A ratio is a comparison of two quantities of the same kind. When you write a : b, you are expressing how many times one quantity contains the other. The key word is same kind — you cannot ratio kilometres to kilograms. The ratio a : b equals a/b, and this fraction is what you work with in every calculation.
Proportion steps in when two ratios are equal: a : b = c : d. This reads as "a is to b as c is to d". The four numbers are called the four terms of the proportion, and the outer pair (a and d) are called extremes, while the inner pair (b and c) are called means. The golden rule is product of extremes = product of means, i.e., ad = bc.
Variation formalises the relationship between two quantities that change together. Think of it this way: if you double the number of workers on a job, the time to finish halves. That is inverse variation. If you double the number of items bought at a fixed price per item, the total cost doubles. That is direct variation.
Here is an analogy that works in the exam hall. Think of a ratio as a recipe — say, 2 cups flour to 1 cup sugar. Proportion says two different batches follow the same recipe. Variation says the recipe itself tells you how ingredients scale when the batch size changes.
One thing CDS loves to test that most students treat as simple: algebraic manipulation of variation. If x ∝ y (x varies directly as y), can you immediately say what x² + y² varies as? This is where the difference between a student who has practised and one who has not shows up cleanly. The entire strategy is to replace the variation statement with an equation — x = ky — and then manipulate that equation algebraically. If you train yourself to do that instinctively, this chapter becomes one of the fastest marks in the paper.
A ratio a : b is in its lowest terms when gcd(a, b) = 1. Always reduce before working. The duplicate ratio of a : b is a² : b². The triplicate ratio is a³ : b³. The sub-duplicate ratio is √a : √b. These are named quantities that CDS setters use directly.
Compound ratio: multiply corresponding terms. The compound ratio of a : b and c : d is ac : bd.
Given a : b = c : d (equivalently, a/b = c/d):
b/a = d/ca/c = b/d(a+b)/b = (c+d)/d(a-b)/b = (c-d)/d(a+b)/(a-b) = (c+d)/(c-d)Componendo-Dividendo is the one you will use the most. If you know the sum-to-difference ratio, you can recover the individual ratio in two lines.
The sum-of-ratios trick: If (a+b) : (b+c) : (c+a) = p : q : r, set each part equal to pk, qk, rk. Then 2(a+b+c) = (p+q+r)k, which gives you a+b+c immediately. From there, subtract each paired sum to isolate variables: c = (a+b+c) - (a+b), and so on. This pattern appears in multiple CDS PYQs.
Direct Variation: x ∝ y means x = ky for some non-zero constant k. The graph is a straight line through the origin.
Inverse Variation: x ∝ 1/y means xy = k. Doubling one halves the other.
Joint Variation: x ∝ yz means x = kyz. This is just direct variation extended to two variables simultaneously.
Partial Variation: x = ay + b where a and b are constants. Here x does not vary directly as y unless b = 0. CDS occasionally tests this subtlety.
This is the move you need to drill. When you see "x varies as..." in a CDS question:
Step 1. Write x = ky (or whatever the variation implies).
Step 2. Substitute into the expression being tested.
Step 3. Factor out constants. Whatever you cannot factor out will tell you what the expression varies as.
Example: x ∝ y, so x = ky. Does x³/y⁴ vary directly as y?
Substitute: x³/y⁴ = (ky)³/y⁴ = k³y³/y⁴ = k³/y.
This is (constant)/y, which means x³/y⁴ ∝ 1/y — inverse variation, not direct. This is exactly what CDS 2026 tested.
A locomotive problem type that recurs: speed decreases by a quantity proportional to some function of load. Setup is always:
Actual speed = Maximum speed − (constant × variation term)
Find the constant from the given data point, then solve for the boundary condition (speed = 0 for "unable to move").
When told P : Q = x : y and given two linear conditions on (P+a)/(Q+a), assign actual values P and Q from the conditions, then read off the ratio. This is faster than parameterising by t when the system is 2×2.
Whenever a CDS variation question asks "what does expression E vary as?", replace the variable with the proportionality equation immediately. Write x = ky, substitute into E, then group all constants together. Whatever pure variable expression remains is what E varies as. This converts a "conceptual" question into pure algebra in one step.
Micro-example: Does (x² + y²) vary as y² given x ∝ y?
Write x = ky: x² + y² = k²y² + y² = (k²+1)y². The factor (k²+1) is a constant, so yes, it varies as y².
Standard method (think from principles): ~40s. Substitute-and-factor: ~12s.
When given (a+b) : (b+c) : (c+a) as a ratio, set them as pk, qk, rk. Add all three: 2(a+b+c) = (p+q+r)k. Once you know a+b+c, subtract each pair:
c = (a+b+c) − (a+b) = (p+q+r)k/2 − pk, and so on.
This pattern gives all three variables in three subtractions. No simultaneous equations needed.
Standard elimination (3 equations): 6–8 steps. Sum-of-pairs shortcut: 3 subtractions, ~20s faster.
If you know (a+b)/(a-b) = r, then a/b = (r+1)/(r-1).
Derivation: let a/b = t. Then (t+1)/(t-1) = r, so t = (r+1)/(r-1).
Micro-example: (a+b)/(a-b) = 5. Then a/b = 6/4 = 3/2.
This avoids writing out the full system. Standard method: 4 steps. Componendo-Dividendo inversion: 1 step, ~25s saved.
When given m:n = 1:2 and p:q = 3:4, just assign m=1, n=2, p=3, q=4 and plug directly into the expression. Do not parameterise with separate multipliers unless the ratios share a variable. The constraint "these are ratios" means any consistent assignment works.
CDS 2024 example: (2m+4p):(n+3q) = (2+12):(2+12) = 14:14 = 1:1. Total time: ~8s.
Standard approach (using unknowns k₁, k₂): ~45s.
To test if PQ ∝ R when P² = aR and Q² = bR: compute P²Q² = abR², so (PQ)² = abR², giving PQ = R√(ab). Since √(ab) is a constant, PQ ∝ R.
This sidesteps taking square roots of P and Q individually (which would require them to be positive and introduce sign issues). Instead, work with PQ as a unit.
Step count without trick: 5 steps (isolate P, isolate Q, multiply, simplify). With trick: 2 steps.
In the exam hall, classify the question in the first five seconds:
Type 1 — Pure ratio/proportion with numerical data (e.g., find x+y : x-y): Assign the smallest consistent integers to the variables immediately. Do not use algebra unless the numbers come out fractional.
Type 2 — Variation statement, algebraic expression (e.g., "x ∝ y, does E vary as...?"): Write the proportionality as an equation first (x = ky), substitute, and factor. If all remaining terms after factoring contain only one variable expression, you have your answer.
Type 3 — Sum-of-pairs ratio (e.g., (a+b):(b+c):(c+a) = p:q:r): Use the sum-of-pairs shortcut. Set equal to pk, qk, rk. Find k from a+b+c, then subtract pairs.
Type 4 — Variation with a given data point (e.g., locomotive problem): Write the general variation equation, substitute the data point to find the constant k, then solve for the boundary condition.
Decision rule for "which statements are correct" type: Test each statement independently using the substitution protocol. Never try to reason about all statements simultaneously — it costs time and invites errors.
Why this question: Tests whether you can apply the substitute-and-factor protocol to two separate expressions under the same variation condition — the core skill of the chapter.
Solving path: Write x = ky. For statement I: x² + y² = k²y² + y² = (k²+1)y². Since (k²+1) is a constant, this varies directly as y². Statement I is correct. For statement II: x³/y⁴ = k³y³/y⁴ = k³/y. This is a constant divided by y, so it varies inversely as y, not directly. Statement II is wrong. Answer: I only.
Why this question: Classic applied variation — speed reduction proportional to a function of load. Tests whether you can find the proportionality constant from given data and then solve for a boundary condition.
Solving path: Model as Speed = 40 - k√n. Substitute Speed = 16, n = 36: 16 = 40 - 6k, so k = 4. Engine stops when 40 - 4√n ≤ 0, giving √n ≥ 10, so n ≥ 100. Smallest integer is 100.
Why this question: Two linear conditions on a ratio — you need to recover actual values, not just the ratio. Also asks for a derived ratio (x+y):(x-y), which catches students who stop one step early.
Solving path: Set up two equations: 10P - 7Q = -3 and 4P - 3Q = -3. Multiply the second by 3 and the first by 3 to eliminate Q: 30P - 21Q = -9 and 28P - 21Q = -21. Subtract: 2P = 12, so P = 6. Back-substitute: Q = 9. Ratio P:Q = 2:3, so x = 2, y = 3. Then (x+y):(x-y) = 5:(-1) = -5.
Why this question: Sum-of-pairs structure with a total constraint — a direct test of whether you know the sum-of-pairs pattern.
Solving path: Set x+y = 3k, y+z = 5k, z+x = 6k. Sum = 2(x+y+z) = 14k. Given x+y+z = 14, so k = 2. Then x+y = 6, y+z = 10, z+x = 12. Isolate: z = 14-6 = 8, x = 14-10 = 4, y = 14-12 = 2. So x²+y²+z² = 16+4+64 = 84.
Why this question: Tests componendo-dividendo in reverse — given the sum-difference ratio, find the squares ratio. Students who don't notice the shortcut waste 90 seconds on algebra.
Solving path: (a+b)/(a-b) = 5/1. Apply componendo-dividendo inversion: a/b = (5+1)/(5-1) = 6/4 = 3/2. Assign a = 3, b = 2. Sum of squares = 9+4 = 13. Difference of squares = 9-4 = 5. Ratio = 13:5.
Not reducing ratios before assigning values. If P:Q = 6:9, the actual values of x and y in x:y form are 2 and 3, not 6 and 9. Confusing the actual numbers with the ratio terms causes wrong answers in problems where (x+y) or (x-y) is computed.
Forgetting that variation means a proportionality equation, not equality. Writing x = y instead of x = ky is the most common variation error. The constant k absorbs the difference in scale, and without it, boundary conditions (like the locomotive stopping) cannot be solved.
Treating "varies with R" as strictly "varies directly as R". In CDS questions, "P² is proportional to R" means P² = aR — the proportionality constant can be any real number, not necessarily 1. This matters when combining two such relations (different constants a and b).
Using the wrong formula for componendo-dividendo. The rule is (a+b)/(a-b) = (c+d)/(c-d) when a/b = c/d. Students sometimes flip the denominator on one side only, especially under time pressure.
Stopping at the ratio instead of computing the derived expression. In problems asking for (x+y):(x-y), many candidates find x:y and report that as the answer. Read the question fully before computing.
Sum-of-pairs: forgetting to divide by 2. When you add (a+b) + (b+c) + (c+a), you get 2(a+b+c), not a+b+c. Dividing by 2 before proceeding is a step that gets skipped under time pressure, leading to a value of k that is half the correct value and cascading wrong answers.