Statistics, at its core, is about summarising a mass of data into a few meaningful numbers. When you have 50 salaries or 200 test scores, you don't want to stare at every individual value — you want to know where the data is centred (central tendency) and how spread out it is (dispersion).
Three measures dominate CDS questions:
Mean is the arithmetic average — sum all values, divide by count. It is the "fair share" number. If you have ₹1,400 to distribute among 5 people, each gets ₹280. That is the mean.
Median is the middle value once the data is sorted. Half the data lies below it, half above. It ignores extreme values, which is why economists prefer it for income data. With an even count, it is the average of the two middle values.
Mode is the value that appears most often. A dataset can have no mode, one mode, or multiple modes. In CDS, mode questions are rare — but recognise mode in context.
Here is the intuition that separates fast solvers from slow ones: the mean is linear. This means:
k to every observation → new mean = old mean + k.k → new mean = old mean × k.That last property is the engine behind a whole class of CDS problems involving "sum of deviations from a number". You will see it repeatedly in PYQs.
Think of the mean as the balance point of a see-saw. Place weights at their respective positions on a number line — the mean is exactly where you would place the fulcrum so the see-saw is perfectly level. Any deviation to the left must be exactly cancelled by a deviation to the right.
Frequency distribution problems ask you to find a missing frequency given that the mean equals some stated value. The method is mechanical: write Σf·x / Σf = given mean, introduce the unknown, and solve. CDS examiners love this format in grouped data.
For a dataset x₁, x₂, ..., xₙ:
Property 1 — Shift invariance: If every xᵢ is replaced by xᵢ + k, the new mean is x̄ + k. This is because:
Property 2 — Scale invariance: If every xᵢ is replaced by c·xᵢ, the new mean is c·x̄.
Property 3 — Zero-deviation property: Σ(xᵢ − x̄) = 0 always.
Property 4 — Deviation from an arbitrary point: If you compute Σ(xᵢ − A), this equals Σxᵢ − nA = n(x̄ − A). So:
This is the master formula for "sum of deviations" problems. Write it on your scratch sheet before reading the question.
CDS frequently gives you two sums of deviations from two different points and asks for the mean. The clean approach:
Let S = Σxᵢ. Then:
A: S − nA = aB: S − nB = bYou now have two equations in two unknowns (S, n). Eliminate S by subtraction:
(S − nA) − (S − nB) = a − b
n(B − A) = a − b
n = (a − b) / (B − A)
Then S = a + nA, and mean = S/n.
Alternatively, if you're given the ratio b/a, substitute directly: (S − nB) = (b/a)(S − nA) and solve for S/n.
Ungrouped: Sort the data. For n observations:
n odd: median = value at position (n+1)/2.n even: median = average of values at positions n/2 and n/2 + 1.Grouped (class intervals): Use the formula:
where L = lower boundary of the median class, F = cumulative frequency before the median class, f = frequency of median class, h = class width. CDS rarely asks you to apply this formula directly — it is more likely to appear in a data-interpretation table where you identify the class.
Given classes with midpoints mᵢ, frequencies fᵢ (one of which is unknown k), and the stated mean:
Cross-multiply, collect k terms on one side, and solve. The arithmetic is the only challenge — do it column by column to avoid errors.
These are statistics problems in disguise. The logic is:
For difference-of-groups questions: find the percentage difference first (no need to compute each group separately), then apply it to the total. That alone saves 30 seconds.
Range = maximum value − minimum value. It is the simplest measure of spread. CDS does not deeply test standard deviation or variance in recent papers, but range appears in data-set questions.
When the problem says "sum of deviations from A is a, sum from B is b, and b/a = r", don't solve two equations. Set up: (x̄ − B) / (x̄ − A) = r directly. This is because Σ(xᵢ − B) / Σ(xᵢ − A) = n(x̄ − B) / n(x̄ − A) = (x̄ − B)/(x̄ − A). Cross-multiply and solve for x̄ in one line.
Example (2023 PYQ): deviations from 10 = a, from 20 = b, b/a = −4. So (x̄ − 20)/(x̄ − 10) = −4 → x̄ − 20 = −4x̄ + 40 → 5x̄ = 60 → x̄ = 12. Standard method (two equations): ~60 seconds. This method: ~15 seconds.
In budget/population problems asking how much MORE or LESS one group is compared to another, compute the percentage difference first, then multiply by the total. Never compute each group's absolute value.
Example (2022 PYQ): A+C = 49%, B+D+E = 51%. Difference = 2%. 2% of ₹300.4 lakhs = ₹6.008 lakhs. Done. Standard method (computing all five groups): ~75 seconds. This method: ~20 seconds.
If a constant k is added to all n observations, the new mean = old mean + k. You do not touch the data. If multiplied by c, new mean = old mean × c.
Use this whenever the question says "each observation is increased/decreased/multiplied/divided by…". Answer by inspection.
Example: mean of 100 observations is 50; 10 is added to each. New mean = 60. Standard route (re-sum 100 numbers): impossible. Pattern recognition: 3 seconds.
In a descending-order sequence with an even count n, the median is the average of the n/2-th and (n/2 + 1)-th terms. Before solving the quadratic, identify which positions the unknowns occupy and confirm they satisfy the ordering constraints to eliminate invalid roots.
Example (2026 PYQ): 12 numbers, median = average of 6th and 7th. Setting up the equation gives x = 4 or x = −12. Plugging x = −12 gives 6th value = −95, which violates the descending order. Eliminate without further calculation. Time saved: ~25 seconds of verification.
In grouped mean problems, write three columns: midpoint m, frequency f, product f·m. Sum the known rows. Let k be the unknown frequency and k·m₀ the unknown product entry. Then: (known Σfm + k·m₀) / (known Σf + k) = given mean. Isolate k. This structured layout prevents arithmetic errors and is faster than mental calculation.
Example (2023 PYQ): known Σfm = 126500, known Σf = 35, m₀ = 3300, mean = 3520. So (126500 + 3300k)/(35 + k) = 3520 → 126500 + 3300k = 123200 + 3520k → 3300 = 220k → k = 15. Setup time vs. solving time: 10s + 20s vs. 90s of unstructured arithmetic.
In the exam hall, categorise the statistics question in the first 10 seconds:
Is it a "sum of deviations" problem?
Yes → Use (x̄ − B)/(x̄ − A) = ratio if a ratio is given. Use two-equation subtraction otherwise. Never expand term by term.
Is it "constant added/multiplied to all observations"? Yes → Answer = old mean ± constant or × constant. No calculation needed.
Is it a percentage-based total/difference problem?
Yes → For total: known value / known % × 100. For difference: compute % gap first, then multiply by total.
Is it a missing frequency in a grouped table? Yes → Build the three-column table (m, f, fm), sum knowns, set up one linear equation, solve for k.
Is it a median with ordering constraints? Yes → Find positions, set up equation, solve algebraically, then verify each root against the ordering. Discard roots that break the sequence.
Is it a simple mean of sub-groups? Yes → Total sum = mean × count. Extract sub-group sums, combine, divide.
Time budget: routine questions should take 90 seconds or less once you recognise the type. If you are past 2 minutes, move on and return.
Why this question: Tests the percentage-to-total conversion — the most common statistics DI format in CDS.
Solving path: Identify the percentage for "below 40" by adding 14 + 29.75 + 26.25 = 70%. You are told 70% of the total = 10.5 lakhs. So total = 10.5 / 0.70 = 15 lakhs. No other computation needed. The 0% entry for 40–44.99 is a deliberate distractor — do not panic, it just means that class is empty.
Why this question: Tests percentage-difference-first shortcut and rewards students who avoid computing each budget head separately.
Solving path: A + C = 40 + 9 = 49%. B + D + E = 18 + 25 + 8 = 51%. Difference = 2%. 2% of ₹300.4 lakhs = 0.02 × 300.4 = ₹6.008 lakhs. Never compute the absolute values of each group — that is the slow path.
Why this question: Classic CDS deviation-ratio question. Tests whether you know the ratio shortcut or are forced into two simultaneous equations.
Solving path: Let mean = x̄. Then sum of deviations from 10 = n(x̄ − 10) = a, and from 20 = n(x̄ − 20) = b. Given b/a = −4: (x̄ − 20)/(x̄ − 10) = −4. Cross-multiply: x̄ − 20 = −4x̄ + 40. So 5x̄ = 60, mean = 12.
Why this question: Standard grouped-frequency mean problem. Tests systematic setup of the three-column product table.
Solving path: Midpoints: 2300, 2800, 3300, 3800, 4300, 4800. Known f·m products: 5×2300 = 11500, 10×2800 = 28000, 8×3800 = 30400, 2×4300 = 8600, 10×4800 = 48000. Sum of known fm = 126500. Sum of known f = 35. Unknown class (3050–3550): midpoint 3300, frequency k. Equation: (126500 + 3300k)/(35 + k) = 3520. Expand: 126500 + 3300k = 123200 + 3520k. So 3300 = 220k, giving k = 15.
Why this question: Tests sub-group mean decomposition — a direct application of "total sum = mean × count" used twice.
Solving path: p+q+r+s+t = 5 × 280 = 1400. p+r+t = 3 × 240 = 720. Therefore q+s = 1400 − 720 = 680. Mean of q and s = 680/2 = 340.
Adding deviations without using the formula. Students write out Σ(xᵢ − A) and try to expand term by term. Use n(x̄ − A) directly — it is a single multiplication, not n subtractions.
Ignoring the ordering constraint in median questions. When the data is in ascending or descending order and you solve a quadratic for an unknown, both roots may satisfy the equation algebraically but only one satisfies the sequence ordering. Always verify every root against the position constraint.
Computing absolute group values in percentage-difference questions. You do not need ₹(A+C) or ₹(B+D+E) separately. The difference in percentages times the total gives the answer directly. Computing all five budget heads wastes 60+ seconds.
Using the wrong midpoint formula. For a class a–b, the midpoint is (a+b)/2. For the class 2050–2550, midpoint is (2050+2550)/2 = 2300, not 2550 or 2050. Students in a hurry sometimes use the upper boundary as the midpoint.
Forgetting that adding a constant shifts the mean, not the median or mode differently. Adding k to every observation shifts mean, median, and mode all by k. Multiplying by c scales all three by c. But if you add k only to some observations, you must recompute from scratch.
Dividing by the wrong n in two-deviation problems. When two sums of deviations give you both n and Σxᵢ, always double-check by substituting back. A sign error in the ratio (positive vs. negative) is the most common arithmetic trap — watch for b/a = −4 (negative ratio) carefully.