Fifteen marks on every GATE paper, needing no subject knowledge, and routinely neglected. The candidates who lose them are almost never the ones who could not do the arithmetic.
Percentages, mixtures, work rates, speeds and scale factors are all the same idea in different clothing. Once a question is set up as a proportion, the arithmetic is trivial; nearly all the errors happen before that, in deciding what the base quantity is.
That result is worth remembering as a caution: a rise followed by an equal fall does not return to the start, because the second change acts on a different base.
Alligation handles mixtures in one line. To find the ratio in which two components at concentrations c and d must be mixed to give a mean C:
Ratios also handle proportional division, partnership profits and scale problems. If a quantity is divided in the ratio 3 : 4 : 5, the parts are 3/12, 4/12 and 5/12 of the whole — write the denominator as the sum of the parts and there is nothing left to get wrong.
Work. Convert everyone to a rate per unit time, then add the rates. Never average the times.
For questions with people joining and leaving, work in fractions of the job completed rather than in days — it keeps the bookkeeping straight, as in worked example 1.
Speed. Average speed is total distance over total time, always. Which mean that reduces to depends on what is held equal:
Relative speed. Add when closing, subtract when moving the same way. For trains or objects with length, the distance covered is the sum of the lengths when passing each other, or one length when passing a fixed point.
Boats and streams are the same idea: downstream speed is boat plus current, upstream is boat minus current, and the two together let you solve for either unknown.
The first decision is always whether order matters. If it does, permutations; if not, combinations.
Then look for a restriction, because that is where the question actually lives. Three techniques cover most of them:
Keep two distinctions clear. Mutually exclusive is not independent — mutually exclusive events cannot both happen, so knowing one occurred tells you a great deal about the other. And with replacement versus without changes whether the probabilities stay constant between draws.
A chart or table with three or four questions attached. The marks are quick, and they are lost to carelessness rather than difficulty.
Read the axes and the units before reading the question — figures given in lakhs, in percentages of a total that itself changes, or on a logarithmic scale are all standard misdirections. Then check whether the question asks for an absolute value or a change, and whether that change is in absolute terms or as a percentage.
The most common trap is the percentage of a moving base: a segment whose share falls from 40 % to 35 % may still have grown in absolute terms, if the total grew faster. Whenever a question mixes shares and totals, convert to absolute numbers first.
Estimate before computing. Most data interpretation options are far enough apart that a rough figure identifies the answer, and the exact calculation is only needed when two options are close.
Roughly half the aptitude section, and the part engineering candidates most often leave to chance.
Reading comprehension and inference. The correct option is the one that cannot be false given the passage. Anything requiring an extra assumption, however reasonable, is wrong. Options that are true in the world but not supported by the passage are the standard distractor.
Word usage. Learn confusable pairs rather than long vocabulary lists — principal and principle, affect and effect, comprise and compose, complement and compliment, discreet and discrete, imply and infer. These recur far more than obscure words do.
Grammar. The recurring items are subject–verb agreement across an intervening phrase, misplaced modifiers, parallel structure in lists, and tense consistency. Read the sentence aloud in your head — most errors are audible.
Critical reasoning. Identify the conclusion first, then the evidence offered for it, then ask what the argument assumes to get from one to the other. A question asking to strengthen or weaken an argument is asking about that assumption, not about the conclusion.
The aptitude section is worth 15 of 100 marks — as much as several core chapters together — and it is the most predictable part of the paper.
These are the cheapest marks on the paper. Treating them as filler is the single most expensive habit a GATE candidate can have.
Three problems in the style GATE sets, each with the wrong answer the question is designed to attract.
A can complete a job in 12 days and B in 18 days. They work together for 4 days, after which A leaves. How long does B take to finish the remaining work?
A alone: 12 days B alone: 18 days Together for 4 days, then B alone
How long does B take to finish the remaining work?
Convert each worker to a rate.
The fraction of the job done per day. This is the whole technique; everything else is arithmetic.
Rates add.
Times do not — a common wrong start is to average 12 and 18.
Work completed in the first four days.
Work remaining.
Time for B alone at his own rate.
Check by totalling: B works 4 + 8 = 12 days at 1/18 per day = 12/18 = 2/3 of the job.
A works 4 days at 1/12 = 1/3. Together 2/3 + 1/3 = 1, the whole job.
Answer8 days
The trap: averaging the times. If A takes 12 days and B takes 18, together they do not take 15 — they take 36/5 = 7.2 days. Rates add; times never do.
(a) A car covers 60 km at 40 km/h and the next 60 km at 60 km/h. Find the average speed. (b) Two trains, 200 m and 160 m long, travel in opposite directions at 54 km/h and 36 km/h. How long do they take to pass each other completely?
(a) 60 km at 40 km/h, then 60 km at 60 km/h (b) Lengths 200 m and 160 m; speeds 54 and 36 km/h, opposite directions
Find the average speed
How long do they take to pass each other completely?
Part (a).
Average speed is total distance over total time — never the average of the two speeds.
Note that it is not 50.
When the two distances are equal, the average speed is the harmonic mean, which is always lower than the arithmetic mean — because the car spends more time at the slower speed.
Worth knowing the companion case.
If the two times were equal instead of the distances, the answer would be the arithmetic mean, 50 km/h. Which mean applies depends on what is held equal.
Part (b).
For trains passing, the distance covered is the sum of both lengths — the rear of one must clear the rear of the other.
Opposite directions means the speeds add.
Convert to m/s with the standard factor 5/18.
Time.
If they were travelling in the same direction.
The relative speed would be the difference, 18 km/h = 5 m/s, and the same pass would take 72 s — five times longer.
Answer(a) 48 km/h (b) 14.4 s
The trap: averaging the two speeds in part (a). It gives 50 km/h, which is one of the options offered precisely because it is what an unthinking answer produces.
A bag contains 5 red and 4 blue balls. Three are drawn at random without replacement. Find the probability that at least one is blue. Then find the probability that exactly two are blue.
5 red, 4 blue, 9 balls total 3 drawn without replacement, order does not matter
Find the probability that at least one is blue
Order does not matter, so this is a combinations problem.
Total number of ways to choose three balls from nine:
For at least one blue, go via the complement.
The opposite of 'at least one blue' is 'no blue at all', which means all three are red — one case instead of three.
Count the all-red selections.
Subtract.
Now exactly two blue.
Choose two of the four blue and one of the five red, and multiply — because every blue pair can be combined with every red choice.
A quick completeness check.
The four possible outcomes must sum to one.
The complement route saved counting three cases instead of one.
Whenever a question says 'at least one', reach for 1 − P(none) before anything else.
AnswerP(at least one blue) = 37/42 ≈ 0.881; P(exactly two blue) = 5/14 ≈ 0.357
The trap: computing 'at least one' by adding the one-blue, two-blue and three-blue cases. It works, but it is three times the arithmetic and three times the chance of a slip — and under time pressure that matters more than elegance.
a + b + ab/100Successive percentage change1/t = 1/t₁ + 1/t₂Combined work rate; together = ab/(a+b)2v₁v₂/(v₁ + v₂)Average speed over equal distances(v₁ + v₂)/2Average speed over equal timeskm/h × 5/18 = m/sAnd m/s × 18/5 = km/hRelative speed = v₁ ± v₂Plus when closing, minus when in the same directionⁿP_r = n!/(n−r)! ⁿC_r = n!/(r!(n−r)!)Order matters, or it does notP(at least one) = 1 − P(none)Nearly always the faster routeP(A or B) = P(A) + P(B) − P(A∩B)Mutually exclusive ≠ independentAlligation: (d − C)/(C − c)Ratio of cheaper to dearer in a mixture