option here satisfies the past condition and fails the present one
• Reversing the past ratio and writing the son's age as twice the father's, which produces a fractional and impossible value of the unknown

Word problems on ages, averages, ratios and simple work or speed relations appear in every quantitative group of this paper, always with small whole-number answers and always with an option set that includes at least one intermediate quantity from the working. The difficulty is almost never algebraic; it lies in translating the sentence into equations correctly and in identifying the quantity actually requested. The most economical preparation is to practise these problems to the point where the translation is automatic, and to adopt two fixed habits: write down in words what the final answer will represent before beginning, and verify the answer by rebuilding the whole situation from it. On age problems in particular, verification is quick enough that it should be treated as part of the solution rather than as an optional check.

No directly related past PYQ was found.

• practice — not a real PYQ

  The present ages of a father and his son are in the ratio 3 : 1. Five years ago the ratio of their ages was 4 : 1. What is the present age of the father in years?

  • (a)40
  • (b)45
  • (c)50
  • (d)60

  Answer(b) 45 — writing the present ages as 3x and x, five years ago they were 3x minus 5 and x minus 5, and the condition that the first is four times the second gives 3x minus 5 equal to 4x minus 20, so x is 15 and the ages are 45 and 15. Checking backwards, five years ago they were 40 and 10, which is 4 : 1 as required, and the constant difference of 30 is the father's age when the son was born.

• practice — not a real PYQ

  The ages of a man and his son are in the ratio 5 : 3, and fifteen years ago the ratio was 2 : 1. What is the son's present age in years?

  • (a)30
  • (b)45
  • (c)60
  • (d)75

  Answer(b) 45 — taking the present ages as 5x and 3x and using the past condition, 5x minus 15 equals twice 3x minus 15, so x is 15 and the ages are 75 and 45. Note how the same working answers two different questions: 45 is the son's present age, 75 the father's, and 30 the difference, which is the father's age at the son's birth. Identifying which of the three is wanted is the whole difficulty of this family of problems.