EPFO APFC 2023 General Studies · Question 36 of 120

Correct — D, (d) Neither 1 nor 2. Both statements are wrecked by the word 'always', and each is true only in a special case that the statement then over-generalises. Statement 1. If A coincides with O, then both chords pass through the centre, so both are diameters. Two diameters cut the circle into four sectors, and a sector's area depends on its angle. If the diameters meet at an angle of θ, the four sectors have angles θ, 180° − θ, θ and 180° − θ, so their areas are equal only when θ = 90°. Take two diameters at 60°: the sectors are 60°, 120°, 60°, 120°, with areas πr²/6, πr²/3, πr²/6, πr²/3 — two distinct values, not one. The four parts are equal only for perpendicular diameters, so the claim that they are always equal is false. Statement 2. If A does not coincide with O, the claim is that two of the four parts must be equal. This is true whenever the configuration is symmetric — if the two chords are equidistant from the centre, the line through O and A is an axis of reflection that exchanges the chords, and it carries one part on to another, forcing those two areas to be equal. But nothing in the statement requires the chords to be equidistant. Take a circle of radius 1 centred at O, one chord along the line at distance 0.2 from the centre and the other perpendicular to it at distance 0.5. They meet at a point that is not the centre, and the four parts have areas of about 0.21, 0.41, 0.97 and 1.56 square units, which sum to π and are all different. One counter-example is enough, so the claim fails. Neither statement survives, and the answer is (d). There is a nice sting in the tail: two of the four parts are always equal in the case statement 2 excludes, because two diameters produce two pairs of equal vertically opposite sectors. The property the statement asserts of the non-coinciding case is exactly the property that does hold in the coinciding one. Read the ask again. It is positive — which of the statements is or are correct — and the negation lives inside statement 2, in the words 'If A and O do not coincide'. On this paper seventeen questions carry a negative ask and the booklet prints the 'not' in bold italic on every one of them; here the 'not' is printed in ordinary weight, because it is part of a statement rather than part of the ask. A candidate trained by those seventeen items to watch for bold gets no warning at all on this one.

• (a)1 only — Choosing '1 only' means accepting the four-equal-parts claim, and it is the commonest error here because the mental picture that comes with 'chords through the centre' is the tidy one of two perpendicular diameters cutting the circle into quarters. That picture is a special case. Coincidence of A with O forces both chords to be diameters but says nothing about the angle between them, and at any angle other than a right angle the four sectors fall into two unequal pairs. The general lesson is that a condition fixes only what it states: 'A and O coincide' fixes the point of intersection, not the inclination of the chords, and everything the statement adds beyond that has to be checked.
• (b)2 only — Choosing '2 only' means rejecting the first statement, which is right, and accepting the second, which is not. The appeal of statement 2 is that a symmetric drawing supports it: sketch two chords of the same length crossing off-centre and two of the four parts really are equal, because the line joining the centre to the intersection point reflects the figure on to itself. But equal-length chords are equidistant from the centre, and the statement imposes no such condition. Chords at different distances from the centre destroy the symmetry, and the four parts then have four different areas. A single free parameter left unconstrained by the hypothesis is exactly where an 'always' claim breaks.
• (c)Both 1 and 2 — 'Both 1 and 2' is the option for a candidate who reads the two statements as a tidy pair — one describing the symmetric case and the other the asymmetric one — and accepts them together because they sound complementary. They are not complementary; they are both over-general. Statement 1 needs the diameters to be perpendicular before its conclusion follows, and statement 2 needs the chords to be equidistant from the centre before its conclusion follows. Neither condition is stated, and in a mathematics item the conclusion may use only what the hypothesis actually gives you.

Two facts about circles do all the work here. First, a chord passes through the centre if and only if it is a diameter, so requiring two chords to intersect at the centre is the same as requiring both to be diameters. Second, the area of a sector is proportional to its central angle: a sector of angle θ in a circle of radius r has area θ/360 × πr², which is why two diameters at an angle θ produce sectors of θ, 180° − θ, θ and 180° − θ, equal in vertically opposite pairs and all four equal only when θ is a right angle. When the chords do not pass through the centre the four regions are no longer sectors but pieces bounded by two chords and an arc, and their areas are computed as combinations of circular segments; the area of a segment cut off by a chord at distance d from the centre of a circle of radius r is r² cos⁻¹(d/r) − d√(r² − d²). Symmetry is the other tool. If a figure has an axis of reflection, regions exchanged by that reflection are equal in area, and two chords equidistant from the centre give exactly such an axis — the line through the centre and the point of intersection. Take the equal distances away and the axis disappears with them. Both statements in this question are applications of these ideas that stop one condition short.

The quantitative block of this paper leans on statement-evaluation items of exactly this type: two claims about a configuration, a positive ask, and the code set '1 only / 2 only / Both 1 and 2 / Neither 1 nor 2', which the guide to this booklet records as occurring on thirteen questions in all. The reason examiners like the format is that it tests judgement rather than computation — you are being asked whether a general claim follows from a stated hypothesis, and the fastest route to 'no' is a counter-example rather than a proof. The discipline that pays is to treat every universal word as a target: 'always', 'shall always have equal areas', 'must'. For each, ask what freedom the hypothesis leaves unconstrained, and then push that freedom to an extreme. Here the free parameters are the angle between the diameters in statement 1 and the two distances of the chords from the centre in statement 2, and pushing either one is enough. The second thing this item teaches is a reading habit. The negation in statement 2 is printed in ordinary weight because it belongs to the statement, not to the ask, and no typographic cue will save a candidate who has learned to scan for bold. Every statement must be read in full, whatever the ask looks like.

• A chord of a circle passes through the centre if and only if it is a diameter, so two chords intersecting at the centre are necessarily two diameters.
• The area of a sector is proportional to its central angle, so two diameters meeting at an angle θ divide the circle into sectors of θ, 180° − θ, θ and 180° − θ; vertically opposite sectors are always equal, and all four are equal only when the diameters are perpendicular.
• Two chords intersecting away from the centre divide the circle into four regions bounded by chord segments and arcs; the area of the circular segment cut off by a chord at distance d from the centre of a circle of radius r is r² cos⁻¹(d/r) − d√(r² − d²).
• If two chords are equidistant from the centre — equivalently, if they are of equal length — the line joining the centre to their point of intersection is an axis of reflection and two of the four regions are equal; if the distances differ, that symmetry is lost and all four regions can have different areas.
• A single counter-example disproves a statement containing 'always'; for a circle of radius 1 with perpendicular chords at distances 0.2 and 0.5 from the centre, the four regions have areas of roughly 0.21, 0.41, 0.97 and 1.56 square units, which sum to π and are all distinct.

• Reading 'chords through the centre' as 'perpendicular diameters'; coincidence of the intersection point with the centre says nothing about the angle, and at any other angle the four sectors form two unequal pairs
• Testing a general claim with the most symmetric drawing you can make; a symmetric figure supports both statements here, and only an asymmetric one exposes them
• Expecting the negation to be flagged: the 'not' in statement 2 is printed in ordinary weight because it belongs to the statement rather than to the ask, unlike the seventeen negative-ask items on this paper where the booklet prints it in bold italic
• Accepting two statements because they look complementary; both describe over-general versions of true special cases, and each has to be judged on its own hypothesis

The quantitative and reasoning blocks of this paper use the two-statement code format repeatedly, and geometry items in that format almost always turn on an unstated condition rather than on a calculation. Expect claims about equal areas, equal lengths, concurrency or similarity that hold in the symmetric case and fail in general, and expect no diagram, so that drawing the figure yourself — twice, once symmetric and once deliberately lopsided — is part of the method. The same approach answers the statistics and reasoning items nearby, where a claim about means or about seating is true for one arrangement and asserted for all.

No directly related past PYQ was found.

• practice — not a real PYQ

  Two diameters of a circle intersect at an angle of 60°. Which of the following statements is/are correct? 1. The circle is divided into four parts of equal area. 2. The four parts consist of two pairs of equal areas. Select the correct answer using the code given below.

  • (a)1 only
  • (b)2 only
  • (c)Both 1 and 2
  • (d)Neither 1 nor 2

  Answer(b) 2 only — two diameters meeting at 60° create sectors of 60°, 120°, 60° and 120°, so the areas are πr²/6, πr²/3, πr²/6 and πr²/3. Vertically opposite sectors are equal, giving two pairs, but the four parts are equal to one another only when the diameters are perpendicular, so the first statement is false and the second is true.

• practice — not a real PYQ

  In a circle of radius r, two chords of equal length intersect at a point other than the centre. Which one of the following must be true of the four parts into which they divide the circle?

  • (a)All four parts have equal areas
  • (b)At least two of the parts have equal areas
  • (c)No two parts can have equal areas
  • (d)Three of the parts have equal areas

  Answer(b) At least two of the parts have equal areas — chords of equal length are equidistant from the centre, so the line joining the centre to their point of intersection is an axis of reflection for the figure, and the reflection carries one region on to another, forcing those two areas to be equal. Equality of all four would require the chords to be diameters as well as perpendicular, which the hypothesis does not give.