A ratio is nothing more than a comparison between two quantities of the same kind using division. When you say the ratio of A to B is 3 : 4, you are saying that for every 3 units of A, there are 4 units of B. The actual values do not matter — only the relative sizes do.
Here is a useful way to think about it: imagine a pizza cut into 7 slices, where A gets 3 slices and B gets 4. You can describe the sharing as 3 : 4. Whether the pizza serves two people or 200 does not change the relative split. That is the essence of ratio — it is scale-independent.
A proportion takes this further. It states that two ratios are equal: a : b = c : d, or equivalently, a/b = c/d. This gives you the cross-multiplication rule: a × d = b × c. This single rule powers more SSC CGL questions than almost anything else in this topic.
A few terms you will encounter:
a : b, a is the antecedent (पूर्वपद) and b is the consequent (उत्तरपद).a : x = x : b, then x is the mean proportional between a and b. Here x = √(ab).a : b = b : x, then x = b²/a is the third proportional.a : b = c : x, then x = bc/a.The practical spin-off of proportion is variation. Direct variation (अनुलोम अनुपात) means both quantities increase or decrease together — if you double the input, you double the output. Inverse variation (व्यत्क्रम अनुपात) means one quantity increases as the other decreases — if you double the speed, you halve the time for the same distance.
Think of variation as the story proportion tells about real-world relationships. You will use it every time you see "more workers, less days" or "more speed, less time" problems.
Given a ratio a : b, you can always write the actual values as a = kx and b = ky for some multiplier k, where x and y are the ratio terms. This multiplier k is what you solve for in most "find the actual number" problems.
Key operations on ratios:
a : b and c : d is ac : bd (multiply across).a : b is a² : b².a : b is a³ : b³.a : b is √a : √b.a : b is b : a.This is where SSC CGL spends real question time. You are given A : B and B : C and asked for A : B : C.
The method: make the value of B identical in both ratios by finding LCM.
A : B = 3 : 4 and B : C = 5 : 6A : B = 15 : 20B : C = 20 : 24A : B : C = 15 : 20 : 24This extends naturally to three-ratio chains (A:B, B:C, C:D) — keep finding the common link and scaling.
Look — this pattern trips up a lot of people, but the method is clean. If 4A = 5B = 6C = k, then:
A = k/4, B = k/5, C = k/6
So A : B : C = 1/4 : 1/5 : 1/6
Multiply through by LCM(4, 5, 6) = 60:
A : B : C = 15 : 12 : 10
The pattern: when you have xA = yB = zC, the ratio A : B : C = 1/x : 1/y : 1/z. Flip the coefficients, then scale by LCM.
The proportion a : b :: c : d (read: a is to b as c is to d) gives the cross-product rule:
a × d = b × c
The product of extremes equals the product of means. This is your go-to when three of four values are known and you need the fourth.
If you divide a total T among A, B, C in the ratio p : q : r:
p + q + rp/(p+q+r) × Tq/(p+q+r) × Tr/(p+q+r) × TNo complicated algebra needed — just identify the share's numerator and divide by the total parts.
Direct proportion: y ∝ x means y = kx for some constant k. Equivalently, y₁/x₁ = y₂/x₂. Use this for speed-distance (constant speed), cost-quantity problems.
Inverse proportion: y ∝ 1/x means xy = k (constant). Equivalently, x₁y₁ = x₂y₂. Use this for time-workers problems, speed-time (constant distance).
The question type "x is added to each term of ratio a : b, new ratio is p : q — find x" uses a simple equation:
(a + x)/(b + x) = p/q
Cross-multiply and solve for x. This is the "4 is added to each" template you see regularly.
When you see pA = qB = rC, write A:B:C = 1/p : 1/q : 1/r, then multiply each term by LCM(p, q, r).
Example: 4A = 5B = 6C → 1/4 : 1/5 : 1/6 → multiply by 60 → 15 : 12 : 10.
Standard method (setting up three equations, solving pairwise): ~6 steps, ~40 seconds. This flip-and-LCM: 3 steps, ~12 seconds.
Instead of finding each person's absolute share and then comparing, directly compute the required share as a fraction of total parts.
Example: Divide ₹1200 in ratio 2 : 3 : 5. You want C's share. C's part = 5, total parts = 10. C gets 5/10 × 1200 = 600. Done.
No need to compute A and B. Standard multi-step approach: 3 calculations. This targeted fraction: 1 calculation, under 8 seconds.
For A:B and B:C, find LCM of B's two values. Scale both ratios so B matches.
Example: A:B = 2:3, B:C = 4:5. B values are 3 and 4. LCM = 12. Scale first ratio ×4 → 8:12. Scale second ×3 → 12:15. Result: A:B:C = 8:12:15.
Mental-arithmetic shortcut: just multiply ratio-1 terms by denominator of ratio-2's B term, and ratio-2 terms by numerator of ratio-1's B term. (Here: ×4 and ×3.) Reduces one step of explicit LCM computation — saves ~10 seconds per question.
After computing shares in a partition question, add all shares. They must equal the original total. If they don't, you made an arithmetic error.
Example: Ratio 3:5:7, total ₹1560. Shares: 390 + 650 + 520 = 1560. Confirmed in 3 seconds.
This eliminates re-reading the question for verification — replaces a 15-second re-check with a 3-second addition.
In mixture ratio problems where only one component changes (water added, milk removed, etc.), keep the unchanged component as the anchor and set up a single equation.
Example: Milk:Water = 5:2. Water added = 7L. New ratio = 5:3. Milk is unchanged. Let milk = 5x. Original water = 2x. New: 5x/(2x+7) = 5/3. One equation, one unknown.
Versus the two-variable approach (milk = 5x, water = 2x, new milk = 5y, new water = 3y + 7): 4 variables, 2 equations, ~50 seconds. Anchor method: 1 equation, ~15 seconds.
Read the question and identify the type in the first five seconds:
1. Find combined ratio (A:B:C from A:B and B:C)? → LCM-bridge on the shared term. Scale and read off.
2. Find actual values from a ratio (numbers in ratio a:b, their sum/difference is given)?
→ Write values as ax and bx. Plug into given condition. Solve for x.
3. Ratio after adding/subtracting a constant?
→ (a + k)/(b + k) = p/q. Cross-multiply. Solve.
4. If pA = qB = rC?
→ Flip coefficients. Scale by LCM. Done.
5. Share from a total in given ratio?
→ Required share = (ratio part / total parts) × total amount. Single step.
6. Mixture ratio with one component changing? → Anchor on the unchanged component. One equation.
7. Direct/inverse proportion? → Direct: ratios equal. Inverse: products equal. Write the equation and solve.
Always do a final sanity check: shares must sum to the total, ratios must simplify to the given form. This costs 3 seconds and catches about 40% of silly errors before you mark the answer.
Why this question: Tests the foundational combined-ratio skill — the LCM-bridge method. If you can do this on autopilot, you can handle any chained-ratio question.
Solving path: A:B = 3:4, B:C = 5:6. B appears as 4 and 5. LCM(4,5) = 20. Multiply first ratio by 5: 15:20. Multiply second by 4: 20:24. Read off: A:B:C = 15:20:24. Match with options — answer is C.
Why this question: The "constant added to both terms" type. This is a guaranteed appearance in CGL mocks and the actual exam. The trap is trying to guess — always set up the equation.
Solving path: Let numbers be 5x and 7x. Equation: (5x+4)/(7x+4) = 3/4. Cross-multiply: 4(5x+4) = 3(7x+4) → 20x + 16 = 21x + 12 → x = 4. Smaller number = 5×4 = 20. Answer: B.
Why this question: Pure partition. Fastest question type in this topic — one multiplication, done.
Solving path: Total parts = 2+3+5 = 10. C's fraction = 5/10. C's share = 5/10 × 1200 = 600. Answer: B. Under 10 seconds.
Why this question: The pA = qB = rC pattern — medium-difficulty but appears more often than it should in CGL. Tests whether you know the flip-and-LCM method.
Solving path: 4A = 5B = 6C = k. So A = k/4, B = k/5, C = k/6. Ratio = 1/4 : 1/5 : 1/6. LCM(4,5,6) = 60. Multiply: 15 : 12 : 10. Answer: C.
Why this question: Mixture problem — the anchor method is critical here. Also tests whether you correctly set up which quantity is unchanged.
Solving path: Let milk = 5x, water = 2x. Only water changes. New ratio: 5x/(2x+7) = 5/3. Cross-multiply: 15x = 10x + 35 → 5x = 35 → x = 7. Milk = 5×7 = 35 litres. Answer: C.
Not making B common before combining ratios. A:B = 3:4 and B:C = 5:6 does NOT give A:B:C = 3:4:6. The B values (4 and 5) are different — you must scale them to the same number first. Skipping this step gives a wrong answer that still looks plausible.
Flipping the inequality direction in pA = qB. If 4A = 5B, then A is the larger quantity (because B needs a bigger multiplier to equal A). Students often read 4A = 5B as "A has coefficient 4, so A is smaller" — this is backwards. A = k/4 > k/5 = B, so A > B.
Treating ratio as an absolute difference. If A:B = 3:4, many students assume A is 3 and B is 4. These are relative values, not absolutes. Always introduce the multiplier k and use the given condition to find it.
Adding the constant to only one term. In "4 is added to each number", the equation must be (5x+4)/(7x+4) = 3/4 — both terms get 4. Writing (5x+4)/(7x) = 3/4 is a frequent error under time pressure.
Confusing direct and inverse proportion. "More workers take less time" is inverse. "More goods cost more" is direct. The tell: if increasing one quantity increases the other, it is direct. If increasing one decreases the other, it is inverse. Write y₁/x₁ = y₂/x₂ for direct and x₁y₁ = x₂y₂ for inverse before solving.
Not cross-checking partition sums. After finding shares, students often mark the answer without verifying the shares add to the total. A 3-second addition check prevents errors from mis-reading the ratio or adding parts incorrectly.