Ratio and proportion — equivalent ratios, dividing in a ratio: Worked examples

The scenario matrix

Each row is a class of problem. The last column tells you which worked example below covers it.

#	Case class	What makes it tricky	Covered by
C1	Plain two-way split of a total	none — the base skill	Example 1
C2	Three-or-more-way split	more parts to add	Example 2
C3	Ratio has a zero term (degenerate)	one share is nothing	Example 3
C4	Ratio given as fractions	must clear denominators first	Example 4
C5	Total not given — a difference is given	"split" logic on a gap	Example 5
C6	Total not given — one share is given	back out the whole	Example 6
C7	Real-world word problem (units, mixture)	translate words → ratio	Example 7
C8	Exam twist: ratio changes after a transfer	two states, solve for unknown	Example 8
C9	Limiting behaviour: ratio $a : b$ as $b \to 0$ or $a = b$	what happens at the edges	Example 9
C10	Ratio not in lowest terms (optional simplify first)	reduce before splitting	Example 10

Everything the topic can throw sits in one of these cells. Let's clear them.

Example 1 — the base case (C1)

Figure s01 shows the bar as one whole total, sliced into $9$ identical cells; the red bracket marks Sam's $5$ cells.

Example 2 — three-way split (C2)

Worked example A prize of

$₹1400$ is split among three teams in the ratio $2 : 3 : 2$ . Forecast: two teams tie for the smaller share and one gets more — predict the middle team's amount before checking.

Step 1 — Add all parts. $2 + 3 + 2 = 7$ . Why this step? The "parts" idea doesn't care how many terms there are — you always add every term to get the total number of equal slices.

Step 2 — One part. $x = \frac{1400}{7} = ₹200$ . Why this step? The whole $₹1400$ is exactly these $7$ equal slices, so one slice is total $\div$ parts — same rule as Example 1, just more parts.

Step 3 — Shares. $2 \times 200 = ₹400$ , $3 \times 200 = ₹600$ , $2 \times 200 = ₹400$ . Why this step? Each team owns its own number of parts, so multiply the one-part value by each team's number in turn.

Verify: $400 + 600 + 400 = 1400$ ✓ and $400 : 600 : 400 = 2 : 3 : 2$ ✓ (divide by $200$ ).

Figure s02 shows the bar split into $7$ cells, with the red block marking the middle team's $3$ cells.

Example 3 — a zero term (C3, degenerate)

Worked example Divide

$₹300$ in the ratio $5 : 0$ . Forecast: what could a " $0$ " part possibly mean? Guess the two shares.

Step 1 — Add the parts. $5 + 0 = 5$ . Why this step? Even a zero counts as a term; adding it changes nothing, which is exactly the point.

Step 2 — One part. $x = \frac{300}{5} = ₹60$ . Why this step? The total is $5$ equal parts, so one part is $300 \div 5$ — the empty term didn't change the count of real parts.

Step 3 — Shares. First $= 5 \times 60 = ₹300$ ; second $= 0 \times 60 = ₹0$ . Why this step? Zero parts means that share is empty — the second person gets nothing, the first gets everything.

Verify: $300 + 0 = 300$ ✓. Sanity check: the ratio $5 : 0$ is not the same as $0 : 5$ — order still matters even with a zero. The only forbidden case is $0 : 0$ , because then total parts $= 0$ and $x = \frac{300}{0}$ is undefined — you cannot divide a real quantity in the ratio $0 : 0$ .

Figure s03 shows all $5$ red cells belonging to the first person; the second person's strip is empty.

Example 4 — a ratio made of fractions (C4)

Worked example Divide

$₹770$ in the ratio $\frac{1}{3} : \frac{1}{4}$ . Forecast: which is bigger, $\frac{1}{3}$ or $\frac{1}{4}$ ? So who gets more?

Step 1 — Clear the denominators. Multiply both terms by the least common multiple (LCM) of the denominators $3$ and $4$ , which is $12$ (the smallest number both $3$ and $4$ divide into): $\frac{1}{3} : \frac{1}{4} = \frac{1}{3} \times 12 : \frac{1}{4} \times 12 = 4 : 3.$ Why this step? You can't easily "add" fractional parts. By the equivalent-ratios definition above, multiplying both terms by $12$ gives the same ratio in whole numbers — which we can add and split. We chose $12$ (the LCM) precisely because it is the smallest multiplier that turns both fractions into whole numbers at once.

Step 2 — Add the parts. $4 + 3 = 7$ . Why this step? Now the terms are whole numbers, so the total splits into $4 + 3$ equal parts.

Step 3 — One part. $x = \frac{770}{7} = ₹110$ . Why this step? Total $\div$ parts, as always.

Step 4 — Shares. $4 \times 110 = ₹440$ and $3 \times 110 = ₹330$ . Why this step? Multiply one-part value by each converted term.

Verify: $440 + 330 = 770$ ✓. And $\frac{1}{3} > \frac{1}{4}$ , so the first person got more — matches ( $440 > 330$ ) ✓.

Figure s04 shows the same bar, first labelled with the awkward fractions, then re-cut into $7$ whole cells after scaling by $12$ .

Example 5 — a difference is given, not the total (C5)

Worked example Two salaries are in the ratio

$7 : 4$ . The first person earns $₹9000$ more than the second. Find both salaries. Forecast: the gap corresponds to how many parts? Guess before Step 1.

Step 1 — Turn the gap into parts. Salaries are $7 x$ and $4 x$ . Their difference is $7 x - 4 x = 3 x$ . Why this step? Whatever fact the problem gives us, we express it in the same currency: parts. Here the difference is $3$ parts.

Step 2 — Match the gap to $₹9000$ . $3 x = 9000 \Rightarrow x = ₹3000$ . Why this step? We were told the $3$ -part gap equals $₹9000$ , so one part $= 9000 \div 3$ .

Step 3 — Build the salaries. First $= 7 \times 3000 = ₹21000$ ; second $= 4 \times 3000 = ₹12000$ . Why this step? Now that one part is known, each salary is its number of parts times the one-part value — the ordinary multiply-back step.

Verify: difference $= 21000 - 12000 = 9000$ ✓ and $21000 : 12000 = 7 : 4$ ✓ (divide by $3000$ ).

Figure s05 shows two bars of $7$ and $4$ cells; the red overhang of $3$ cells is the $₹9000$ gap.

Figure s06 shows the red block as the known middle share of $5$ cells; the rest of the bar is rebuilt from it.

Example 7 — real-world mixture, watch the units (C7)

Worked example A drink is made by mixing syrup and water in the ratio

$2 : 9$ (by volume). How much syrup is in a $1.1$ litre bottle? Forecast: syrup is the small number ( $2$ of $11$ parts) — guess roughly what fraction of the bottle it is.

Step 1 — Same units first. Both quantities are volumes (litres), so they're of the same kind — the ratio is valid. Convert the total to a single unit: $1.1$ L $= 1100$ mL. Why this step? A ratio only compares like with like; and mixing L with mL silently would wreck the arithmetic. Pick one unit and stick to it.

Step 2 — Total parts. $2 + 9 = 11$ . Why this step? The bottle is chopped into $2 + 9$ equal volumes, so we add the terms to count the slices.

Step 3 — One part. $x = \frac{1100}{11} = 100$ mL. Why this step? The whole $1100$ mL equals $11$ equal parts, so one part is total $\div$ parts.

Step 4 — Syrup share. $2 \times 100 = 200$ mL $= 0.2$ L. Why this step? Syrup owns $2$ of the parts, so multiply the one-part volume by $2$ ; then convert back to litres to answer in the asked unit.

Verify: syrup $200$ mL $+$ water $900$ mL $= 1100$ mL $= 1.1$ L ✓. Syrup is $\frac{2}{11} \approx 0.18$ of the bottle — matches the forecast that it's a small fraction.

Figure s07 shows the bottle bar as $11$ cells; the red $2$ cells at the bottom are the syrup.

Example 8 — exam twist: ratio changes after a transfer (C8)

Worked example A and B share money in the ratio

$5 : 3$ . A gives $₹40$ to B, and now they hold equal amounts. How much did each start with? Forecast: equal-at-the-end means A must have started with more — how much more, in parts?

Step 1 — Name the two states in parts. Start: A $= 5 x$ , B $= 3 x$ . Why this step? Using one unknown $x$ for the part-size captures the whole ratio in a single letter, so we get one equation, not two.

Step 2 — Apply the transfer. After A gives $₹40$ : A has $5 x - 40$ , B has $3 x + 40$ . Why this step? We translate the event ("gives $₹40$ ") directly into the algebra — subtract from the giver, add to the receiver.

Step 3 — Use the "equal now" condition. $5 x - 40 = 3 x + 40 \Rightarrow 2 x = 80 \Rightarrow x = 40.$ Why this step? "Now equal" is the missing fact that pins down $x$ . Setting the two after-amounts equal gives a single linear equation — one unknown, no fractions — which we solve by collecting the $x$ terms on one side and the numbers on the other. This is the same "turn the condition into an easy equation" idea the parent note used for cross-multiplication.

Step 4 — Recover the starting amounts. A $= 5 \times 40 = ₹200$ ; B $= 3 \times 40 = ₹120$ . Why this step? Now that the part-size $x = 40$ is known, each start is its number of parts times $x$ .

Verify: after transfer A has $200 - 40 = 160$ , B has $120 + 40 = 160$ — equal ✓. And $200 : 120 = 5 : 3$ ✓ (divide by $40$ ).

Figure s08 shows two bars before ( $5$ vs $3$ cells) and after the $₹40$ arrow, ending level.

Example 9 — the two limiting edges (C9)

Worked example What happens to "divide

$T$ in ratio $a : b$ " at the extremes $a = b$ and $b \to 0$ ? Forecast: for $a = b$ predict the two shares; for a tiny $b$ predict who gets almost everything.

Edge 1 — Equal ratio $a = b$ (say $1 : 1$ ). One part $= \frac{T}{1 + 1} = \frac{T}{2}$ , so each share $= \frac{T}{2}$ . Why this matters: $a : b$ with $a = b$ is a fair half-and-half split — the ratio machinery still works and just returns equal halves.

Edge 2 — One part shrinks, $b \to 0$ . The symbol $b \to 0$ reads " $b$ heads toward zero" — imagine holding $a = 1$ fixed while $b$ marches through a sequence of ever-smaller values. Watch the small share $\frac{b}{a + b} T$ along that sequence: $1 : 1 \to \frac{1}{2} T, 3 : 1 \to \frac{1}{4} T, 99 : 1 \to \frac{1}{100} T, 999 : 1 \to \frac{1}{1000} T, \dots$ Each ratio in this list ( $1 : 1, 3 : 1, 99 : 1, 999 : 1$ ) is just one snapshot on the way to $b = 0$ ; the ratios $99 : 1$ and $999 : 1$ are not the endpoint, they are stepping-stones that show the trend. As $b$ keeps shrinking, the big share $\to T$ and the small share $\to 0$ — smoothly approaching the degenerate $a : 0$ case of Example 3. Why this matters: the formula $\frac{a}{a + b} T$ is continuous — it doesn't "break" near the edge, it just glides into the all-or-nothing split.

Figure s09 shows three snapshots ( $1 : 1$ , $3 : 1$ , $99 : 1$ ) along that shrinking sequence; as the divider slides right, the red (small) share smoothly vanishes.

Verify (with $T = ₹800$ , ratio $99 : 1$ ): big $= \frac{99}{100} \times 800 = 792$ , small $= \frac{1}{100} \times 800 = 8$ ; sum $792 + 8 = 800$ ✓.

Example 10 — ratio not in lowest terms (C10)

Worked example Divide

$₹500$ in the ratio $40 : 60$ . Forecast: those numbers look big — but is $40 : 60$ really different from a much simpler ratio? Guess the simplest form first.

Step 1 — Simplify first (optional but wise). Divide both terms by the greatest common divisor (GCD) of $40$ and $60$ , which is $20$ (the largest number dividing both exactly): $40 : 60 = \frac{40}{20} : \frac{60}{20} = 2 : 3.$ Why this step? By the equivalent-ratios definition (Example 4), dividing both terms by the same number leaves the ratio unchanged — but the arithmetic below is far lighter with small numbers. You could split with $40 : 60$ directly; you'd just carry bigger numbers for the same answer.

Step 2 — Add the parts. $2 + 3 = 5$ . Why this step? The total is chopped into $2 + 3$ equal parts.

Step 3 — One part. $x = \frac{500}{5} = ₹100$ . Why this step? Total $\div$ parts.

Step 4 — Shares. $2 \times 100 = ₹200$ and $3 \times 100 = ₹300$ . Why this step? Multiply one-part value by each term.

Verify: $200 + 300 = 500$ ✓. Cross-check the un-simplified route: with $40 : 60$ , total parts $= 100$ , one part $= 500/100 = ₹5$ , shares $= 40 \times 5 = 200$ and $60 \times 5 = 300$ — identical ✓, proving simplifying first changed nothing but the effort.

Figure s10 shows the bulky $40 : 60$ bar collapsing onto the tidy $2 : 3$ bar — same red proportion, fewer cells.

Recall

Recall Which cell? Match the disguise

A problem gives a difference of $₹9000$ , ratio $7 : 4$ — what's the first move? ::: Express the difference in parts ( $7 x - 4 x = 3 x$ ), set $3 x = 9000$ . A ratio is $\frac{1}{3} : \frac{1}{4}$ — what must you do before adding parts? ::: Multiply both terms by the LCM (least common multiple) of the denominators ( $12$ ) to get whole numbers ( $4 : 3$ ). Divide $₹300$ in $5 : 0$ — is this allowed? ::: Yes; second share is $₹0$ . Only $0 : 0$ is forbidden (division by zero). One share is $₹250$ for the " $5$ " of $2 : 5 : 6$ — first step? ::: One part $= 250/5 = ₹50$ , then total $= 13 \times 50$ . A ratio $40 : 60$ — smartest first move before splitting? ::: Simplify to $2 : 3$ by dividing both terms by the GCD (greatest common divisor) $= 20$ ; the split is identical but lighter.

Connections

Fractions and simplification — Example 4 clears fractional ratio terms.
Highest Common Factor (GCD) — Example 10 reduces $40 : 60$ using the GCD.
Unitary method — "value of one part" is the unitary idea in every example here.
Direct and inverse proportion — Example 8's changing state is a proportion condition.
Percentages — Example 9's $\frac{99}{100}$ share is a percentage in disguise.
Similar figures — the same "parts" logic scales matching sides.