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
Answer
Why
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.
Why the others are wrong
- (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 in this figure.
- (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.
Concept
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 in the figure: 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 in this figure, 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 in the south-east quadrant, the figure 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.
Key facts
- 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.
Study next
Common traps
- 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.
Related PYQs
Ram goes North, turns right, then goes right again and then goes to left. In which direction Ram is now?
- (a) EAST
- (b) SOUTH
- (c) NORTH
- (d) WEST
Answer(a) EAST
The same family at its other end. The 2025 item gives turns and no distances, so it resolves by tracking the facing direction (North, right to East, right again to South, left back to East) while this 2023 item gives distances and a diagonal and has to be resolved in coordinates. Between them they mark the two forms BPSC alternates: a compass direction as the answer, or a number in kilometres.
Practice
- 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.