The length of candle B is 3 times the length of candle A. The speed of burning of Candle B is 4 times the speed of burning of Candle A. A party begins with the lighting of these two candles. The party ends when the heights of these candles become equal. If the numerical value of length (in metre) of the burnt away portion of Candle A and the speed (in metre per hour) of its burning are same, how long, in hours, did the party continue ?
- (a)1/2
- (b)1
- (c)1 1/4
- (d)1 1/2
Answer
Why
Correct — B, (b) 1. One hour, and the step that decides it is the last sentence of the stem, not the ratios at the beginning.
Set up the symbols. Let candle A have length L metres and burn at v metres per hour. Then candle B has length 3L and burns at 4v.
WHEN DO THE HEIGHTS BECOME EQUAL? After t hours, A stands at L − vt and B at 3L − 4vt. The party ends when those are equal: L − vt = 3L − 4vt 3vt = 2L t = 2L / (3v).
THE CONDITION THAT FIXES THE ANSWER. The burnt away portion of candle A is its speed multiplied by the time, that is vt metres. The stem says this length, in metres, has the same numerical value as the speed, in metres per hour. So vt = v, and therefore t = 1.
The party lasted one hour. Notice what this condition really does: it does not tell us how long or how fast the candles are, and it does not need to. It compares distance travelled with speed, and the only way those two numbers can agree is if the time is one hour.
CHECK THAT THE WHOLE STORY HOLDS TOGETHER. Putting t = 1 into t = 2L/(3v) gives L = 3v/2, so the two conditions are consistent rather than contradictory. Take v = 2 metres per hour: then A is 3 metres long and B is 9 metres long, and B burns at 8 metres per hour. After one hour A stands at 3 − 2 = 1 metre and B at 9 − 8 = 1 metre — equal, as required. The burnt portion of A is 2 metres and its speed is 2 metres per hour, the same number, as required. Every condition in the stem is satisfied and the duration is one hour, whatever value of v is chosen.
The lesson for this kind of problem: when a word problem gives a relation between a DISTANCE and a SPEED as numbers, it is really giving you the time, because distance divided by speed is time.
Why the others are wrong
- (a)1/2 — Half an hour would require the burnt length of candle A to be half its hourly rate, not equal to it. A candidate who reaches this has usually formed the equation for equal heights, obtained t = 2L/(3v), and then estimated a value for L in terms of v rather than using the condition given. The condition is not an estimate; it fixes the time exactly.
- (c)1 1/4 — An hour and a quarter fits none of the conditions. Values of this shape appear when the ratios three and four are combined arithmetically — for instance by taking a fraction such as five-fourths out of the burning rates — instead of being used to write the two heights and set them equal. The ratios describe the candles; they do not by themselves produce the answer.
- (d)1 1/2 — An hour and a half is what comes out if the equal-height equation is solved as t = 3L/(2v) — that is, with the two-thirds inverted — and L and v are then taken to be equal. Both steps go wrong at once. From 3vt = 2L the time is two-thirds of L over v, not three-halves of it, and the numerical condition in the stem relates the BURNT LENGTH to the speed, not the whole length.
Concept
Uniform burning is a distance-speed-time problem in disguise: a candle burning at a constant rate loses height exactly as a vehicle covers distance, so height remaining equals initial height minus rate times time. Problems of this family are solved by writing an expression for each quantity at a general time t, imposing the condition described in words, and solving. Where the quantities are given only as ratios — three times as long, four times as fast — carrying them as multiples of a single unknown keeps the algebra clean, and the unknown usually cancels. Where the problem instead compares the numerical VALUES of two quantities measured in related units, as here with metres and metres per hour, that comparison is itself an equation, and it is generally the one that determines the answer.
The quantitative block of an EPFO general ability paper rewards candidates who read the last sentence of a word problem as carefully as the first. This item spends four sentences setting a scene of ratios and one sentence stating the condition that actually decides the answer. A candidate who works only with the ratios will spend a long time and arrive at nothing determinate, since the ratios alone leave both L and v free.
The stem is five sentences long, of which four set a scene and one states the condition that decides the answer. That distribution is deliberate. Candidates who begin calculating at the first sentence spend their time on ratios that cannot by themselves produce a number, and candidates who read to the end first see immediately that a length has been equated numerically with a speed, which is a statement about TIME. Where a word problem gives quantities only as ratios, the determining datum is almost always a separate numerical condition, and it is usually placed last.
Key facts
- For a uniformly burning candle, height remaining equals initial height minus burning rate multiplied by time.
- Setting two such expressions equal gives the moment at which the heights coincide.
- Here that gives t = 2L/(3v), where L and v are candle A's length and burning rate.
- Burnt length equals rate times time, so equating burnt length numerically with rate gives t = 1 directly.
- Distance divided by speed is time — a numerical equality between a distance and a speed therefore fixes the time at one unit.
- The two conditions together give L = 3v/2, so the data are consistent rather than over-determined.
- Concrete check: with v = 2 m/h, A is 3 m and B is 9 m; after one hour both stand at 1 m.
- Quantities given only as ratios should be carried as multiples of one unknown, which usually cancels.
Study next
Common traps
- Working only with the ratios and ignoring the numerical condition, which alone determines the answer.
- Inverting the fraction when solving 3vt = 2L.
- Equating the whole length of candle A with its speed rather than the burnt portion.
- Assuming a problem needs actual values of every quantity; here they cancel.
These papers set two or three word problems of this shape per paper. They typically supply data as ratios and then hide the deciding condition in the final clause. The reliable method is to write every quantity in terms of one unknown, translate each sentence into an equation as you read it, and only then look at the options.
Related PYQs
EPFO_APFC_2023_Q53A train starting from rest with a uniform acceleration attains a speed of 108 km/h in 5 minutes. The distance covered by the train in attaining the speed is
- (a) 9000 m
- (b) 4500 m
- (c) 355 m
- (d) 108 m
Answer(b) 4500 m
A distance-speed-time item from the earlier paper built on the same relation, where a train's uniform acceleration and final speed are used to find the distance covered.
Practice
- practice — not a real PYQ
A candle 24 cm long burns uniformly at 3 cm per hour and a second candle 30 cm long burns uniformly at 5 cm per hour. Both are lit together. After how many hours are the two candles of equal height ?
- (a)2
- (b)3
- (c)4
- (d)6
Answer(b) 3
- practice — not a real PYQ
A candle burns uniformly at r centimetres per hour. If the length burnt away, in centimetres, is numerically twice the burning rate, for how long has it been burning ?
- (a)Half an hour
- (b)One hour
- (c)Two hours
- (d)Four hours
Answer(c) Two hours