In a desert, X and Y are two spots that are separated by 100 km. At point X, are placed two pillars, one black and one white. Along the line joining X and Y, a white pillar is placed for every 180 m and a black pillar is placed for every 350 m. How many times can one find black and white pillars together while traveling from X to Y ?
- (a)14
- (b)15
- (c)16
- (d)17
Correct — C, (c) 16. A black and a white pillar stand together only where a multiple of 180 m coincides with a multiple of 350 m, so the coincidences occur at the common multiples of the two spacings — that is, at multiples of their LCM. Step 1, the LCM. 180 = 2² × 3² × 5 and 350 = 2 × 5² × 7, so LCM(180, 350) = 2² × 3² × 5² × 7 = 4 × 9 × 25 × 7 = 6300 m. The two kinds of pillar meet every 6300 metres, and nowhere else. Step 2, how many such points lie on the journey. The distance is 100 km = 1,00,000 m. Dividing, 1,00,000 ÷ 6300 = 15.87, so the multiples of 6300 that fit are 6300 × 1 = 6300 up to 6300 × 15 = 94,500. That is 15 meeting points after leaving X. (The next one would be 6300 × 16 = 1,00,800 m, which is past Y.) Step 3, the point that decides the answer. The stem states that at X itself two pillars are placed, one black and one white. X is the start of the journey and it already has a black and a white pillar together — a coincidence at distance zero, which is also a multiple of 6300. Counting it gives 15 + 1 = 16. That second sentence of the stem is not scene-setting. It is the whole question, and it is why the answer is 16 rather than 15. It is also worth checking the far end: 1,00,000 is divisible by neither 180 nor 350, so no pillar of either colour stands exactly at Y and there is no ambiguity about whether the endpoint counts. (The booklet prints 'traveling' with a single l; the stem is reproduced as printed.)
- (a)14 — Two short. There is no reading of the problem that produces 14 — it requires dropping both the pair at X and one of the fifteen interior coincidences. It is present as the outer edge of the option cluster, which is how a set of four consecutive integers is usually built: two plausible answers in the middle and one padding value on each side.
- (b)15 — This is the trap, and it is the answer a candidate reaches by doing the mathematics correctly and reading the stem carelessly. Fifteen is the number of coincidences strictly after X — the multiples 6300 through 94,500. It ignores the pair the stem expressly places at X itself. Any question that goes out of its way to tell you what is at the starting point is telling you the starting point counts.
- (d)17 — One too many. Seventeen requires counting both X and a further coincidence at or beyond Y, but 6300 × 16 = 1,00,800 m lies 800 m past Y, and no pillars of either colour stand exactly at Y in any case, since 1,00,000 is a multiple of neither 180 nor 350. This is the over-inclusive mirror of option (b).
This is an LCM problem wearing a story. Whenever two events repeat at fixed intervals — pillars every 180 m and every 350 m, bells every so many seconds, buses every so many minutes — they coincide exactly at the common multiples of the two intervals, and the first coincidence after the start is at the LCM. The count over a finite stretch is then the floor of the total length divided by the LCM, plus one if the starting point itself counts as a coincidence. That last clause is where these questions are won and lost, and it is why the fence-post distinction — intervals versus posts — is worth naming explicitly: fifteen gaps of 6300 m have sixteen endpoints.
The quantitative block of the EO/AO paper favours problems whose arithmetic is easy and whose reading is not. Here the LCM is a routine computation, but the option set offers four consecutive integers, which is the examiner's signal that the answer turns on a boundary decision rather than on the calculation. The habit rewarded is finishing the arithmetic and then re-reading the stem for what it says about the endpoints.
- Two repeating patterns with periods a and b coincide at the common multiples of a and b, that is, at multiples of LCM(a, b).
- 180 = 2² × 3² × 5; 350 = 2 × 5² × 7; LCM = 2² × 3² × 5² × 7 = 6300.
- 100 km = 1,00,000 m; 1,00,000 ÷ 6300 = 15.87, giving 15 complete multiples of 6300 within the distance.
- The stem places one black and one white pillar at X, so the starting point is itself a coincidence and must be counted: 15 + 1 = 16.
- 1,00,000 is divisible by neither 180 nor 350, so no pillar stands exactly at Y and the far endpoint raises no boundary question.
- Fence-post principle: n equal gaps along a line have n + 1 endpoints, so a count of positions is one more than a count of intervals.
- LCM(a, b) × HCF(a, b) = a × b, a useful check: HCF(180, 350) = 10, and 6300 × 10 = 63,000 = 180 × 350.
- Forgetting the coincidence at the starting point when the stem has explicitly placed pillars there.
- Counting a multiple that lies beyond the endpoint. 6300 × 16 exceeds 1,00,000.
- Adding rather than multiplying the spacings, or using the HCF instead of the LCM.
- Unit slips — the distance is given in kilometres and the spacings in metres, so 100 km must become 1,00,000 m before anything else happens.
The EO/AO quantitative block runs to about fifteen items across the paper, and the recurring types are LCM and HCF applications, time and work, ratio and proportion, averages, percentages and simple probability. They are testable in under a minute each once the underlying family is recognised, so the scoring skill is classification rather than calculation.
No directly related past PYQ was found.
- practice — not a real PYQ
Along a straight road 2 km long, a lamp post is placed every 40 m and a milestone every 60 m, both starting from the beginning of the road. At how many points does a lamp post stand together with a milestone ?
- (a)16
- (b)17
- (c)18
- (d)34
Answer(b) 17
- practice — not a real PYQ
Three bells ring at intervals of 9, 12 and 15 minutes respectively. If they ring together at 8 a.m., at what time will they next ring together ?
- (a)9 a.m.
- (b)10 a.m.
- (c)11 a.m.
- (d)12 noon
Answer(c) 11 a.m.