Exercises — Ratio and proportion — equivalent ratios, dividing in a ratio

What we want: the comparison red $:$ blue as small whole numbers. Step 1 — write it raw. $12:8$. Step 2 — find the largest common divisor. $\gcd (12,8)=4$ (largest number dividing both). Why the largest? So no further reduction is possible. In the figure, $4$ is the biggest bundle-size that splits both colours evenly. Step 3 — divide both terms by $4$. $12÷4:8÷4=3:2$. This counts the bundles: $3$ red bundles to $2$ blue bundles. Check: $\frac{12}{8}=1.5=\frac{3}{2}$ ✓. Answer: $\boxed{3:2}$.

Idea: "parts" are the equal slices; the ratio numbers tell how many slices each person takes. Step — add the terms. $5+3=8$. Why add? Because the total is made of every slice, and the two people together take all of them. Answer: $\boxed{8\text{ parts}}$.

Step 1 — total parts. $5+4=9$ (the number of cells in the bar). Step 2 — value of one part. x = \dfrac{72}{9} = \8$.\ast Whydivideby$9$?\ast Eachofthe$9$equalcellsmusthavethesamevalue,andtheyaddto$$72$.\ast \ast Step3—multiplyback.\ast \ast Amy$=5\times 8 = $40$(5cells),Ben$=4\times 8 = $32$(4cells).\ast \ast Check:\ast \ast$40+32 = 72$✓and$40:32 = 5:4$(dividebothby$8$)✓.\ast \ast Answer:\ast \ast$\boxed{\text{Amy}=$40,\ \text{Ben}=$32}$.

Step 1 — total parts. $2+3+4=9$. Why extend so easily? The "parts" bar just gets more colours — add all the numbers. Step 2 — one part. x = \dfrac{180}{9} = \20$.\ast \ast Step3—shares.\ast \ast$2\times 20=$40,\quad 3\times 20=$60,\quad 4\times 20=$80$.\ast \ast Check:\ast \ast$40+60+80 = 180$✓.\ast \ast Answer:\ast \ast$\boxed{$40,\ $60,\ $80}$.

Tool: cross-multiplication. For $\frac{a}{b}=\frac{c}{d}$, multiplying both sides by $bd$ gives $ad=bc$ — this kills the fractions (as the intuition box shows, it's a same-area statement) and leaves a simple linear equation. Step 1 — cross-multiply. $x\times 5=15\times 8$, so $5x=120$. Step 2 — solve. $x=\frac{120}{5}=24$. Check: $24:15=8:5$? Divide both by $3$: $24÷3:15÷3=8:5$ ✓. Answer: $\boxed{x=24}$.

What's known: the $7$-part share equals \105$.\ast \ast Wha{t}^{\prime }shidden:\ast \ast thevalueofonepart.\ast \ast Step1—onepart.\ast \ast Thelargershareisthe$7$-partside,so$7$\ast equal\ast partstogethermake$$105$.Tofind\ast \ast one\ast \ast ofthoseequalpartsweundothe"times$7$":divideby$7$.\ast Whydivideby$7$?\ast Becausethe$7$partsareallthesamesizeandaddto$$105$;sharing$$105$equallyamong$7$identicalpartsgiveseachpart$\dfrac{105}{7} = $15$.\ast \ast Step2—totalparts.\ast \ast$3+7 = 10$.\ast \ast Step3—totalmoney.\ast \ast$10 \times 15 = $150$.\ast \ast Check:\ast \ast shareswouldbe$45$and$105$;$45+105=150$✓and$45:105 = 3:7$✓.\ast \ast Answer:\ast \ast$\boxed{$150}$.

Step 1 — one part. Total parts $=7+3=10$, so one part =\dfrac{96}{10}=\9.60$.\ast Whydivideby$10$?\ast The$10$equalpartsaddto$$96$,soeachis$96/10$.\ast \ast Step2—thegapinparts.\ast \ast Larger$-$smaller$= 7-3 = 4$parts.\ast Whyworkwiththedifferenceinparts?\ast Wedo{n}^{\prime }tevenneedbothshares—theexcessis$4$partsdirectly.\ast \ast Step3—valueofthegap.\ast \ast$4 \times 9.60 = $38.40$.\ast \ast Check:\ast \ast shares$= 67.20$and$28.80$;difference$67.20-28.80 = 38.40$✓.\ast \ast Answer:\ast \ast$\boxed{$38.40}$.

Step 1 — read the two percentages. Girls $=60\%$, so boys $=100\%-60\%=40\%$. Step 2 — write as a ratio. girls $:$ boys $=60:40$. Step 3 — simplify. $\gcd (60,40)=20$, so $60÷20:40÷20=3:2$. Check: $\frac{60}{40}=1.5=\frac{3}{2}$ ✓. Answer: $\boxed{3:2}$.

Step 1 — find current green. $2$ parts $=20$ red $\Rightarrow$ one part $=\frac{20}{2}=10$. Green $=5$ parts $=5\times 10=50$. Step 2 — set up the new condition. Red stays $20$. Let $g$ green be added, so green $=50+g$. We need $\frac{20}{50+g}=\frac{1}{2}$. Step 3 — cross-multiply. $20\times 2=1\times (50+g)\Rightarrow 40=50+g\Rightarrow g=-10$. That negative means we'd have to remove green, not add it. So $1:2$ can't be reached by adding — instead we take $10$ green away: green becomes $50-10=40$, and $20:40=1:2$. Check: $20:40=1:2$ ✓. Answer: $\boxed{g=-10,\text{i.e. remove }10\text{ green marbles}}$.

Step 1 — total parts. $2+9=11$. Step 2 — one part. $x=\frac{3.3}{11}=0.3$ litre. Why divide by $11$? The $11$ equal parts fill the $3.3$-litre bottle. Step 3 — syrup share. $2\times 0.3=0.6$ litre. Check: water $=9\times 0.3=2.7$; $0.6+2.7=3.3$ ✓ and $0.6:2.7=2:9$ ✓. Answer: $\boxed{0.6\text{ litre of syrup}}$.

Step 1 — call one part $x$. Cara $=4x$, Dan $=5x$. Step 2 — write the "after" condition. Cara gains $6$, Dan loses $6$, and they become equal: $4x+6=5x-6$ Why this equation? "Equal amounts" means the two new totals are the same number. Step 3 — solve. $6+6=5x-4x\Rightarrow 12=x$. Step 4 — original amounts. Cara $=4\times 12=48$, Dan $=5\times 12=60$. Check: after transfer $48+6=54$ and $60-6=54$ — equal ✓, and $48:60=4:5$ ✓. Answer: $\boxed{\text{Cara }48,\text{Dan }60}$.

Problem: $B$ appears as $3$ in the first ratio but $4$ in the second — they must be made the same before we can chain. Tool: equivalent ratios. Scale each ratio so $B$ matches. We choose $\text{lcm}(3,4)=12$ for $B$. Why the lcm and not just any common multiple? Any common multiple (like $24$ or $36$) also works, but the smallest one keeps every term as small as possible, so the final triple comes out already in lowest terms — no extra $\gcd$ step needed. Step 1 — scale the first. $A:B=2:3=8:12$ (multiply both by $4$). Step 2 — scale the second. $B:C=4:5=12:15$ (multiply both by $3$). Step 3 — combine. Now $B=12$ in both, so $A:B:C=8:12:15$. Check: $A:B=8:12=2:3$ ✓ and $B:C=12:15=4:5$ ✓; $\gcd (8,12,15)=1$ (no whole number bigger than $1$ divides all three) so it's simplest — confirming the lcm gave us lowest terms. Answer: $\boxed{8:12:15}$.

Match the figure to the algebra. In the drawing, the green rectangle is the small one, labelled width = 6 cm along its base and 4 cm up its left edge. The lavender rectangle is the large one: its base is labelled width = 15 cm and its left edge (in coral) carries the unknown height = ? we are solving for. The butter arrow between them is stamped "$\times 5/2$" — that arrow is the scale factor we compute in Step 1, showing every side of the green shape stretching into the lavender one. Step 1 — the length ratio (scale factor). Widths: $6\to 15$, so scale $=\frac{15}{6}=\frac{5}{2}$. This is the number on the butter arrow. Step 2 — apply to the height. Follow the coral edge: height $=4\times \frac{5}{2}=10\text{ cm}$. Why the same factor? In similar figures every side is multiplied by the one scale factor — the whole green rectangle grows uniformly into the lavender one. Step 3 — area ratio. Areas: small $=6\times 4=24$, large $=15\times 10=150$. Ratio $=24:150$. Simplify by $\gcd (24,150)=6$: $4:25$. Notice: $4:25=2^2:5^2$ — the area ratio is the square of the side ratio $2:5$ (the two boxes differ in area far more than in side, which the figure makes visually obvious). Answer: $\boxed{\text{height}=10\text{ cm},\text{area ratio}=4:25}$.

Step 1 — express the gap in parts. Z gets $8$ parts and X gets $3$ parts, so Z $-$ X $=8-3=5$ parts. Why parts, not dollars yet? We don't know a dollar value yet, but we do know the difference is exactly $5$ equal parts. Step 2 — value of one part. Those $5$ equal parts are worth \60$altogether,so\ast one\ast part$=\dfrac{60}{5}=$12$.\ast Whydivideby$5$?\ast The$5$identicalpartssharethe$$60$gapequally.\ast \ast Step3—totalparts,thentotalmoney.\ast \ast Totalparts$=3+5+8 = 16$;totalmoney$=16\times12 = $192$.\ast \ast Check:\ast \ast shares$= 3\times12,,5\times12,,8\times12 = 36,,60,,96$;$96-36 = 60$✓and$36+60+96 = 192$✓.\ast \ast Answer:\ast \ast$\boxed{$192}$.

(Below, the answer is written after each ::: — cover it, answer, then reveal.) Value of one part when splitting $T$ in $a:b:c$ ::: $\frac{T}{a+b+c}$ To chain $A:B$ and $B:C$, first make the two $B$ values equal — scale to their ::: $\text{lcm}$ For similar figures, the area ratio is the ___ of the side ratio ::: square Difference of shares in parts for $3:5:8$ between Z and X ::: $8-3=5$ parts Before writing "one part $=T/(a+b)$" you must check the sum of parts is not ::: zero