SSC CGL Maths · Ratio, Proportion and Variation

Combining and Changing Ratios

Two ratios that share a term join into one once that term is made equal in both, and a ratio that changes after items are added or removed gives one equation for the ratio unit.

All Ratio, Proportion and Variation notes

Why this matters

Nineteen questions in the bank, sixteen of them EASY: ten join two, three or four ratios through a common term and nine change a ratio by adding or removing items. Both have a fast route that skips the algebra, and both need care about which term is shared or unchanged.

Concept 1 of 2: Joining ratios through a common term

A : B = 4 : 5 and B : C = 3 : 7 describe B in two different units, 5 parts in one and 3 parts in the other. Before the ratios can be joined, B must be counted in the same unit in both, which means scaling each ratio until the B terms match.

Definition

Make the shared term equal: scale A : B and B : C so that B becomes the LCM of its two terms, then read A : B : C.
Fast route (the N rule): for A : B = $a_1 : b_1$ and B : C = $b_2 : c_2$ , A : B : C = $a_1b_2 : b_1b_2 : b_1c_2$ . Reduce at the end.
A longer chain: join two at a time, or for only the end terms multiply the fractions, $\frac{A}{D} = \frac{A}{B} \times \frac{B}{C} \times \frac{C}{D}$ .
Reduce each given ratio first (6 : 8 is 3 : 4). The numbers stay small and the answer is already in lowest terms more often.

Joining ratios

A : B : C = a_1b_2 : b_1b_2 : b_1c_2 \qquad \frac{A}{D} = \frac{A}{B} \times \frac{B}{C} \times \frac{C}{D}

a_1 : b_1the ratio A : B
b_2 : c_2the ratio B : C

Worked example

If P : Q = 4 : 5 and Q : R = 3 : 7, find P : Q : R.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 1 · Q51Easy

Example 1 · Ratio, Proportion and Variation · Combining and Changing Ratios

If A : B = 7 : 5, B : C = 2 : 3, and C : D = 4 : 5, find the ratio A : D.

Writing the ratios side by side

A : B = 3 : 4 and B : C = 5 : 6 do not give A : B : C = 3 : 4 : 6. B is 4 parts in one and 5 in the other; matched at 20, the answer is 15 : 20 : 24.

Drill 9 more on joining ratios through a common term

Concept 2 of 2: A ratio after items are added or removed

Before the change the quantities are ak and bk for one unknown unit k. After the change they are known expressions in k, and their ratio is given, so one equation finds k. If one quantity does not change at all, it is a fixed ruler, and the other quantity's change can be read off it in parts.

Definition

Full method: write the quantities as $ak$ and $bk$ , apply the change, and set $\frac{ak + x}{bk + y} = \frac{m}{n}$ . Cross-multiply and solve for k.
Fast route (the unchanged term): if one quantity does not change, scale the old and new ratios so its term is the same in both. Then the change in the other term, in parts, is the number of items added or removed.
The fast route is safe only after that scaling. Reading parts straight off two ratios whose unchanged terms differ gives a wrong unit.
The same number added to both (or taken from both): $\frac{ak + x}{bk + x} = \frac{m}{n}$ , then the full method.
Read the question again before answering: it may ask for the number before the change, after it, or the total.

Changed ratio

\frac{ak + x}{bk + y} = \frac{m}{n}

a : bthe ratio before the change
x, ythe numbers added (negative if removed)
m : nthe ratio after the change

Worked example

Cars and bikes in a car park are in the ratio 3 : 5. After 12 more cars arrive, the ratio becomes 9 : 10. How many bikes are there?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 12 Sep 2025, Shift 2 · Q61Easy

Example 2 · Ratio, Proportion and Variation · Combining and Changing Ratios

The ratio of boys to girls in a school is 7:5. If 20 more girls join the school, the new ratio of boys to girls becomes 7:6. What is the total number of students in the school initially?

Reading parts before matching the unchanged term

Girls and boys are 2 : 3, and after some girls join they are 3 : 4. The boys did not change, but their terms are 3 and 4, so first write 8 : 12 and 9 : 12. Only then does the change in parts (8 to 9) give the number who joined.

Answering for the wrong moment

Many stems ask for the number BEFORE the change, but the equation is solved for the unit k. Finish by working out the quantity the question names, then check it against both ratios.

Drill 8 more on a ratio after items are added or removed

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Joining ratios through a common term
Joining ratios
$A : B : C = a_1b_2 : b_1b_2 : b_1c_2 \qquad \frac{A}{D} = \frac{A}{B} \times \frac{B}{C} \times \frac{C}{D}$
A ratio after items are added or removed
Changed ratio
$\frac{ak + x}{bk + y} = \frac{m}{n}$

Watch out for (3)

Writing the ratios side by side
→ Joining ratios through a common term
Reading parts before matching the unchanged term
→ A ratio after items are added or removed
Answering for the wrong moment
→ A ratio after items are added or removed

Test yourself on Ratio, Proportion and Variation

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Ratio, Proportion and Variation chapter test Drill all 19 questions