EPFO EO/AO 2023 General Ability Test · Question 105 of 120

Correct — B, (b) 3 km. The stem deliberately refuses to name the first direction. It is the closing condition, that he ends up east of where he started, which pins it down. Treat the walk as two displacements added together. The second leg is 4 km due south, so it moves him 4 km southward and not at all east or west. If the final position is due east of the start, then the north-south accounts must balance exactly: his total northward movement must equal his total southward movement, otherwise he would finish north-east or south-east of the start rather than due east. The only source of northward movement is the first leg, so the first leg must carry him exactly 4 km north. The first leg is 5 km long. Resolve it into its north component and its east component. The north component is 4 km, and by Pythagoras the east component is the square root of 5 squared minus 4 squared, which is the square root of 25 - 16 = 9, so 3 km. That first leg is therefore a slanting walk 3 km east and 4 km north, the familiar 3-4-5 right triangle in disguise. Adding the two legs: east 3 + 0 = 3 km, north 4 - 4 = 0 km. He finishes 3 km due east of his starting point, so the distance is 3 km. Two wrong readings are worth naming because both produce numbers that are not on the list, and their absence is itself a clue. Adding the two distances gives 5 + 4 = 9 km, which is not offered. Applying Pythagoras to the two legs as though they were perpendicular gives the square root of 25 + 16, about 6.4 km, which is also not offered. The option set contains neither, which tells a candidate that the closing condition is not decoration and has to be used.

• (a)2 km — For the final distance to be 2 km due east, the first leg would need an east component of 2 km, and then its north component would be the square root of 25 - 4, about 4.58 km. After walking 4 km south he would still be about 0.58 km north of the starting line, so he would finish north-east of the start and not due east, contradicting the stem. The option is there for a candidate who has taken the difference of the two given distances in the wrong direction, or who has guessed at a small number without resolving the slanting leg into components.
• (c)4 km — Four kilometres is the length of the second leg, and it is the answer a candidate reaches by carrying a number straight down from the stem. Tested properly it fails: an east component of 4 km on a 5 km leg leaves a north component of 3 km, so after the 4 km walk south he would be 1 km south of the starting line and 4 km east of it, which is south-east rather than due east, and his distance from the start would be the square root of 16 + 1, about 4.12 km. The stem's closing condition rules that arrangement out. Being alert to an option that simply repeats a figure from the question is a useful reflex in direction problems.
• (d)5 km — Five kilometres is the length of the first leg, offered for the same reason as (c). If the whole first leg were due east, he would be 5 km east of the start, and the 4 km walk south would then leave him 5 km east and 4 km south, which is not due east at all; his distance from the start would be the square root of 25 + 16, about 6.4 km. The only way the walk can end due east is for the first leg to have a northward component of exactly 4 km, and a leg that is entirely eastward has none. The number is a decoy drawn from the stem, and the closing condition is what disqualifies it.

Direction-and-distance questions are vector addition written in plain English. Set the starting point at the origin, treat north and east as the positive directions, and record every leg of the journey as a pair of components, one east-west and one north-south. Legs along a named compass direction contribute to one component only. A leg described merely as 'a certain direction' has both components unknown, and something later in the stem must supply the missing information; here it is the statement that the walk ends due east, which forces the north-south components to cancel. Once every leg is resolved, the final position is the sum of the east components and the sum of the north components, and the straight-line distance from the start is the square root of the sum of their squares. Two facts make these questions quick. First, 'due east of the start' means the net north-south displacement is zero, and 'due north of the start' means the net east-west displacement is zero; each is one equation. Second, examiners build the slanting legs out of Pythagorean triples, so recognising 3-4-5, 5-12-13, 8-15-17 and 7-24-25 on sight saves the whole calculation. A rough sketch, even a bad one, prevents almost every error in this family, because it makes an impossible configuration visible immediately.

The reasoning blocks of the EO/AO General Ability Test regularly include one direction-sense item, and its difficulty is controlled by how much of the route is left unnamed. An easy version names every direction and asks only for the resultant. A harder version, like this one, hides a direction and supplies instead a condition about where the walker finishes, which the candidate has to convert into an equation. The habit rewarded is to read the last sentence of the stem as data rather than as a restatement of the question, and to draw the route before computing anything.

• Resolve every leg into an east-west component and a north-south component, then add the components separately.
• 'Ends up due east of the start' means the net north-south displacement is zero.
• 'Ends up due north of the start' means the net east-west displacement is zero.
• A 5 km leg with a 4 km north component has an east component of 3 km, since 3, 4 and 5 form a right triangle.
• The straight-line distance from the start is the square root of the sum of the squares of the net components.
• Useful Pythagorean triples for this family: 3-4-5, 5-12-13, 8-15-17 and 7-24-25.
• Here the totals are 3 km east and 0 km north, so the answer is 3 km due east.

• Adding the leg lengths instead of the components; distance travelled is not displacement.
• Applying Pythagoras to the two legs as though they were at right angles when one of them is slanting.
• Ignoring the closing condition about where the walk ends, which is the only thing fixing the first direction.
• Choosing an option that simply repeats a number from the stem, such as 4 km or 5 km here.
• Confusing a turn to the right, which depends on the way the walker faces, with an absolute direction such as east.

EO/AO papers set direction sense in three shapes. The first is a chain of named legs ending in 'how far and in which direction is he from the start', solved by adding components. The second, used here, hides one direction and supplies a condition about the final position instead. The third involves turns to the left or right, which must be interpreted relative to the direction the walker currently faces, and where a small sketch is not optional. Practise resolving into components rather than visualising, and keep the common triples in mind so that the arithmetic never becomes the obstacle.

No directly related past PYQ was found.

• practice — not a real PYQ

  A man walks 13 km in a certain direction and then walks 5 km due south. If he ends up due east of his starting point, how far is he from the starting point ?

  • (a)8 km
  • (b)10 km
  • (c)12 km
  • (d)14 km

  Answer(c) 12 km

• practice — not a real PYQ

  A person walks 6 km north, then 8 km east, and then 6 km south. How far is he from his starting point, and in which direction ?

  • (a)8 km, east
  • (b)10 km, north-east
  • (c)14 km, east
  • (d)2 km, south

  Answer(a) 8 km, east