Proportional Reasoning – 1

www.akankshaclasses.com
आ
CLASS VIII Mathematics ~4–5 marks Ch 7 of 14
Proportional Reasoning – 1

Class 8 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • A ratio $a:b$ compares two quantities — for every $a$ units of the first there are $b$ units of the second. The numbers $a,b$ are its terms.
  • Two ratios are proportional ($a:b::c:d$) when their simplest forms are equal, i.e. each term changes by the same factor. The quick test is cross multiplication: $a:b::c:d \Leftrightarrow ad=bc$.
  • Rule of Three (Trairāśika): given three of four proportional quantities, the fourth is $d=\dfrac{b\,c}{a}$ — ancient India (Āryabhaṭa, 199 CE) called this $pram\bar{a}\dfrac{}{}\!na$, $phala$, $ichchh\bar a$.
  • Sharing in a ratio $m:n$: split $x$ into $m+n$ equal groups; the parts are $m\times\dfrac{x}{m+n}$ and $n\times\dfrac{x}{m+n}$.
  • Watch out: equal/proportional change is by multiplication, not addition. And not everything is direct — e.g. higher speed means less time, so the Rule of Three does not apply there.
  • Board weightage: ~4–5 marks — usually a proportion/simplest-form question and a Rule-of-Three or sharing word problem.
Detailed notes

1. Where this chapter begins — similar pictures

The chapter opens with five photographs of a tiger, all of different sizes. Images A, C and D look similar (same shape, just scaled); images B and E look distorted — the tiger is stretched in B and squashed in E. Measuring their rectangles explains why:

A: $60\times40$,   B: $40\times20$,   C: $30\times20$,   D: $90\times60$,   E: $60\times60$ (width $\times$ height, mm)

Compare C with A: width $30$ is half of $60$, and height $20$ is half of $40$ — both changed by the same factor $\tfrac12$, so C looks similar. For B, the height halved but the width did not, so B looks wrong. When width and height change by the same factor, the changes are proportional. Key idea: equal-looking change means change by the same multiplying factor, not the same subtraction.

2. Ratios and their terms (§7.2)

A ratio represents such a relationship. The ratio of width to height of image A is written $60:40$. The numbers $60$ and $40$ are the terms of the ratio.

In a ratio of the form $a:b$, for every $a$ units of the first quantity there are $b$ units of the second.

So $60:40$ says: for every $60$ mm of width there are $40$ mm of height. Multiplying both terms of $60:40$ by $\tfrac12$ gives $30:20$ — the ratio of image C — which is why A and C are proportional. The terms must change by the same factor.

3. Simplest form and proportion (§7.3)

To compare ratios cleanly, reduce each to its simplest form by dividing both terms by their HCF.

  • $60:40$ → divide by HCF $20$ → $\mathbf{3:2}$ (image A).
  • $90:60$ → divide by HCF $30$ → $\mathbf{3:2}$ (image D). Same simplest form, so A and D are proportional.
  • $40:20=\mathbf{2:1}$ (image B) and $60:60=\mathbf{1:1}$ (image E) — different, so B and E are not proportional to A, C, D.
When two ratios have the same simplest form, they are in proportion (proportional). We write $a:b::c:d$ (read "$a$ is to $b$ as $c$ is to $d$").
So $60:40::30:20$ and $60:40::90:60$.

4. Solving with proportional reasoning (§7.4)

Most problems give three quantities and ask for a fourth, keeping the ratio proportional. Find the factor of change in one pair of terms and apply the same factor to the other.

NCERT Example 1 — are $3:4$ and $72:96$ proportional?

$3:4$ is already simplest. $72:96$ — divide by HCF $24$ → $3:4$. Same simplest form, so yes, they are proportional.

NCERT Example 2 — Kesang's lemonade

$6$ glasses use $10$ spoons sugar, ratio $6:10$. For $18$ glasses, model as $6:10::18:?$. The first term went $6\to18$, factor $=18\div6=3$. Apply the same factor to $10$: $10\times3=30$. So she needs $30$ spoons of sugar for the same sweetness.

NCERT Example 3 — equally strong walls

Nitin: $60$ ft wall, $3$ bags $\Rightarrow 60:3=20:1$. Hari: $40$ ft wall, $2$ bags $\Rightarrow 40:2=20:1$. Same simplest form, so the walls are equally strong — Nitin needn't worry. (More wall simply needs more cement in the same ratio.)

NCERT Example 6 — Neelima's age (addition is NOT proportional)

At $3$ years, mother is $10\times3=30$, ratio $3:30=1:10$. Nine years later Neelima is $12$, mother is $39$, ratio $12:39=4:13$ — different. Adding the same number to both terms changes the ratio, so it is not necessarily proportional to the original.

NCERT Example 7 — missing terms proportional to $14:21$

(i) $\_:42$ — second term $42=2\times21$, so first $=2\times14=28 \Rightarrow 28:42$. (ii) $6:\_$ — first term $14\to6$ means factor $\tfrac{6}{14}=\tfrac37$, so $21\times\tfrac37=9 \Rightarrow 6:9$. (iii) $2:\_$ — divide $14$ by HCF $7$ to get $2$, so divide $21$ by $7$ → $3 \Rightarrow 2:3$.

5. Cross multiplication & the Rule of Three / Trairāśika

If $a:b::c:d$ then $c=fa$ and $d=fb$ for one common factor $f$. Dividing, $\dfrac{c}{a}=\dfrac{d}{b}=f$, so $\dfrac{c}{a}=\dfrac{d}{b}$. Multiplying both sides by $ab$ gives the master rule:

$$a:b::c:d \;\Longleftrightarrow\; ad=bc \qquad\text{(cross multiplication)}, \qquad d=\dfrac{b\,c}{a}.$$

This is the ancient Indian Rule of Three (Trairāśika). Āryabhaṭa (199 CE) named the three known numbers pramāṇa (measure $=a$), phala (fruit $=b$), ichchhā (requisition $=c$), and the unknown ichchhāphala (yield $=d$), with the rule "multiply the phala by the ichchhā and divide by the pramāṇa": $\;ichchh\bar aphala=\dfrac{phala\times ichchh\bar a}{pram\bar a\dfrac{}{}\!na}$.

NCERT Example 8 — mid-day meal rice

$120$ students need $15$ kg rice; only $80$ came. $120:15::80:?$. Factor in first term $=\dfrac{80}{120}=\dfrac23$. So rice $=15\times\dfrac23=\mathbf{10}$ kg, no food wasted.

NCERT Example 9 — distance of a car (watch the units!)

$90$ km in $150$ min; distance in $4$ hours? First convert $4$ h $=240$ min (same unit). $150:90::240:x$. Cross multiply: $150x=240\times90$, so $x=\dfrac{240\times90}{150}=\mathbf{144}$ km.

NCERT Example 10 — which tea is dearer?

Himachal: $200$ g for ₹$200$ → $200:200=1:1$. Meghalaya: $1$ kg for ₹$800$ = $1000$ g for ₹$800$ → $5:4$. Different simplest forms, so not proportional. Compare per kg: Meghalaya ₹$800$/kg; Himachal $\tfrac15 x=200\Rightarrow x=$ ₹$1000$/kg. So Himachal tea is more expensive.

Important caution (Puneeth's father): a journey takes $2$ h at $50$ km/h; at $75$ km/h it takes less time. This is inverse behaviour, so it cannot be written as $50:2::75:?$ — the Rule of Three only fits quantities that grow together (direct proportion).

6. Sharing, but not equally! (§7.5)

To divide a whole $x$ in the ratio $m:n$, think in groups: the first share gets $m$ groups, the second gets $n$ groups, so there are $m+n$ equal groups in all.

Size of each group $=\dfrac{x}{m+n}$. Then $\;\text{first part}=m\times\dfrac{x}{m+n},\quad \text{second part}=n\times\dfrac{x}{m+n}.$

Example (from text): sharing $12$ counters in $3:1$ → groups $=3+1=4$, each group $=12\div4=3$, so parts are $3\times3=9$ and $1\times3=3$. Sharing $42$ in $4:3$ → groups $=7$, each $=42\div7=6$, parts $24$ and $18$.

NCERT Example 11 — sharing profit

Prashanti invests ₹$75000$, Bhuvan ₹$25000$; profit ₹$4000$ shared in investment ratio. $75000:25000=3:1$, groups $=4$, each group $=4000\div4=1000$. So Prashanti gets $3\times1000=$ ₹$3000$ and Bhuvan $1\times1000=$ ₹$1000$.

NCERT Example 12 — adjusting a mixture

$40$ kg of sand:cement $=3:1$ → sand $=\dfrac{3}{4}\times40=30$ kg, cement $=\dfrac14\times40=10$ kg. New ratio sand:cement $=5:2$ with sand still $30$: $5:2::30:?$, so cement $=\dfrac25\times30=12$ kg. Already have $10$ kg, so add $2$ kg of cement.

7. Unit conversions (§7.6)

Proportion problems often need a unit conversion first (as in Examples 9 and 10). Handy conversions from the chapter:

  • Length: $1$ metre $=3.281$ feet.
  • Area: $1\ \text{m}^2=10.764\ \text{ft}^2$; $1$ acre $=43{,}560\ \text{ft}^2$; $1$ hectare $=10{,}000\ \text{m}^2=2.471$ acres.
  • Volume: $1$ mL $=1$ cc; $1$ litre $=1000$ mL $=1000$ cc.
  • Temperature: $\text{F}=\dfrac95\times\text{C}+32$ and $\text{C}=\dfrac59(\text{F}-32)$; e.g. $25^\circ\text{C}=77^\circ\text{F}$.

Golden rule: before forming a proportion, make both quantities in a ratio use the same unit (minutes with minutes, grams with grams) — otherwise the cross multiplication is meaningless.

8. "Figure it Out" — solved exercises

Proportion check (true ones). Using $ad=bc$: $4:7::12:21$? $4\cdot21=84,\;7\cdot12=84$ ✔. $8:3::24:6$? $48\ne72$ ✘. $7:12::12:7$? $49\ne144$ ✘. $21:6::35:10$? $210=210$ ✔. $12:18::28:12$? $144\ne504$ ✘. $24:8::9:3$? $72=72$ ✔. So (i), (iv), (vi) are true.

Three ratios proportional to $4:9$: $8:18,\ 12:27,\ 16:36$ (multiply both terms by $2,3,4$).

Missing terms proportional to $18:24$ (simplest $3:4$): $3:\mathbf{4}$;   $12:\mathbf{16}$;   $20:\mathbf{?}$ → $20=3\times\tfrac{20}{3}$, term $=4\times\tfrac{20}{3}=\dfrac{80}{3}=26\tfrac23$;   $27:\mathbf{36}$ (factor $9$).

Divide ₹$4500$ in $2:3$: groups $=5$, each $=900$; parts ₹$1800$ and ₹$2700$.

Acid:water $=1:5$ in $240$ mL: groups $=6$, each $=40$ mL; acid $=40$ mL, water $=200$ mL.

Blue:yellow $=3:5$, make $40$ mL green: groups $=8$, each $=5$ mL; blue $=15$ mL, yellow $=25$ mL. Add $20$ mL more yellow → yellow $=45$, ratio $15:45=\mathbf{1:3}$.

Rice:urad dal $=2:1$, need $6$ cups: groups $=3$, each $=2$ cups; rice $=4$ cups, urad dal $=2$ cups.

Earth's orbit: $940$ million km in a year ($52$ weeks) → per week $\dfrac{940}{52}\approx18.08$ million km.

Buses: $3$ buses carried $162$ → $54$ per bus; $204\div54\approx3.78$, so 4 buses are needed (the buses won't all be full).

Orange : apple juice $=600:900=\mathbf{2:3}$ in simplest form.

9. Common mistakes to avoid

  • Treating equal difference as proportional — proportion needs the same multiplying factor (Example 6, image B).
  • Forming a ratio with different units — convert first (Example 9: $4$ h $=240$ min).
  • Forcing the Rule of Three on inverse situations (speed–time): more speed → less time, so $50:2::75:?$ is wrong.
  • In sharing, forgetting groups $=m+n$ (not just $m$ or $n$), so each group $=\dfrac{x}{m+n}$.
  • Not reducing to simplest form before declaring two ratios proportional.

10. Quick revision checklist

  • Ratio $a:b$ — for every $a$ of the first there are $b$ of the second; terms $a,b$.
  • Reduce by HCF to simplest form; equal simplest forms ⇒ proportional $a:b::c:d$.
  • Cross multiply test: $ad=bc$; find the missing fourth as $d=\dfrac{bc}{a}$ (Rule of Three).
  • Share $x$ in $m:n$: each group $=\dfrac{x}{m+n}$, parts $m,n$ groups.
  • Convert units before comparing; proportion fits direct relations, not inverse ones.
Practice MCQs
1. The simplest form of the ratio $90:60$ is:
  1. $9:6$
  2. $3:2$
  3. $2:3$
  4. $30:20$
Answer: (B) divide both terms by HCF $30$ → $3:2$.
2. Which pair of ratios is proportional?
  1. $8:3$ and $24:6$
  2. $7:12$ and $12:7$
  3. $21:6$ and $35:10$
  4. $12:18$ and $28:12$
Answer: (C) $21\times10=210=6\times35$, so $ad=bc$.
3. The terms of the ratio $60:40$ are:
  1. $6$ and $4$
  2. $60$ and $40$
  3. $3$ and $2$
  4. $20$
Answer: (B) the terms are the two numbers $60$ and $40$ as written.
4. If $6:10::18:x$ (same sweetness), then $x=$
  1. $22$
  2. $28$
  3. $30$
  4. $36$
Answer: (C) factor $=18\div6=3$, so $x=10\times3=30$.
5. A car covers $90$ km in $150$ min. At the same speed, distance in $240$ min is:
  1. $120$ km
  2. $144$ km
  3. $150$ km
  4. $240$ km
Answer: (B) $150:90::240:x\Rightarrow x=\dfrac{240\times90}{150}=144$ km.
6. $12$ counters shared in the ratio $3:1$ give the two shares:
  1. $6$ and $6$
  2. $8$ and $4$
  3. $9$ and $3$
  4. $10$ and $2$
Answer: (C) groups $=4$, each $=12\div4=3$, so $3\times3=9$ and $1\times3=3$.
7. Sand and cement in $40$ kg are mixed $3:1$. The weight of cement is:
  1. $10$ kg
  2. $12$ kg
  3. $20$ kg
  4. $30$ kg
Answer: (A) cement $=\dfrac{1}{3+1}\times40=10$ kg.
8. Adding the same number to both terms of a ratio:
  1. always keeps it proportional
  2. need not keep it proportional
  3. halves the ratio
  4. makes it $1:1$
Answer: (B) proportion needs the same multiplying factor; adding (Example 6) generally changes the ratio.
9. Puneeth's father travels at $50$ km/h in $2$ h; at $75$ km/h the time will be:
  1. more than $2$ h
  2. exactly $2$ h
  3. less than $2$ h
  4. $3$ h
Answer: (C) higher speed ⇒ less time (inverse) — the Rule of Three does not apply here.
10. $600$ mL orange juice mixed with $900$ mL apple juice — the ratio in simplest form is:
  1. $6:9$
  2. $3:2$
  3. $2:3$
  4. $1:2$
Answer: (C) $600:900$, divide by $300$ → $2:3$.
11. If $a:b::c:d$, then which relation is always true?
  1. $a+d=b+c$
  2. $ad=bc$
  3. $ac=bd$
  4. $a-b=c-d$
Answer: (B) cross multiplication: $ad=bc$.
12. ₹$4500$ divided in the ratio $2:3$ gives the parts:
  1. ₹$2000$ and ₹$2500$
  2. ₹$1800$ and ₹$2700$
  3. ₹$1500$ and ₹$3000$
  4. ₹$900$ and ₹$3600$
Answer: (B) groups $=5$, each $=900$; $2\times900=1800$, $3\times900=2700$.
Assertion–Reason
A: $72:96$ is proportional to $3:4$.   R: Two ratios are proportional when their simplest forms are equal.
Answer: Both A and R are true, and R explains A — $72:96$ divided by HCF $24$ is $3:4$.
A: A trip at higher speed can be solved by $50:2::75:?$.   R: Speed and time for a fixed distance are directly proportional.
Answer: Both A and R are false — for fixed distance, speed and time are inverse, so the Rule of Three does not apply (Puneeth's father).
Exam-style questions
Q1. A car travels $90$ km in $150$ minutes. At the same speed, how far will it go in $4$ hours? (3 marks)
Solution: $4$ h $=240$ min. $150:90::240:x \Rightarrow 150x=240\times90 \Rightarrow x=\dfrac{21600}{150}=\mathbf{144\ \text{km}}$.
Q2. Prashanti and Bhuvan invest ₹$75{,}000$ and ₹$25{,}000$ and earn ₹$4{,}000$ profit, shared in the investment ratio. Find each share. (3 marks)
Solution: $75000:25000=3:1$, groups $=4$, each group $=4000\div4=1000$. Prashanti $=3\times1000=$ ₹$3000$, Bhuvan $=1\times1000=$ ₹$1000$.
Q3. A mixture of $40$ kg has sand and cement in $3:1$. How much cement must be added to make sand : cement $=5:2$? (3 marks)
Solution: sand $=\tfrac34\times40=30$ kg, cement $=\tfrac14\times40=10$ kg. For $5:2$ with sand $30$: cement $=\tfrac25\times30=12$ kg. Already $10$ kg present ⇒ add $2$ kg cement.
Q4. For $120$ students a cook makes $15$ kg rice. On a rainy day only $80$ students come. How much rice should be cooked so none is wasted? (2 marks)
Solution: $120:15::80:x$. Factor $=\dfrac{80}{120}=\dfrac23$, so $x=15\times\dfrac23=\mathbf{10\ \text{kg}}$.
Want personal coaching in Dwarka?
Book a free demo class
More Class 8 Mathematics chapters
A Square and A Cube · Power Play · A Story of Numbers · Quadrilaterals · Number Play · We Distribute, Yet Things Multiply · Fractions in Disguise · The Baudhayana-Pythagoras Theorem
Chat with us