Chapter 11 Direct and Inverse Proportions

Solution:

The cost of a particular quality of cloth is directly proportional to its length. This means that as the length of the cloth increases, the cost also increases by the same factor.

Let the length of the cloth be $x$ metres and the cost of the cloth be $y$ rupees.

In the case of direct proportion, the ratio $\frac{y}{x}$ is a constant.

Given that the cost of 5 metres of cloth is $\textsf{₹}$ 210.

So, when $x = 5$ metres, $y = 210$ rupees.

The constant of proportionality, $k$, is given by:

$k = \frac{y}{x}$

$k = \frac{210}{5}$

$k = 42$

Thus, the relationship between the cost ($y$) and the length ($x$) of the cloth is $y = 42x$.

Now, we can use this relation to find the cost for the required lengths of cloth:

For a length of 2 metres:

$y = 42 \times 2 = 84$

The cost of 2 metres of cloth is $\textsf{₹}$ 84.

For a length of 4 metres:

$y = 42 \times 4 = 168$

The cost of 4 metres of cloth is $\textsf{₹}$ 168.

For a length of 10 metres:

$y = 42 \times 10 = 420$

The cost of 10 metres of cloth is $\textsf{₹}$ 420.

For a length of 13 metres:

$y = 42 \times 13$

$y = 546$

The cost of 13 metres of cloth is $\textsf{₹}$ 546.

The costs for the specified lengths are tabulated below:

Solution:

Under similar conditions (same time of day, same location), the ratio of the height of an object to the length of its shadow is constant. This is a problem involving direct proportion.

Given:

Height of electric pole ($H_p$) = 14 metres

Length of shadow of pole ($S_p$) = 10 metres

Length of shadow of tree ($S_t$) = 15 metres

To Find:

Height of the tree ($H_t$)

Solution:

Since the height of an object is directly proportional to the length of its shadow under similar conditions, the ratio $\frac{\text{Height}}{\text{Shadow}}$ is constant for both the pole and the tree.

So, we can set up the proportion:

$\frac{\text{Height of Pole}}{\text{Shadow of Pole}} = \frac{\text{Height of Tree}}{\text{Shadow of Tree}}$

Substituting the given values:

$\frac{14}{10} = \frac{H_t}{15}$

To find the value of $H_t$, we can multiply both sides of the equation by 15:

$H_t = \frac{14}{10} \times 15$

$H_t = \frac{\cancel{14}^{7}}{\cancel{10}_{2}} \times 15$

$H_t = \frac{7}{2} \times 15$

$H_t = \frac{7 \times 15}{2}$

$H_t = \frac{105}{2}$

$H_t = 52.5$

Wait, let's recheck the calculation $\frac{14}{10} \times 15$.

$H_t = \frac{14 \times 15}{10}$

$H_t = \frac{210}{10}$

$H_t = 21$

The height of the tree is 21 metres.

The height of the tree is $\mathbf{21}$ metres.

Solution:

This is a problem of direct proportion, as the weight of the paper increases with the number of sheets.

Given:

Number of sheets ($N_1$) = 12 sheets

Weight of 12 sheets ($W_1$) = 40 grams

Required weight ($W_2$) = $2\frac{1}{2}$ kilograms

To Find:

Number of sheets ($N_2$) weighing $2\frac{1}{2}$ kilograms.

Solution:

First, convert the required weight to grams so that the units are consistent.

$2\frac{1}{2}$ kg = 2.5 kg

Since 1 kilogram = 1000 grams,

$W_2 = 2.5 \text{ kg} = 2.5 \times 1000 \text{ grams} = 2500 \text{ grams}$.

Let the number of sheets weighing 2500 grams be $N_2$.

In a direct proportion, the ratio of the corresponding quantities is constant.

$\frac{\text{Number of sheets}}{\text{Weight}} = \text{constant}$

So, we have:

$\frac{N_1}{W_1} = \frac{N_2}{W_2}$

Substitute the given values:

$\frac{12}{40} = \frac{N_2}{2500}$

To find $N_2$, multiply both sides by 2500:

$N_2 = \frac{12}{40} \times 2500$

Simplify the fraction $\frac{12}{40}$ by dividing both numerator and denominator by their greatest common divisor, which is 4:

$N_2 = \frac{\cancel{12}^{3}}{\cancel{40}_{10}} \times 2500$

$N_2 = \frac{3}{10} \times 2500$

$N_2 = 3 \times \frac{2500}{10}$

$N_2 = 3 \times 250$

$N_2 = 750$

Therefore, 750 sheets of the same paper would weigh $2\frac{1}{2}$ kilograms.

The number of sheets is 750.

Solution:

Since the train is moving at a uniform speed, the distance traveled is directly proportional to the time taken.

Given:

Uniform speed of the train = 75 km/hour

This means the train travels 75 km in 1 hour.

(i) How far will it travel in 20 minutes?

First, convert the time into hours so that the units are consistent with the speed unit (km/hour).

20 minutes = $\frac{20}{60}$ hours = $\frac{1}{3}$ hours

Let the distance traveled in 20 minutes be $d$ km.

Using the relationship between distance, speed, and time: distance = speed $\times$ time.

$d = \text{speed} \times \text{time}$

$d = 75 \text{ km/hour} \times \frac{1}{3} \text{ hours}$

$d = \frac{75}{3}$ km

$d = 25$ km

Alternatively, using direct proportion: Let $t_1 = 1$ hour = 60 minutes, $d_1 = 75$ km. Let $t_2 = 20$ minutes, $d_2 = d$ km.

$\frac{d_1}{t_1} = \frac{d_2}{t_2}$

$\frac{75}{60} = \frac{d}{20}$

Multiply both sides by 20:

$d = \frac{75}{60} \times 20$

$d = \frac{75}{\cancel{60}_{3}} \times \cancel{20}^{1}$

$d = \frac{75}{3}$

$d = 25$

The train will travel $\mathbf{25}$ km in 20 minutes.

(ii) Find the time required to cover a distance of 250 km.

Let the time required to cover 250 km be $T$ hours.

Using the relationship between distance, speed, and time: time = $\frac{\text{distance}}{\text{speed}}$.

$T = \frac{250 \text{ km}}{75 \text{ km/hour}}$

$T = \frac{250}{75}$ hours

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 25:

$T = \frac{\cancel{250}^{10}}{\cancel{75}_{3}}$ hours

$T = \frac{10}{3}$ hours

We can express this in hours and minutes. $\frac{10}{3}$ hours = $3\frac{1}{3}$ hours.

$\frac{1}{3}$ hours = $\frac{1}{3} \times 60$ minutes = 20 minutes.

So, $T = 3$ hours and 20 minutes.

Alternatively, using direct proportion: Let $d_1 = 75$ km, $t_1 = 1$ hour. Let $d_2 = 250$ km, $t_2 = T$ hours.

$\frac{t_1}{d_1} = \frac{t_2}{d_2}$

$\frac{1}{75} = \frac{T}{250}$

Multiply both sides by 250:

$T = \frac{1}{75} \times 250$

$T = \frac{250}{75}$

$T = \frac{10}{3}$ hours

The time required is $\mathbf{\frac{10}{3}}$ hours or 3 hours 20 minutes.

Solution:

The scale of a map represents a direct proportion between the distance on the map and the corresponding actual distance on the ground. A scale of $1:30000000$ means that $1$ unit of distance on the map corresponds to $30000000$ units of the same distance in reality.

Given:

Map scale = $1:30000000$

Distance between two cities on the map = $4$ cm

To Find:

The actual distance between the two cities.

Solution:

According to the map scale, $1$ cm on the map represents $30000000$ cm in actual distance.

Since the cities are $4$ cm apart on the map, the actual distance will be $4$ times the distance represented by $1$ cm on the map.

Actual distance = Map distance $\times$ Scale factor

Actual distance = $4 \text{ cm} \times 30000000$

Actual distance = $120000000$ cm

The actual distance is $120000000$ cm.

We can convert this distance from centimetres to a more convenient unit like kilometres.

We know the conversion factors: $1$ metre ($m$) = $100$ centimetres ($cm$) and $1$ kilometre ($km$) = $1000$ metres ($m$).

Combining these, $1$ km = $1000 \times 100$ cm = $100000$ cm.

To convert centimetres to kilometers, we divide the number of centimetres by the number of centimetres in a kilometer ($100000$).

Actual distance in km = $\frac{120000000 \text{ cm}}{100000 \text{ cm/km}}$

Actual distance in km = $\frac{120000000}{100000}$

Cancel out the zeros from the numerator and the denominator:

Actual distance in km = $\frac{1200\cancel{00000}}{\cancel{100000}}$

Actual distance in km = $1200$

The actual distance between the two cities is $1200$ km.

The actual distance between the two cities is 1200 km.

Solution:

For two quantities to be in direct proportion, the ratio of corresponding values must be constant.

Let the parking time be $T$ (in hours) and the parking charges be $C$ (in $\textsf{₹}$). If $T$ and $C$ are in direct proportion, then $\frac{C}{T}$ should be a constant value for all given pairs of time and charges.

Given Parking Charges for different times:

• 4 hours: $\textsf{₹}$ 60
• 8 hours: $\textsf{₹}$ 100
• 12 hours: $\textsf{₹}$ 140
• 24 hours: $\textsf{₹}$ 180

Calculate the ratio $\frac{C}{T}$ for each case:

Case 1: For 4 hours, Charges = $\textsf{₹}$ 60

Ratio = $\frac{60}{4} = 15$

Case 2: For 8 hours, Charges = $\textsf{₹}$ 100

Ratio = $\frac{100}{8} = \frac{25}{2} = 12.5$

Case 3: For 12 hours, Charges = $\textsf{₹}$ 140

Ratio = $\frac{140}{12} = \frac{35}{3} \approx 11.67$

Case 4: For 24 hours, Charges = $\textsf{₹}$ 180

Ratio = $\frac{180}{24} = \frac{15}{2} = 7.5$

Comparing the ratios calculated:

$15 \neq 12.5 \neq \frac{35}{3} \neq 7.5$

Since the ratio of parking charges to the parking time is not constant for all given data points, the parking charges are not in direct proportion to the parking time.

Solution:

The problem describes a paint mixture where the parts of red pigment and the parts of base are in direct proportion.

This means that the ratio of the parts of base to the parts of red pigment is constant.

Given that 1 part of red pigment is mixed with 8 parts of base.

Let $x$ represent the parts of red pigment and $y$ represent the parts of base.

According to the problem, when $x=1$, $y=8$.

The constant of proportionality $k$ is given by $\frac{y}{x}$.

$k = \frac{8}{1} = 8$

So, the relationship between the parts of base and red pigment is $y = 8x$.

We need to find the parts of base ($y$) for the given parts of red pigment ($x$) in the table.

For $x = 4$: $y = 8 \times 4 = 32$

For $x = 7$: $y = 8 \times 7 = 56$

For $x = 12$: $y = 8 \times 12 = 96$

For $x = 20$: $y = 8 \times 20 = 160$

The completed table is as follows:

The parts of base needed are 32, 56, 96, and 160 for 4, 7, 12, and 20 parts of red pigment, respectively.

Solution:

This problem deals with the direct proportion between the amount of red pigment and the amount of base required to make a mixture.

Given:

1 part of red pigment requires 75 mL of base.

Amount of base available = 1800 mL.

To Find:

The amount of red pigment (in parts) to mix with 1800 mL of base.

Solution:

Let the number of parts of red pigment be $P$ and the amount of base in mL be $B$. Since the quantities are in direct proportion, the ratio $\frac{P}{B}$ is constant.

We are given that $P=1$ when $B=75$.

So, the constant ratio is $\frac{1}{75}$.

We need to find the value of $P$ when $B=1800$. Let this unknown number of parts be $x$.

Using the direct proportion relation:

$\frac{P_1}{B_1} = \frac{P_2}{B_2}$

Substitute the given values:

$\frac{1}{75} = \frac{x}{1800}$

To solve for $x$, multiply both sides of the equation by 1800:

$x = \frac{1}{75} \times 1800$

$x = \frac{1800}{75}$

We can simplify the fraction by cancelling common factors.

$x = \frac{\cancel{1800}^{24}}{\cancel{75}_{1}}$ (Dividing both numerator and denominator by 75)

$x = 24$

So, 24 parts of red pigment should be mixed with 1800 mL of base.

The amount of red pigment to mix is 24 parts.

Solution:

Assuming the machine works at a uniform rate, the number of bottles filled is directly proportional to the time taken.

Given:

Number of bottles filled ($B_1$) = 840 bottles

Time taken ($T_1$) = 6 hours

New time ($T_2$) = 5 hours

To Find:

Number of bottles filled ($B_2$) in 5 hours.

Solution:

In a direct proportion, the ratio of the corresponding quantities is constant.

$\frac{\text{Number of bottles}}{\text{Time}} = \text{constant}$

So, we have:

$\frac{B_1}{T_1} = \frac{B_2}{T_2}$

Substitute the given values:

$\frac{840}{6} = \frac{B_2}{5}$

To find $B_2$, multiply both sides by 5:

$B_2 = \frac{840}{6} \times 5$

Simplify the fraction $\frac{840}{6}$:

$\frac{840}{6} = \frac{\cancel{840}^{140}}{\cancel{6}_{1}} = 140$

So, $B_2 = 140 \times 5$

$B_2 = 700$

Therefore, the machine will fill 700 bottles in five hours.

The number of bottles filled in five hours is 700.

Solution:

This problem involves direct proportion because the enlarged length of the bacteria photograph is directly proportional to the number of times it is enlarged. The actual length of the bacteria is a constant value.

Given:

Enlargement Factor 1 ($F_1$) = 50,000 times

Enlarged Length at $F_1$ ($L_{enlarged,1}$) = 5 cm

Enlargement Factor 2 ($F_2$) = 20,000 times

To Find:

1. The actual length of the bacteria ($L_{actual}$).

2. The enlarged length when enlarged 20,000 times ($L_{enlarged,2}$).

Solution - Part 1: Actual length of the bacteria

The relationship between the enlarged length, actual length, and enlargement factor is:

Enlarged Length = Actual Length $\times$ Enlargement Factor

$L_{enlarged,1} = L_{actual} \times F_1$

Substitute the given values:

$5 \text{ cm} = L_{actual} \times 50000$

Solve for $L_{actual}$:

$L_{actual} = \frac{5 \text{ cm}}{50000}$

$L_{actual} = \frac{1}{10000} \text{ cm}$

We can express this in standard form and also in meters.

$L_{actual} = \frac{1}{10^4} \text{ cm} = 10^{-4} \text{ cm}$

To convert centimetres to meters, we know that 1 cm $= 10^{-2}$ m.

$L_{actual} = 10^{-4} \times 10^{-2} \text{ m}$

$L_{actual} = 10^{-4 + (-2)} \text{ m}$

$L_{actual} = 10^{-6} \text{ m}$

In standard form, the actual length is $1 \times 10^{-6}$ m.

Note that $10^{-6}$ meters is equal to 1 micrometer ($\mu$m).

The actual length of the bacteria is $\mathbf{10^{-4}}$ cm or $\mathbf{1 \times 10^{-6}}$ m (or 1 micrometer).

Solution - Part 2: Enlarged length at 20,000 times enlargement

We can use the actual length calculated in Part 1 and the new enlargement factor $F_2 = 20000$.

$L_{enlarged,2} = L_{actual} \times F_2$

$L_{enlarged,2} = \frac{1}{10000} \text{ cm} \times 20000$

$L_{enlarged,2} = \frac{20000}{10000} \text{ cm}$

$L_{enlarged,2} = 2 \text{ cm}$

Alternatively, using direct proportion $\frac{L_{enlarged}}{F} = \text{constant}$:

$\frac{L_{enlarged,1}}{F_1} = \frac{L_{enlarged,2}}{F_2}$

$\frac{5}{50000} = \frac{L_{enlarged,2}}{20000}$

Multiply both sides by 20000 to solve for $L_{enlarged,2}$:

$L_{enlarged,2} = \frac{5}{50000} \times 20000$

$L_{enlarged,2} = 5 \times \frac{20000}{50000}$

$L_{enlarged,2} = 5 \times \frac{2}{5}$

$L_{enlarged,2} = \cancel{5} \times \frac{2}{\cancel{5}}$

$L_{enlarged,2} = 2$

The enlarged length would be 2 cm.

The enlarged length when the photograph is enlarged 20,000 times is 2 cm.

Solution:

This problem involves the concept of scale. The dimensions of the model ship are in direct proportion to the corresponding dimensions of the actual ship.

Given:

Height of the mast in the model ($H_{mast, model}$) = 9 cm

Height of the mast in the actual ship ($H_{mast, actual}$) = 12 m

Length of the actual ship ($L_{actual}$) = 28 m

To Find:

Length of the model ship ($L_{model}$).

Solution:

First, determine the scale factor of the model. The scale factor is the ratio of a dimension in the model to the corresponding dimension in the actual ship.

We are given the heights of the masts for both the model and the actual ship. Let's use these to find the scale factor.

It's important to have consistent units. Let's convert the actual mast height from meters to centimetres:

$H_{mast, actual} = 12 \text{ m} = 12 \times 100 \text{ cm} = 1200 \text{ cm}$.

The scale factor ($k$) can be found by taking the ratio of the model mast height to the actual mast height:

$k = \frac{H_{mast, model}}{H_{mast, actual}}$

$k = \frac{9 \text{ cm}}{1200 \text{ cm}}$

$k = \frac{9}{1200}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 3:

$k = \frac{\cancel{9}^{3}}{\cancel{1200}_{400}}$

$k = \frac{3}{400}$

The scale of the model is $3:400$. This means that every 3 units on the model represent 400 units on the actual ship.

Now we can use this scale factor to find the length of the model ship.

Let $L_{model}$ be the length of the model ship. We are given the length of the actual ship, $L_{actual} = 28$ m.

We can set up a proportion using the scale factor:

$\frac{L_{model}}{L_{actual}} = k$

$\frac{L_{model}}{28 \text{ m}} = \frac{3}{400}$

To find $L_{model}$, multiply both sides by $28$ m:

$L_{model} = \frac{3}{400} \times 28 \text{ m}$

$L_{model} = \frac{3 \times 28}{400} \text{ m}$

$L_{model} = \frac{84}{400} \text{ m}$

We can simplify the fraction $\frac{84}{400}$ by dividing the numerator and denominator by their greatest common divisor, which is 4:

$L_{model} = \frac{\cancel{84}^{21}}{\cancel{400}_{100}} \text{ m}$

$L_{model} = \frac{21}{100} \text{ m}$

$L_{model} = 0.21 \text{ m}$

It is standard to express model dimensions in centimetres, so let's convert the length from meters to centimetres:

$L_{model} = 0.21 \text{ m} \times 100 \text{ cm/m}$

$L_{model} = 21 \text{ cm}$

The length of the model ship is $\mathbf{21}$ cm.

Solution:

The number of sugar crystals is directly proportional to the weight of the sugar.

Given:

Weight of sugar ($W_1$) = 2 kg

Number of crystals in $W_1$ ($N_1$) = $9 \times 10^6$ crystals

To Find:

(i) Number of crystals in 5 kg of sugar.

(ii) Number of crystals in 1.2 kg of sugar.

Solution:

In a direct proportion, the ratio of the number of crystals to the weight of sugar is constant.

$\frac{\text{Number of crystals}}{\text{Weight of sugar}} = \text{constant}$

So, for any two weights and their corresponding number of crystals, we have:

$\frac{N_1}{W_1} = \frac{N_2}{W_2}$

(i) Find the number of sugar crystals in 5 kg of sugar.

Let the number of crystals in 5 kg of sugar be $N_{5kg}$.

Here, $W_1 = 2$ kg, $N_1 = 9 \times 10^6$, and $W_2 = 5$ kg.

Using the proportion:

$\frac{9 \times 10^6}{2} = \frac{N_{5kg}}{5}$

To find $N_{5kg}$, multiply both sides by 5:

$N_{5kg} = \frac{9 \times 10^6}{2} \times 5$

$N_{5kg} = \frac{9 \times 5}{2} \times 10^6$

$N_{5kg} = \frac{45}{2} \times 10^6$

$N_{5kg} = 22.5 \times 10^6$

We can express this in standard form. Move the decimal point one place to the left to get 2.25. This means we multiply by $10^1$.

$N_{5kg} = 2.25 \times 10^1 \times 10^6$

Using the law $a^m \times a^n = a^{m+n}$:

$N_{5kg} = 2.25 \times 10^{1+6}$

$N_{5kg} = 2.25 \times 10^{7}$

The number of sugar crystals in 5 kg of sugar is $\mathbf{2.25 \times 10^7}$.

(ii) Find the number of sugar crystals in 1.2 kg of sugar.

Let the number of crystals in 1.2 kg of sugar be $N_{1.2kg}$.

Here, $W_1 = 2$ kg, $N_1 = 9 \times 10^6$, and $W_2 = 1.2$ kg.

Using the proportion:

$\frac{9 \times 10^6}{2} = \frac{N_{1.2kg}}{1.2}$

To find $N_{1.2kg}$, multiply both sides by 1.2:

$N_{1.2kg} = \frac{9 \times 10^6}{2} \times 1.2$

$N_{1.2kg} = 9 \times 10^6 \times \frac{1.2}{2}$

$N_{1.2kg} = 9 \times 10^6 \times 0.6$

$N_{1.2kg} = (9 \times 0.6) \times 10^6$

$N_{1.2kg} = 5.4 \times 10^6$

The number of sugar crystals in 1.2 kg of sugar is $\mathbf{5.4 \times 10^6}$.

Solution:

The map scale represents a direct proportion between the distance on the map and the corresponding actual distance on the ground.

Given:

Map scale: 1 cm on the map represents 18 km in actual distance.

Actual distance covered = 72 km

To Find:

The distance covered on the map.

Solution:

Let the distance on the map be $d_{map}$ (in cm) and the actual distance be $d_{actual}$ (in km).

According to the scale, the ratio $\frac{d_{map}}{d_{actual}}$ is constant.

From the scale, we have $\frac{1 \text{ cm}}{18 \text{ km}}$. This is our constant ratio.

We are given an actual distance of 72 km, and we need to find the corresponding distance on the map. Let this distance be $x$ cm.

Using the direct proportion:

$\frac{d_{map,1}}{d_{actual,1}} = \frac{d_{map,2}}{d_{actual,2}}$

Substitute the given values:

$\frac{1 \text{ cm}}{18 \text{ km}} = \frac{x \text{ cm}}{72 \text{ km}}$

To find $x$, multiply both sides by 72:

$x = \frac{1}{18} \times 72$

$x = \frac{72}{18}$

Divide 72 by 18:

$x = 4$

So, the distance covered on the map is 4 cm.

The distance covered in the map is 4 cm.

Solution:

Under similar conditions (same time of day), the ratio of the height of an object to the length of its shadow is constant. This is a problem involving direct proportion.

Given initial pole and shadow:

Height of the pole ($H_1$) = 5 m 60 cm

Length of the shadow ($S_1$) = 3 m 20 cm

First, convert all measurements to the same unit, let's use centimetres.

1 metre = 100 centimetres

$H_1 = 5 \text{ m} + 60 \text{ cm} = 5 \times 100 \text{ cm} + 60 \text{ cm} = 500 \text{ cm} + 60 \text{ cm} = 560 \text{ cm}$

$S_1 = 3 \text{ m} + 20 \text{ cm} = 3 \times 100 \text{ cm} + 20 \text{ cm} = 300 \text{ cm} + 20 \text{ cm} = 320 \text{ cm}$

The ratio of height to shadow length for the first pole is $\frac{H_1}{S_1} = \frac{560}{320}$.

Simplify this ratio:

$\frac{560}{320} = \frac{56}{32}$ (Dividing by 10)

$\frac{56}{32} = \frac{\cancel{56}^{7}}{\cancel{32}_{4}}$ (Dividing by 8)

The constant ratio is $\frac{7}{4}$.

So, $\frac{\text{Height}}{\text{Shadow}} = \frac{7}{4}$.

(i) Find the length of the shadow cast by another pole 10 m 50 cm high.

Let the height of the second pole be $H_2$ and the length of its shadow be $S_2$.

$H_2 = 10 \text{ m} + 50 \text{ cm} = 10 \times 100 \text{ cm} + 50 \text{ cm} = 1000 \text{ cm} + 50 \text{ cm} = 1050 \text{ cm}$

Using the constant ratio:

$\frac{H_2}{S_2} = \frac{7}{4}$

$\frac{1050}{S_2} = \frac{7}{4}$

To solve for $S_2$, we can cross-multiply:

$7 \times S_2 = 1050 \times 4$

$7 S_2 = 4200$

Divide by 7:

$S_2 = \frac{4200}{7}$

$S_2 = 600 \text{ cm}$

Convert the shadow length back to meters and centimetres:

$600 \text{ cm} = 6 \text{ m}$

The length of the shadow cast by the 10 m 50 cm pole is $\mathbf{6}$ m.

(ii) Find the height of a pole which casts a shadow 5 m long.

Let the height of this pole be $H_3$ and the length of its shadow be $S_3$.

$S_3 = 5 \text{ m} = 5 \times 100 \text{ cm} = 500 \text{ cm}$

Using the constant ratio:

$\frac{H_3}{S_3} = \frac{7}{4}$

$\frac{H_3}{500} = \frac{7}{4}$

To solve for $H_3$, multiply both sides by 500:

$H_3 = \frac{7}{4} \times 500 \text{ cm}$

$H_3 = 7 \times \frac{500}{4} \text{ cm}$

$H_3 = 7 \times 125 \text{ cm}$

$H_3 = 875 \text{ cm}$

Convert the height back to meters and centimetres:

$875 \text{ cm} = 800 \text{ cm} + 75 \text{ cm} = 8 \text{ m} + 75 \text{ cm}$

The height of the pole is 8 m 75 cm.

The height of the pole is 8 m 75 cm.

Solution:

This is a problem of direct proportion because the distance traveled at a uniform speed is directly proportional to the time taken. If the speed is constant, as time increases, the distance traveled also increases proportionally.

Given:

Distance traveled ($d_1$) = 14 km

Time taken ($t_1$) = 25 minutes

New time ($t_2$) = 5 hours

To Find:

Distance traveled ($d_2$) in 5 hours.

Solution:

First, we need to ensure that the units of time are consistent. The speed is constant, so the ratio of distance to time is constant. We have time in minutes and hours, so we convert 5 hours into minutes.

1 hour = 60 minutes

5 hours = $5 \times 60$ minutes = 300 minutes.

So, $t_2 = 300$ minutes.

Since the distance is directly proportional to the time at a constant speed, the ratio of distance to time is constant:

$\frac{\text{Distance}_1}{\text{Time}_1} = \frac{\text{Distance}_2}{\text{Time}_2}$

Substitute the known values:

$\frac{14 \text{ km}}{25 \text{ minutes}} = \frac{d_2 \text{ km}}{300 \text{ minutes}}$

To find $d_2$, multiply both sides of the equation by 300:

$d_2 = \frac{14}{25} \times 300$

Calculate the value by simplifying the fraction:

$d_2 = 14 \times \frac{\cancel{300}^{12}}{\cancel{25}_{1}}$

$d_2 = 14 \times 12$

$d_2 = 168$

So, the truck can travel 168 km in 5 hours.

The distance the truck can travel in 5 hours is 168 km.

Solution:

This is a problem of inverse proportion. The number of pipes and the time taken to fill a tank are inversely proportional. This means that if you increase the number of pipes, the time taken will decrease, and if you decrease the number of pipes, the time taken will increase, assuming all pipes are of the same type and flow rate.

Given:

Number of pipes ($N_1$) = 6 pipes

Time taken by $N_1$ pipes ($T_1$) = 1 hour 20 minutes

New number of pipes ($N_2$) = 5 pipes

To Find:

Time taken by $N_2$ pipes ($T_2$).

Solution:

First, convert the initial time into a single unit, such as minutes.

1 hour = 60 minutes

$T_1 = 1 \text{ hour } + 20 \text{ minutes} = 60 \text{ minutes } + 20 \text{ minutes} = 80 \text{ minutes}$

In inverse proportion, the product of the number of items and the time taken is constant.

$N_1 \times T_1 = N_2 \times T_2$

Substitute the given values into the equation:

$6 \text{ pipes} \times 80 \text{ minutes} = 5 \text{ pipes} \times T_2 \text{ minutes}$

$480 = 5 \times T_2$

To find $T_2$, divide both sides by 5:

$T_2 = \frac{480}{5}$

$T_2 = 96$ minutes

The time taken is 96 minutes.

We can convert this back to hours and minutes.

96 minutes = 60 minutes + 36 minutes = 1 hour 36 minutes.

It will take 1 hour 36 minutes for 5 pipes to fill the tank.

Solution:

This is a problem of inverse proportion. The number of students and the duration for which the food provisions will last are inversely proportional. If the number of students increases, the provisions will last for fewer days, and vice versa, assuming the consumption rate per student is constant.

Given:

Initial number of students ($N_1$) = 100 students

Duration of provisions for $N_1$ students ($D_1$) = 20 days

Number of students joining = 25 students

To Find:

The duration for which the provisions will last if 25 more students join ($D_2$).

Solution:

Calculate the new number of students ($N_2$).

$N_2 = \text{Initial number of students} + \text{Number of students joining}$

$N_2 = 100 + 25 = 125$ students

In inverse proportion, the product of the number of students and the duration of provisions is constant.

$N_1 \times D_1 = N_2 \times D_2$

Substitute the given values into the equation:

$100 \text{ students} \times 20 \text{ days} = 125 \text{ students} \times D_2 \text{ days}$

$2000 = 125 \times D_2$

To find $D_2$, divide both sides by 125:

$D_2 = \frac{2000}{125}$

We can simplify this fraction. Divide both numerator and denominator by 25:

$D_2 = \frac{\cancel{2000}^{80}}{\cancel{125}_{5}}$

Now divide 80 by 5:

$D_2 = \frac{80}{5} = 16$

The provisions will last for 16 days.

The provisions will last for 16 days if 25 more students join the group.

Solution:

This is a problem of inverse proportion. The number of workers and the time taken to build a wall are inversely proportional. If you increase the number of workers, the time taken will decrease, and vice versa, assuming all workers work at the same rate.

Given:

Number of workers ($W_1$) = 15 workers

Time taken by $W_1$ workers ($T_1$) = 48 hours

New time required ($T_2$) = 30 hours

To Find:

Number of workers required to complete the work in 30 hours ($W_2$).

Solution:

In inverse proportion, the product of the number of workers and the time taken to complete the same amount of work is constant.

$W_1 \times T_1 = W_2 \times T_2$

Substitute the given values into the equation:

$15 \text{ workers} \times 48 \text{ hours} = W_2 \text{ workers} \times 30 \text{ hours}$

$15 \times 48 = W_2 \times 30$

To find $W_2$, divide both sides by 30:

$W_2 = \frac{15 \times 48}{30}$

Simplify the expression by cancelling common factors.

$W_2 = \frac{\cancel{15}^{1} \times 48}{\cancel{30}_{2}}$

$W_2 = \frac{48}{2}$

$W_2 = 24$

Therefore, 24 workers will be required to do the same work in 30 hours.

The number of workers required is 24.

Solution:

Two quantities are in inverse proportion if the product of their corresponding values is constant. This means that as one quantity increases, the other quantity decreases proportionally.

Let's analyse each case:

(i) The number of workers on a job and the time to complete the job.

Assuming all workers work at the same rate and the total amount of work is fixed, if you increase the number of workers, the time taken to complete the job will decrease. If you decrease the number of workers, the time taken will increase. The product of the number of workers and the time taken (which represents the total work done) remains constant.

Therefore, this is an example of inverse proportion.

(ii) The time taken for a journey and the distance travelled in a uniform speed.

Assuming the speed is uniform and positive, if you increase the time taken, the distance travelled will also increase proportionally. If you decrease the time, the distance will decrease. The ratio of distance to time is constant (which is the speed).

Therefore, this is an example of direct proportion.

(iii) Area of cultivated land and the crop harvested.

Assuming the yield per unit area is constant and other factors like weather, soil quality, etc., are the same, if you increase the area of cultivated land, the amount of crop harvested will also increase proportionally. If you decrease the area, the crop harvested will decrease.

Therefore, this is an example of direct proportion.

(iv) The time taken for a fixed journey and the speed of the vehicle.

Assuming the distance of the journey is fixed and the speed is positive, if you increase the speed of the vehicle, the time taken to cover the fixed distance will decrease. If you decrease the speed, the time taken will increase. The product of speed and time is constant (which is the fixed distance).

Therefore, this is an example of inverse proportion.

(v) The population of a country and the area of land per person.

Assuming the total area of the country is fixed, if the population increases, the area of land available per person will decrease. If the population decreases, the area of land per person will increase. The product of the population and the area of land per person is constant (which is the total area of the country).

Therefore, this is an example of inverse proportion.

The following are in inverse proportion:

(i) The number of workers on a job and the time to complete the job.

(iv) The time taken for a fixed journey and the speed of the vehicle.

(v) The population of a country and the area of land per person.

Solution:

The total prize money is fixed at $\textsf{₹}$ 1,00,000, and it is to be divided equally amongst the winners. Let the number of winners be $N$ and the prize money for each winner be $P$. The total prize money is the product of the number of winners and the prize money for each winner, i.e., $N \times P$.

Given that the total prize money is $\textsf{₹}$ 1,00,000.

So, $N \times P = 1,00,000$.

This equation shows that the product of the number of winners and the prize money per winner is a constant (1,00,000). This is the definition of inverse proportion.

Therefore, the prize money given to an individual winner is inversely proportional to the number of winners.

To complete the table, we use the relationship $N \times P = 1,00,000$, or $P = \frac{100000}{N}$.

For $N=1$, $P = \frac{100000}{1} = 100000$. (Given)

For $N=2$, $P = \frac{100000}{2} = 50000$. (Given)

For $N=4$, $P = \frac{100000}{4} = 25000$.

For $N=5$, $P = \frac{100000}{5} = 20000$.

For $N=8$, $P = \frac{100000}{8} = 12500$.

For $N=10$, $P = \frac{100000}{10} = 10000$.

For $N=20$, $P = \frac{100000}{20} = 5000$.

The completed table is as follows:

The prize money given to an individual winner is inversely proportional to the number of winners.

Solution:

When spokes are fixed in a wheel such that the angles between consecutive spokes are equal, the total angle around the center is $360^\circ$. If there are $N$ spokes, there are $N$ equal angles between consecutive spokes. Let $A$ be the angle between a pair of consecutive spokes.

The relationship is $N \times A = 360^\circ$.

Since the product $N \times A$ is a constant ($360^\circ$), the number of spokes ($N$) and the angle between a pair of consecutive spokes ($A$) are in inverse proportion.

(i) Check if the number of spokes and the angles are in inverse proportion.

We calculate the product of the number of spokes and the angle for the given values in the table:

For 4 spokes: $4 \times 90^\circ = 360^\circ$

For 6 spokes: $6 \times 60^\circ = 360^\circ$

Since the product is constant ($360^\circ$), the number of spokes and the angles formed between the pairs of consecutive spokes are indeed in inverse proportion.

Complete the table:

The relationship is $N \times A = 360^\circ$, so $A = \frac{360^\circ}{N}$.

For $N=8$: $A = \frac{360^\circ}{8} = 45^\circ$

For $N=10$: $A = \frac{360^\circ}{10} = 36^\circ$

For $N=12$: $A = \frac{360^\circ}{12} = 30^\circ$

The completed table is:

(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.

Let $N = 15$. We need to find the angle $A$ when $N=15$.

Using the inverse proportion relationship $N \times A = 360^\circ$:

$15 \times A = 360^\circ$

$A = \frac{360^\circ}{15}$

$A = 24^\circ$

The angle between a pair of consecutive spokes on a wheel with 15 spokes is 24$^\circ$.

(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?

Let $A = 40^\circ$. We need to find the number of spokes $N$ when $A=40^\circ$.

Using the inverse proportion relationship $N \times A = 360^\circ$:

$N \times 40^\circ = 360^\circ$

$N = \frac{360^\circ}{40^\circ}$

$N = 9$

9 spokes would be needed if the angle between a pair of consecutive spokes is 40°.

The number of spokes needed is 9.

Solution:

The total number of sweets in the box is fixed. When the sweets are divided equally among the children, the number of sweets each child gets is inversely proportional to the number of children. If the number of children decreases, the number of sweets each child gets will increase, and vice versa.

Given:

Initial number of children ($C_1$) = 24 children

Number of sweets each child gets ($S_1$) = 5 sweets

Number of children reduced by = 4

To Find:

Number of sweets each child would get if the number of children is reduced by 4 ($S_2$).

Solution:

First, calculate the total number of sweets in the box.

Total number of sweets = Number of children $\times$ Sweets per child

Total sweets = $24 \times 5 = 120$ sweets

This total number of sweets remains constant.

Next, calculate the new number of children ($C_2$).

New number of children = Initial number of children - Number of children reduced

$C_2 = 24 - 4 = 20$ children

Now, divide the total number of sweets equally among the new number of children to find the number of sweets each child gets ($S_2$).

Number of sweets per child ($S_2$) = $\frac{\text{Total number of sweets}}{\text{New number of children}}$

$S_2 = \frac{120}{20}$

$S_2 = 6$ sweets

Alternatively, using inverse proportion: The product of the number of children and the number of sweets per child is constant (equal to the total number of sweets).

$C_1 \times S_1 = C_2 \times S_2$

Substitute the known values:

$24 \times 5 = (24-4) \times S_2$

$120 = 20 \times S_2$

To find $S_2$, divide both sides by 20:

$S_2 = \frac{120}{20}$

$S_2 = 6$

If the number of children is reduced by 4, each child would get 6 sweets.

Each child would get 6 sweets.

Solution:

This is a problem of inverse proportion. The number of animals and the duration for which the food will last are inversely proportional, assuming the consumption rate per animal is constant and the total amount of food is fixed. As the number of animals increases, the food will last for fewer days.

Given:

Initial number of animals ($N_1$) = 20 animals

Duration the food lasts for $N_1$ animals ($D_1$) = 6 days

Number of additional animals = 10 animals

To Find:

The duration for which the food will last if there are 10 more animals ($D_2$).

Solution:

First, calculate the new number of animals ($N_2$).

New number of animals = Initial number of animals + Number of additional animals

$N_2 = 20 + 10 = 30$ animals

In an inverse proportion, the product of the number of items and the duration is constant.

$N_1 \times D_1 = N_2 \times D_2$

Substitute the given values into the equation:

$20 \text{ animals} \times 6 \text{ days} = 30 \text{ animals} \times D_2 \text{ days}$

$120 = 30 \times D_2$

To find $D_2$, divide both sides by 30:

$D_2 = \frac{120}{30}$

$D_2 = \frac{12}{3} = 4$

The food would last for 4 days if there were 10 more animals.

The food would last for 4 days.

Solution:

This is a problem of inverse proportion. The number of persons working on a job and the time taken to complete the job are inversely proportional, assuming all persons work at the same rate and the total amount of work is fixed. If the number of persons increases, the time taken will decrease.

Given:

Initial number of persons ($P_1$) = 3 persons

Time taken by $P_1$ persons ($D_1$) = 4 days

New number of persons ($P_2$) = 4 persons

To Find:

Time taken by $P_2$ persons ($D_2$).

Solution:

In inverse proportion, the product of the number of persons and the time taken to complete the same work is constant.

$P_1 \times D_1 = P_2 \times D_2$

Substitute the given values into the equation:

$3 \text{ persons} \times 4 \text{ days} = 4 \text{ persons} \times D_2 \text{ days}$

$12 = 4 \times D_2$

To find $D_2$, divide both sides by 4:

$D_2 = \frac{12}{4}$

$D_2 = 3$

Therefore, if the contractor uses 4 persons instead of 3, they should take 3 days to complete the job.

They should take 3 days to complete the job.

Solution:

This problem involves the concept of inverse proportion. The total number of bottles in the batch is fixed. If you increase the number of bottles in each box, the number of boxes required to pack the same batch will decrease, and vice versa.

Given:

Initial number of boxes ($B_1$) = 25 boxes

Number of bottles per box in the initial packing ($N_1$) = 12 bottles

New number of bottles per box ($N_2$) = 20 bottles

To Find:

The number of boxes that would be filled with 20 bottles in each box ($B_2$).

Solution:

First, calculate the total number of bottles in the batch. This quantity is constant regardless of how the bottles are packed.

Total number of bottles = Number of boxes $\times$ Bottles per box

Total bottles = $25 \times 12 = 300$ bottles

Now, we need to find how many boxes are needed if 20 bottles are packed in each box. Let the number of boxes be $B_2$.

Total number of bottles = New number of boxes $\times$ New bottles per box

$300 = B_2 \times 20$

To find $B_2$, divide both sides by 20:

$B_2 = \frac{300}{20}$

$B_2 = \frac{30}{2} = 15$

So, 15 boxes would be filled if the same batch is packed using 20 bottles in each box.

Alternatively, using inverse proportion: The product of the number of boxes and the number of bottles per box is constant (equal to the total number of bottles).

$B_1 \times N_1 = B_2 \times N_2$

Substitute the known values:

$25 \times 12 = B_2 \times 20$

$300 = B_2 \times 20$

Divide by 20:

$B_2 = \frac{300}{20}$

$B_2 = 15$

Therefore, 15 boxes would be filled.

The number of boxes that would be filled is 15.

Solution:

This is a problem of inverse proportion. The number of machines working at the same rate and the time taken to produce a fixed number of articles are inversely proportional. If you increase the number of machines, the time taken will decrease, and vice versa.

Given:

Initial number of machines ($M_1$) = 42 machines

Time taken by $M_1$ machines ($D_1$) = 63 days

New time required ($D_2$) = 54 days

To Find:

Number of machines required to produce the same number of articles in 54 days ($M_2$).

Solution:

The total amount of work (producing the given number of articles) is constant. The work done is proportional to the number of machines multiplied by the number of days they work.

In inverse proportion, the product of the number of machines and the time taken is constant for the same amount of work.

$M_1 \times D_1 = M_2 \times D_2$

Substitute the given values into the equation:

$42 \text{ machines} \times 63 \text{ days} = M_2 \text{ machines} \times 54 \text{ days}$

$42 \times 63 = M_2 \times 54$

To find $M_2$, divide both sides by 54:

$M_2 = \frac{42 \times 63}{54}$

Simplify the expression by cancelling common factors.

We can divide 42 and 54 by their greatest common divisor, which is 6:

$M_2 = \frac{\cancel{42}^{7} \times 63}{\cancel{54}_{9}}$

$M_2 = \frac{7 \times 63}{9}$

Now, divide 63 by 9:

$M_2 = 7 \times \frac{\cancel{63}^{7}}{\cancel{9}_{1}}$

$M_2 = 7 \times 7$

$M_2 = 49$

Therefore, 49 machines would be required to produce the same number of articles in 54 days.

The number of machines required is 49.

Solution:

This problem involves the concept of inverse proportion. For a fixed distance (the distance to the destination), the speed of the vehicle and the time taken to cover that distance are inversely proportional. If the speed increases, the time taken will decrease, and vice versa.

Given:

Initial speed ($S_1$) = 60 km/h

Time taken at $S_1$ ($T_1$) = 2 hours

New speed ($S_2$) = 80 km/h

To Find:

Time taken at $S_2$ ($T_2$).

Solution:

First, calculate the distance to the destination. Since the distance is fixed, it will be the same in both cases.

Distance = Speed $\times$ Time

Distance = $60 \text{ km/h} \times 2 \text{ hours} = 120 \text{ km}$

The distance to the destination is 120 km.

Now, we need to find the time taken to cover the same distance (120 km) at a new speed of 80 km/h. Let the time taken be $T_2$ hours.

Distance = New Speed $\times$ New Time

$120 \text{ km} = 80 \text{ km/h} \times T_2 \text{ hours}$

To find $T_2$, divide both sides by 80:

$T_2 = \frac{120}{80}$ hours

Simplify the fraction by cancelling common factors:

$T_2 = \frac{12}{8}$ hours (Dividing by 10)

$T_2 = \frac{3}{2}$ hours (Dividing by 4)

So, $T_2 = \frac{3}{2} = 1.5$ hours.

This can also be expressed as 1 hour and 0.5 hours. 0.5 hours = $0.5 \times 60$ minutes = 30 minutes.

So, $T_2 = 1$ hour 30 minutes.

Alternatively, using inverse proportion: The product of speed and time for a fixed distance is constant.

$S_1 \times T_1 = S_2 \times T_2$

Substitute the known values:

$60 \text{ km/h} \times 2 \text{ hours} = 80 \text{ km/h} \times T_2 \text{ hours}$

$120 = 80 \times T_2$

Divide by 80:

$T_2 = \frac{120}{80}$

$T_2 = \frac{3}{2}$ hours

Therefore, it will take $\frac{3}{2}$ hours or 1 hour 30 minutes.

It will take $\frac{3}{2}$ hours or 1 hour 30 minutes.

Solution:

This problem involves inverse proportion. The number of persons working on a job and the time taken to complete the job are inversely proportional, assuming all persons work at the same rate and the total amount of work is fixed.

Given:

Initial number of persons ($P_1$) = 2 persons

Time taken by $P_1$ persons ($D_1$) = 3 days

In inverse proportion, the product of the number of persons and the time taken is constant for the same amount of work.

$P_1 \times D_1 = P_2 \times D_2 = \text{Constant work units}$

The total work units = $2 \text{ persons} \times 3 \text{ days} = 6 \text{ person-days}$.

(i) One of the persons fell ill before the work started. How long would the job take now?

If one person fell ill, the new number of persons ($P_2$) is $2 - 1 = 1$ person.

Let the time taken by 1 person be $D_2$ days.

Using the inverse proportion relationship:

$P_1 \times D_1 = P_2 \times D_2$

$2 \times 3 = 1 \times D_2$

$6 = D_2$

The job would take 6 days if only 1 person works.

The job would take 6 days.

(ii) How many persons would be needed to fit the windows in one day?

We want the time taken ($D_3$) to be 1 day.

Let the number of persons needed be $P_3$.

Using the inverse proportion relationship:

$P_1 \times D_1 = P_3 \times D_3$

$2 \times 3 = P_3 \times 1$

$6 = P_3$

So, 6 persons would be needed to fit the windows in one day.

The number of persons needed is 6.

Solution:

This problem assumes that the total duration of school hours per day is fixed. The duration of each period is inversely proportional to the number of periods in the day. If the number of periods increases, the duration of each period must decrease to maintain the same total school time.

Given:

Initial number of periods per day ($N_1$) = 8 periods

Duration of each period for $N_1$ periods ($D_1$) = 45 minutes

New number of periods per day ($N_2$) = 9 periods

Total school hours per day remain the same.

To Find:

Duration of each period if there are 9 periods a day ($D_2$).

Solution:

The total duration of school hours is the product of the number of periods and the duration of each period. Since the total duration is constant, we have the inverse proportion relationship:

$N_1 \times D_1 = N_2 \times D_2$

Substitute the given values into the equation:

$8 \text{ periods} \times 45 \text{ minutes} = 9 \text{ periods} \times D_2 \text{ minutes}$

$8 \times 45 = 9 \times D_2$

Calculate the total duration on the left side:

$360 = 9 \times D_2$

To find $D_2$, divide both sides by 9:

$D_2 = \frac{360}{9}$

$D_2 = 40$

The duration of each period would be 40 minutes if the school has 9 periods a day.

Each period would be 40 minutes long.