Aptitude · Ratio and Proportion · Model 3

Two ratios, one shared term, one chain

A : B = 2 : 3 and B : C = 4 : 5 describe the same B at two different sizes. Scale both ratios until the shared term matches and the two comparisons become one three-term ratio — and there is a shortcut that skips the scaling entirely.

Join two ratios on their shared term →

A : B = 2 : 3 and B : C = 4 : 5 gives A : B : C = 8 : 12 : 15. Fill each gap with the number beside it, then multiply down the columns.

01 The idea

The bridge term

A is to B as 2 is to 3. B is to C as 4 is to 5. Both statements are about the same B, but one calls it 3 and the other calls it 4. You cannot write A : B : C as 2 : 3 : 5 or 2 : 4 : 5, because neither of those respects both statements. B is the bridge, and the bridge has to be one number.

Fixing it uses the golden rule from the first lesson. Multiplying both terms of a ratio by the same number leaves it unchanged, so scale each ratio until the two Bs agree. The LCM of 3 and 4 is 12: multiply 2 : 3 by 4 to get 8 : 12, and multiply 4 : 5 by 3 to get 12 : 15. Now both say B = 12, and the chain reads 8 : 12 : 15.

There is a faster route that never mentions the LCM. Write the two ratios as two rows and fill each blank cell with the number sitting next to it, then multiply down each column: A = 2 × 4 = 8, B = 3 × 4 = 12, C = 3 × 5 = 15. Same answer, no division, and it extends to four or five terms without changing.

Two things a paper will do to make this harder. It may give you the second ratio backwards — “D : C = 5 : 6” when your chain needs C : D — and the fix is to flip it to 6 : 5 before you start. Or it may give the ratios as fractions, like A : B = 1/2 : 3/8, and the fix is to cross-multiply: 1 × 8 : 2 × 3 = 8 : 6 = 4 : 3. Both are one extra line, and both are worth marks precisely because people skip them.

Scale each ratio until the shared term is the same number in both, then read the chain straight off. Or fill each gap with its neighbour and multiply down the columns.

Bridge termThe quantity that appears in both ratios — B in A : B and B : C. It is the only term you have to do any work on.

Neighbour fillA : B : C = ac : bc : bd for A : B = a : b and B : C = c : d. Fill each empty cell with the number beside it and multiply down.

Cleaning a fractional ratio1/2 : 3/8 becomes 1×8 : 2×3 = 8 : 6 = 4 : 3. Cross-multiplying diagonally clears both denominators in one move, with no LCM needed.

02 Worked example

A : B = 2 : 3 and B : C = 4 : 5, joined two ways

One chain runs the whole lesson. If A : B = 2 : 3 and B : C = 4 : 5, find A : B : C. Do it once with the LCM to see why it works, then once with the shortcut you will actually use.

Spot the bridge and its two sizesB is the term the two ratios share. It is written as 3 on the left and 4 on the right, and until those agree the chain cannot be written down.A : B = 2 : 3 B : C = 4 : 5 ⇒ B is 3 and also 4

Scale each ratio to the LCM of the two BsThe LCM of 3 and 4 is 12. Multiply the first ratio by 4 and the second by 3. Neither ratio changes value, because both terms move together.(2 : 3) × 4 = 8 : 12 (4 : 5) × 3 = 12 : 15

MergeB is now 12 in both, so the two ratios overlap exactly and become one chain. There is nothing left to cancel here.A : B : C = 8 : 12 : 15

Now the shortcut, with no LCMTwo rows, each gap filled with the number beside it, then multiply down the columns. This is the version worth having under exam conditions.A = 2×4 = 8, B = 3×4 = 12, C = 3×5 = 15 ✓

Read the pairs back out as a checkThe answer has to contain both of the ratios you started with. Reduce the first two terms, then the last two.8 : 12 = 2 : 3 ✓ 12 : 15 = 4 : 5 ✓

That last check is also the fastest way through a multiple-choice version of this question. The correct option must contain 2 : 3 in its first two terms, and usually only one option does — so you can answer without joining anything. If two options survive, check the last pair against 4 : 5, and if both still survive, take the one in simplest form, because an unsimplified ratio is not the answer to a ratio question.

03 The method

The general form, and the three ways a paper complicates it

One formula covers the whole model, and it extends to any length of chain. The complications are all in the setup, never in the joining.

A : B = a : b and B : C = c : d gives A : B : C = ac : bc : bd. For four terms, A : B = a : b, B : C = c : d and C : D = e : f gives ace : bce : bde : bdf — the same neighbour-fill rule with one more row.

Three setup fixes, each one line. If a ratio arrives reversed, flip it: “D : C = 5 : 6” is C : D = 6 : 5, and joining it unflipped is the single most common error in this model. If a ratio arrives as fractions, cross-multiply diagonally: 1/2 : 3/8 is 8 : 6 = 4 : 3. If it arrives as an equation, swap the coefficients first: 2A = 3B is A : B = 3 : 2. Do all three fixes before you join anything, and the joining itself never varies.

Given	Setup fix	Chain
A:B = 2:3, B:C = 4:5	`none needed`	`8 : 12 : 15`
P:Q = 3:4, Q:R = 8:9	`Q already divides: ×2 on the left`	`6 : 8 : 9`
2A = 3B, 4B = 5C	`swap each: 3:2 and 5:4`	`15 : 10 : 8`
A:B = 1/2 : 3/8, B:C = 1/3 : 5/9	`cross-multiply: 4:3 and 3:5`	`4 : 3 : 5`
A:B = 1:2, B:C = 3:2, C:D = 1:3	`three rows of neighbour fill`	`3 : 6 : 4 : 12`
A:B = 1:2, B:C = 3:4, D:C = 5:6	`flip D:C to C:D = 6:5`	`9 : 18 : 24 : 20`
Joining D:C without flipping it	`the classic error`	gives a chain that fails its own check

05 Cheat sheet

Model 3 on one page

One formula, three setup fixes, and the check that lets you answer a multiple-choice version without joining anything.

Case	Rule	On 2 : 3 and 4 : 5
Two ratios sharing a term	`ac : bc : bd`	8 : 12 : 15
LCM route	`scale both to LCM(b, c)`	B becomes 12 in both
Four-term chain	`ace : bce : bde : bdf`	1:2, 3:2, 1:3 → 3 : 6 : 4 : 12
Fractional ratio	`cross-multiply diagonally`	1/2 : 3/8 → 4 : 3
Ratio given as an equation	`swap the coefficients`	2A = 3B → 3 : 2
Ratio given backwards	`flip it before joining`	D : C = 5 : 6 means C : D = 6 : 5
The answer check	`both pairs must survive`	8 : 12 = 2 : 3 and 12 : 15 = 4 : 5

The chain contains both originalsReduce the first two terms of your answer and you must recover the first ratio; reduce the last two and you must recover the second. Any answer that fails this is wrong, and the test takes about five seconds.

Option elimination is usually enoughThe correct option must show 2 : 3 in its leading pair. In a four-option question that typically leaves one survivor, so you can answer without ever joining the ratios.

Simplest form breaks tiesIf two options both satisfy every pair check, one is a multiple of the other and the simplified one is the answer. A ratio question always wants the simplified ratio.

06 Where & why

Where this shows up

Connecting ratios is almost never the whole question. It is the step that turns two given ratios into the one ratio the rest of the question needs.

TCS NQT · Bank Prelims

“If A:B = 2:3 and B:C = 4:5, find A:B:C”

Asked directly and worth a mark. Neighbour fill answers it in about five seconds, and option elimination answers it in three.

Model 2, one step later

“… now divide ₹1,400 among them”

Join the ratios to 8 : 12 : 15, which is 35 parts, so one part is ₹40 and B’s share is ₹480. The joining is the setup; the split is the question.

SSC CGL Tier 2

Four-term chains and reversed ratios

“A:B = 1:2, B:C = 3:4, D:C = 5:6” needs the last ratio flipped before anything else. Marks here are lost in the setup, never in the arithmetic.

Partnerships and mixtures

Three-way ratios from pairwise data

Capital contributions or ingredient ratios often arrive as two overlapping pairs. This model is how they become the single ratio those chapters need.

Do the setup fixes first — flip reversed ratios, clear fractions, swap coefficients — then neighbour fill, then check both pairs. That sequence never varies, and following it makes this the most mechanical model in the chapter.

07 Interview questions

What gets asked

Ten. The first four are the model; the rest are the setups that make it look harder than it is.

If A : B = 2 : 3 and B : C = 4 : 5, find A : B : C.

8 : 12 : 15. B is 3 in one ratio and 4 in the other, so scale both to 12: 2 : 3 becomes 8 : 12 and 4 : 5 becomes 12 : 15. Check it by reducing the pairs — 8 : 12 is 2 : 3 and 12 : 15 is 4 : 5.

Why can you not just write 2 : 3 : 5?

Because that says B is 3 while the second ratio says B is 4. Both ratios describe the same quantity, so the bridge term has to be one number, and getting it there is the whole model.

Explain the neighbour-fill shortcut.

Write the two ratios as two rows with the missing cells filled by the number beside them, then multiply down each column. For a : b and c : d that gives ac : bc : bd. It is the LCM method with the division already cancelled out, so it needs no LCM.

How do you check a three-term answer quickly?

Reduce its first two terms and its last two terms. Both must return the ratios you were given. On 8 : 12 : 15 that is 2 : 3 and 4 : 5, so the answer stands.

If P : Q = 3 : 4 and Q : R = 8 : 9, find P : Q : R.

6 : 8 : 9. Here Q’s two sizes are 4 and 8, and 4 divides 8, so only the first ratio needs scaling — multiply it by 2. Neighbour fill also works and gives 24 : 32 : 36, which reduces to the same thing.

The question gives D : C = 5 : 6 and you need C : D. What do you do?

Flip it to C : D = 6 : 5 before joining anything. Joining it in the given order produces a chain that fails its own pair check, and it is the most frequently made mistake in this model.

If A : B = 1/2 : 3/8, what is A : B in whole numbers?

4 : 3. Cross-multiply diagonally: 1 × 8 : 2 × 3 = 8 : 6, which reduces to 4 : 3. This clears both denominators in one move; taking the LCM of 2 and 8 gets the same answer with more writing.

If 2A = 3B and 4B = 5C, find A : B : C.

15 : 10 : 8. Swap each equation first: 2A = 3B gives A : B = 3 : 2, and 4B = 5C gives B : C = 5 : 4. Then neighbour fill: 3×5 : 2×5 : 2×4. A quick shortcut on a multiple-choice paper is that A : B = 3 : 2 means A must exceed B, which kills any option whose first term is smaller than its second.

If A : B = 1 : 2, B : C = 3 : 2 and C : D = 1 : 3, find A : B : C : D.

Hard

If A : B = 2 : 1, B : C = 3 : 2 and C : D = 4 : 3: (a) find A : B : C : D. (b) Find A^2 : BD. (c) Explain why you may substitute the four chain values directly into (b) instead of carrying an unknown multiplier through, and give one example of a question about the same four quantities where you could not.

Follow-up

Part (c) is the real content. A^2 : BD is homogeneous of degree two on both sides, so the unknown scale factor cancels and direct substitution is exact rather than approximate. Ask for A + D in rupees instead and it fails immediately, because that needs a real total.

Show the hint

For (b) substitute your chain values as if they were the actual quantities, then simplify. For (c) look at what happens if you replace every quantity by k times itself.