Chapter 7: Proportional Reasoning-1

Two images look similar when one looks like a scaled (enlarged or reduced) copy of the other — the shape stays the same even though the size is different.

Comparing the five pictures by eye, Images A, C and D show the same picture at different sizes: none of them is stretched or squashed. They just look like bigger or smaller versions of one another.

Image B looks flatter (stretched sideways) and Image E looks more square-like (stretched top-to-bottom). So they do not look like scaled copies of A, C or D.

Hence, Images A, C and D look similar to each other, while Images B and E look different from the rest.

No. Images A, C and D are just scaled (bigger or smaller) versions of the same picture, so they all keep the same shape.

Image B looks stretched out sideways (it is wider than it should be for its height), and Image E looks squashed vertically or stretched to be almost square. Because their shapes have been changed and not merely their sizes, B and E do not look like the other three images.

Let us find the width-to-height ratio of each image in its simplest form.

Image A: $$60 : 40 = \dfrac{60}{40} = \dfrac{3}{2} = 3 : 2$$.

Image B: $$40 : 20 = \dfrac{40}{20} = \dfrac{2}{1} = 2 : 1$$.

Image C: $$30 : 20 = \dfrac{30}{20} = \dfrac{3}{2} = 3 : 2$$.

Image D: $$90 : 60 = \dfrac{90}{60} = \dfrac{3}{2} = 3 : 2$$.

Image E: $$60 : 60 = \dfrac{60}{60} = \dfrac{1}{1} = 1 : 1$$.

Images A, C and D all have the same width-to-height ratio $$3 : 2$$. That is why one looks like a scaled copy of the others — the shape is preserved.

Images B $$(2 : 1)$$ and E $$(1 : 1)$$ have different width-to-height ratios, so their shapes are different from A, C and D and from each other.

Image A has width $$60$$ mm and height $$40$$ mm. Image D has width $$90$$ mm and height $$60$$ mm.

Factor by which width changed:

$$\text{width factor} = \dfrac{\text{width of D}}{\text{width of A}} = \dfrac{90}{60} = \dfrac{3}{2}.$$

Factor by which height changed:

$$\text{height factor} = \dfrac{\text{height of D}}{\text{height of A}} = \dfrac{60}{40} = \dfrac{3}{2}.$$

Both the width and height are multiplied by the same factor $$\dfrac{3}{2}$$. That is why Image D looks like a proper enlargement of Image A.

Both terms of the ratio must be multiplied by the same number. Let that factor be $$k$$. Then

$$60 \times k = 90 \quad\Rightarrow\quad k = \dfrac{90}{60} = \dfrac{3}{2}.$$

Check the other term:

$$40 \times \dfrac{3}{2} = 60. \checkmark$$

So multiplying each term of $$60 : 40$$ by $$\dfrac{3}{2}$$ gives $$90 : 60$$.

The regular coffee has coffee : milk $$= 15 : 35 = 3 : 7$$. As a decimal, coffee is $$\dfrac{3}{7} \approx 0.4286$$ times the milk.

A mix is stronger than regular if $$\dfrac{\text{coffee}}{\text{milk}} > \dfrac{3}{7}$$ (i.e. more coffee per unit of milk), lighter if $$\dfrac{\text{coffee}}{\text{milk}} < \dfrac{3}{7}$$, and regular if the ratio equals $$\dfrac{3}{7}$$.

We compare each row's ratio with $$\dfrac{3}{7}$$.

• $$300 : 600 = \dfrac{300}{600} = \dfrac{1}{2} = 0.5 > \dfrac{3}{7}$$  →  Stronger.
• $$150 : 500 = \dfrac{150}{500} = \dfrac{3}{10} = 0.3 < \dfrac{3}{7}$$  →  Lighter.
• $$200 : 400 = \dfrac{200}{400} = \dfrac{1}{2} = 0.5 > \dfrac{3}{7}$$  →  Stronger.
• $$24 : 56 = \dfrac{24}{56} = \dfrac{3}{7}$$  →  Regular (proportional to $$15 : 35$$).
• $$100 : 300 = \dfrac{100}{300} = \dfrac{1}{3} \approx 0.333 < \dfrac{3}{7}$$  →  Lighter.

Strength depends on how much coffee decoction there is compared to milk. Compare the given ratio with the regular one.

Regular coffee: $$15 : 35 = \dfrac{15}{35} = \dfrac{3}{7} \approx 0.43$$.

Given mix: $$20 : 30 = \dfrac{20}{30} = \dfrac{2}{3} \approx 0.67$$.

Since $$\dfrac{2}{3} > \dfrac{3}{7}$$, for every mL of milk there is more coffee decoction in the new mix than in the regular one. More decoction per unit milk gives a stronger flavour, so the coffee tastes stronger.

Compare with the regular ratio $$15 : 35 = \dfrac{3}{7} \approx 0.43$$.

Given mix: $$10 : 40 = \dfrac{10}{40} = \dfrac{1}{4} = 0.25$$.

Since $$\dfrac{1}{4} < \dfrac{3}{7}$$, there is less coffee decoction per mL of milk than in the regular coffee. That is why this coffee tastes lighter (milkier).

Yes. A real human body has fixed relationships between the sizes of its parts — for example the head is about a certain fraction of the torso, the arms are about a certain fraction of the torso, and so on.

When your drawing keeps the same ratios (head : torso, torso : arms, torso : legs) as your friend's actual body, every part is a scaled copy of the real part. So the whole figure looks like a shrunken version of your friend and appears realistic.

If instead you draw, say, a very long head with tiny legs, the ratios are no longer proportional to real ones — the figure looks like a cartoon or a caricature, not a real person.

This is a hands-on activity — the answer will depend on your dish and your family size. Here is one worked example so you can follow the same method for your own recipe.

Chosen dish: Vegetable pulao for a family of 4.

Now we want to cook for $$4 + 15 = 19$$ people. The scaling factor is

$$k = \dfrac{\text{new number of people}}{\text{original number of people}} = \dfrac{19}{4}.$$

Multiply every quantity by $$k = \dfrac{19}{4}$$.

For your own recipe, list your ingredients, note the number of people it feeds, compute $$k = \dfrac{\text{total guests}}{\text{original people}}$$, and multiply every quantity by $$k$$.

First find the distance between Lucknow and Kanpur.

$$\text{Distance} = \text{speed} \times \text{time} = 50 \;\text{km/h} \times 2 \;\text{h} = 100 \;\text{km}.$$

The distance does not change. If he now drives at $$75$$ km/h, the time taken is

$$\text{time} = \dfrac{\text{distance}}{\text{speed}} = \dfrac{100}{75} = \dfrac{4}{3} \;\text{hours} = 1\;\text{hour }20\;\text{minutes}.$$

Since he is going faster, he reaches Kanpur in less time.

Can it be written as $$50 : 2 :: 75 : \square$$? No — that proportion would give

$$\square = \dfrac{2 \times 75}{50} = 3 \;\text{hours},$$

which means he takes more time when he goes faster. That is the wrong relationship. Speed and time (for a fixed distance) are inversely proportional, not directly proportional. The correct proportion is

$$50 \times 2 = 75 \times \square,$$

giving $$\square = \dfrac{100}{75} = \dfrac{4}{3}$$ hours.

Let us compute the price per mL for each container.

The price per mL is different for every container, so volume and price are not in the same proportion.

Notice also that the sachet costs $$\dfrac{2}{6} \approx \text{₹}0.33$$ per mL — cheaper than the small and medium bottles — while the large bottle is the cheapest per mL. The reasons are:

• Fixed costs: The price of a container includes the packing, printing, transport, shopkeeper's margin and GST. These costs do not double when the volume doubles.
• Discounts on bulk: Companies deliberately sell bigger bottles at a lower price per mL to encourage customers to buy larger quantities.
• Small units for affordability: Sachets are priced low so that customers who cannot afford a bottle can still buy the product, even though the price per mL is not the cheapest.

Because price = (cost of shampoo) + (fixed costs) + (profit), and fixed costs do not scale with volume, the price is not directly proportional to the volume.

This is a hands-on activity — the two of you can split 12 counters in many different ways. Any pair of whole numbers that adds up to $$12$$ is a possible sharing.

Here are all the ways of dividing 12 counters between two people (partner : you), together with the ratio of the two shares in simplest form:

Try a few of these physically and check that in each case the two shares add up to $$12$$.

There are $$12$$ counters in all and they are divided equally between the two of you. Each of you therefore gets

$$\dfrac{12}{2} = 6 \text{ counters.}$$

The ratio of your counters to your partner's counters is

$$6 : 6 = \dfrac{6}{6} = \dfrac{1}{1} = 1 : 1.$$

The two of you together have $$12$$ counters, and your partner takes $$5$$. So you get

$$12 - 5 = 7 \text{ counters.}$$

The ratio of your partner's counters to your counters is

$$5 : 7.$$

Since $$5$$ and $$7$$ have no common factor greater than $$1$$, the ratio is already in its simplest form.

The ratio $$3 : 1$$ splits the $$12$$ counters into $$3 + 1 = 4$$ equal groups. Then one person takes $$3$$ groups and the other takes $$1$$ group.

Size of each group:

$$\dfrac{12}{4} = 3 \text{ counters.}$$

Larger share $$= 3$$ groups $$= 3 \times 3 = 9$$ counters.
Smaller share $$= 1$$ group $$= 1 \times 3 = 3$$ counters.

Check: $$9 + 3 = 12$$ and $$9 : 3 = 3 : 1$$. ✓

The ratio $$4 : 3$$ means we split the counters into $$4 + 3 = 7$$ equal groups. Your partner takes $$4$$ groups and you take $$3$$ groups.

Size of one group:

$$\dfrac{42}{7} = 6 \text{ counters.}$$

Partner's share $$= 4 \times 6 = 24$$ counters.
Your share $$= 3 \times 6 = 18$$ counters.

Check: $$24 + 18 = 42$$ and $$24 : 18 = 4 : 3$$. ✓

The total number of groups is $$4 + 3 = 7$$. The $$42$$ counters are distributed equally among these $$7$$ groups, so the size of each group is

$$\dfrac{42}{7} = 6 \text{ counters.}$$

Two ratios $$a : b$$ and $$c : d$$ are proportional if $$\dfrac{a}{b} = \dfrac{c}{d}$$, i.e. $$a \times d = b \times c$$.

Method 1 — reduce the second ratio.

$$\dfrac{72}{96} = \dfrac{72 \div 24}{96 \div 24} = \dfrac{3}{4}.$$

So $$72 : 96 = 3 : 4$$. The two ratios are equal, hence proportional.

Method 2 — cross-multiply.

$$3 \times 96 = 288 \quad\text{and}\quad 4 \times 72 = 288.$$

The two cross-products are equal, so $$3 : 4 :: 72 : 96$$.

For the same sweetness, glasses of lemonade and spoons of sugar must be in the same ratio throughout.

She has already used $$6$$ glasses with $$10$$ spoons of sugar. She now wants to make $$18$$ more glasses. Let the sugar needed for these $$18$$ glasses be $$x$$ spoons.

Set up the proportion glasses : sugar

$$6 : 10 :: 18 : x.$$

Cross-multiplying,

$$6 \times x = 10 \times 18,$$
$$6x = 180,$$
$$x = \dfrac{180}{6} = 30.$$

Alternative reasoning: $$18$$ glasses is $$3$$ times $$6$$ glasses, so the sugar must also be tripled: $$10 \times 3 = 30$$ spoons.

Check: $$18 : 30 = \dfrac{18}{30} = \dfrac{3}{5}$$ and $$6 : 10 = \dfrac{6}{10} = \dfrac{3}{5}$$. ✓

Strength depends on how much cement is used per unit length of wall, not on the total number of bags. Let us find the cement-per-foot for each wall.

Nitin's wall: $$3$$ bags for $$60$$ ft, so $$\dfrac{3}{60} = \dfrac{1}{20}$$ bag per foot.

Hari's wall: $$2$$ bags for $$40$$ ft, so $$\dfrac{2}{40} = \dfrac{1}{20}$$ bag per foot.

Both walls have the same cement per foot. Equivalently, the ratio of cement to length is

$$3 : 60 = 1 : 20 \quad\text{and}\quad 2 : 40 = 1 : 20,$$

which are proportional. So the two walls are equally strong; Hari simply needed less cement because her wall was shorter.

This question needs data from your own school. The method is:

Step 1. Count the number of teachers $$T$$ and the number of students $$S$$ in your school. Write the ratio $$T : S$$ and reduce it to simplest form by dividing both numbers by their HCF.

Step 2. Reduce the given ratio $$5 : 170$$ to simplest form:

$$5 : 170 = \dfrac{5}{170} = \dfrac{1}{34} = 1 : 34.$$

So one teacher looks after $$34$$ students.

Step 3. Compare $$T : S$$ (in simplest form) with $$1 : 34$$. They are proportional exactly when $$\dfrac{T}{S} = \dfrac{1}{34}$$, i.e. when $$S = 34\, T$$ or equivalently $$34\, T = S$$.

Illustration. Suppose your school has $$8$$ teachers and $$240$$ students. Then $$T : S = 8 : 240 = 1 : 30$$. Since $$1 : 30 \neq 1 : 34$$, your ratio is not proportional to $$5 : 170$$. Cross-check: $$5 \times 240 = 1200$$ and $$170 \times 8 = 1360$$; the two products differ, confirming the ratios are not proportional.

If instead you had $$10$$ teachers and $$340$$ students, then $$10 : 340 = 1 : 34$$, which is proportional to $$5 : 170$$.

Use a metre-scale and measure the blackboard's width $$W$$ and height $$H$$ in centimetres. Reduce $$W : H$$ to its simplest form. A typical school blackboard might be $$W = 240$$ cm and $$H = 120$$ cm, giving

$$W : H = 240 : 120 = \dfrac{240}{120} = 2 : 1.$$

Drawing a proportional rectangle. Pick any scale factor $$k$$ and draw a rectangle of width $$k\,W$$ and height $$k\,H$$. For example, with $$k = \dfrac{1}{20}$$ we get $$12 \text{ cm} \times 6 \text{ cm}$$.

Do the rectangles look the same? All the rectangles that keep the ratio $$2 : 1$$ have the same shape, but they can have different sizes because different classmates may pick different scale factors $$k$$. So they all look like scaled copies of the blackboard, not identical rectangles.

When Neelima is 3. Her mother is $$10 \times 3 = 30$$ years old. So

$$\text{Neelima} : \text{Mother} = 3 : 30 = \dfrac{3}{30} = \dfrac{1}{10} = 1 : 10.$$

Age difference between them. Mother is older by $$30 - 3 = 27$$ years. This difference stays the same forever.

When Neelima is 12. Her mother is $$12 + 27 = 39$$ years old. So

$$\text{Neelima} : \text{Mother} = 12 : 39 = \dfrac{12}{39} = \dfrac{12 \div 3}{39 \div 3} = \dfrac{4}{13} = 4 : 13.$$

Since $$\dfrac{1}{10} \neq \dfrac{4}{13}$$, the ratio has changed.

Would it remain the same? No. Both ages increase by the same number of years each year, but a ratio changes when the same number is added to both terms (unless they were already equal). As Neelima grows up, the ratio Neelima : Mother keeps increasing towards $$1 : 1$$.

First simplify $$14 : 21$$:

$$14 : 21 = \dfrac{14}{21} = \dfrac{2}{3} = 2 : 3.$$

So a ratio $$a : b$$ is proportional to $$14 : 21$$ iff $$\dfrac{a}{b} = \dfrac{2}{3}$$, i.e. $$3a = 2b$$.

(i) $$\square : 42$$. Here $$b = 42$$, so $$a = \dfrac{2}{3} \times 42 = 28$$. Answer: $$28 : 42$$.

(ii) $$6 : \square$$. Here $$a = 6$$, so $$b = \dfrac{3}{2} \times 6 = 9$$. Answer: $$6 : 9$$.

(iii) $$2 : \square$$. Here $$a = 2$$, so $$b = \dfrac{3}{2} \times 2 = 3$$. Answer: $$2 : 3$$.

Check each: $$28 : 42 = 2 : 3$$, $$6 : 9 = 2 : 3$$, $$2 : 3 = 2 : 3$$. All are proportional to $$14 : 21$$. ✓

The rice needed is directly proportional to the number of students. Let $$x$$ kg of rice be needed for $$80$$ students.

$$120 \text{ students} : 15 \text{ kg} :: 80 \text{ students} : x \text{ kg}.$$

Cross-multiplying,

$$120 \times x = 15 \times 80,$$
$$120 x = 1200,$$
$$x = \dfrac{1200}{120} = 10.$$

Alternative reasoning: rice per student $$= \dfrac{15}{120} = \dfrac{1}{8}$$ kg. For $$80$$ students the cook needs $$80 \times \dfrac{1}{8} = 10$$ kg.

First convert both time measurements to the same unit. Take minutes.

$$4 \text{ hours} = 4 \times 60 = 240 \text{ minutes.}$$

At a constant speed, distance is directly proportional to time. Let $$x$$ km be the distance covered in $$240$$ minutes.

$$90 \text{ km} : 150 \text{ min} :: x \text{ km} : 240 \text{ min},$$

so

$$\dfrac{90}{150} = \dfrac{x}{240} \quad\Rightarrow\quad x = \dfrac{90 \times 240}{150} = \dfrac{21600}{150} = 144.$$

Alternative check via speed:

$$\text{Speed} = \dfrac{90 \text{ km}}{150 \text{ min}} = \dfrac{90}{150} = 0.6 \text{ km/min} = 36 \text{ km/h}.$$

$$\text{Distance in }4\text{ h} = 36 \times 4 = 144 \text{ km}. \checkmark$$

Bring both weights to the same unit. Take grams.

$$1 \text{ kg} = 1000 \text{ g}.$$

Himachal: weight : price $$= 200 \text{ g} : \text{₹}200$$, so price per gram $$= \dfrac{200}{200} = \text{₹}1$$.

Meghalaya: weight : price $$= 1000 \text{ g} : \text{₹}800$$, so price per gram $$= \dfrac{800}{1000} = \text{₹}0.80$$.

Are the ratios proportional? Reduce each:

$$200 : 200 = 1 : 1, \qquad 1000 : 800 = 5 : 4.$$

Since $$1 : 1 \neq 5 : 4$$, the two weight-to-price ratios are not proportional.

Comparing prices per gram, Himachal tea costs ₹$$1$$ per gram while Meghalaya tea costs ₹$$0.80$$ per gram. So Himachal tea is more expensive.

First find the ratio of investments in its simplest form.

$$\text{Prashanti} : \text{Bhuvan} = 75000 : 25000 = \dfrac{75000}{25000} : 1 = 3 : 1.$$

The ratio $$3 : 1$$ splits the profit into $$3 + 1 = 4$$ equal parts. So one part is

$$\dfrac{\text{₹}4000}{4} = \text{₹}1000.$$

Prashanti's share $$= 3 \times \text{₹}1000 = \text{₹}3000$$.
Bhuvan's share $$= 1 \times \text{₹}1000 = \text{₹}1000$$.

Check: $$\text{₹}3000 + \text{₹}1000 = \text{₹}4000$$ and $$3000 : 1000 = 3 : 1$$. ✓

Step 1 — find how much sand and cement are already there.

Ratio $$3 : 1$$ means $$3 + 1 = 4$$ parts in $$40$$ kg. One part is $$\dfrac{40}{4} = 10$$ kg.

Sand $$= 3 \times 10 = 30$$ kg. Cement $$= 1 \times 10 = 10$$ kg.

Step 2 — set up the new ratio.

Only cement is added; sand stays at $$30$$ kg. Let the cement added be $$x$$ kg, so the new cement weight is $$(10 + x)$$ kg. We want

$$\dfrac{\text{sand}}{\text{cement}} = \dfrac{30}{10 + x} = \dfrac{5}{2}.$$

Step 3 — solve. Cross-multiplying,

$$5\,(10 + x) = 2 \times 30,$$
$$50 + 5x = 60,$$
$$5x = 10,$$
$$x = 2.$$

Check: new cement $$= 10 + 2 = 12$$ kg. Ratio $$= 30 : 12 = \dfrac{30}{12} = \dfrac{5}{2} = 5 : 2$$. ✓

Ratios proportional to $$4 : 9$$ are obtained by multiplying (or dividing) both terms by the same non-zero number. Multiplying by $$2, 3, 4$$:

$$4 \times 2 : 9 \times 2 = 8 : 18,$$
$$4 \times 3 : 9 \times 3 = 12 : 27,$$
$$4 \times 4 : 9 \times 4 = 16 : 36.$$

Check each ratio by cross-multiplying with $$4 : 9$$: $$4 \times 18 = 72 = 9 \times 8$$, $$4 \times 27 = 108 = 9 \times 12$$, $$4 \times 36 = 144 = 9 \times 16$$. ✓

Other correct answers include $$20 : 45$$, $$40 : 90$$, $$100 : 225$$, and so on.

First simplify $$18 : 24$$:

$$18 : 24 = \dfrac{18}{24} = \dfrac{3}{4} = 3 : 4.$$

So each ratio $$a : b$$ must satisfy $$\dfrac{a}{b} = \dfrac{3}{4}$$, i.e. $$b = \dfrac{4a}{3}$$.

(i) $$3 : \square$$. $$b = \dfrac{4 \times 3}{3} = 4$$. Answer: $$3 : 4$$.

(ii) $$12 : \square$$. $$b = \dfrac{4 \times 12}{3} = 16$$. Answer: $$12 : 16$$.

(iii) $$20 : \square$$. $$b = \dfrac{4 \times 20}{3} = \dfrac{80}{3}$$. Answer: $$20 : \dfrac{80}{3}$$ (i.e. $$60 : 80$$).

(iv) $$27 : \square$$. $$b = \dfrac{4 \times 27}{3} = 36$$. Answer: $$27 : 36$$.

Check each pair reduces to $$3 : 4$$. ✓

Two rectangles are similar iff their width : height ratios are equal.

Measure the width $$w$$ and height $$h$$ of each rectangle A, B, C, D, E in your book with a scale. For each, compute the simplified ratio $$w : h$$.

Group rectangles with the same simplified ratio — those form one similarity class.

Sample working with the sizes printed in the NCERT textbook:

Since $$A$$ and $$D$$ both have ratio $$2 : 1$$, they are similar. $$B$$ and $$E$$ both have ratio $$3 : 1$$, so they are similar. Rectangle $$C$$ (ratio $$3 : 2$$) does not match any other, so it is not similar to the rest.

Use your own measured values, but the method is the same — rectangles with equal width-to-height ratios are similar.

Measure the given rectangle. Suppose its width is $$w$$ cm and its height is $$h$$ cm. In simplest form let the ratio be $$w : h$$ (for example, if $$w = 8$$ and $$h = 4$$, then $$w : h = 2 : 1$$).

To draw another rectangle with the same $$w : h$$ ratio, choose a scale factor $$k$$ and draw a rectangle of width $$kw$$ and height $$kh$$:

• Smaller: pick $$k < 1$$. E.g. $$k = \dfrac{1}{2}$$ gives $$4 \text{ cm} \times 2 \text{ cm}$$.
• Bigger: pick $$k > 1$$. E.g. $$k = \dfrac{3}{2}$$ gives $$12 \text{ cm} \times 6 \text{ cm}$$.

Are all the classmates' drawings the same? No — different classmates will pick different scale factors, so their rectangles will be of different sizes. But every rectangle drawn with the correct ratio will look like a scaled version of the original — the shape is the same.

Are they wrong? No. All rectangles with the same $$w : h$$ ratio are equally correct — they are all similar to the given rectangle, just different in size.

This is an activity — the exact numbers depend on your friend's actual measurements. Here is a worked example so you can copy the method.

Suppose you measure your friend and get

• Head $$= 24$$ cm
• Torso $$= 60$$ cm
• Arms $$= 66$$ cm
• Legs $$= 90$$ cm

Then

$$\text{head : torso} = 24 : 60 = \dfrac{24}{60} = \dfrac{2}{5} = 2 : 5,$$

$$\text{torso : arms} = 60 : 66 = \dfrac{60}{66} = \dfrac{10}{11} = 10 : 11,$$

$$\text{torso : legs} = 60 : 90 = \dfrac{60}{90} = \dfrac{2}{3} = 2 : 3.$$

Drawing the figure. Choose a scale factor $$k$$ (say $$k = \dfrac{1}{6}$$, so $$1$$ cm on your friend becomes $$\dfrac{1}{6}$$ cm on paper). Multiply every real length by $$k$$:

• Head: $$24 \times \dfrac{1}{6} = 4$$ cm
• Torso: $$60 \times \dfrac{1}{6} = 10$$ cm
• Arms: $$66 \times \dfrac{1}{6} = 11$$ cm
• Legs: $$90 \times \dfrac{1}{6} = 15$$ cm

Because you have multiplied every length by the same $$k$$, the ratios head : torso, torso : arms, and torso : legs stay the same as those of your friend's real body — so the drawn figure looks like a proper scaled copy.

At a (roughly) constant speed, distance is proportional to time. There are $$52$$ weeks in a year, so let the distance travelled in one week be $$x$$ million kilometres.

$$\dfrac{\text{distance in 1 year}}{\text{weeks in 1 year}} = \dfrac{940}{52}.$$

Therefore

$$x = \dfrac{940}{52} \approx 18.08 \text{ million km.}$$

Being a bit more careful, a year is $$365.25$$ days $$= 52.18$$ weeks, giving $$x \approx \dfrac{940}{52.18} \approx 18.02$$ million km, essentially the same answer.

So in one week the Earth covers about $$18$$ million kilometres, i.e. approximately $$1.8 \times 10^{7}$$ km.

Since all walls have the same height and thickness, the number of bricks is directly proportional to the total length of wall to be built. So first find the total wall length.

Step 1 — Outer perimeter of the L-shape.

Place the big rectangle so it is $$12$$ ft wide and $$15$$ ft tall. Attach the small rectangle ($$9$$ ft $$\times$$ $$6$$ ft) so that its $$9$$ ft side lies along the bigger rectangle (this makes the shared boundary the $$9$$-ft inner wall). Walking round the L, the sides are

$$18 + 9 + 6 + 6 + 12 + 15 = 66 \text{ ft.}$$

Check by another method: Perimeter of big + perimeter of small $$- 2 \times$$ (shared side) $$= 2(12+15) + 2(9+6) - 2(9) = 54 + 30 - 18 = 66$$ ft. ✓

Step 2 — Add the inner wall. The inner wall between the two rooms is $$9$$ ft.

$$\text{Total wall length} = 66 + 9 = 75 \text{ ft.}$$

Step 3 — Bricks by proportion. $$10$$ ft of wall needs $$1450$$ bricks. Let $$x$$ bricks be needed for $$75$$ ft:

$$10 : 1450 :: 75 : x \quad\Rightarrow\quad x = \dfrac{1450 \times 75}{10} = 145 \times 75.$$

Now $$145 \times 75 = 145 \times 75 = 10{,}875$$. So he needs approximately $$10{,}875$$ bricks.

The ratio $$2 : 3$$ splits the money into $$2 + 3 = 5$$ equal parts.

$$\text{Value of one part} = \dfrac{\text{₹}4500}{5} = \text{₹}900.$$

First part $$= 2 \times 900 = \text{₹}1800$$.
Second part $$= 3 \times 900 = \text{₹}2700$$.

Check: $$1800 + 2700 = 4500$$ and $$1800 : 2700 = 2 : 3$$. ✓

The ratio $$1 : 5$$ divides the solution into $$1 + 5 = 6$$ equal parts.

$$\text{Volume of one part} = \dfrac{240}{6} = 40 \text{ mL.}$$

Acid $$= 1 \times 40 = 40$$ mL.
Water $$= 5 \times 40 = 200$$ mL.

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

Step 1 — Split 40 mL in the ratio $$3 : 5$$.

Total parts $$= 3 + 5 = 8$$. One part $$= \dfrac{40}{8} = 5$$ mL.

Blue $$= 3 \times 5 = 15$$ mL.
Yellow $$= 5 \times 5 = 25$$ mL.

Step 2 — Add 20 mL of yellow.

New yellow $$= 25 + 20 = 45$$ mL. Blue is unchanged at $$15$$ mL.

Step 3 — Simplify the new ratio.

$$\text{Blue : Yellow} = 15 : 45 = \dfrac{15}{45} = \dfrac{1}{3} = 1 : 3.$$

The ratio $$2 : 1$$ divides the mixture into $$2 + 1 = 3$$ equal parts.

$$\text{Volume of one part} = \dfrac{6}{3} = 2 \text{ cups.}$$

Rice $$= 2 \times 2 = 4$$ cups.
Urad dal $$= 1 \times 2 = 2$$ cups.

Check: $$4 + 2 = 6$$ cups and $$4 : 2 = 2 : 1$$. ✓

Suppose one bucket holds $$V$$ litres. The orange bucket has red and yellow in the ratio $$3 : 5$$, split into $$3 + 5 = 8$$ equal parts of $$\dfrac{V}{8}$$ each.

$$\text{Red in orange bucket} = 3 \times \dfrac{V}{8} = \dfrac{3V}{8}.$$
$$\text{Yellow in orange bucket} = 5 \times \dfrac{V}{8} = \dfrac{5V}{8}.$$

Now we add one full bucket of yellow, i.e. $$V$$ litres of yellow.

$$\text{New yellow} = \dfrac{5V}{8} + V = \dfrac{5V + 8V}{8} = \dfrac{13V}{8}.$$

Red is unchanged at $$\dfrac{3V}{8}$$. The new ratio is

$$\text{Red : Yellow} = \dfrac{3V}{8} : \dfrac{13V}{8} = 3 : 13.$$

Ratio of orange juice to apple juice:

$$600 : 900 = \dfrac{600}{900}.$$

Divide the numerator and denominator by their HCF. $$\text{HCF}(600, 900) = 300$$, so

$$\dfrac{600}{900} = \dfrac{600 \div 300}{900 \div 300} = \dfrac{2}{3} = 2 : 3.$$

Step 1 — Seating capacity of one bus. Since $$3$$ full buses seated $$162$$ people,

$$\text{seats per bus} = \dfrac{162}{3} = 54.$$

Step 2 — Buses needed for 204 students.

$$\dfrac{204}{54} = 3\dfrac{42}{54} = 3\dfrac{7}{9}.$$

Three buses can seat only $$3 \times 54 = 162$$ students, which is less than $$204$$. So the school must hire a fourth bus.

Step 3 — Will all the buses be full?

$$4 \text{ buses} \times 54 = 216 \text{ seats.}$$

But there are only $$204$$ students. Three buses will be full ($$162$$ students) and the fourth bus will carry $$204 - 162 = 42$$ students, leaving $$54 - 42 = 12$$ seats empty.

Crowdedness is measured by population density, i.e. the number of people per square kilometre. A city with more people per square km is more crowded.

Delhi:

$$\text{density} = \dfrac{\text{population}}{\text{area}} = \dfrac{30{,}000{,}000}{1484} \approx 20{,}215 \text{ people/sq km.}$$

Mumbai:

$$\text{density} = \dfrac{20{,}000{,}000}{550} \approx 36{,}364 \text{ people/sq km.}$$

Since $$36{,}364 > 20{,}215$$, Mumbai has more people packed into every square kilometre.

Delhi has more people in total, but it also has a much bigger area, so the people are spread out. Mumbai has fewer people but is squeezed into less than half the area, so it is more crowded.

Let the neck be $$4x$$ and the rest of the body be $$6x$$. Then

$$\text{total height} = 4x + 6x = 10x.$$

So $$\text{neck} : \text{total height} = 4 : 10 = 2 : 5$$, i.e. the neck is $$\dfrac{2}{5}$$ of the total height.

Check with the crane: total $$= 155$$ cm, so $$x = \dfrac{155}{10} = 15.5$$. Neck $$= 4 \times 15.5 = 62$$ cm and rest $$= 6 \times 15.5 = 93$$ cm. Sum $$= 155$$ cm. ✓

For yourself. If your height is $$H$$ cm, then

$$\text{your neck} = \dfrac{2}{5} \times H = \dfrac{2H}{5} \text{ cm.}$$

Example: if $$H = 150$$ cm, your neck would be $$\dfrac{2 \times 150}{5} = 60$$ cm. (For comparison, a real human neck is only about $$10\text{–}15$$ cm — so a crane's neck is proportionally much longer than a person's.)

The weight of saffron is directly proportional to the money spent. Let $$x$$ palas be the quantity that can be bought for $$9$$ niskas. Then

$$\dfrac{2\tfrac{1}{2} \text{ palas}}{\tfrac{3}{7} \text{ niskas}} = \dfrac{x \text{ palas}}{9 \text{ niskas}}.$$

Writing $$2\tfrac{1}{2} = \dfrac{5}{2}$$ and cross-multiplying,

$$x \times \dfrac{3}{7} = 9 \times \dfrac{5}{2},$$
$$\dfrac{3x}{7} = \dfrac{45}{2},$$
$$3x = \dfrac{45}{2} \times 7 = \dfrac{315}{2},$$
$$x = \dfrac{315}{6} = \dfrac{105}{2} = 52\tfrac{1}{2}.$$

Check by unit-rate method. For $$\dfrac{3}{7}$$ niskas you get $$\dfrac{5}{2}$$ palas, so for $$1$$ niska you get

$$\dfrac{5/2}{3/7} = \dfrac{5}{2} \times \dfrac{7}{3} = \dfrac{35}{6} \text{ palas.}$$

For $$9$$ niskas: $$9 \times \dfrac{35}{6} = \dfrac{315}{6} = 52\tfrac{1}{2}$$ palas. ✓

The difference of their ages stays the same forever: brother is always older by

$$5 - 1 = 4 \text{ years.}$$

Let Harmain's age be $$x$$ years when the ratio Harmain : brother is $$1 : 2$$. Then brother's age is $$2x$$, and the difference is

$$2x - x = x = 4.$$

So Harmain will be $$4$$ years old. At that time, brother will be $$2 \times 4 = 8$$ years old, and $$4 : 8 = 1 : 2$$. ✓

(This happens $$4 - 1 = 3$$ years from now.)

For the same volume $$V$$, mass of gold : mass of water $$= 37 : 2$$. Let the mass of $$1$$ litre of gold be $$m$$ kg. Since $$1$$ litre of water is $$1$$ kg,

$$\dfrac{m}{1} = \dfrac{37}{2}.$$

Therefore

$$m = \dfrac{37}{2} = 18.5 \text{ kg.}$$

(Indeed, gold has density about $$19.3$$ g/cm³, so this is close to the real value.)

Step 1 — Area of the plot in square feet.

$$\text{Area} = 200 \text{ ft} \times 500 \text{ ft} = 1{,}00{,}000 \text{ sq ft.}$$

Step 2 — Convert to acres. From the unit-conversions section, $$1 \text{ acre} = 43{,}560 \text{ sq ft}$$. Therefore

$$\text{Area in acres} = \dfrac{1{,}00{,}000}{43{,}560} \approx 2.296 \text{ acres.}$$

Step 3 — Apply the proportion. Manure needed is directly proportional to area:

$$\text{manure} = 2.296 \times 10 \approx 22.96 \text{ tonnes.}$$

So the farmer should buy about $$23$$ tonnes of cow manure.

Same tap ⇒ same flow rate. Time is directly proportional to the volume filled.

Convert the bucket to mL: $$10$$ L $$= 10 \times 1000 = 10{,}000$$ mL.

Let $$t$$ be the time (in seconds) to fill the bucket.

$$500 : 15 :: 10{,}000 : t.$$

Cross-multiplying,

$$500 \times t = 15 \times 10{,}000,$$
$$500 t = 1{,}50{,}000,$$
$$t = \dfrac{1{,}50{,}000}{500} = 300 \text{ seconds.}$$

$$300 \text{ seconds} = \dfrac{300}{60} = 5 \text{ minutes.}$$

Alternative: flow rate $$= \dfrac{500 \text{ mL}}{15 \text{ s}} = \dfrac{100}{3}$$ mL/s. Time to fill $$10{,}000$$ mL $$= \dfrac{10{,}000}{100/3} = 300$$ s. ✓

Cost is directly proportional to area, but the two areas are in different units. Convert them to the same unit — square feet.

$$1 \text{ acre} = 43{,}560 \text{ sq ft.}$$

So $$43{,}560$$ sq ft costs ₹$$15{,}00{,}000$$. Let $$C$$ be the cost of $$2{,}400$$ sq ft. Then

$$43{,}560 : 15{,}00{,}000 :: 2{,}400 : C,$$

i.e.

$$C = \dfrac{15{,}00{,}000 \times 2{,}400}{43{,}560}.$$

Compute in steps:

$$C = \dfrac{15{,}00{,}000}{43{,}560} \times 2{,}400 \approx 34.4353 \times 2{,}400 \approx 82{,}644.6.$$

So the cost of $$2{,}400$$ sq ft is approximately ₹$$82{,}645$$ (roughly ₹$$82{,}650$$).

Oxen. Time is directly proportional to area to be ploughed.

$$1 \text{ acre} \rightarrow 6 \text{ hours,} \quad 20 \text{ acres} \rightarrow 20 \times 6 = 120 \text{ hours.}$$

Tractor. "4 times faster" means the tractor takes $$\dfrac{1}{4}$$ of the time the oxen take for the same area. So

$$\text{tractor time} = \dfrac{120}{4} = 30 \text{ hours.}$$

(Equivalently, if the oxen take $$6$$ h per acre, the tractor takes $$\dfrac{6}{4} = 1.5$$ h per acre, so $$20$$ acres $$\rightarrow 20 \times 1.5 = 30$$ h.)

Step 1 — Mass of each metal in the coin.

Ratio $$3 : 1$$ splits the total mass into $$3 + 1 = 4$$ equal parts.

$$\text{one part} = \dfrac{7.74}{4} = 1.935 \text{ g.}$$

Copper $$= 3 \times 1.935 = 5.805$$ g.
Nickel $$= 1 \times 1.935 = 1.935$$ g.

Step 2 — Cost per gram of each metal.

$$1 \text{ kg} = 1000 \text{ g,}$$

$$\text{copper cost per g} = \dfrac{906}{1000} = \text{₹}0.906,$$

$$\text{nickel cost per g} = \dfrac{1341}{1000} = \text{₹}1.341.$$

Step 3 — Cost of the metals in one coin.

Cost of copper $$= 5.805 \times 0.906 \approx \text{₹}5.26$$.
Cost of nickel $$= 1.935 \times 1.341 \approx \text{₹}2.59$$.

$$\text{Total} = 5.26 + 2.59 \approx \text{₹}7.85.$$

So the copper and nickel in one ₹$$10$$ coin together are worth about ₹$$7.85$$ (a little less than the face value of the coin).