Ratio, proportion, and percentage are three faces of the same idea: comparing quantities. Once you see that connection clearly, the three topics stop feeling like separate things to memorize.
Ratio is a bare comparison — a : b tells you how many times one quantity is relative to another. It has no unit. The ratio 3 : 5 is identical to 6 : 10 or 0.6 : 1. The moment you fix the total, a ratio becomes actual numbers.
Proportion is the claim that two ratios are equal: a : b :: c : d, or equivalently a/b = c/d. The single most useful fact here is the cross-multiplication rule: if a/b = c/d, then ad = bc. This one identity solves virtually every "add/subtract x to make four numbers proportional" question that CTET loves to set.
Percentage is ratio with a denominator fixed at 100. Saying "35%" is just saying the ratio 35 : 100. Because the base is fixed, percentage makes comparisons across different-sized wholes easy. That's why shopkeepers, banks, and census reports all use it.
Here is the analogy that connects all three for your Class 6–8 students: imagine cutting a pizza. Ratio is "I got 3 slices and you got 5 slices." Proportion is "Every time I get 3 slices, you get 5 — that holds across every pizza." Percentage is "I got 37.5% of the pizza" — useful when someone asks how your share compares to the whole.
For CTET Paper II, you need two things: the mechanical skill to solve these in under 90 seconds each, and a conceptual understanding of why each method works — because the paper sometimes asks you to identify a misconception a student might hold, or to pick the right teaching approach.
A ratio a : b in lowest terms means both a and b share no common factor. To compare or combine ratios, bring them to a common consequent (the second term). If A : B = 3 : 4 and B : C = 5 : 6, find A : B : C by making B the same in both — B = LCM(4, 5) = 20. So A : B = 15 : 20 and B : C = 20 : 24, giving A : B : C = 15 : 20 : 24.
The "equal multiples" trick: When 4A = 6B = 9C = k, write each variable as k divided by its coefficient: A = k/4, B = k/6, C = k/9. Then A : B : C = 1/4 : 1/6 : 1/9. Multiply through by LCM(4, 6, 9) = 36 to get A : B : C = 9 : 6 : 4. This is faster than solving simultaneous equations and appears almost verbatim in CTET PYQs.
CTET Paper II has repeatedly asked: "Add x to (or subtract x from) four numbers so that they are in proportion." The setup is always:
(a + x) / (b + x) = (c + x) / (d + x)
Cross-multiply immediately:
(a + x)(d + x) = (b + x)(c + x)
Expand both sides. The x² terms cancel — they always do, which is why the problem has a unique solution. Collect x terms on one side:
ad + (a + d)x = bc + (b + c)x
ad - bc = (b + c - a - d)x
x = (ad - bc) / (b + c - a - d)
You don't need to memorize this formula — just remember that cross-multiplying, expanding, and cancelling x² is the mechanical path. The algebra takes about 40 seconds once you're practiced.
Every percentage question is one of three types:
(P/100) × Q(Q/R) × 100Q / (P/100) = Q × (100/P)Identify the type first. Misidentifying is the most common error — more on that in Common Mistakes.
Successive percentage changes appear in discount and population problems. If a quantity changes by +a% then by +b%, the net effect is:
Net % change = a + b + (ab/100)
When one change is positive and one is negative (say +10% then -10%), the net is always a loss of (10 × 10)/100 = 1%. Never zero. This is a classic trap.
Mark-up and successive discounts: Start with CP = 100 (always use this base). Apply the mark-up multiplicatively: CP × (1 + mark-up%). Apply each discount multiplicatively: × (1 - d₁%) × (1 - d₂%). The final figure minus 100 is the net profit or loss percentage directly. No formula to memorize — just sequential multiplication.
When you add a pure substance to a mixture, only the ratio of that substance changes; the other stays fixed. Set the unchanged component as kx in both the before and after situations, then equate and solve. This gives you x directly.
Example pattern: milk : water = 7 : 3 originally. Add 10 L water to get 7 : 5. Milk is unchanged at 7x. Before: water = 3x. After: water = 3x + 10. Set up 7x / (3x + 10) = 7/5. Solve: x = 5, milk = 35 L.
When you see mA = nB = pC, flip the coefficients and use LCM as the multiplier. Write A : B : C = 1/m : 1/n : 1/p, then multiply by LCM(m, n, p). No algebra needed.
Micro-example: 4A = 6B = 9C. LCM(4,6,9) = 36. Ratios: 36/4 : 36/6 : 36/9 = 9 : 6 : 4. Done in 10 seconds vs. 45 seconds of simultaneous equations.
Two discounts d₁% and d₂% applied in sequence give a net discount of d₁ + d₂ - (d₁ × d₂)/100 percent. This is always less than d₁ + d₂. Use it to find the effective selling price in one step.
Micro-example: 10% then 15% off a marked price of 140. Net discount = 10 + 15 - (10×15)/100 = 25 - 1.5 = 23.5%. SP = 140 × (1 - 0.235) = 140 × 0.765 = 107.1. Profit = 7.1%. Standard sequential method: 3 multiplications (~50s). This formula: 1 subtraction + 1 multiplication (~20s).
Any quantity increased by r% then decreased by r% (or vice versa) always results in a net loss of r²/100 percent. The result is never the original value — this is a trap question at every exam.
Micro-example: Population +10%, then -10%. Net = -(10²/100)% = -1%. If end population is 29700, original = 29700 / 0.99 = 30000. Knowing the net effect is exactly -1% saves you from setting up the two-step equation (~30s saved).
In any "add x / subtract x to four numbers to make them proportional" question, cross-multiply, expand, and immediately cancel x² from both sides. You are guaranteed to get a linear equation. This eliminates the instinct to use the quadratic formula or factor — which wastes 60-90 seconds and introduces sign errors.
Micro-example: (14+x)(30+x) = (12+x)(34+x). Expand: 420 + 44x + x² = 408 + 46x + x². Cancel x²: 420 + 44x = 408 + 46x → x = 6. 3 steps, under 25 seconds.
When a substance is added to a mixture without removing anything, identify the component that does NOT change. Let that component = kx (the same kx in both ratios). Set up one equation, solve for x. Never set up two variables for both components.
Micro-example: milk:water = 7:3, add 10L water → 7:5. Milk unchanged at 7x. Equation: 7x/(3x+10) = 7/5. One unknown, one equation. Solve in 20 seconds vs. 45 seconds with two variables.
Read the question and classify it in the first 5 seconds:
Is it a proportion problem (four numbers, add/subtract x)? → Cross-multiply → expand → cancel x² → solve linear equation.
Is it an equal-product ratio (mA = nB = pC)? → Write 1/m : 1/n : 1/p → multiply by LCM → done.
Is it a mixture problem? → Identify the unchanged component → set as kx → one equation → solve.
Is it a percentage problem? → Identify type (find %, is what % of, is % of what?) → CP = 100 base for profit/discount → apply multipliers sequentially.
Is it successive same-rate change (+r% then -r%)? → Net = -r²/100%. Use directly. Do not set up two steps.
If the question involves a radical or square root at the end (like √(12x + 9)), solve for x first using the proportion method, then substitute. Never try to work backwards from the answer options through the radical — it wastes time.
Why this question: The classic "add x to make proportional" format, with a twist — you must evaluate a radical after finding x. CTET uses this to test whether you can chain proportion-solving with arithmetic.
Solving path: Set up (14+x)/(12+x) = (34+x)/(30+x). Cross-multiply and expand both sides: 420 + 44x + x² = 408 + 46x + x². Cancel x² immediately: 12 = 2x → x = 6. Now evaluate: √(12×6 + 9) = √(72 + 9) = √81 = 9. Answer: 9.
Why this question: The "subtract x" variant of the same pattern. The algebra is identical in structure but students sometimes make sign errors when distributing negatives during expansion.
Solving path: Set (50-x)/(61-x) = (92-x)/(117-x). Cross-multiply: (50-x)(117-x) = (61-x)(92-x). Expand left: 5850 - 167x + x². Expand right: 5612 - 153x + x². Cancel x²: 5850 - 167x = 5612 - 153x → 238 = 14x → x = 17. Answer: 17.
Why this question: The equal-product ratio type where LCM-flipping is far faster than solving simultaneous equations. CTET tests this exact structure in division/sharing problems.
Solving path: 4A = 6B = 9C means A : B : C = 1/4 : 1/6 : 1/9. LCM(4,6,9) = 36. Multiply: 9 : 6 : 4. Sum = 19. A = (9/19) × 1710 = 810. C = (4/19) × 1710 = 360. Difference = 450. Answer: ₹450.
Why this question: Ratio-transfer problem — fruits moved from one basket to another. Teaches careful tracking of what increases and what decreases. The total is conserved, which is a useful check.
Solving path: Let A = 7k, B = 9k. After transfer: A = 7k - 6, B = 9k + 6. Ratio = 1:3, so 3(7k-6) = 9k+6 → 21k - 18 = 9k + 6 → 12k = 24 → k = 2. Total = 7(2) + 9(2) = 14 + 18 = 32. Note: the verified answer from the equation is 32; use the cross-multiplication check 8:24 = 1:3 to confirm before moving on.
Why this question: Mixture ratio problem where milk is the fixed component. Tests whether you correctly identify what stays constant when water is added.
Solving path: Original: milk = 7x, water = 3x. Add 10 L water. New ratio: 7x / (3x+10) = 7/5. Cross-multiply: 35x = 21x + 70 → 14x = 70 → x = 5. Milk = 35 L. Answer: 35 litres.
Why this question: Mark-up plus successive discounts — a three-step percentage chain. The multiplicative approach (CP = 100 base) is cleanest here.
Solving path: Let CP = 100. MP = 140. After 10% discount: 140 × 0.9 = 126. After 15% discount: 126 × 0.85 = 107.1. Net profit = 7.1%. Answer: 7.1% profit.
Why this question: The symmetric +r% / -r% trap. The net is never zero — it is always a small loss. This question is a direct test of that principle.
Solving path: P × 1.10 × 0.90 = P × 0.99 = 29700. So P = 29700 / 0.99 = 30000. Alternatively: net change = -(10²/100)% = -1%, so original is 29700 / 0.99 = 30000. Answer: 30000.
Forgetting that x² cancels in proportion problems. Students set up the cross-multiplication correctly, then panic when they see x² and reach for the quadratic formula. The x² terms are identical on both sides and always cancel. If they don't cancel, re-check your expansion — you've made a sign error.
Treating +r% then -r% as a wash. A 10% increase followed by a 10% decrease does not return to the original. The net is always -(r²/100)%. In the population question above, the net is -1%, not 0%. This exact error appears in student work at the Class 8 level and CTET tests it.
Using sum of discounts instead of multiplicative discounts. Two successive discounts of 10% and 15% are not equal to a single 25% discount. The correct equivalent single discount is 10 + 15 - (10×15)/100 = 23.5%. Adding them directly over-estimates the discount and gives a wrong selling price.
Choosing the wrong component as fixed in mixture problems. When only water is added, milk is fixed — not water. When nothing is removed, the component you are not adding stays the same. Always identify the unchanged component before setting up the equation.
Dividing the sum in a ratio without adding the ratio parts first. In the A:B:C = 9:6:4 problem, the total parts are 9+6+4 = 19, not 9 or 4. Students sometimes divide the sum by just one ratio part, or by the total of only two. Add all parts in the ratio before dividing.
Confusing "percentage of" with "percentage more than". If A is 20% more than B, then A = 1.2B — not A = 0.2B. And B is not 20% less than A; B is (20/120) × 100 = 16.67% less than A. These are different calculations, and CTET word problems exploit this asymmetry.