Chapter 3: Proportional Reasoning 2

Taste depends on the ratio of rice to urad dal, not on the absolute amounts used. So we compare the two ratios.

Viswanath's ratio of rice to urad dal is $$6 : 3$$. Dividing both terms by their common factor $$3$$, this simplifies to $$2 : 1$$.

Puneet's ratio of rice to urad dal is $$4 : 2$$. Dividing both terms by $$2$$, this also simplifies to $$2 : 1$$.

Since both mixes have the same simplified ratio $$2 : 1$$, the proportion of rice to urad dal is identical in the two batters. When cooked in the same way, the two batches of idlis will taste the same.

The letters RF stand for Representative Fraction. It is the fixed ratio that tells us how a length measured on the map is related to the corresponding real distance on the ground.

The statement $$\text{RF} = 1 : 60{,}00{,}000$$ means:

$$\text{Distance on the map} : \text{Actual distance on the ground} = 1 : 60{,}00{,}000.$$

In other words, every $$1$$ unit measured on the map represents $$60{,}00{,}000$$ of the same units on the ground. So $$1$$ cm on the map stands for $$60{,}00{,}000$$ cm on the ground, and $$1$$ inch on the map would stand for $$60{,}00{,}000$$ inches on the ground.

Because the ratio is the same for every pair of points, we can use it (with a ruler) to work out the real geographical distance between any two places shown on the map.

We use the standard unit conversions:

$$1 \text{ m} = 100 \text{ cm}, \qquad 1 \text{ km} = 1000 \text{ m} = 1{,}00{,}000 \text{ cm}.$$

So dividing centimetres by $$1{,}00{,}000$$ gives kilometres:

$$60{,}00{,}000 \text{ cm} = \dfrac{60{,}00{,}000}{1{,}00{,}000} \text{ km} = 60 \text{ km}.$$

Therefore an RF of $$1:60{,}00{,}000$$ means that $$1$$ cm on the map represents $$60$$ km on the ground.

The scale of the map is $$\text{RF} = 1:60{,}00{,}000$$, and from the previous question we know that this means

$$1 \text{ cm on the map} \;=\; 60 \text{ km on the ground}.$$

Bengaluru — Chennai. Measure the straight-line distance between the two cities on the map with a ruler. Typically this comes out to about $$5.8$$ cm. Multiplying by the scale:

$$\text{Actual distance} = 5.8 \times 60 \text{ km} \approx 348 \text{ km}.$$

So the geographical distance between Bengaluru and Chennai is about $$350$$ km.

Mangaluru — Chennai. Now measure the distance between Mangaluru and Chennai on the same map. This comes out to roughly $$11.7$$ cm. Applying the scale:

$$\text{Actual distance} = 11.7 \times 60 \text{ km} \approx 702 \text{ km}.$$

So the geographical distance between Mangaluru and Chennai is about $$700$$ km.

(Your own measurements with a ruler may differ by a few millimetres, so answers between roughly $$340$$–$$360$$ km and $$680$$–$$720$$ km respectively are all reasonable.)

Take a different map of India (or of southern India) whose RF is, say, $$1 : 1{,}00{,}00{,}000$$ — here $$1$$ cm on the map stands for $$100$$ km on the ground.

Measure the same two pairs of cities (Bengaluru–Chennai and Mangaluru–Chennai) on this new map. Because the scale is smaller, the ruler readings will now be different — the distances on the paper will be about $$3.5$$ cm and $$7$$ cm respectively. Converting with the new scale:

$$\text{Bengaluru–Chennai} \approx 3.5 \times 100 \text{ km} = 350 \text{ km},$$

$$\text{Mangaluru–Chennai} \approx 7 \times 100 \text{ km} = 700 \text{ km}.$$

These are essentially the same values we obtained in the previous question. This is exactly what we should expect: the geographical distance between two fixed places on the Earth does not depend on which map we use. Different maps only change the paper measurement, but the scale is designed to compensate for that, so the actual distance always comes out (approximately) the same. Small differences arise only from measurement error, from projection distortions, and from whether we measure a straight line or follow the road.

To keep the taste unchanged, Puneet must keep the proportions of the four ingredients exactly the same as in Viswanath's recipe, namely

$$\text{coriander seeds : red chillies : toor dal : fenugreek seeds} = 8 : 4 : 2 : 1.$$

In Viswanath's recipe there are $$4$$ red chillies. Puneet has only $$2$$ red chillies, which is half of $$4$$. So Puneet must scale every ingredient down by the same factor

$$k = \dfrac{2}{4} = \dfrac{1}{2}.$$

Multiplying each quantity by $$\tfrac{1}{2}$$:

Check. The new ratio is $$4 : 2 : 1 : \tfrac{1}{2}$$; multiplying every term by $$2$$ gives $$8 : 4 : 2 : 1$$, the same as before, so the flavour will indeed be the same.

The three components of the purple paint are $$4$$ litres of red, $$6$$ litres of blue and $$10$$ litres of white. Adding them together gives the total volume of the mixture:

$$\text{Total volume} = 4 + 6 + 10 = 20 \text{ litres}.$$

Consistency check. The proportion of the three colours in this total is $$4 : 6 : 10$$; dividing every term by $$2$$ this becomes $$2 : 3 : 5$$, which is the ratio the recipe asked for. So the shade of purple is correct.

The recipe ratio is

$$\text{Red} : \text{Blue} : \text{White} = 2 : 3 : 5.$$

This says that for every $$5$$ parts of white paint we need $$2$$ parts of red and $$3$$ parts of blue. Yasmin has $$10$$ litres of white paint, which corresponds to the $$5$$ parts in the ratio. So one part must be

$$1 \text{ part} = \dfrac{10}{5} \text{ L} = 2 \text{ L}.$$

Therefore

$$\text{Red} = 2 \text{ parts} = 2 \times 2 = 4 \text{ L},$$

$$\text{Blue} = 3 \text{ parts} = 3 \times 2 = 6 \text{ L}.$$

Check. $$4 : 6 : 10 = 2 : 3 : 5$$ ✓. Yasmin should mix $$4$$ L red and $$6$$ L blue with her $$10$$ L white.

The mixture ratio is

$$\text{Cement} : \text{Sand} : \text{Gravel} = 1 : 1.5 : 3.$$

Here $$1$$ part of cement gives a certain amount of concrete. We are given $$3$$ bags of cement, which is $$3$$ times the $$1$$ part in the recipe. So we scale every ingredient by the factor $$3$$:

$$\text{Sand} = 1.5 \times 3 = 4.5 \text{ bags},$$

$$\text{Gravel} = 3 \times 3 = 9 \text{ bags}.$$

The total amount of concrete produced (measured in bag-equivalents) is the sum of the three quantities:

$$\text{Total concrete} = 3 + 4.5 + 9 = 16.5 \text{ bags}.$$

Check. The proportion of cement, sand and gravel in the mixture is $$3 : 4.5 : 9$$. Dividing every term by $$3$$ gives $$1 : 1.5 : 3$$, which matches the required ratio ✓.

The ratio in which the three ingredients are mixed is

$$\text{Cement} : \text{Sand} : \text{Gravel} = 1 : 1.5 : 3.$$

The total number of parts in the mixture is

$$1 + 1.5 + 3 = 5.5 \text{ parts}.$$

These $$5.5$$ parts must together give $$110$$ units of concrete, so the size of one part is

$$1 \text{ part} = \dfrac{110}{5.5} = 20 \text{ units}.$$

Multiplying each ratio number by $$20$$:

$$\text{Cement} = 1 \times 20 = 20 \text{ units},$$

$$\text{Sand} = 1.5 \times 20 = 30 \text{ units},$$

$$\text{Gravel} = 3 \times 20 = 60 \text{ units}.$$

Check. $$20 + 30 + 60 = 110$$ units ✓, and $$20 : 30 : 60 = 1 : 1.5 : 3$$ ✓.

The recipe ratio is

$$\text{Red} : \text{Blue} : \text{White} = 2 : 3 : 5.$$

The total number of parts in this ratio is

$$2 + 3 + 5 = 10 \text{ parts}.$$

These $$10$$ parts together must make $$50$$ ml of paint, so

$$1 \text{ part} = \dfrac{50}{10} = 5 \text{ ml}.$$

Multiplying each ratio number by $$5$$:

$$\text{Red} = 2 \times 5 = 10 \text{ ml},$$

$$\text{Blue} = 3 \times 5 = 15 \text{ ml},$$

$$\text{White} = 5 \times 5 = 25 \text{ ml}.$$

Check. $$10 + 15 + 25 = 50$$ ml ✓ and $$10 : 15 : 25 = 2 : 3 : 5$$ ✓.

Step 1: Find the three angles.

The sum of the angles of any triangle is $$180^{\circ}$$. The angles are in the ratio $$1:3:5$$, so the total number of parts is

$$1 + 3 + 5 = 9 \text{ parts}.$$

Hence one part equals

$$1 \text{ part} = \dfrac{180^{\circ}}{9} = 20^{\circ}.$$

Multiplying each ratio number by $$20^{\circ}$$:

$$\text{Angles} = 1 \times 20^{\circ},\; 3 \times 20^{\circ},\; 5 \times 20^{\circ} = 20^{\circ},\; 60^{\circ},\; 100^{\circ}.$$

Check: $$20^{\circ} + 60^{\circ} + 100^{\circ} = 180^{\circ}$$ ✓.

Step 2: Construct the triangle (ASA method).

1. Draw a line segment $$BC$$ of any convenient length (say $$6$$ cm). This will be one side of the triangle.
2. At $$B$$, use a protractor to draw a ray making an angle of $$60^{\circ}$$ with $$BC$$.
3. At $$C$$, use a protractor to draw a ray making an angle of $$20^{\circ}$$ with $$CB$$, on the same side of $$BC$$ as the previous ray.
4. Mark the point where the two rays meet as $$A$$. Triangle $$ABC$$ is the required triangle.

The third angle at $$A$$ is automatically $$180^{\circ} - 60^{\circ} - 20^{\circ} = 100^{\circ}$$, and the three angles are in the required ratio $$1 : 3 : 5$$. Because we picked $$BC$$ freely, there are infinitely many such triangles — they are all similar to one another, but not congruent unless the side lengths are chosen the same.

The four activities occupy time in the ratio

$$\text{Warm-up/cool-down} : \text{Batting} : \text{Bowling} : \text{Fielding} = 3 : 4 : 3 : 5.$$

The total number of parts is

$$3 + 4 + 3 + 5 = 15 \text{ parts}.$$

The whole session is $$150$$ minutes, so one part corresponds to

$$1 \text{ part} = \dfrac{150}{15} = 10 \text{ minutes}.$$

Multiplying each ratio number by $$10$$ minutes:

$$\text{Warm-up/cool-down} = 3 \times 10 = 30 \text{ min},$$

$$\text{Batting} = 4 \times 10 = 40 \text{ min},$$

$$\text{Bowling} = 3 \times 10 = 30 \text{ min},$$

$$\text{Fielding} = 5 \times 10 = 50 \text{ min}.$$

Check. $$30 + 40 + 30 + 50 = 150$$ min ✓.

The number of books in the three languages is in the ratio

$$\text{Odia} : \text{Hindi} : \text{English} = 3 : 2 : 1.$$

Odia takes up $$3$$ parts, and the actual number of Odia books is $$288$$. So one part equals

$$1 \text{ part} = \dfrac{288}{3} = 96 \text{ books}.$$

Now use this value of one part for the other two languages:

$$\text{Hindi} = 2 \text{ parts} = 2 \times 96 = 192 \text{ books},$$

$$\text{English} = 1 \text{ part} = 1 \times 96 = 96 \text{ books}.$$

Check. $$288 : 192 : 96 = 3 : 2 : 1$$ (divide every term by $$96$$) ✓.

Step 1: Find how many coins of each denomination there are.

The number of coins is split in the ratio

$$\text{₹10} : \text{₹5} : \text{₹2} : \text{₹1} = 4 : 3 : 2 : 1.$$

Total parts $$= 4 + 3 + 2 + 1 = 10$$. Since there are $$100$$ coins in all,

$$1 \text{ part} = \dfrac{100}{10} = 10 \text{ coins}.$$

Multiplying each ratio number by $$10$$:

$$\text{₹10 coins} = 4 \times 10 = 40,\quad \text{₹5 coins} = 3 \times 10 = 30,\quad \text{₹2 coins} = 2 \times 10 = 20,\quad \text{₹1 coins} = 1 \times 10 = 10.$$

Step 2: Compute the total value in rupees.

Value contributed by each denomination = (number of coins) × (value of one coin):

$$\text{₹10 coins} \to 40 \times 10 = ₹400,$$

$$\text{₹5 coins} \to 30 \times 5 = ₹150,$$

$$\text{₹2 coins} \to 20 \times 2 = ₹40,$$

$$\text{₹1 coins} \to 10 \times 1 = ₹10.$$

Total money

$$= 400 + 150 + 40 + 10 = ₹600.$$

Step 1: Choose actual side lengths.

The ratio $$3 : 4 : 5$$ only fixes the relative lengths of the three sides. We must pick a positive number $$k$$ (called the scale factor) and take the sides to be

$$3k,\; 4k,\; 5k.$$

For a convenient drawing take $$k = 1$$ cm, giving sides of $$3$$ cm, $$4$$ cm and $$5$$ cm.

Step 2: Triangle inequality check.

We must verify the sum of any two sides exceeds the third:

$$3 + 4 = 7 > 5,\quad 3 + 5 = 8 > 4,\quad 4 + 5 = 9 > 3.\quad\checkmark$$

So the triangle is constructible.

Step 3: Construction (SSS method).

1. Draw the longest side $$BC$$ of length $$5$$ cm.
2. With centre $$B$$ and radius $$3$$ cm, draw an arc on one side of $$BC$$.
3. With centre $$C$$ and radius $$4$$ cm, draw a second arc that cuts the first arc. Call the intersection point $$A$$.
4. Join $$BA$$ and $$CA$$. Triangle $$ABC$$ is the required triangle.

(In fact $$3^2 + 4^2 = 9 + 16 = 25 = 5^2$$, so by the converse of Pythagoras' theorem this is a right-angled triangle with the right angle at $$A$$.)

Step 4: Are all such triangles congruent?

No. Choosing a different scale factor $$k$$ produces a different-sized triangle. For example, with $$k = 2$$ cm the sides are $$6$$ cm, $$8$$ cm, $$10$$ cm, and with $$k = 10$$ cm they are $$30$$ cm, $$40$$ cm, $$50$$ cm. These triangles all have the same shape (i.e. equal corresponding angles $$-$$ they are similar), but their sides have different actual lengths, so they are not congruent. Two triangles are congruent only when their corresponding sides are equal in length, which is a stronger condition than being in the same ratio.

Take the three sides to be $$k, 3k, 5k$$ for some positive length $$k$$.

A triangle can only be constructed if it satisfies the triangle inequality: the sum of the two shorter sides must be strictly greater than the longest side. Check this on the two shortest sides:

$$k + 3k = 4k.$$

Comparing with the longest side $$5k$$:

$$4k \;<\; 5k.$$

So the sum of the two shorter sides is smaller than the longest side. The two shorter sides could never meet to close the triangle — one arc would fall short of the other. Since the failure comes from a strict inequality on the ratio itself, no choice of $$k$$ will fix it.

Therefore, no triangle can be constructed with sides in the ratio $$1 : 3 : 5$$.

A pie chart represents the whole (here $$40$$ students) by a full circle whose central angle at the centre is $$360^{\circ}$$. Each grade must occupy a slice whose angle is the same fraction of $$360^{\circ}$$ as that grade's fraction of the total. So the angle for each grade is

$$\text{angle} = \dfrac{\text{number of students in that grade}}{40} \times 360^{\circ}.$$

Applying this formula:

Check. $$108^{\circ} + 90^{\circ} + 72^{\circ} + 54^{\circ} + 36^{\circ} = 360^{\circ}$$ ✓.

Drawing the pie chart. Draw a circle. Draw one radius, then measure the first angle ($$108^{\circ}$$) from that radius with a protractor and draw the second radius; this marks the slice for Grade A. Continue from the last radius drawn, measuring $$90^{\circ}, 72^{\circ}, 54^{\circ}, 36^{\circ}$$ in turn for Grades B, C, D, E. Label each slice with its grade.

To reduce a ratio to its simplest form we divide every term by the highest common factor (HCF) of all the terms. The five terms here are $$12, 10, 8, 6, 4$$.

Their factors:

$$12 = 2^2 \times 3,\quad 10 = 2 \times 5,\quad 8 = 2^3,\quad 6 = 2 \times 3,\quad 4 = 2^2.$$

The only common prime factor is $$2$$, and its lowest power in the list is $$2^1$$. So

$$\gcd(12,10,8,6,4) = 2.$$

Dividing every term by $$2$$:

$$12 : 10 : 8 : 6 : 4 \;=\; 6 : 5 : 4 : 3 : 2.$$

Since $$\gcd(6,5,4,3,2) = 1$$ (the number $$5$$ shares no factor greater than $$1$$ with the rest), the ratio $$6 : 5 : 4 : 3 : 2$$ is in simplest form.

Note. Reducing does not change the pie chart. The sum of the reduced parts is $$6 + 5 + 4 + 3 + 2 = 20$$, so each part corresponds to $$\tfrac{360^{\circ}}{20} = 18^{\circ}$$, giving angles $$108^{\circ}, 90^{\circ}, 72^{\circ}, 54^{\circ}, 36^{\circ}$$ — exactly the same as before.

Step 1: Find how many people liked each season.

Total people $$= 360$$. Summer $$= 90$$, Rainy $$= 120$$, and the remaining people liked winter:

$$\text{Winter} = 360 - (90 + 120) = 360 - 210 = 150.$$

Step 2: Compute the central angle for each season.

For a category with $$n$$ votes out of $$N$$ total, the pie-chart angle is $$\dfrac{n}{N} \times 360^{\circ}$$.

Because the total is exactly $$360$$ people, each person corresponds to $$\dfrac{360^{\circ}}{360} = 1^{\circ}$$. So the angle in degrees is numerically equal to the number of people!

Check. $$90^{\circ} + 120^{\circ} + 150^{\circ} = 360^{\circ}$$ ✓.

Step 3: Construct the pie chart. Draw a circle and mark its centre. Draw a starting radius. Using a protractor, mark off a $$90^{\circ}$$ sector for Summer, then continue with a $$120^{\circ}$$ sector for Rainy, and finally a $$150^{\circ}$$ sector for Winter. Colour or label each slice with the season name and, if desired, its percentage ($$25\%$$, $$33\tfrac{1}{3}\%$$ and $$41\tfrac{2}{3}\%$$ respectively).

The whole $$100\%$$ has to fill the full circle of $$360^{\circ}$$. So each $$1\%$$ corresponds to

$$\dfrac{360^{\circ}}{100} = 3.6^{\circ}.$$

The central angle of a slice for a category with percentage $$p$$ is therefore $$p \times 3.6^{\circ}$$ (equivalently $$\dfrac{p}{100} \times 360^{\circ}$$). Applying this to each channel type:

Check. $$180^{\circ} + 90^{\circ} + 54^{\circ} + 36^{\circ} = 360^{\circ}$$ ✓.

Construction. Draw a circle and one starting radius. Using a protractor, mark off sectors of $$180^{\circ}$$ (Entertainment — this is a semicircle), $$90^{\circ}$$ (Sports — a right-angle sector), $$54^{\circ}$$ (News) and $$36^{\circ}$$ (Information) one after the other, and label each slice with its name and percentage.

This is a classroom-activity question, so the numbers will change from class to class. The method is the same in every case; here is a fully worked example using an illustrative set of counts for a class of $$40$$ students.

Step 1: Collect the data. Ask each of your classmates to name their single favourite subject and tally the votes. Suppose the tally comes out like this:

Total $$= 4 + 3 + 2 + 5 + 6 + 10 + 10 = 40$$ students.

Step 2: Compute the central angle for each subject. The angle is $$\dfrac{\text{students in subject}}{40} \times 360^{\circ}$$, and here each student corresponds to $$\dfrac{360^{\circ}}{40} = 9^{\circ}$$.

Check. $$36^{\circ} + 27^{\circ} + 18^{\circ} + 45^{\circ} + 54^{\circ} + 90^{\circ} + 90^{\circ} = 360^{\circ}$$ ✓.

Step 3: Draw the pie chart. Draw a circle, mark the centre, and draw one radius. Starting from that radius and using a protractor, mark off consecutive sectors of $$36^{\circ}, 27^{\circ}, 18^{\circ}, 45^{\circ}, 54^{\circ}, 90^{\circ}, 90^{\circ}$$. Colour the sectors differently and label each with the subject name (and, if you like, the count and percentage).

For your own class, replace the sample counts by your actual tally, use the same formula $$\text{angle} = \dfrac{\text{students}}{\text{total students}} \times 360^{\circ}$$, and draw the resulting chart.

All workers work at the same rate, so the number of bricks moved in a day is proportional to the number of workers. This is a direct proportion.

Method 1 (Unitary method).

Bricks moved by $$1$$ worker in a day:

$$= \dfrac{4500}{5} = 900 \text{ bricks per worker}.$$

To move $$18000$$ bricks, the number of workers required is

$$\dfrac{18000}{900} = 20 \text{ workers}.$$

Method 2 (Ratio). Let $$x$$ be the required number of workers. Since bricks $$\propto$$ workers,

$$\dfrac{4500}{5} = \dfrac{18000}{x} \;\Rightarrow\; 4500\, x = 5 \times 18000 \;\Rightarrow\; x = \dfrac{5 \times 18000}{4500} = 20.$$

Both methods give the same answer: $$20$$ workers are needed.

Step 1: Find the distance between Lucknow and Kanpur. Using distance $$=$$ speed $$\times$$ time on the motorcycle trip:

$$\text{Distance} = 30 \; \mathrm{km/h} \times 3 \; \text{h} = 90 \; \text{km}.$$

This distance does not change, no matter what vehicle is used.

Step 2: Compute the new travel time. With the car at $$60 \; \mathrm{km/h}$$,

$$\text{Time} = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{90 \; \text{km}}{60 \; \mathrm{km/h}} = 1.5 \; \text{h}.$$

So the trip takes $$1.5$$ hours, i.e. $$1$$ hour $$30$$ minutes.

Consistency with inverse proportion. For a fixed distance, time is inversely proportional to speed: $$\text{speed} \times \text{time}$$ is constant. Here $$30 \times 3 = 90$$ and $$60 \times 1.5 = 90$$, so the product is indeed the same. The speed doubled, so the time halved.

No, the statement $$30 : 60 :: 3 : x$$ is not correct for this problem. That statement asserts that speed and time are in direct proportion. If it were true, we would compute

$$\dfrac{30}{60} = \dfrac{3}{x} \;\Rightarrow\; x = \dfrac{3 \times 60}{30} = 6 \; \text{hours},$$

which would mean the car takes longer than the motorcycle — clearly wrong.

Why? For a fixed distance, a faster vehicle covers it in less time, not more. As speed increases, travel time decreases. So speed and time are in inverse proportion, not direct proportion. The correct relation is

$$\text{speed} \times \text{time} = \text{distance (constant)},$$

which gives $$30 \times 3 = 60 \times x$$ and so $$x = \dfrac{90}{60} = 1.5$$ hours, matching Example 2.

Compare walking with bicycling. The speed goes from $$5$$ km/h to $$15$$ km/h, so it is multiplied by

$$\dfrac{15}{5} = 3.$$

The time goes from $$18$$ hours down to $$6$$ hours, so it is multiplied by

$$\dfrac{6}{18} = \dfrac{1}{3}.$$

So when the speed becomes $$3$$ times, the time becomes $$\tfrac{1}{3}$$ times. The two factors are reciprocals of one another — that is exactly the signature of an inverse proportion. Equivalently, the product speed $$\times$$ time stays the same:

$$5 \times 18 = 90 = 15 \times 6.$$

So yes, the time decreases in exactly the same factor by which the speed increases.

The two things to check are (a) the product speed $$\times$$ time is the same for every mode of transport, and (b) whenever the speed is multiplied by some number, the time gets multiplied by its reciprocal.

(a) Product speed $$\times$$ time:

The product is the same value $$90$$ (which is the distance in km) for every row, confirming inverse proportion.

(b) Reciprocal factors:

• Bicycle vs Motorcycle: speed becomes $$\dfrac{30}{15} = 2$$ times; time becomes $$\dfrac{3}{6} = \dfrac{1}{2}$$ times. Reciprocals. ✓
• Motorcycle vs Car: speed becomes $$\dfrac{60}{30} = 2$$ times; time becomes $$\dfrac{1.5}{3} = \dfrac{1}{2}$$ times. Reciprocals. ✓
• Walk vs Car: speed becomes $$\dfrac{60}{5} = 12$$ times; time becomes $$\dfrac{1.5}{18} = \dfrac{1}{12}$$ times. Reciprocals. ✓

So the same behaviour holds for every pair of modes: whenever the speed is multiplied by a factor $$k$$, the time is multiplied by $$\dfrac{1}{k}$$. This is exactly what inverse proportion means.

Since $$x$$ and $$y$$ are in inverse proportion, the product $$xy$$ takes the same value $$k$$ in every column. The first column gives

$$k = xy = 16 \times 9 = 144.$$

Now fill each missing cell using $$xy = 144$$.

Second column ($$x = 12$$, $$y = ?$$):

$$y = \dfrac{144}{12} = 12.$$

Third column ($$x = ?$$, $$y = 48$$):

$$x = \dfrac{144}{48} = 3.$$

Fourth column ($$x = 36$$, $$y = ?$$):

$$y = \dfrac{144}{36} = 4.$$

Filled table:

Check. $$16\times 9 = 144, \; 12\times 12 = 144, \; 3\times 48 = 144, \; 36\times 4 = 144$$ ✓.

The amount of work (laying the same road) is fixed. If every worker works at the same rate, then more workers finish faster and fewer workers take longer. So the number of workers and the number of days needed are in inverse proportion:

$$(\text{workers}) \times (\text{days}) = \text{constant}.$$

Using the given data,

$$20 \times 4 = 10 \times x \;\Rightarrow\; 80 = 10\, x \;\Rightarrow\; x = 8.$$

So $$10$$ workers will take $$8$$ days to lay the same road.

Sense check. The number of workers has been halved, so — for the same job — the time needed should double. $$4 \to 8$$ ✓.

Every pump fills water at the same steady rate, and the tank has a fixed size, so the number of pumps and the time to fill are in inverse proportion:

$$(\text{pumps}) \times (\text{time}) = \text{constant}.$$

Originally there are $$2$$ pumps and it takes $$18$$ hours. After adding $$2$$ more pumps of the same kind, there are $$2 + 2 = 4$$ pumps. Let $$x$$ be the new filling time. Then

$$2 \times 18 = 4 \times x \;\Rightarrow\; 36 = 4\, x \;\Rightarrow\; x = 9 \; \text{hours}.$$

Sense check. The number of pumps doubled ($$2 \to 4$$), so the time should halve ($$18 \to 9$$) ✓.

Assume every student eats the same amount of food each day. The total food is fixed, so more students will finish it in fewer days: the number of students and the number of days for which the food lasts are in inverse proportion:

$$(\text{students}) \times (\text{days}) = \text{constant} = \text{total student-days of food}.$$

Originally $$80$$ students can be fed for $$15$$ days, so the total supply is

$$80 \times 15 = 1200 \; \text{student-days of food}.$$

After $$20$$ more students join, the number of students becomes $$80 + 20 = 100$$. Let $$x$$ be the number of days the food will now last. Since the total supply is unchanged,

$$100 \times x = 1200 \;\Rightarrow\; x = \dfrac{1200}{100} = 12 \; \text{days}.$$

So the provisions will last $$12$$ days for $$100$$ students.

Step 1: Convert each person's speed to a rate per hour.

Ram finishes the whole job in $$1$$ hour, so his rate is

$$\text{Ram's rate} = \dfrac{1 \; \text{job}}{1 \; \text{h}} = 1 \; \text{job/hour}.$$

Shyam finishes the same job in $$1.5$$ hours, so his rate is

$$\text{Shyam's rate} = \dfrac{1 \; \text{job}}{1.5 \; \text{h}} = \dfrac{2}{3} \; \text{job/hour}.$$

Step 2: Add the rates when they work together.

$$\text{Combined rate} = 1 + \dfrac{2}{3} = \dfrac{3}{3} + \dfrac{2}{3} = \dfrac{5}{3} \; \text{job/hour}.$$

Step 3: Compute the time to finish one job together.

$$\text{Time} = \dfrac{\text{Work}}{\text{Rate}} = \dfrac{1}{\;5/3\;} = \dfrac{3}{5} \; \text{hour}.$$

Convert to minutes: $$\dfrac{3}{5} \times 60 = 36 \; \text{minutes}.$$

Sense check. Working together, they should finish the job in less time than the faster person alone. $$\tfrac{3}{5} \; \text{h} = 36 \; \text{min} < 60 \; \text{min}$$ ✓.

All the pencils cost the same amount each. So the total cost is proportional to the number of pencils bought — this is a direct proportion, not an inverse one.

Method 1 (Unitary method). Cost of one pencil:

$$\dfrac{₹120}{24} = ₹5 \text{ per pencil}.$$

Cost of $$50$$ pencils:

$$50 \times 5 = ₹250.$$

Method 2 (Ratio). Let $$x$$ be the cost of $$50$$ pencils. Then

$$\dfrac{24}{120} = \dfrac{50}{x} \;\Rightarrow\; 24\, x = 120 \times 50 = 6000 \;\Rightarrow\; x = 250.$$

Either way, $$50$$ pencils cost $$₹250$$.

Assumptions.

• Every family uses the same amount of water per day.
• Each family's daily usage does not change when new families move in.
• The tank is not refilled during the period under consideration.

Under these assumptions the total water in the tank is fixed, and the number of families and the number of days the water lasts are in inverse proportion:

$$(\text{families}) \times (\text{days}) = \text{total family-days of water} = \text{constant}.$$

Step 1: Compute the constant.

$$20 \times 6 = 120 \; \text{family-days of water}.$$

Step 2: New number of families. After $$10$$ more families move in, the number of families becomes $$20 + 10 = 30$$.

Step 3: New number of days. Let $$x$$ be the number of days the water lasts for $$30$$ families. Then

$$30 \times x = 120 \;\Rightarrow\; x = \dfrac{120}{30} = 4 \; \text{days}.$$

Sense check. The number of families has become $$\tfrac{30}{20} = 1.5$$ times, so the water should last for $$\tfrac{1}{1.5} = \tfrac{2}{3}$$ of the previous time: $$\tfrac{2}{3} \times 6 = 4$$ days ✓.

The pie chart next to each animal shows what fraction of the $$24$$-hour day the animal spends asleep. To read off the sleep hours, measure that shaded fraction $$\dfrac{\text{sleep-angle}}{360^{\circ}}$$ and multiply by $$24$$:

$$\text{Sleep hours per day} = \dfrac{\text{shaded angle}}{360^{\circ}} \times 24.$$

For example, a chart shaded exactly a quarter shows $$\dfrac{1}{4} \times 24 = 6$$ h; a chart shaded half shows $$\dfrac{1}{2} \times 24 = 12$$ h; and so on.

Matching each chart to the closest value in the given list $$\{15, 2.5, 20, 8, 3.5, 13, 10.5, 18\}$$ (each value used exactly once, using well-known biological averages) gives the following:

Where does $$2.5$$ go? The remaining value in the list is $$2.5$$ hours, which is used for the animal whose chart shows the smallest shaded slice. Depending on the illustrations in your textbook, this may correspond to the elephant (some biology sources give elephants only about $$2.5$$ hours of sleep per day rather than $$3.5$$, in which case the elephant takes $$2.5$$ and the value $$3.5$$ is not used in that particular book).

Method to apply to your own book. For each pie chart, read the shaded fraction of the circle (using the marked angle if given, or by comparing to a full circle). Multiply that fraction by $$24$$ hours to obtain the sleep hours. Match to the closest value in the list.

Assumptions.

• Every worker paints at the same steady rate, so their outputs simply add up.
• Adding a new worker does not change the pace of the others (nobody gets in anybody's way).
• The size and difficulty of the fence stays the same.

Under these assumptions the fence is a fixed job, and the number of workers and the number of days needed to finish it are in inverse proportion:

$$(\text{workers}) \times (\text{days}) = \text{constant} = \text{total worker-days needed}.$$

Step 1: Total work in worker-days.

$$3 \times 4 = 12 \; \text{worker-days}.$$

Step 2: New number of workers. One more worker joins, so the team is now $$3 + 1 = 4$$ workers.

Step 3: New number of days. Let $$x$$ be the number of days needed. Then

$$4 \times x = 12 \;\Rightarrow\; x = \dfrac{12}{4} = 3 \; \text{days}.$$

Sense check. The number of workers has become $$\tfrac{4}{3}$$ times, so the time should become $$\tfrac{3}{4}$$ times: $$\tfrac{3}{4} \times 4 = 3$$ days ✓.

Only one pump is used all through, and it works at a steady rate. So the total time is proportional to the total amount of water pumped, i.e. to the number of tanks: this is a direct proportion.

Method (Unitary). Time to fill $$1$$ tank:

$$\dfrac{6 \; \text{h}}{2} = 3 \; \text{h per tank}.$$

Time to fill $$5$$ tanks:

$$5 \times 3 = 15 \; \text{hours}.$$

Method (Ratio). $$\dfrac{6}{2} = \dfrac{x}{5}$$, so $$x = \dfrac{6 \times 5}{2} = 15$$ hours.

The total number of chairs does not change when they are rearranged. Since

$$(\text{rows}) \times (\text{chairs per row}) = \text{total chairs} = \text{constant},$$

the number of rows and the number of chairs per row are in inverse proportion.

Step 1: Total number of chairs.

$$25 \times 12 = 300 \; \text{chairs}.$$

Step 2: New number of rows. With $$20$$ chairs per row, let $$x$$ be the number of rows. Then

$$20 \times x = 300 \;\Rightarrow\; x = \dfrac{300}{20} = 15 \; \text{rows}.$$

Check. $$15 \times 20 = 300$$ ✓.

The total number of minutes of school time in a day is fixed. Since

$$(\text{number of periods}) \times (\text{length of one period}) = \text{total school minutes} = \text{constant},$$

the number of periods and the length of one period are in inverse proportion.

Step 1: Total school minutes per day.

$$8 \times 45 = 360 \; \text{minutes per day} \; (=\; 6 \; \text{hours}).$$

Step 2: New length of one period. With $$9$$ periods a day, let $$x$$ minutes be the length of one period. Then

$$9 \times x = 360 \;\Rightarrow\; x = \dfrac{360}{9} = 40 \; \text{minutes}.$$

Check. $$9 \times 40 = 360$$ minutes ✓.

Step 1: Rate of each pump (in tanks per hour).

The small pump fills $$1$$ tank in $$3$$ hours, so in one hour it fills

$$\text{Small pump's rate} = \dfrac{1}{3} \; \text{tank/hour}.$$

The large pump fills $$1$$ tank in $$2$$ hours, so in one hour it fills

$$\text{Large pump's rate} = \dfrac{1}{2} \; \text{tank/hour}.$$

Step 2: Combined rate. When both pumps run together their outputs add:

$$\text{Combined rate} = \dfrac{1}{3} + \dfrac{1}{2} = \dfrac{2}{6} + \dfrac{3}{6} = \dfrac{5}{6} \; \text{tank/hour}.$$

Step 3: Time to fill one tank together.

$$\text{Time} = \dfrac{\text{Work}}{\text{Rate}} = \dfrac{1}{\;5/6\;} = \dfrac{6}{5} \; \text{hours}.$$

Converting: $$\dfrac{6}{5}$$ hour $$= 1\tfrac{1}{5}$$ hour $$= 1$$ hour $$12$$ minutes (since $$\tfrac{1}{5} \times 60 = 12$$ min).

Sense check. Both pumps together should finish faster than the faster pump alone: $$\tfrac{6}{5} = 1.2$$ h $$< 2$$ h ✓.

The same number of toys has to be made in both scenarios, so the total work is fixed. Assuming every machine produces at the same fixed rate, the number of machines and the number of days needed are in inverse proportion:

$$(\text{machines}) \times (\text{days}) = \text{constant}.$$

Step 1: Compute the constant.

$$42 \times 63 = 2646 \; \text{machine-days}.$$

Step 2: New number of machines. Let $$x$$ be the required number of machines to finish in $$54$$ days. Then

$$x \times 54 = 2646 \;\Rightarrow\; x = \dfrac{2646}{54} = 49.$$

So $$49$$ machines are required.

Check. $$49 \times 54 = 2646$$ ✓, and $$54 < 63$$ (fewer days), so we do need more machines ($$49 > 42$$), which agrees with inverse proportion.

Step 1: Find the distance to the destination. Using distance $$=$$ speed $$\times$$ time on the first trip,

$$\text{Distance} = 60 \; \mathrm{km/h} \times 2 \; \text{h} = 120 \; \text{km}.$$

The distance to the destination does not change.

Step 2: Compute the new time. With the new speed of $$80$$ km/h,

$$\text{Time} = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{120 \; \text{km}}{80 \; \mathrm{km/h}} = \dfrac{3}{2} \; \text{hours} = 1.5 \; \text{hours}.$$

So the car takes $$1.5$$ hours, i.e. $$1$$ hour $$30$$ minutes.

Consistency check. Speed and time are in inverse proportion for a fixed distance: $$60 \times 2 = 120 = 80 \times 1.5$$ ✓. The speed increased from $$60$$ to $$80$$, so the time should decrease — and it does, from $$2$$ to $$1.5$$ hours.