The original lay of a rectangular plot ABCD on open ground is 80 m long along AB, and 60 m wide along BC. Concreted pathways are intended to be laid on the inside of the plot all around the sides. The pathways along BC and DA are each 4 m wide. The pathways along AB and DC will mutually be of equal widths such that the un-concreted internal plot will measure three-fourth of the original area of the plot ABCD. What will be the width of each of these pathways along AB and DC ?

Correct — C, (c) 5 m.

FIRST, A NOTE ON READING THIS QUESTION. No figure is printed with it. The plot is described entirely in words, and nothing has been omitted — the corner labels, the two dimensions and the four pathways are all the information there is, and they are enough. The one thing a reader must supply for himself is the convention that the WIDTH of a path laid along a side is measured PERPENDICULAR to that side. Once that is fixed, the problem solves itself.

SET THE RECTANGLE UP. ABCD is a rectangle with its corners named in order, so AB and DC are the pair of opposite sides of one length and BC and DA the pair of the other. The stem gives AB as 80 m and BC as 60 m, so the plot is 80 m by 60 m and its area is

80 × 60 = 4800 square metres.

THE FIRST PAIR OF PATHS. The pathways along BC and DA are each 4 m wide. BC and DA are the 60-metre sides, one at each END of the 80-metre direction, so each of these strips is 60 m long and 4 m wide, and each eats 4 m out of the 80-metre dimension. Together they take 8 m:

inner length = 80 − 4 − 4 = 72 m

THE SECOND PAIR OF PATHS. The pathways along AB and DC are each of an unknown equal width w. AB and DC are the 80-metre sides, one at each end of the 60-metre direction, so each strip is 80 m long and w metres wide, and each eats w out of the 60-metre dimension:

inner width = 60 − 2w

THE CONDITION. The un-concreted inner plot is three-fourths of the original area:

72 × (60 − 2w) = (3/4) × 4800 = 3600 60 − 2w = 3600 ÷ 72 = 50 2w = 10 w = 5 metres

So each of the pathways along AB and DC is 5 m wide, which is option (c).

THE ELEGANT ROUTE, using factors instead of areas. The length has been cut from 80 to 72, a factor of 72/80 = 0·9. Area factors MULTIPLY, so if the two dimensions are reduced by factors f and g the area is reduced by f × g. We need the product to be 0·75, so

0·9 × g = 0·75 → g = 0·75 / 0·9 = 5/6

The width must therefore fall to five-sixths of 60, that is to 50 m, and the 10 m removed is shared equally between two paths at 5 m each. This route is quicker and it makes the structure visible: a path all round reduces each dimension by TWICE its width, and the area falls by the product of the two reduction factors.

A CHECK THAT ALSO SETTLES THE READING. Suppose the 4-metre paths were taken instead to eat into the 60-metre dimension. The inner width would be 60 − 8 = 52, and the condition would give an inner length of 3600 ÷ 52 = 69·23 m, so the remaining pathways would each be (80 − 69·23)/2 = 5·38 m — a figure that appears nowhere on the option list. Only the reading set out above produces one of the printed answers, which confirms that the width of a path along a side is measured across that side and not along it.

The stem prints "un-concreted" with a hyphen, spells "three-fourth" out in words rather than as a fraction, and uses "lay" rather than "layout", all as the booklet sets them.

Why the others are wrong

• (a)3 m — Three metres would leave an inner width of 60 − 6 = 54 m and an inner rectangle of 72 × 54 = 3888 square metres. As a fraction of the original 4800 that is 0·81, or eighty-one per cent — the paths would take only nineteen per cent of the plot, not the twenty-five per cent the stem requires. The option is the natural landing place for a candidate who has grasped that the answer must be reasonably close to the 4-metre width of the other pair but has guessed on the low side, or who has divided the required 10-metre reduction unevenly. It is worth noting how quickly back-substitution disposes of it. Each option can be tested in a single multiplication: reduce 60 by twice the trial width, multiply by 72, and compare with 3600. That check takes a few seconds per option and is the fastest way through the whole item for a candidate who is not confident in setting up the equation.
• (b)4 m — Four metres is the width given in the stem for the OTHER pair of pathways, and it is the strongest trap on this list. Questions of this shape supply one dimension and ask for a second, and the supplied dimension is almost never the answer, because if the two widths were equal the stem would have said the paths were of uniform width all round and asked a much easier question. The whole point of the phrase "will mutually be of equal widths such that" is that this second width is determined by the area condition rather than given. Test it: at 4 m the inner rectangle would be 72 × 52 = 3744 square metres, which is 0·78 of the original — a good deal more than three-quarters. To bring the area down to exactly three-quarters the second pair of paths must be wider than the first, and the arithmetic makes them 5 m. A uniform 4 m path all round would leave more than the stem allows.
• (d)6 m — Six metres would leave an inner width of 60 − 12 = 48 m and an inner rectangle of 72 × 48 = 3456 square metres, which is 0·72 of the original — the paths would swallow twenty-eight per cent of the plot instead of twenty-five. It is the mirror image of the lowest option, an overshoot rather than an undershoot, and it catches a candidate who has correctly worked out that the second pair must be wider than the first but has not pinned down by how much. There is a quick way to see that six is too many. The first pair of paths, at 4 m each, has already removed a tenth of the plot by cutting the length from 80 to 72. Only fifteen per cent of the original area remains to be taken by the second pair, and fifteen per cent of 4800 is 720 square metres; spread over two strips each 72 m long once the first pair is allowed for, that gives 720 ÷ (2 × 72) = 5 m apiece.

PATHS AND BORDERS INSIDE A RECTANGLE are a standard mensuration topic, and everything in it follows from one rule: a strip of width t laid along one side of a rectangle, on the inside, reduces the dimension PERPENDICULAR to that side by t. A strip on each of two opposite sides therefore reduces that dimension by 2t.

So for a rectangle L by B with an internal border of uniform width t all round, the inner rectangle is (L − 2t) by (B − 2t), and the area of the border is LB − (L − 2t)(B − 2t). The commonest error in the topic is subtracting t once instead of twice, and it comes from picturing the path on one side only.

THIS PROBLEM GENERALISES THAT SETTING by allowing the two pairs of paths to have different widths, which is why it names the sides individually. The inner rectangle is then (L − 2s) by (B − 2t), where s is the width of the pair on the B-sides and t the width of the pair on the L-sides. Keeping track of which width attaches to which dimension is the only real difficulty, and the reliable way to do it is to name the strips by what they eat rather than by where they lie: the paths along the short sides eat into the LONG dimension, and the paths along the long sides eat into the SHORT one.

AREA FACTORS MULTIPLY, and this is the idea worth carrying beyond mensuration. If one dimension is scaled by f and the other by g, the area is scaled by f × g. Here f = 72/80 = 0·9, and the requirement that the product be 0·75 forces g = 5/6. The same principle explains why halving a radius quarters a circle's area, why a photograph enlarged by 20 per cent in both directions gains 44 per cent in area, and why two successive percentage changes never simply add. Setting a mensuration problem up in factors rather than in absolute areas often removes the arithmetic altogether.

A RELATED FAMILY worth recognising: a path laid OUTSIDE the rectangle adds 2t to each dimension instead of subtracting it; a path of uniform width running through the middle of a field, in a cross, is handled by adding the two strips and subtracting their overlap, which is the inclusion-exclusion principle in geometric dress; and a border of unknown width whose AREA is given leads to a quadratic in t, of which only the root smaller than half the shorter side is admissible.

READING A GEOMETRY QUESTION SET ENTIRELY IN WORDS is a skill in itself. Name the corners in order, fix which pair of sides is which length, and write down the two dimensions before touching the conditions. A rectangle ABCD has AB opposite DC and BC opposite DA, and that ordering is what lets a stem describe a shape precisely without drawing it.

Key facts

• The plot ABCD is 80 m along AB and 60 m along BC, so its area is 4800 square metres and the inner plot must be three-quarters of that, namely 3600.
• The width of a path laid along a side is measured perpendicular to that side, so a pair of paths on opposite sides reduces that dimension by twice their width.
• The 4 m paths lie along BC and DA, the 60 m sides, so they eat into the 80 m dimension and leave an inner length of 80 − 8 = 72 m.
• The unknown paths lie along AB and DC, the 80 m sides, so they eat into the 60 m dimension and leave an inner width of 60 − 2w.
• The condition 72 × (60 − 2w) = 3600 gives 60 − 2w = 50, hence w = 5 metres.
• By factors: the length falls to 72/80 = 0·9 of its value, so the width must fall to 0·75 ÷ 0·9 = 5/6 of 60, that is to 50 m, and 10 m is shared between two paths.
• Back-substitution checks the option list quickly — widths of 3, 4 and 6 metres leave inner areas of 3888, 3744 and 3456 square metres, that is 81, 78 and 72 per cent of the plot.
• Reading the 4 m paths as eating into the 60 m dimension instead would make the unknown width about 5·38 m, which matches no option and so confirms the intended reading.
• No figure is printed with this item; the geometry is given entirely in words and the corner labels carry the information a drawing would otherwise supply.

Common traps

• Subtracting a path's width once instead of twice. Two opposite strips each take their width from the same dimension.
• Attaching the 4 m width to the wrong pair of sides. The paths along BC and DA reduce the 80 m dimension, not the 60 m one.
• Answering with 4 m, the width supplied in the stem for the other pair of paths. A supplied dimension is almost never what is being asked for.
• Assuming both pairs of paths must be equally wide. The stem makes the second pair equal to EACH OTHER, not equal to the first pair.
• Looking for a figure that is not there. The description is complete, and the corner labels do the work a drawing would.

• practice — not a real PYQ

  A rectangular lawn measuring 40 m by 30 m has a path of uniform width 2 m running inside it all round. What is the area of the path ?

  • (a)264 square metres
  • (b)272 square metres
  • (c)280 square metres
  • (d)296 square metres

  Answer(a) 264 square metres — a 2 m path on each of two opposite sides takes 4 m from each dimension, so the inner rectangle is 36 m by 26 m with an area of 936. The path is the difference, 1200 − 936 = 264 square metres.

• practice — not a real PYQ

  A rectangular plot measures 50 m by 40 m. A path of uniform width is laid inside it all round, and the inner rectangle that remains has an area of 1344 square metres. What is the width of the path ?

  • (a)2 m
  • (b)3 m
  • (c)4 m
  • (d)5 m

  Answer(c) 4 m — the inner rectangle is (50 − 2w) by (40 − 2w), and setting that product equal to 1344 gives w² − 45w + 164 = 0, whose admissible root is 4. Checking, 42 × 32 = 1344, and the other root of 41 is impossible since it exceeds half the shorter side.