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CSIR-NET Part A — The Aptitude Section Worth Practising

By Aniket Bhardwaj · 12 September 2026 · CSIR-NET Chemistry

The CSIR-NET chemical sciences paper is set in three parts: Part A, a general aptitude section, and Parts B and C, which test the chemistry syllabus. For the number of questions in each part, the marks, the time allowed and any negative marking, read the current official notification — those details are the examining body's to state, they have changed over the years, and no coaching page (this one included) should be trusted for them.

What this page is for is different, and more useful: Part A is the only section where a chemistry postgraduate is competing on skills nobody taught them during MSc. Most aspirants spend all their preparation on spectroscopy and thermodynamics and then walk into a set of number-series and data-interpretation questions cold. That is a mistake, because aptitude is the most trainable part of the whole paper — the question families are finite, the methods are short, and a few weeks of honest practice moves the needle far faster than a fifth revision of coordination chemistry.

The question families

General-aptitude sets across Indian national examinations draw on the same well-defined pool. Confirm the definitive list against the official syllabus, but you will be preparing along these lines.

FamilyWhat it testsTypical sub-types
Number and letter seriesPattern recognitionDifferences, ratios, squares/cubes, alternating series
Arithmetic reasoningPercentages, ratio, averageSuccessive change, mixtures and alligation, profit and loss, interest
Time-based problemsRatesTime and work, time–speed–distance, trains, pipes and cisterns
Counting and chanceCombinatoricsPermutations, combinations, elementary probability
Data interpretationReading quantitative displaysTables, bar and pie representations, growth rates, averages
Basic statisticsSummarising dataMean, median, mode, standard deviation, simple correlation
Geometry and mensurationShapes and spaceAreas, volumes, scaling arguments
Logical puzzlesStructured deductionSeating and ordering, blood relations, direction sense, clocks and calendars
Numerical estimationOrder-of-magnitude sensePowers of ten, approximation, unit reasoning

The last family is the one where chemists have an unfair advantage and often fail to use it. You already manipulate 10⁻²³ and 10⁻³⁴ every day. Treat estimation questions as dimensional-analysis questions and they become the fastest marks in the section.

Ten worked examples, with the reasoning shown

1. Number series. Find the next term: 3, 8, 15, 24, 35, ?

Method A — differences: 8−3 = 5, 15−8 = 7, 24−15 = 9, 35−24 = 11. The differences rise by 2, so the next difference is 13 and the term is 35 + 13 = 48.
Method B — closed form: 4−1 = 3, 9−1 = 8, 16−1 = 15, 25−1 = 24, 36−1 = 35, so the rule is n² − 1 and the sixth term is 49 − 1 = 48. Two independent routes agreeing is the fastest confidence check available in this section.

2. Successive percentage change. A sample loses 20% of its mass, then loses 25% of what remains. What is the total loss?

Start with 100 units. After the first loss: 100 × 0.80 = 80. After the second: 80 × 0.75 = 60. Total loss = 100 − 60 = 40%, not 45%.

The general rule: net factor = 0.80 × 0.75 = 0.60, so the net change is −40%. Percentages multiply, they do not add — this single idea covers discounts, dilutions and yields.

3. Time and work. A completes a task in 12 days, B in 18 days. Working together, how long?

Work per day: A = 1/12, B = 1/18. Common denominator 36: 3/36 + 2/36 = 5/36 of the task per day.
Time = 36/5 = 7.2 days.

Speed method: take the total work as the LCM of 12 and 18, i.e. 36 units. Then A does 3 units/day and B does 2, together 5, so 36/5 = 7.2 days — no fractions at any stage.

4. Speed from a crossing time. A train 150 m long passes a stationary pole in 9 s. Find its speed in km h⁻¹.

Passing a pole means covering its own length: v = 150 ÷ 9 = 16.667 m s⁻¹.
Convert: 16.667 × 18/5 = 16.667 × 3.6 = 60 km h⁻¹.

Remember the distinction: past a pole the distance is the train's length; past a platform it is train length + platform length.

5. Combinations and probability. A committee of 3 is drawn at random from 5 chemists and 4 physicists. What is the probability that at least 2 are chemists?

Favourable cases: exactly 2 chemists = C(5,2) × C(4,1) = 10 × 4 = 40; exactly 3 chemists = C(5,3) = 10. Total favourable = 40 + 10 = 50.
Total cases = C(9,3) = (9 × 8 × 7)/(3 × 2 × 1) = 504/6 = 84.
P = 50/84 = 25/42 ≈ 0.595.

Check by complement: fewer than 2 chemists means 1 chemist (C(5,1) × C(4,2) = 5 × 6 = 30) or 0 chemists (C(4,3) = 4), total 34. And 50 + 34 = 84. The two counts close, so the answer stands.

6. Mean, median and standard deviation. For the readings 12, 15, 11, 18, 14, find the mean, the median and the standard deviation.

Sum = 12 + 15 + 11 + 18 + 14 = 70, so mean = 70/5 = 14.
Sorted: 11, 12, 14, 15, 18 — the middle value is 14.
Deviations from the mean: −2, +1, −3, +4, 0. Squares: 4, 1, 9, 16, 0; sum = 30.

Now be careful, because two conventions exist and both are correct in their own context. Treating the five values as the whole population, divide by n = 5: variance = 6 and σ = √6 = 2.449. Treating them as a sample of a larger population, divide by n − 1 = 4: variance = 7.5 and s = √7.5 = 2.739. If the question does not say, state which you used.

7. Clock angle. What is the angle between the hands at 3:40?

The minute hand moves 360°/60 = 6° per minute: 40 × 6 = 240° from 12.
The hour hand moves 30° per hour plus 0.5° per minute: 3 × 30 + 40 × 0.5 = 90 + 20 = 110° from 12.
Difference = 240 − 110 = 130°.

The forgotten 0.5°/min on the hour hand is the single commonest error in clock questions. If the difference exceeds 180°, subtract it from 360° to report the smaller angle.

8. Compound versus simple interest. ₹10,000 is invested for 2 years at 10% per annum. Find the compound interest, and how much it exceeds the simple interest.

Amount = 10000 × (1.10)² = 10000 × 1.21 = ₹12,100, so CI = ₹2,100.
Simple interest = 10000 × 0.10 × 2 = ₹2,000.
Difference = ₹100, which equals the interest earned on the first year's interest: 1000 × 0.10 = 100. That identity is a fast way to verify any two-year CI question.

9. Mixtures (alligation). In what ratio must a 20% solution be mixed with a 50% solution to obtain a 30% solution?

Ratio of the two components = (50 − 30) : (30 − 20) = 20 : 10 = 2 : 1, with the larger share going to the weaker solution.
Verify with 2 L and 1 L: solute = 2(0.20) + 1(0.50) = 0.40 + 0.50 = 0.90 in 3 L, giving 0.90/3 = 0.30 = 30%. Correct.

10. Order-of-magnitude estimation. Evaluate (6.4 × 10⁵) ÷ (1.6 × 10⁻³) without a calculator.

Split the mantissas and the powers: 6.4 ÷ 1.6 = 4, and 10⁵ ÷ 10⁻³ = 10⁵⁻⁽⁻³⁾ = 10⁸.
Answer = 4 × 10⁸.

The sign trap is the exponent subtraction: dividing by 10⁻³ adds 3 to the power. Any chemist who has ever computed [H⁺] from pH has made this error at least once.

How to practise it without stealing time from chemistry

Aptitude rewards short, frequent contact far more than long sessions. A workable pattern for a chemistry aspirant looks like this:

One thing worth saying plainly: aptitude marks count the same as chemistry marks. A student who treats Part A as beneath them is choosing to leave the easiest marks in the paper on the table.

Traps that cost marks

  • Adding percentages instead of multiplying factors. A 20% fall followed by a 25% fall is 40%, not 45%.
  • Ignoring the hour hand's own movement in clock problems.
  • Answering the wrong quantity. The question may ask for the difference, the ratio, or the value after the change — read the last line twice.
  • Not stating the standard-deviation convention. n and n − 1 give genuinely different numbers.
  • Forgetting to convert m s⁻¹ to km h⁻¹ (multiply by 18/5) or the reverse (multiply by 5/18).
  • Over-committing to one hard puzzle. Marks are per question, not per minute spent.
  • Assuming a data-interpretation "percentage increase" is from the first year. Growth is usually year-on-year unless stated otherwise.

Where aptitude sits in the wider picture

ExamIs there a general aptitude component?
CSIR-NET chemical sciencesYes — Part A, alongside subject Parts B and C
GATE ChemistryYes — a general aptitude component alongside the subject paper
IIT-JAM ChemistryNo separate aptitude section; the paper is subject-only
CUET-PG ChemistrySubject-focused; check the current notification for the exact composition

Because two of these examinations carry an aptitude component, the practice is not wasted effort confined to one paper. Confirm the structure of each from its own official notification before you plan your time.

Use the calculator to check, not to solve. Aptitude is a mental-arithmetic skill, so work each question by hand first and verify afterwards — the suite's scientific calculator, nPr/nCr tool and mean/median/standard-deviation tool between them cover every numerical family listed above. This article explains a whole section rather than one formula, so the button opens the suite itself rather than a single tool.

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Preparing for CSIR-NET, GATE, IIT-JAM or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and fully online for students across India — details at abcchemistry.in.