Four branches of a company are located at M, N , O and P . M is in the North of N at a distance of 4 km; P is in the South of O at a distance of 2 km; N is in the Southeast of O by 1 km. What is the distance between M and P in km?
- (a)5·34
- (b)6·74
- (c)28·5
- (d)None of the above
Correct — A, 5·34. Set coordinates rather than sketching: put O at the origin, with East along the positive x-axis and North along the positive y-axis, and measure everything in kilometres. P is 2 km due South of O, so P = (0, −2). N is 1 km Southeast of O, and in these items Southeast always means the exact 45-degree bisector, so that 1 km splits equally into an eastward and a southward component of 1 ÷ √2 ≈ 0·707 km each: N = (0·707, −0·707). M is 4 km due North of N, which changes only the y-coordinate: M = (0·707, −0·707 + 4) = (0·707, 3·293). The two points wanted are now fixed. Their horizontal separation is 0·707 km — the whole of it inherited from N's eastward offset, because M sits directly above N and P directly below O. Their vertical separation is 3·293 − (−2) = 5·293 km. Pythagoras gives MP = √(0·707² + 5·293²) = √(0·5 + 28·01) = √28·51 ≈ 5·34 km. In exact form MP = √(37 − 6√2), which evaluates to 5·3399, so the printed 5·34 is the correctly rounded value.
- (b)6·74 — This is what a North-for-South slip on the diagonal produces. Place N a kilometre to the north-east of O instead of the south-east and M rises to y = 4 + 0·707 = 4·707, so the vertical leg becomes 6·707 and √(0·5 + 44·99) ≈ 6·74. The stem says Southeast, which carries N below the east-west line through O and pulls M down by that same 0·707 km, not up.
- (c)28·5 — 28·51 is MP squared, offered to the candidate who does everything right and then forgets the final square root. It can also be rejected without any arithmetic at all: the three distances the question supplies are 4, 2 and 1 km, so even if every branch were strung out in a straight line no two of them could be more than 7 km apart. A separation of 28·5 km is impossible.
- (d)None of the above — Only correct if none of the three numeric options matched, and 5·34 does match to the two decimal places printed. It is the honest choice for a candidate who cannot resolve a diagonal into components — but the resolution is standard: a displacement of d km along any of the four intercardinal directions is d ÷ √2 in each of the two cardinal directions it lies between.
Direction-and-distance items place points relative to one another using the compass and ask for a separation, a final bearing or a net displacement. Two conventions make them solvable and both are fixed. North is taken up the page and East to the right, so a right turn from North faces East. And the four intercardinal directions — north-east, south-east, south-west, north-west — mean the exact 45-degree bisectors, never a vague quadrant; a journey of d km to the south-east is therefore d ÷ √2 km east together with d ÷ √2 km south. Once those are accepted the whole family collapses into coordinate geometry: give one point the origin, convert every instruction into a change in x or y or both, and finish with the distance formula. Where the legs are perpendicular the last step is a Pythagoras triple often enough — 3-4-5, 5-12-13, 8-15-17 — that recognising one saves time, though not here.
The instruction to read first is the one that fixes a shared reference. Here O is named twice, in the clauses about P and about N, so O is the natural origin; starting from M or P instead means carrying an unknown through the working. After that the order does not matter. The step that decides the question is resolving the single diagonal: N is a kilometre south-east of O, which is 0·707 km east and 0·707 km south, and that eastward 0·707 is the only horizontal displacement anywhere — M is due north of N and P is due south of O, so both of them simply inherit the x-coordinate of the point they were measured from. Recognising that turns a four-point problem into one right-angled triangle with legs 1 ÷ √2 and 6 − 1 ÷ √2. Two exam-hall habits pay here. Carry a bound: the legs given total 4 + 1 + 2 = 7 km, so any option above 7 is not a distance, which kills 28·5 in a second. And check the last operation: a squared quantity offered as a length is the standard final trap, and 28·5 is exactly MP². One caveat worth stating plainly — the answer depends on Southeast meaning precisely 45 degrees. Read loosely as anywhere would be under-determined and MP could be anything from 5 km, if N were due south of O, to about 6·08 km, if N were due east; the 45-degree convention is what makes the question well posed, and it is the convention every setter of these items uses.
- Intercardinal convention: a displacement of d km to the south-east is d ÷ √2 km east and d ÷ √2 km south — for d = 1 km that is about 0·707 km in each direction
- Coordinates with O at the origin, East as +x and North as +y: O (0, 0), P (0, −2), N (0·707, −0·707), M (0·707, 3·293)
- MP has a horizontal leg of 1 ÷ √2 ≈ 0·707 km and a vertical leg of 6 − 1 ÷ √2 ≈ 5·293 km, so MP = √(37 − 6√2) = 5·3399 ≈ 5·34 km
- Sanity bound: the three given legs total 4 + 1 + 2 = 7 km, so no two branches in this figure can be further apart than 7 km — which disqualifies 28·5 as a distance before any calculation
- 28·51 is MP², and 6·74 is what you get from √(0·5 + 6·707²) after placing N to the north-east instead of the south-east
The only horizontal displacement in the whole figure is N's 0·707 km eastward offset: M sits directly above N and P directly below O, so each inherits its x-coordinate unchanged.
- Treating an intercardinal direction as a rough quadrant rather than as an exact 45-degree bisector, which leaves the figure under-determined
- Confusing north-east with south-east on the diagonal leg — the error that produces 6·74
- Stopping at the squared value: 28·51 is MP², and the option list prints 28·5 to catch exactly that
BPSC has put a direction item in the reasoning block of successive General Studies papers and moves it along a spectrum — turns only, with the answer a compass direction, in the 2025 block, and a full coordinate problem with distances and a decimal answer here in 2023. The decimal answer is the signal that a diagonal has to be resolved rather than merely tracked. UPSC keeps direction sense in CSAT Paper II under general mental ability, usually as a turns-and-facing puzzle rather than as a distance calculation.
No directly related past PYQ was found.
- practice — not a real PYQ
A cyclist rides 2√2 km to the north-east from point X and then rides 2 km due south. How far is he from X?
- (a)2 km
- (b)2√2 km
- (c)4 km
- (d)6 km
Answer(a) 2 km — 2√2 km to the north-east is 2 km east and 2 km north, taking him to (2, 2); riding 2 km south brings him to (2, 0), which is 2 km due east of X.
- practice — not a real PYQ
A man starts from point X, walks 4 km north, then 3 km east, then 8 km south. How far is he from X?
- (a)5 km
- (b)7 km
- (c)15 km
- (d)25 km
Answer(a) 5 km — he ends at (3, −4), so the distance is √(3² + 4²) = 5 km; 15 km is the total walked and 25 km is the square of the answer.