Proportional Reasoning-1

Given: Regular coffee ratio = 20 mL coffee : 80 mL milk = $20:80 = 1:4$.

A coffee is stronger if the ratio of coffee decoction to milk is greater than $1:4$ (more coffee per unit milk).
A coffee is lighter if the ratio of coffee decoction to milk is less than $1:4$.
A coffee is regular if the ratio equals $1:4$.

We compare each ratio to $1:4$ by writing it in the form $\text{coffee}:\text{milk}$ and simplifying.

Row 1: 300 mL coffee, 600 mL milk
$$300:600 = 1:2$$
Compare $1:2$ with $1:4$. Since \frac{1}{2} > \frac{1}{4}, there is more coffee per unit milk.
$\Rightarrow$ Strong

Row 2: 150 mL coffee, 500 mL milk
$$150:500 = 3:10$$
Compare $\frac{3}{10} = 0.3$ with $\frac{1}{4} = 0.25$. Since 0.3 > 0.25, more coffee per unit milk.
$\Rightarrow$ Strong

Row 3: 200 mL coffee, 400 mL milk
$$200:400 = 1:2$$
Same as Row 1. \frac{1}{2} > \frac{1}{4}.
$\Rightarrow$ Strong

Row 4: 24 mL coffee, 56 mL milk
$$24:56 = 3:7$$
Compare $\frac{3}{7} \approx 0.429$ with $\frac{1}{4} = 0.25$. Since 0.429 > 0.25, more coffee per unit milk.
$\Rightarrow$ Strong

Row 5: 100 mL coffee, 300 mL milk
$$100:300 = 1:3$$
Compare $\frac{1}{3} \approx 0.333$ with $\frac{1}{4} = 0.25$. Since 0.333 > 0.25, more coffee per unit milk.
$\Rightarrow$ Strong

| Coffee Decoction (mL) | Milk (mL) | Regular/Strong/Light |
|---|---|---|
| 300 | 600 | Strong |
| 150 | 500 | Strong |
| 200 | 400 | Strong |
| 24 | 56 | Strong |
| 100 | 300 | Strong |

Concept: Two ratios $a:b$ and $c:d$ are in proportion if $ad = bc$ (cross-multiplication rule), or equivalently if both ratios simplify to the same fraction.

(i) $4:7 :: 12:21$
Check: $4 \times 21 = 84$ and $7 \times 12 = 84$.
Since $84 = 84$, this is TRUE. ✓

(ii) $8:3 :: 24:6$
Check: $8 \times 6 = 48$ and $3 \times 24 = 72$.
Since $48 \neq 72$, this is FALSE. ✗

(iii) $7:12 :: 12:7$
Check: $7 \times 7 = 49$ and $12 \times 12 = 144$.
Since $49 \neq 144$, this is FALSE. ✗

(iv) $21:6 :: 35:10$
Check: $21 \times 10 = 210$ and $6 \times 35 = 210$.
Since $210 = 210$, this is TRUE. ✓

(v) $12:18 :: 28:12$
Check: $12 \times 12 = 144$ and $18 \times 28 = 504$.
Since $144 \neq 504$, this is FALSE. ✗

(vi) $24:8 :: 9:3$
Check: $24 \times 3 = 72$ and $8 \times 9 = 72$.
Since $72 = 72$, this is TRUE. ✓

Answer: Statements (i), (iv), and (vi) are true proportions.

Concept: Ratios proportional to $4:9$ are obtained by multiplying both terms by the same non-zero number.

$$4:9 = \frac{4 \times 2}{9 \times 2} = 8:18$$
$$4:9 = \frac{4 \times 3}{9 \times 3} = 12:27$$
$$4:9 = \frac{4 \times 4}{9 \times 4} = 16:36$$

Answer: Three ratios proportional to $4:9$ are $\mathbf{8:18}$, $\mathbf{12:27}$, and $\mathbf{16:36}$.

(Any other multiples such as $20:45$, $40:90$, etc. are also correct.)

Concept: Ratios proportional to $18:24$ must have the same simplified form. Simplify $18:24$:
$$18:24 = 3:4 \quad (\text{dividing both by } 6)$$
So for any ratio $a:b$ proportional to $18:24$, we need $\frac{b}{a} = \frac{4}{3}$, i.e., $b = \frac{4}{3} \times a$.

$3 : ?$
$$b = \frac{4}{3} \times 3 = 4$$
Answer: $3 : \mathbf{4}$

$12 : ?$
$$b = \frac{4}{3} \times 12 = 16$$
Answer: $12 : \mathbf{16}$

$20 : ?$
$$b = \frac{4}{3} \times 20 = \frac{80}{3} \approx 26.67$$
Answer: $20 : \mathbf{\dfrac{80}{3}}$ (or approximately $26.\overline{6}$)

$27 : ?$
$$b = \frac{4}{3} \times 27 = 36$$
Answer: $27 : \mathbf{36}$

Summary: $3:4$, $12:16$, $20:\dfrac{80}{3}$, $27:36$.

Note: The actual figure is not visible in the OCR. The method to solve this is described below.

Method:
1. Measure the width ($w$) and height ($h$) of each rectangle using a ruler.
2. Write the ratio $w:h$ for each rectangle.
3. Simplify each ratio to its lowest terms.
4. Rectangles whose simplified $w:h$ ratios are equal are similar to each other (their dimensions are proportional).

Example approach: If Rectangle A has $w:h = 4:6 = 2:3$ and Rectangle C has $w:h = 6:9 = 2:3$, then A and C are similar.

Conclusion: Measure each rectangle, compute and simplify the ratio $w:h$, and group those with equal ratios as similar rectangles.

Note: The figure is not visible in the OCR. The method is described below.

Step 1: Measure the width ($w$) and height ($h$) of the given rectangle and find the ratio $w:h$ in its simplest form. For example, if $w = 6$ cm and $h = 4$ cm, then $w:h = 3:2$.

Step 2 — Smaller rectangle: Multiply both dimensions by a factor less than 1 (or divide by a whole number). E.g., $3:2 \Rightarrow 3\times 1 : 2\times 1 = 3$ cm $\times$ 2 cm.

Step 3 — Bigger rectangle: Multiply both dimensions by a factor greater than 1. E.g., $3:2 \Rightarrow 9$ cm $\times$ 6 cm.

Are classmates' drawings the same? They may have different actual sizes but the ratio $w:h$ will be the same. They are not wrong — there are infinitely many rectangles with the same ratio but different sizes. All such rectangles are similar to the original.

Conclusion: Yes, the drawings look different in size but are all correct as long as the ratio $w:h$ is maintained. This illustrates the concept of similar figures.

Note: The actual figure is not visible in the OCR. The method to solve this is described below.

Method:
1. Count the number of grey bricks in one repeating unit of the pattern. Call this $g$.
2. Count the number of coloured bricks in the same repeating unit. Call this $c$.
3. Write the ratio $g:c$ and simplify by dividing both by their HCF.

Example: If in one repeating unit there are 6 grey bricks and 2 coloured bricks:
$$g:c = 6:2 = 3:1$$

General instruction for students: Identify the repeating pattern unit in the figure, count grey and coloured bricks in that unit, and express the ratio in simplest form by dividing both terms by their HCF.

This is an activity-based question. A sample solution is provided.

Step 1 — Measure (sample values):
- Head length = 20 cm
- Torso length = 50 cm
- Arm length = 45 cm
- Leg length = 80 cm

Step 2 — Write ratios:
$$\text{head : torso} = 20:50 = 2:5$$
$$\text{torso : arms} = 50:45 = 10:9$$
$$\text{torso : legs} = 50:80 = 5:8$$

Step 3 — Draw a scaled figure:
Choose a scale, e.g., divide all measurements by 10:
- Head = 2 cm, Torso = 5 cm, Arms = 4.5 cm, Legs = 8 cm

Draw the figure using these scaled measurements. The ratios remain the same:
$$2:5 = 2:5, \quad 5:4.5 = 10:9, \quad 5:8 = 5:8$$

Does it look realistic? Yes! When the ratios of body parts are proportional to actual human proportions, the drawing looks more realistic because it faithfully represents the relative sizes of body parts. If the ratios are not maintained, the figure will look distorted (e.g., too-long arms or too-small head).

Given:
- Distance travelled in 1 year = 940 million km
- 1 year = 52 weeks (approximately)

Concept: Distance is proportional to time (at constant speed).

$$\text{Distance in 1 week} = \frac{940 \text{ million km}}{52}$$

$$= \frac{940}{52} \text{ million km}$$

$$= 18.08 \text{ million km (approximately)}$$

$$\approx 1{,}80{,}76{,}923 \text{ km}$$

Answer: The Earth travels approximately $18.08$ million km (about $1.81 \times 10^7$ km) in a week.

Note: The figure is not fully visible in the OCR. A standard approach is described. Assume the house dimensions from a typical version of this problem: outer walls form a rectangle of 30 ft × 20 ft, and there is one inner wall of 20 ft.

Step 1 — Find total wall length:
- Outer perimeter = $2 \times (30 + 20) = 100$ ft
- Inner wall = $20$ ft
- Total wall length = $100 + 20 = 120$ ft

Step 2 — Use proportion:
$$10 \text{ ft wall} \rightarrow 1450 \text{ bricks}$$
$$120 \text{ ft wall} \rightarrow ?$$

$$\text{Bricks needed} = \frac{1450}{10} \times 120 = 145 \times 120 = 17{,}400 \text{ bricks}$$

Answer: The mason would need approximately 17,400 bricks.

*(Note: The exact answer depends on the dimensions shown in the figure. Students should measure the total wall length from the diagram and apply the same proportion: $\text{bricks} = \frac{1450}{10} \times \text{total length in feet}$.)*

Given: Total amount = ₹4,500; Ratio = $2:3$.

Concept: If a quantity $x$ is divided in ratio $m:n$, then:
$$\text{First part} = \frac{m}{m+n} \times x, \quad \text{Second part} = \frac{n}{m+n} \times x$$

Step 1: Total parts = $2 + 3 = 5$.

Step 2:
$$\text{First part} = \frac{2}{5} \times 4500 = \frac{9000}{5} = ₹1800$$

$$\text{Second part} = \frac{3}{5} \times 4500 = \frac{13500}{5} = ₹2700$$

Verification: $1800 + 2700 = 4500$ ✓ and $1800:2700 = 2:3$ ✓

Answer: The two parts are ₹1,800 and ₹2,700.

Given: Ratio of acid to water = $1:5$; Total solution = 240 mL.

Step 1: Total parts = $1 + 5 = 6$.

Step 2:
$$\text{Acid} = \frac{1}{6} \times 240 = 40 \text{ mL}$$

$$\text{Water} = \frac{5}{6} \times 240 = 200 \text{ mL}$$

Verification: $40 + 200 = 240$ mL ✓ and $40:200 = 1:5$ ✓

Answer: The bottle contains 40 mL of acid and 200 mL of water.

Part 1 — Finding quantities for 40 mL of green paint:

Given ratio: Blue : Yellow = $3:5$; Total = 40 mL.

Total parts = $3 + 5 = 8$.

$$\text{Blue} = \frac{3}{8} \times 40 = 15 \text{ mL}$$

$$\text{Yellow} = \frac{5}{8} \times 40 = 25 \text{ mL}$$

Verification: $15 + 25 = 40$ mL ✓

Part 2 — New ratio after adding 20 mL of yellow:

- Blue remains = 15 mL
- New Yellow = $25 + 20 = 45$ mL

$$\text{New ratio} = 15:45 = 1:3 \quad (\text{dividing both by 15})$$

Answer:
- Blue needed = 15 mL, Yellow needed = 25 mL.
- New ratio of Blue to Yellow = 1 : 3.

Given: Ratio of rice to urad dal = $2:1$; Total mixture = 6 cups.

Step 1: Total parts = $2 + 1 = 3$.

Step 2:
$$\text{Rice} = \frac{2}{3} \times 6 = 4 \text{ cups}$$

$$\text{Urad dal} = \frac{1}{3} \times 6 = 2 \text{ cups}$$

Verification: $4 + 2 = 6$ cups ✓ and $4:2 = 2:1$ ✓

Answer: You will need 4 cups of rice and 2 cups of urad dal.

Given: Original mixture has red : yellow = $3:5$.
Let the original bucket contain $3k$ parts red and $5k$ parts yellow, so total = $8k$.

A second bucket of yellow paint is added. The second bucket is the same size as the first, so it contains $8k$ parts of yellow.

New amounts:
- Red = $3k$
- Yellow = $5k + 8k = 13k$

$$\text{New ratio} = 3k : 13k = 3:13$$

Answer: The new ratio of red paint to yellow paint is 3 : 13.

Given: Orange juice = 600 mL; Apple juice = 900 mL.

$$\text{Ratio} = 600:900$$

Simplify by dividing both terms by their HCF. $\text{HCF}(600, 900) = 300$.

$$600:900 = \frac{600}{300} : \frac{900}{300} = 2:3$$

Answer: The ratio of orange juice to apple juice in its simplest form is $\mathbf{2:3}$.

Given:
- Last year: 162 people, 3 buses (all full).
- This year: 204 people.

Step 1 — Find capacity of one bus:
$$\text{Capacity per bus} = \frac{162}{3} = 54 \text{ people}$$

Step 2 — Find number of buses needed:
$$\text{Buses needed} = \frac{204}{54} = 3.\overline{7}$$

Since we cannot hire a fraction of a bus, we need 4 buses.

Step 3 — Will all buses be full?
$$4 \times 54 = 216 \text{ seats available}$$
But only 204 students are going.
$$216 - 204 = 12 \text{ empty seats}$$

Answer: We will need 4 buses, but they will not all be full — the last bus will have 12 empty seats.

Concept: Crowdedness is measured by population density = $\dfrac{\text{Population}}{\text{Area}}$.

Delhi:
$$\text{Density} = \frac{30{,}000{,}000}{1484} \approx 20{,}215 \text{ people per sq. km}$$

Mumbai:
$$\text{Density} = \frac{20{,}000{,}000}{550} \approx 36{,}364 \text{ people per sq. km}$$

Comparison: 36{,}364 > 20{,}215

Answer: Mumbai is more crowded than Delhi because it has a higher population density — approximately 36,364 people per sq. km compared to Delhi's 20,215 people per sq. km. Even though Delhi has a larger population, Mumbai's much smaller area means more people are packed into less space.

Given: Crane's total height = 155 cm; Neck : Rest of body = $4:6$.

Step 1 — Find the crane's neck length:
Total parts = $4 + 6 = 10$.
$$\text{Neck of crane} = \frac{4}{10} \times 155 = 62 \text{ cm}$$

Step 2 — Apply to your own height:
Let your height = $H$ cm (measure your own height).
$$\text{Your neck (if same ratio)} = \frac{4}{10} \times H = 0.4 \times H \text{ cm}$$

Example: If your height is 150 cm:
$$\text{Neck} = 0.4 \times 150 = 60 \text{ cm}$$

Answer: If your height is $H$ cm, your neck would be $\mathbf{0.4 \times H}$ cm long under the crane's ratio. (Substitute your actual height to get the numerical answer.)

Given:
- $2\frac{1}{2}$ palas of saffron costs $\dfrac{3}{7}$ niskas.
- Find: quantity of saffron for 9 niskas.

Step 1 — Set up proportion:
$$\frac{3}{7} \text{ niskas} \rightarrow 2\frac{1}{2} \text{ palas}$$
$$9 \text{ niskas} \rightarrow ? \text{ palas}$$

Step 2 — Find the factor of change:
$$\text{Factor} = \frac{9}{\frac{3}{7}} = 9 \times \frac{7}{3} = \frac{63}{3} = 21$$

Step 3 — Find the quantity:
$$\text{Saffron} = 2\frac{1}{2} \times 21 = \frac{5}{2} \times 21 = \frac{105}{2} = 52\frac{1}{2} \text{ palas}$$

Answer: For 9 niskas, one can buy $\mathbf{52\dfrac{1}{2}}$ palas of saffron.

Given:
- Harmain's current age = 1 year
- Brother's current age = 5 years
- Find: Harmain's age when her age : brother's age = $1:2$.

Step 1 — Set up equation:
Let $x$ years pass. Then:
- Harmain's age = $1 + x$
- Brother's age = $5 + x$

$$\frac{1+x}{5+x} = \frac{1}{2}$$

Step 2 — Solve:
$$2(1+x) = 1(5+x)$$
$$2 + 2x = 5 + x$$
$$2x - x = 5 - 2$$
$$x = 3$$

Step 3 — Find Harmain's age:
$$\text{Harmain's age} = 1 + 3 = 4 \text{ years}$$

Verification: Brother's age = $5 + 3 = 8$ years. Ratio = $4:8 = 1:2$ ✓

Answer: Harmain will be 4 years old when the ratio of her age to her brother's age is $1:2$.

Given:
- Mass of equal volumes: gold : water = $37:2$
- Mass of 1 litre of water = 1 kg

Step 1 — Set up proportion:
$$\frac{\text{mass of gold}}{\text{mass of water}} = \frac{37}{2}$$

Step 2 — Find mass of 1 litre of gold:
$$\text{Mass of 1 litre of gold} = \frac{37}{2} \times \text{mass of 1 litre of water}$$
$$= \frac{37}{2} \times 1 = 18.5 \text{ kg}$$

Answer: The mass of 1 litre of gold is 18.5 kg.

Given:
- Manure required = 10 tonnes per acre
- Plot size = 200 ft × 500 ft

Step 1 — Find area of plot in square feet:
$$\text{Area} = 200 \times 500 = 1{,}00{,}000 \text{ sq. ft}$$

Step 2 — Convert to acres:
From unit conversions: 1 acre = 43,560 sq. ft.
$$\text{Area in acres} = \frac{1{,}00{,}000}{43{,}560} \approx 2.296 \text{ acres}$$

Step 3 — Find manure required:
$$\text{Manure} = 10 \times 2.296 \approx 22.96 \text{ tonnes}$$

Answer: The farmer should buy approximately 23 tonnes of cow manure (more precisely, about 22.96 tonnes).

Given:
- Time to fill 500 mL mug = 15 seconds
- Bucket capacity = 10 litres = 10,000 mL

Concept: Time is proportional to volume (at constant flow rate).

Step 1 — Find flow rate:
$$\text{Flow rate} = \frac{500 \text{ mL}}{15 \text{ s}} = \frac{100}{3} \text{ mL/s}$$

Step 2 — Find time for bucket:
$$\text{Time} = \frac{10{,}000}{\frac{100}{3}} = 10{,}000 \times \frac{3}{100} = 300 \text{ seconds}$$

$$300 \text{ seconds} = 5 \text{ minutes}$$

Answer: The tap will take 300 seconds (5 minutes) to fill the 10-litre bucket.

Given:
- Cost of 1 acre = ₹15,00,000
- 1 acre = 43,560 sq. ft
- Find: cost of 2,400 sq. ft

Step 1 — Find cost per square foot:
$$\text{Cost per sq. ft} = \frac{15{,}00{,}000}{43{,}560} \approx ₹34.44 \text{ per sq. ft}$$

Step 2 — Find cost of 2,400 sq. ft:
$$\text{Cost} = 34.44 \times 2400 \approx ₹82{,}645$$

Alternatively using proportion:
$$\frac{15{,}00{,}000}{43{,}560} = \frac{x}{2400}$$
$$x = \frac{15{,}00{,}000 \times 2400}{43{,}560} = \frac{36{,}00{,}00{,}000}{43{,}560} \approx ₹82{,}645$$

Answer: The cost of 2,400 square feet of land is approximately ₹82,645.

Given:
- Oxen: 6 hours per acre
- Tractor is 4 times faster than oxen
- Total field = 20 acres

Part 1 — Time with oxen:
$$\text{Time} = 6 \text{ hours/acre} \times 20 \text{ acres} = 120 \text{ hours}$$

Part 2 — Time with tractor:
Since the tractor is 4 times faster:
$$\text{Tractor time per acre} = \frac{6}{4} = 1.5 \text{ hours/acre}$$

$$\text{Total time with tractor} = 1.5 \times 20 = 30 \text{ hours}$$

Answer:
- With a pair of oxen: 120 hours
- With a tractor: 30 hours

Given:
- Copper : Nickel = $3:1$
- Total mass of coin = 7.74 g
- Cost of copper = ₹906 per kg
- Cost of nickel = ₹1,341 per kg

Step 1 — Find mass of copper and nickel in the coin:
Total parts = $3 + 1 = 4$.

$$\text{Mass of copper} = \frac{3}{4} \times 7.74 = 5.805 \text{ g}$$

$$\text{Mass of nickel} = \frac{1}{4} \times 7.74 = 1.935 \text{ g}$$

Step 2 — Convert costs to per gram:
$$\text{Cost of copper per gram} = \frac{906}{1000} = ₹0.906 \text{ per g}$$

$$\text{Cost of nickel per gram} = \frac{1341}{1000} = ₹1.341 \text{ per g}$$

Step 3 — Find cost of metals in the coin:
$$\text{Cost of copper} = 5.805 \times 0.906 = ₹5.26 \text{ (approx.)}$$

$$\text{Cost of nickel} = 1.935 \times 1.341 = ₹2.59 \text{ (approx.)}$$

$$\text{Total cost of metals} = 5.26 + 2.59 = ₹7.85 \text{ (approx.)}$$

Detailed calculation:
- Copper: $5.805 \times 0.906 = 5.259..\approx ₹5.26$
- Nickel: $1.935 \times 1.341 = 2.594..\approx ₹2.59$
- Total $\approx$ ₹7.85

Answer: The cost of metals in a ₹10 coin is approximately ₹5.26 worth of copper and ₹2.59 worth of nickel, totalling approximately ₹7.85.