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GATE General Aptitude — How Not to Lose the Easy Marks

By Aniket Bhardwaj · 12 September 2026 · GATE Chemistry

General Aptitude is a section common to every GATE paper, whatever your discipline. For a chemistry candidate it is strange territory — nothing in an MSc syllabus prepares you for a seating-arrangement puzzle — and that is exactly why it gets neglected until the last week. The marks in this section are the cheapest in the whole paper: no derivation, no mechanism, no spectrum to assign. This guide covers what the section contains, eight worked examples of the recurring numerical types, and the habits that stop you leaking marks here.

Before anything else: the number of General Aptitude questions, the mark split between one-mark and two-mark questions, and the negative-marking scheme are all stated in the official GATE information brochure for the year you are appearing in. Read that document yourself. Do not plan your attempt strategy on figures quoted in a coaching handout or a blog, including this one — patterns are revised from time to time, and the official notification is the only source that is current by definition.

What the section actually covers

Recent GATE General Aptitude syllabi have grouped the questions into four areas. Check the syllabus PDF for your year, but the shape has been stable:

AreaTypical contentHow to prepare
Verbal aptitudeWord usage, sentence completion, grammar, reading comprehension, inferenceRead English non-fiction daily; keep a small notebook of confusable word pairs
Quantitative aptitudeRatio, percentage, averages, speed–time, work–rate, mensuration, probability, simple statisticsClass 8–10 arithmetic, practised for speed rather than depth
Analytical aptitudeLogic, deduction, number series, puzzles, data interpretation from tables and graphsTimed practice; learn to draw the grid or table quickly
Spatial aptitudeRotation, folding, mirror images, assembling shapes, pattern completionPractise with actual paper folding; visualise, do not calculate

Question formats across the paper are multiple choice, multiple select and numerical answer type. The rule that matters for guessing — which formats carry negative marking and which do not — is in the official brochure. Confirm it before the exam and build your guessing policy around what it says that year.

The eight numerical patterns that keep coming back

Almost all the quantitative and data-interpretation questions reduce to one of these. Each one is worked completely below, because seeing the full arithmetic once is worth more than reading a shortcut.

Worked example 1 — successive percentage change

A quantity increases by 20 % and then decreases by 20 %. What is the net change?

Multiply the factors, never add the percentages:
final ÷ initial = 1.20 × 0.80 = 0.96
Net change = 0.96 − 1 = −0.04 → a 4 % decrease.

The intuition worth carrying: an equal percentage rise and fall never cancels, and the result is always a net loss. A 50 % rise followed by a 50 % fall gives 1.5 × 0.5 = 0.75, a 25 % loss.

Worked example 2 — keeping expenditure constant

The price of a commodity rises by 25 %. By what percentage must consumption be reduced so that expenditure stays the same?

Expenditure = price × consumption. Let both start at 1.
New price factor = 1.25
Required consumption factor = 1 ÷ 1.25 = 0.80
Reduction = 1 − 0.80 = 0.20 → 20 %

Cross-check with the standard formula for a price rise of r %:
reduction % = 100r ÷ (100 + r) = (100 × 25) ÷ 125 = 2500 ÷ 125 = 20 %. The two agree.

The trap is answering 25 %. A 25 % rise needs only a 20 % cut, because the cut is taken on a larger base.

Worked example 3 — work and rate

A can finish a job in 12 days and B in 18 days. Working together, how long do they take?

Work with rates, not days.
A's rate = 1/12 job per day, B's rate = 1/18 job per day
Combined rate = 1/12 + 1/18. LCM of 12 and 18 is 36:
= 3/36 + 2/36 = 5/36 job per day
Time = 1 ÷ (5/36) = 36/5 = 7.2 days

Sanity check: the answer must be less than the faster worker's 12 days and more than half of it (6 days). 7.2 sits in that window, so it is plausible. Adding or averaging the days — 15 days — fails this check instantly.

Worked example 4 — average speed over equal distances

A car covers the first half of a journey at 40 km/h and the second half at 60 km/h. What is the average speed for the whole journey?

Average speed = total distance ÷ total time. Let each half be d km.
time for first half = d/40 h, time for second half = d/60 h
total time = d/40 + d/60 = 3d/120 + 2d/120 = 5d/120 = d/24 h
total distance = 2d km
average speed = 2d ÷ (d/24) = 2 × 24 = 48 km/h

Cross-check with the harmonic mean formula for equal distances:
2 × 40 × 60 ÷ (40 + 60) = 4800 ÷ 100 = 48 km/h. The two agree.

The answer is never 50, the arithmetic mean, because more time is spent at the slower speed. Note the condition: equal distances give the harmonic mean; equal times would give the arithmetic mean, 50 km/h. Read which one the question states.

Worked example 5 — mixtures and ratios

A 40-litre mixture contains milk and water in the ratio 3 : 1. How much water must be added to make the ratio 1 : 1?

Split the original mixture into parts. Total parts = 3 + 1 = 4, so one part = 40 ÷ 4 = 10 L.
milk = 3 × 10 = 30 L, water = 1 × 10 = 10 L
Adding x litres of water leaves the milk unchanged at 30 L, so for a 1 : 1 ratio:
30 = 10 + x → x = 20 litres

Check: the new mixture is 60 L with 30 L milk and 30 L water — ratio 1 : 1. The habit that makes mixture questions easy is to fix on the quantity that does not change (here the milk) and work from it.

Worked example 6 — trains, platforms and unit conversion

A 150 m long train travelling at 72 km/h crosses a 250 m platform. How long does it take?

Step 1 — convert the speed. To go from km/h to m/s, multiply by 1000/3600 = 5/18:
72 × 1000 ÷ 3600 = 72 000 ÷ 3600 = 20 m/s

Step 2 — the distance is the sum, not the platform alone. The train has crossed only when its last coach leaves the platform, so it travels its own length plus the platform length:
distance = 150 + 250 = 400 m

Step 3 — time.
400 ÷ 20 = 20 seconds

Forgetting to add the train's own length is the standard error here, and it gives 12.5 s.

Worked example 7 — data interpretation with mean, median and spread

Five readings are 12, 15, 18, 20 and 25. Find the mean, the median and the standard deviation.

Mean. Sum = 12 + 15 + 18 + 20 + 25 = 90; mean = 90 ÷ 5 = 18

Median. Already in ascending order; with five values the middle one is the third = 18

Deviations and their squares.
12 − 18 = −6 → 36
15 − 18 = −3 → 9
18 − 18 = 0 → 0
20 − 18 = 2 → 4
25 − 18 = 7 → 49
The deviations sum to −6 − 3 + 0 + 2 + 7 = 0, which is the built-in check that the mean is right. Sum of squares = 36 + 9 + 0 + 4 + 49 = 98

Standard deviation — and here two conventions genuinely differ.
Treating the five numbers as the whole population, divide by n = 5:
variance = 98 ÷ 5 = 19.6, σ = √19.6 = 4.43
Treating them as a sample from a larger population, divide by n − 1 = 4:
variance = 98 ÷ 4 = 24.5, s = √24.5 = 4.95

Both are correct answers to different questions. If the question says "population" or gives you all the data there is, use n. If it says "sample", use n − 1. If it says neither and the options contain only one of the two values, the options are telling you which convention is intended.

Worked example 8 — number series and probability

Series. Find the next term: 2, 6, 12, 20, 30, ?

Route 1 — first differences. 6 − 2 = 4, 12 − 6 = 6, 20 − 12 = 8, 30 − 20 = 10. The differences increase by 2, so the next difference is 12 and the next term is 30 + 12 = 42.

Route 2 — closed form. 2 = 1×2, 6 = 2×3, 12 = 3×4, 20 = 4×5, 30 = 5×6, so the pattern is n(n + 1) and the sixth term is 6 × 7 = 42. Two independent routes agreeing is what makes a series answer safe.

Probability. Two fair dice are rolled. What is the probability that the sum is 8?

Total outcomes = 6 × 6 = 36.
Favourable outcomes: (2,6), (3,5), (4,4), (5,3), (6,2) — five of them. Note that (2,6) and (6,2) are different outcomes because the dice are distinguishable, but (4,4) occurs only once.
P = 5/36 ≈ 0.139

Attempt strategy that actually works

Common mistakes that cost marks

  • Adding percentages instead of multiplying factors. +20 % then −20 % is not zero; it is −4 %.
  • Averaging speeds. For equal distances the answer is the harmonic mean, not the arithmetic mean.
  • Averaging days in work problems. Add the rates, then invert.
  • Forgetting the train's own length when it crosses a platform, bridge or another train.
  • Missing unit conversions. km/h to m/s is × 5/18; m/s to km/h is × 18/5. Write the conversion down rather than doing it in your head under pressure.
  • Treating (4,4) as two outcomes in dice problems, or forgetting that (2,6) and (6,2) are two.
  • Leaving the section till the last month. It is the one part of the paper where a chemistry graduate has no built-in advantage, so it needs its own small slot in the timetable.
  • Assuming last year's pattern still holds. Verify question counts, mark splits and negative marking from the current official notification every single year.

How to use these patterns while revising

PatternThe one thing to get right
Percentage changeMultiply the factors; never add the percentages
Price and consumptionThe reduction is taken on the new, larger base
Work and rateAdd rates, then invert; sanity-check against the faster worker
Average speedEqual distances → harmonic mean; equal times → arithmetic mean
MixturesAnchor on the component whose amount does not change
Trains and platformsDistance = own length + object length
Mean, median, spreadDeviations must sum to zero; state n or n − 1 convention
Series and probabilityConfirm the pattern by a second route before answering

Practise the arithmetic with a tool that shows its steps. The ABC Chemistry Calculator Suite bundles the statistics, permutation and combination, regression and unit conversion tools that these aptitude patterns lean on, alongside the chemistry calculators — useful for checking your own working when you self-review a mock test.

Open the ABC Chemistry Calculator Suite →

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