Aptitude · Percentages · Model 1

Delete the hundreds and there are no percentages left

“If 50% of P equals 25% of Q…” looks like a percentage question and is really one line of algebra. The hundreds cancel on both sides, and what remains is a ratio you can read straight off.

Set the two percentages and watch the hundreds cancel →

a% of P = b% of Q means aP = bQ. The hundreds cancel, and the ratio P : Q = b : a falls out immediately.

01 The idea

The percentages are decoration

If 50% of P is equal to 25% of Q, what per cent of Q is P? Written out, the equation is 50P/100 = 25Q/100. Both sides have a hundred underneath, so multiply through and it is gone: 50P = 25Q, or 2P = Q.

So P is half of Q, which is 50% of Q. The whole question was one cancellation and one division, and nothing about percentages survived past the first line. That is the pattern for every question in this model.

The variants only change what you are asked to read off the ratio. Sometimes it is a percentage, sometimes the ratio P : Q itself, sometimes an expression like (2A − 3B)/(A + B). Once you have A = 5B or 2A = 3B, all of those are substitution.

The most useful habit is to reduce to the simplest relation and then choose convenient numbers. If A = 5B, set B = 1 and A = 5. Now any expression in A and B is arithmetic rather than algebra, and there is nothing left to slip on.

Get to a single relation like A = 5B, then pick B = 1. Every expression the question asks for becomes arithmetic.

The cancellationa% of P = b% of Q becomes aP = bQ. Both sides carry a factor of 1/100 and it divides out.

Reading the ratioFrom aP = bQ, the ratio P : Q is b : a — note the swap.

Convenient valuesOnce you have a relation, assign the smallest whole numbers that satisfy it. Legitimate because the answer is a ratio or percentage and does not depend on the actual sizes.

02 Worked example

50% of P = 25% of Q, and then a nastier one

One easy case to fix the method, then the shape that looks hard. (i) If 50% of P equals 25% of Q, find P as a percentage of Q. (ii) If 20% of (P + Q) equals 50% of (P − Q), find P : Q.

Write part (i) as an equationBoth sides get a hundred underneath, so both hundreds go.50P/100 = 25Q/100 ⇒ 50P = 25Q

Simplify and read it offDivide by 25. P is half of Q.2P = Q ⇒ P = Q/2 = 50% of Q

Now part (ii) — same first moveThe brackets change nothing about the cancellation.20(P + Q) = 50(P − Q) ⇒ 2(P + Q) = 5(P − Q)

Expand and collectMultiply out and gather P on one side, Q on the other.2P + 2Q = 5P − 5Q ⇒ 7Q = 3P

Read the ratioFrom 3P = 7Q, the ratio P : Q is 7 : 3 — the coefficients swap over.P/Q = 7/3 ⇒ P : Q = 7 : 3

Watch the swap in the last step, because it is the one place this model catches people. From 3P = 7Q the ratio is 7 : 3, not 3 : 7 — the larger coefficient belongs to the smaller quantity. Check it with numbers: if P = 7 and Q = 3 then 3 × 7 = 21 and 7 × 3 = 21, so the relation holds.

03 The method

The variants, and the substitution habit

The first move never changes. What varies is what the question wants once you have the relation.

a% of P = b% of Q ⇒ aP = bQ ⇒ P : Q = b : a, and hence P = (b/a) × 100 per cent of Q.

Assign numbers as soon as you have a relation. If the question ends up wanting (2A − 3B)/(A + B) and you know A = 5B, set B = 1 and A = 5: the expression is (10 − 3)/6 = 7/6, done. Chained conditions work the same way — from 60A = 30B and B = 40% of C you get B = 2A and C = 5A, so C is 500% of A, without ever handling a decimal.

The question gives	It reduces to	Then
50% of P = 25% of Q	`2P = Q`	P is 50% of Q
20(P+Q) = 50(P−Q)	`3P = 7Q`	P : Q = 7 : 3
90% of A = 30% of B	`B = 3A`	B is 300% of A
20% of A = 80% of B	`A = 4B`	substitute and evaluate
8% of X = 4% of Y	`Y = 2X`	20% of X is 10% of Y
Chained conditions	`reduce one at a time`	60A=30B, B=40%C → C=500% of A
Reading aP = bQ as P:Q = a:b	`wrong, it swaps`	P : Q = b : a

05 Cheat sheet

Model 1 on one page

One first move, then whatever the question wants from the relation.

Step	Do this	On 50% of P = 25% of Q
1. Cancel the hundreds	`a% of P = b% of Q → aP = bQ`	50P = 25Q
2. Simplify	`divide by the hcf`	2P = Q
3. Read the ratio	`P : Q = b : a`	1 : 2
4. As a percentage	`(b/a) × 100`	P is 50% of Q
5. For an expression	`assign B = 1`	arithmetic, not algebra
Chained conditions	`reduce one at a time`	60A=30B, B=40%C → C=500%A
Swapping the ratio	`the usual slip`	3P = 7Q gives 7 : 3

The hundreds always cancelBoth sides of the equation carry a factor of 1/100, so the percentages vanish in the first line. There is no percentage arithmetic in this model at all.

The coefficients swapFrom aP = bQ the ratio P : Q is b : a. The larger coefficient belongs to the smaller quantity, which is worth checking with numbers once.

Assign convenient valuesOnce you have A = 5B, set B = 1. Legitimate because the answer is a ratio, and it turns every expression into arithmetic.

06 Where & why

Where this shows up

A short, mechanical model that appears as a warm-up and as a component of harder questions.

Bank PO · SSC CGL

“If a% of P = b% of Q…”

Set directly, usually asking for a ratio or a percentage. Ten seconds once the cancellation is instinct.

TCS NQT · Infosys

Expression evaluation

“Find (2A − 3B)/(A + B).” Reduce to a relation, assign numbers, evaluate. No algebra needed beyond the first line.

Chained conditions

Three quantities linked in sequence

“60% of A = 30% of B, B = 40% of C, C = X% of A.” Reduce one link at a time and the chain unwinds.

Ratio and proportion

The same relations in other clothing

This model is really ratio questions phrased with percentages. Recognising that makes both chapters shorter.

The only thing to remember here is the first line. Cancel the hundreds, and a question that looked like percentages turns out to be a ratio you can read off.

07 Interview questions

What gets asked

Nine, and the ratio swap in the middle is the one place this model bites.

If 50% of P equals 25% of Q, what per cent of Q is P?

Fifty per cent. Writing it out, 50P/100 = 25Q/100, so the hundreds cancel and 50P = 25Q, giving 2P = Q. So P is half of Q, which is 50% of Q.

Why do the percentages disappear so quickly?

Because both sides of the equation carry a factor of 1/100, and multiplying through by 100 removes them together. After that first line there is no percentage arithmetic left — only algebra.

From 3P = 7Q, what is the ratio P : Q?

7 : 3. The coefficients swap over, because P = (7/3)Q. Check it with numbers: P = 7 and Q = 3 gives 3 × 7 = 21 on the left and 7 × 3 = 21 on the right. Reading it as 3 : 7 is the commonest error in this model.

If 20% of (P + Q) equals 50% of (P − Q), find P : Q.

7 : 3. Cancel to 2(P + Q) = 5(P − Q), expand to 2P + 2Q = 5P − 5Q, and collect: 7Q = 3P. So P : Q = 7 : 3. Brackets change nothing about the method.

If 20% of A equals 80% of B, find (B + A)/(B − A).

−5/3. The relation is A = 4B, so set B = 1 and A = 4. Then (1 + 4)/(1 − 4) = 5/(−3). Assigning numbers once you have the relation is what makes these expression questions trivial.

How do you handle chained conditions?

Reduce one link at a time. From 60% of A = 30% of B you get B = 2A. Then B = 40% of C gives 2A = 0.4C, so C = 5A, which is 500% of A. Never try to combine all the links in one step.

If 8% of X equals 4% of Y, then 20% of X is what per cent of Y?

Ten per cent. First 8X = 4Y gives Y = 2X, so X = Y/2. Then 20% of X is 0.2 × Y/2 = 0.1Y, which is 10% of Y. Two short steps, and neither involves a decimal if you keep it as halves.

Is it legitimate to just pick numbers for the variables?

Hard

(a) Show that if a% of P equals b% of Q then P : Q = b : a, and state why the coefficients swap. (b) Hence show that P is (b/a) × 100 per cent of Q, and that Q is (a/b) × 100 per cent of P. (c) Verify that these two percentages multiply to 10,000 rather than adding to 200, and explain what that tells you about the relationship between the two readings.

Follow-up

Part (c) is the interesting one. The two percentages are reciprocals scaled by 10,000, not complements — so “P is 400% of Q” pairs with “Q is 25% of P”, and 400 × 25 = 10,000. Students who expect them to sum to 200 have imported an intuition from the more/less comparison rule, where it also fails.

Show the hint

For (c), multiply (b/a)×100 by (a/b)×100 and see what cancels.