Aptitude · Ratio and Proportion · Model 4

The rule you cannot break is the model you get paid for

Adding a fixed amount to both terms of a ratio changes it. That is the one thing you may not do to a ratio — and it is exactly why a paper can tell you the ratio before and after a change and expect you to recover the actual numbers.

Set a ratio, a shift and a target, and solve for one part →

5 : 7 becomes 7 : 11 when nine leave each group, and that pins one part at 6. The shift changes the ratio; the change is the information.

01 The idea

Why a broken rule is useful

Lesson one said you may multiply or divide both terms of a ratio by the same number but never add to them. That was not a prohibition — it was a measurement. Adding to both terms moves a ratio by a definite amount, and if you know where it started and where it ended, the size of the move tells you the actual quantities.

A class has boys and girls in the ratio 5 : 7. Nine of each leave, and the ratio becomes 7 : 11. You were never told how many students there were, and you do not need to be. Call one part x, so there are 5x boys and 7x girls. After the change there are 5x − 9 and 7x − 9, and those are in the ratio 7 : 11. Cross-multiply, solve, and x = 6 — so 30 boys and 42 girls.

The equation is always linear. Cross-multiplying (5x − 9)/(7x − 9) = 7/11 gives 55x − 99 = 49x − 63, and the x terms sit on both sides at the first power. There is no quadratic to worry about, so every question in this model comes down to one collection of like terms and one division.

Here is the part worth carrying away. When both groups move by the same amount, the difference between them is untouched — 42 − 30 = 12 before, and 33 − 21 = 12 after — while the ratio moves. Scaling does the reverse: it keeps the ratio and changes the difference. Ratios and differences are two independent facts about a pair of numbers, and the two operations each preserve exactly one of them.

Call one part x, write both quantities as multiples of it, apply the shift, then cross-multiply. One linear equation, one division, done.

One part, xThe unknown scale factor. With no total given, 5 : 7 becomes 5x and 7x, and the whole model is solving for x.

The shiftThe fixed amount added to or subtracted from each quantity. The two shifts need not be equal — “50 boys leave and 50 girls join” is −50 and +50.

The uniform-shift shortcutIf every ratio term rises by the same number of parts, that part count equals the shift. 2 : 3 : 5 → 4 : 5 : 7 is +2 parts each, so 2 parts = the shift and no algebra is needed.

02 Worked example

5 : 7 becomes 7 : 11 when nine leave each group

One class runs the whole lesson. In a classroom the boys and girls are in the ratio 5 : 7. If 9 students are removed from each group, the ratio becomes 7 : 11. Find the difference between the number of boys and the number of girls.

Attach an unknown to the ratioNo total is given, so the ratio alone cannot fix the numbers. Let one part be x; there are 5x boys and 7x girls.boys = 5x, girls = 7x

Apply the change to bothNine leave each group, so subtract 9 from each expression. The unknown x rides through untouched.boys → 5x − 9, girls → 7x − 9

Set the new ratio and cross-multiplyThe shifted pair is in the ratio 7 : 11. Cross-multiplying turns that into a linear equation — the x terms are first power on both sides.11(5x − 9) = 7(7x − 9) ⇒ 55x − 99 = 49x − 63

Collect and solveBring the x terms to one side and the numbers to the other. One division finishes it.6x = 36 ⇒ x = 6 ⇒ boys 30, girls 42

Answer the question, then checkThe question asked for the difference, which is 2x. Then verify: after the change the numbers are 21 and 33, and those must reduce to 7 : 11.difference = 42 − 30 = 12 check: 21 : 33 = 7 : 11 ✓

The difference of 12 is worth staring at. Before the change it was 42 − 30; after the change it was 33 − 21; both are 12, because subtracting the same 9 from both numbers cannot alter the gap between them. That is the mirror image of the golden rule, and it is also a free check: in any equal-shift question, the difference before and after must match.

03 The method

The general equation, and the shortcut that avoids it

One setup covers every question in this model. The shortcut covers the subset where every term shifts by the same number of parts, and that subset is set surprisingly often.

For a : b becoming c : d after shifts of p and q: (ax + p)/(bx + q) = c/d, so x = (cq − dp)/(ad − cb). Substituting back gives the two quantities. For 5 : 7 → 7 : 11 with p = q = −9 that is x = (−63 + 99)/(55 − 49) = 36/6 = 6.

If every term rises by the same number of parts, the algebra disappears. Three classes are in the ratio 2 : 3 : 5; forty students join each, and the ratio becomes 4 : 5 : 7. Look at the terms: 2→4, 3→5, 5→7 — each up by exactly 2 parts. So 2 parts = 40 students, one part = 20, and the original total is 10 parts = 200. Confirm it the long way once: (2x+40)/(3x+40) = 4/5 gives 10x + 200 = 12x + 160, so x = 20. The shortcut is only valid when the part increase is the same for every term, so check all of them before you use it.

Question shape	Setup	Answer
Ratio 5 : 7, minus 9 each, becomes 7 : 11	`(5x−9)/(7x−9) = 7/11`	x = 6 → 30 and 42
Ratio 3 : 5, minus 9 each, becomes 12 : 23	`(3x−9)/(5x−9) = 12/23`	x = 11 → 33 and 55
Ratio 5 : 3, −50 and +50, becomes 9 : 7	`(5x−50)/(3x+50) = 9/7`	x = 100 → 500 and 300
Ratio 4 : 5, −10 and +10, becomes 1 : 2	`(4x−10)/(5x+10) = 1/2`	x = 10 → total 90
Ratio 1 : 2, plus 7 each, becomes 3 : 5	`(x+7)/(2x+7) = 3/5`	x = 14 → 14 and 28
Ratio 2 : 3 : 5, plus 40 each, becomes 4 : 5 : 7	`every term up 2 parts`	2 parts = 40 → total 200
Total given as well as the ratio	`no x needed at all`	504 in 13 : 11 → 273 and 231

05 Cheat sheet

Model 4 on one page

The setup, the shortcut, and the two checks that between them catch every error this model can produce.

Case	Rule	On 5 : 7 → 7 : 11, −9 each
No total given	`write ax and bx`	5x and 7x
The equation	`(ax+p)/(bx+q) = c/d`	(5x−9)/(7x−9) = 7/11
Solving	`x = (cq − dp)/(ad − cb)`	36/6 = 6
Equal shifts	`the difference is preserved`	12 before, 12 after
Unequal shifts	`gap changes by \|q − p\|`	−50 and +50 moves it by 100
Every term up the same parts	`skip the algebra`	2:3:5 → 4:5:7 is +2 parts each
Total given as well	`no x at all`	504 in 13:11 gives 273 and 231 directly

The equation is always linearCross-multiplying two ratios in x gives x terms at the first power on both sides. If a quadratic appears, you multiplied two unknown quantities together somewhere and should re-read the setup.

Equal shifts preserve the differenceSubtracting 9 from both 30 and 42 leaves the gap at 12. So in any equal-shift question, checking that the before and after differences match costs nothing and catches a sign error immediately.

A given total removes the algebra“504 students in the ratio 13 : 11, then 12 girls join” needs no x: 24 parts of 504 is 21, so 273 boys and 231 girls, and the new ratio is 273 : 243 = 91 : 81. Reaching for x when the total is given is wasted work.

06 Where & why

Where this shows up

Model 4 is the most algebraic model in the chapter, which is why it is the one that separates candidates in a hard paper.

Bank PO · SSC CGL

“If 9 are removed from each, the ratio becomes …”

The standard form, with the total deliberately withheld so you have to introduce x. Two or three marks and about ninety seconds.

TCS NQT · Infosys

Opposite shifts

“50 boys leave and 50 girls join.” Same equation, but the signs differ between the two groups, and a sign slip produces a plausible wrong answer rather than an obvious one.

Three-term versions

“40 students are added to each of three classes”

Check whether every term rises by the same number of parts. When it does — 2 : 3 : 5 to 4 : 5 : 7 — the whole question is one division.

Multi-stage word problems

Two changes, one after the other

“The ratio becomes 2 : 1 when 15 girls leave; then 1 : 5 when 45 boys also leave.” Two equations in two unknowns, solved by substitution. This is the hardest shape the chapter has.

One habit decides this model: read whether the total is given. If it is, find the actual numbers first and never introduce x at all. If it is not, introduce x immediately and expect one linear equation.

07 Interview questions

What gets asked

Ten. The second question is the one that explains why this model exists at all.

Boys and girls are in the ratio 5 : 7. Nine leave each group and the ratio becomes 7 : 11. Find the difference between the two numbers.

12. Write the groups as 5x and 7x, apply the shift, and set (5x − 9)/(7x − 9) = 7/11. Cross-multiplying gives 6x = 36, so x = 6 and the groups are 30 and 42. The difference is 2x = 12.

Adding to both terms is the one thing you cannot do to a ratio. So why is there a whole model about it?

Because the change is measurable. Adding a constant moves a ratio by a definite amount, so knowing the ratio before and after pins down the scale. The rule is not a prohibition, it is the mechanism the model runs on.

Does subtracting the same number from both quantities change their difference?

No. Subtracting 9 from 30 and 42 gives 21 and 33, and the gap is 12 either way. Equal shifts preserve the difference and destroy the ratio; scaling preserves the ratio and destroys the difference. Each operation keeps exactly one of the two facts.

Why does the equation never come out quadratic?

Because cross-multiplying two ratios that are each linear in x gives x terms at the first power on both sides. The x² terms would only appear if you multiplied two unknown quantities together, which this setup never does.

Three classes are in the ratio 2 : 3 : 5. Forty students join each and the ratio becomes 4 : 5 : 7. Find the original total.

200. Each ratio term rises by exactly 2 — 2 to 4, 3 to 5, 5 to 7 — so 2 parts equal 40 students and one part is 20. The original total is 2 + 3 + 5 = 10 parts, so 200. No algebra needed, but check all three terms before trusting the shortcut.

When can you skip introducing x altogether?

When the total is given. “504 students in the ratio 13 : 11” means 24 parts of 504, so one part is 21 and the groups are 273 and 231 straight away. Then apply the change to the actual numbers. Introducing x here is extra work with extra chances to slip.

A ratio 5 : 3 becomes 9 : 7 when 50 boys leave and 50 girls join. Find the number of boys.

500. The shifts have opposite signs: (5x − 50)/(3x + 50) = 9/7, so 35x − 350 = 27x + 450, giving 8x = 800 and x = 100. Boys are 5x = 500, girls 300, and after the change 450 and 350, which is 9 : 7.

What number must be added to both terms of 7 : 13 to make it 2 : 3?

5. Here the unknown is the shift rather than the scale: (7 + x)/(13 + x) = 2/3, so 21 + 3x = 26 + 2x and x = 5. Check: 12 : 18 = 2 : 3. Note that the answer moves the ratio towards 1 : 1, as adding to both terms always does.

What is the fastest check on an answer in this model?

Hard

In a group, the ratio of boys to girls becomes 2 : 1 when 15 girls leave. Afterwards, when 45 boys also leave, the ratio becomes 1 : 5. (a) Find the original number of girls and boys. (b) Verify both stated ratios against your answer. (c) Explain why this problem needs two equations rather than one, and what feature of the wording tells you that in advance.

Follow-up

Two events in sequence means two unknowns and two equations, solved by substitution rather than by a single cross-multiplication. Part (c) is the transferable skill: the phrase “afterwards” is what signals that the second ratio applies to the already-changed group, not to the original one.

Show the hint

Let the original numbers be x boys and y girls. The first sentence relates x to y − 15; the second relates x − 45 to the same y − 15.