A rectangular garden is to be twice as long as its width. If 360 m of fencing including gates will totally enclose this garden, what is the length of the garden ?
- (a)120 m
- (b)130 m
- (c)140 m
- (d)150 m
Answer
Why
Correct — A, (a) 120 m.
FENCING IS PERIMETER. That is the first translation to make, and it is the one the phrase 'including gates' is there to settle: the 360 metres runs all the way round the plot and nothing is to be deducted for the openings. So the perimeter of the rectangle is exactly 360 m.
SET IT UP AND SOLVE IT.
Let the width be w. 'Twice as long as its width' makes the length 2w. Perimeter = 2 x (length + width) = 2 x (2w + w) = 6w 6w = 360, so w = 60 Length = 2w = 120 m
The garden is 60 m by 120 m; its perimeter is 2 x (120 + 60) = 360 m, as required.
THE ONE-LINE VERSION, WHICH IS THE INSIGHT WORTH KEEPING. When the length is twice the width, the perimeter consists of two lengths and two widths, that is 2w + 2w + w + w — SIX EQUAL PARTS, each equal to the width. The length is two of those six parts, so
length = one third of the perimeter = 360 / 3 = 120 m
and the width is one sixth, 60 m. Any rectangle in a stated ratio can be handled this way: express the perimeter as a number of equal parts and read off whichever side is asked for. With a length-to-width ratio of 3 : 1 the perimeter is eight parts; with 3 : 2 it is ten.
READ WHICH SIDE IS BEING ASKED FOR. The width, 60 m, is the number that comes out of the division first, and it is the natural place to stop. The question asks for the LENGTH, so one more step is needed. Notice that the paper does not even offer 60 among the choices — a candidate who stops early finds nothing to mark and is sent back to the working, which is a mercy rather than a trap.
CHECKING BACKWARDS IS FASTER THAN SOLVING FORWARDS HERE. Take any option as the length, halve it for the width, and add up the perimeter: only 120 m returns 360 m. The other three give 390, 420 and 450 metres, each demanding more fence than the question supplies.
Why the others are wrong
- (b)130 m — Test it against the data and it fails at once. If the length were 130 m then the width, being half of it, would be 65 m, and the fence needed would be 2 x (130 + 65) = 390 m — thirty metres more than the 360 m the question allows. There is no reading of the stem on which the extra fence appears from anywhere; 'including gates' means the gates are part of the 360 m rather than an addition to it. This option and the two after it are simply the correct answer walked upward in steps of ten, and each step of ten metres in the length adds thirty metres to the fence, because a longer side and a wider side both grow. The general lesson is that a mensuration answer can always be checked by substitution in a single line, and on an item with four numerical choices that check is usually quicker than a fresh solution.
- (c)140 m — A length of 140 m implies a width of 70 m and a perimeter of 2 x (140 + 70) = 420 m, which is sixty metres beyond what the question provides. It is worth asking where a number like this could come from, because that is what makes an option instructive rather than mere padding. No standard mis-step in this problem produces 140: dividing 360 by 6 gives the width, 60; multiplying by 2 gives the answer, 120; halving the perimeter gives 180, the sum of a length and a width; dividing by 4 gives 90, the side of a square with the same perimeter. None of those is 140. When an option cannot be produced by any plausible error, that is a reason to check the arithmetic that DID produce your own answer rather than to worry about which error the option represents.
- (d)150 m — Here the width would be 75 m and the fence 2 x (150 + 75) = 450 m, a quarter more than is available. This is the largest of the three wrong values, and it is worth noticing what would have to change in the question for it to be right: if only THREE sides were being fenced, say because one side ran along a wall, the fencing would be 2w + l = 2w + 2w = 4w, giving w = 90 and a length of 180 m — still not 150. If the 360 m were the SEMI-perimeter, the length would be 240 m. So no natural variation of the problem lands here either. The one habit that protects against all of these is the same: convert the words to an equation before touching the numbers, and state in that equation exactly which lengths the fence covers.
Concept
THIS IS A PERIMETER PROBLEM DISGUISED AS A FENCING PROBLEM, and almost all of the work is in the translation from words to algebra.
THE MENSURATION FACTS NEEDED, for a rectangle of length l and width w:
PERIMETER = 2(l + w) — the distance all the way round, and therefore the length of fence, boundary wall, edging or ribbon required. AREA = l x w — the surface enclosed, and therefore the quantity of grass, tiles, paint or crop. DIAGONAL = square root of (l squared + w squared), by Pythagoras.
KEEPING PERIMETER AND AREA APART is the single most useful discipline in this chapter, because the words of a problem point at one or the other and the two behave completely differently. Fencing, running, edging and walking round all mean perimeter. Turfing, tiling, painting and cropping all mean area. And they scale differently: double every side of a rectangle and the perimeter doubles while the area quadruples.
TRANSLATING THE ENGLISH is where marks are actually won or lost:
'twice as long as its width' → l = 2w 'the length exceeds the width by 5' → l = w + 5 'in the ratio 3 : 2' → l = 3k, w = 2k 'a square' → l = w
THE RATIO METHOD, which turns most of these into mental arithmetic. If the sides are in the ratio a : b, write them as a x k and b x k. The perimeter is then 2(a + b) x k, a fixed number of equal parts, and any side asked for is a known fraction of the perimeter. For 2 : 1 the perimeter is six parts, so the length is one third of it and the width one sixth. For 3 : 1 it is eight parts; for 3 : 2, ten parts.
THE VARIATIONS EXAMINERS BUILD ON THIS BASE. Fencing three sides only, because a wall or a river closes the fourth. A path of uniform width laid inside or outside a plot, where the trick is to write down the outer and inner rectangles and subtract. A gate or gap that must be deducted from the fencing — which is precisely what this question rules out by saying the 360 metres includes the gates. And cost problems, where the fencing rate per metre or the turfing rate per square metre is applied afterwards, and the only real question remains whether the quantity wanted is a perimeter or an area.
Quantitative aptitude is the largest strand on this APFC paper — roughly a third of Part B — and mensuration is a steady part of it, appearing here as a rectangle, as a staircase-shaped figure, as circles, and as a plot with concreted pathways laid inside it. None of these needs a formula beyond school level; what they need is the discipline of setting up the right equation and answering the question actually asked.
This item is at the easy end, and it is best treated as a timing opportunity rather than a puzzle. In a section this large, an item that resolves in two lines should be finished in well under a minute so that the time goes to the ones that genuinely need it. The two lines are: fencing means perimeter, and the length is one third of it.
The habit that most improves accuracy across the whole quantitative section is to underline the quantity being asked for before starting. Word problems characteristically make you compute an intermediate value — the width here — that is not the answer, and stopping at it is one of the commonest ways of losing an item that was solved correctly. A useful routine is to write down what the unknown letter stands for at the top of the working, and to check it again at the end.
The second habit is the back-substitution check. Wherever four numerical options are offered and the relation is simple, testing an option against the data takes one line and confirms the answer outright.
Key facts
- Fencing, walling and edging all mean PERIMETER; turfing, tiling and painting all mean AREA. The words of the problem decide which.
- Perimeter of a rectangle = 2(length + width); area = length x width.
- If the length is twice the width, the perimeter is six equal parts of the width, so the length is one third of the perimeter and the width one sixth.
- Here 6w = 360 gives w = 60 m and length = 120 m; the check is 2 x (120 + 60) = 360 m.
- The phrase 'including gates' means nothing is to be deducted from the fencing for the openings — the whole 360 m is the boundary.
- In general, if the sides are in the ratio a : b, the perimeter is 2(a + b) equal parts and any side is a known fraction of it.
- Doubling every side doubles the perimeter but quadruples the area — the two do not scale alike.
- When only three sides are fenced, because a wall or river closes the fourth, the fencing is 2w + l rather than 2(l + w).
- With four numerical options, back-substitution is usually the fastest confirmation: here 130, 140 and 150 m give perimeters of 390, 420 and 450 m.
Study next
Common traps
- Answering the intermediate value. The width comes out first; the question asks for the length.
- Using area when the problem is about fencing, or perimeter when it is about turfing.
- Deducting for gates when the stem says the fencing includes them.
- Writing the perimeter as l + w rather than 2(l + w), which halves every answer.
- Reading 'twice as long as its width' backwards and making the width twice the length.
- Skipping the one-line check by substitution, which would catch every one of these errors.
Mensuration on EPFO papers comes as short word problems with four close numerical options, and the arithmetic is deliberately light — the examiner is testing the translation from English into an equation and the discipline of answering the question asked. Rectangles and circles dominate, usually with one relation given between the sides or the radii and one total given for the perimeter, the area or a cost. Where a figure is involved the dimensions are stated in the stem, so nothing depends on measuring. Items of this kind are best done early and quickly, with a substitution check at the end, so that the harder data-interpretation and statistics questions in the same section get the time they need.
Related PYQs
EPFO_APFC_2016_Q109The 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 ?
- (a) 3 m
- (b) 4 m
- (c) 5 m
- (d) 6 m
Answer(c) 5 m
A rectangular plot 80 m by 60 m with concreted pathways laid inside it, asking for the path width — the same rectangle mensuration one step harder, and the classic inner-and-outer rectangle subtraction.
EPFO_APFC_2016_Q19What is the perimeter of the figure shown below ? AJ = 10 cm, JI = 12 cm, AB = x, CD = x + 1, EF = x + 2, GH = x + 3, BC = DE = FG = HI = y
- (a) 44 cm
- (b) 48 cm
- (c) 54 cm
- (d) 58 cm
Answer(a) 44 cm
The perimeter of a staircase-shaped figure on this same paper — perimeter asked where the sides must first be reorganised, and the item that shows why perimeter reasoning is worth practising.
EPFO_APFC_2016_Q113If the radius of a circle is reduced by 50%, its area will be reduced by
- (a) 30%
- (b) 50%
- (c) 60%
- (d) 75%
Answer(d) 75%
By how much the area of a circle falls when the radius is halved — the scaling half of mensuration, where area and linear measure behave differently.
Practice
- practice — not a real PYQ
The length of a rectangle is three times its width. If the perimeter of the rectangle is 96 cm, its length is
- (a)24 cm
- (b)30 cm
- (c)36 cm
- (d)48 cm
Answer(c) 36 cm — with length 3w and width w the perimeter is 8w, so 8w = 96 gives w = 12 cm and length 36 cm. In ratio terms the perimeter is eight equal parts and the length is three of them, that is three eighths of 96.
- practice — not a real PYQ
A rectangular field is twice as long as it is wide, and its area is 512 square metres. Its perimeter is
- (a)96 m
- (b)128 m
- (c)144 m
- (d)192 m
Answer(a) 96 m — with width w the area is 2w squared, so 2w squared = 512 gives w = 16 m and length 32 m, and the perimeter is 2 x (32 + 16) = 96 m. Note that the same ratio of sides is given here but the total supplied is an area rather than a perimeter, so the first step is a square root rather than a division.