11.2-Direct Proportion

11.2-Direct Proportion Important Formulae

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11.2 - Direct Proportion
  • In direct proportion, as one quantity increases, the other also increases at a constant rate.
  • Mathematically, if two quantities, $x$ and $y$, are in direct proportion, then $y \propto x$ or $y = kx$, where $k$ is a constant.
  • The constant $k$ is called the constant of proportionality.
  • If $x_1, y_1$ and $x_2, y_2$ are two pairs of values in direct proportion, then $ \frac{y_1}{x_1} = \frac{y_2}{x_2}$.
  • Example: If $y = 3x$, when $x = 2$, $y = 6$.

11.2-Direct Proportion

In this subtopic, we will understand the concept of Direct Proportion, one of the important relationships in mathematics, particularly in the context of ratios and proportions.

Definition: When two quantities increase or decrease in such a way that they always maintain the same ratio, they are said to be in direct proportion. In other words, the ratio between two variables remains constant. This means that as one quantity increases, the other also increases in the same proportion, or as one quantity decreases, the other decreases in the same proportion.

Mathematical Representation: If two quantities, say $x$ and $y$, are in direct proportion, we can express this relationship as:

x \propto y or y = kx

Where $k$ is a constant, called the constant of proportionality. This means that the ratio of $x$ to $y$ is always equal to $k$. Therefore, we can write the equation as:

\frac{x}{y} = k

Example 1: If $x = 2$ and $y = 4$, and we know that $x$ and $y$ are in direct proportion, we can find the constant $k$ as:

k = \frac{x}{y} = \frac{2}{4} = 0.5

Now, if $x = 6$, we can use the formula $y = kx$ to find the value of $y$:

y = 0.5 \times 6 = 3

Properties of Direct Proportion:

  • The ratio of the two quantities is always constant.
  • If one quantity increases, the other increases proportionally.
  • If one quantity decreases, the other decreases proportionally.
  • The graph of two quantities in direct proportion is a straight line passing through the origin (0,0).

Example 2: If the cost of 5 apples is ₹50, we can say the cost of apples is in direct proportion to the number of apples. Let the cost of 1 apple be $C$, then:

Cost = k \times \text{number of apples}

We know that when the number of apples is 5, the cost is ₹50, so:

50 = k \times 5 \Rightarrow k = \frac{50}{5} = 10

Thus, the cost of 1 apple is ₹10. Therefore, the cost of $n$ apples would be:

Cost = 10n

Formulae for Direct Proportion:

  • If $x$ and $y$ are in direct proportion, then $y = kx$.
  • The constant of proportionality is $k = \frac{x}{y}$.
  • If $x_1$ and $y_1$ are related by direct proportion, and $x_2$ and $y_2$ are another set of values, then:
    • x_1 / y_1 = x_2 / y_2
    • x_1 y_2 = x_2 y_1

Direct proportion is a fundamental concept used in solving problems related to speed, cost, time, and many other areas of daily life. The key takeaway is that the relationship between the quantities in direct proportion is always represented by a constant ratio.

11.2 - Direct Proportion

प्रत्यक्ष अनुपात (Direct Proportion) तब होता है जब दो मात्राएँ एक दूसरे के साथ सीधे संबंध में होती हैं, यानी एक राशि के बढ़ने पर दूसरी राशि भी बढ़ती है, और एक राशि के घटने पर दूसरी राशि भी घटती है। इसे "सिधा अनुपात" भी कहा जाता है।

यदि दो राशियाँ $x$ और $y$ प्रत्यक्ष अनुपात में हैं, तो यह निम्नलिखित तरीके से व्यक्त किया जाता है:

अर्थात, यदि $x$ और $y$ प्रत्यक्ष अनुपात में हैं, तो:

$x \propto y$ या $\frac{x}{y} = k$

जहाँ $k$ एक स्थिरांक (constant) होता है, जिसे अनुपात का स्थिरांक कहा जाता है।

इसका मतलब है कि $x$ और $y$ के बीच अनुपात हमेशा स्थिर रहता है।

यदि $x_1$ और $x_2$ दो विभिन्न मान हों और $y_1$ और $y_2$ उनके संबंधित मान हों, तो:

$\frac{x_1}{y_1} = \frac{x_2}{y_2}$

यह नियम हमें यह बताता है कि दो राशियाँ हमेशा एक दूसरे के साथ प्रत्यक्ष अनुपात में रहती हैं, अगर उनका अनुपात एक स्थिरांक के बराबर होता है।

उदाहरण: मान लीजिए, एक व्यक्ति 3 घंटे में 180 किलोमीटर की यात्रा करता है। अगर हम जानना चाहते हैं कि वह 5 घंटे में कितनी दूरी तय करेगा, तो हम इसे प्रत्यक्ष अनुपात के आधार पर हल कर सकते हैं।

यहाँ, दूरी और समय दो राशियाँ हैं जो प्रत्यक्ष अनुपात में हैं। यदि 3 घंटे में 180 किलोमीटर तय किया गया, तो:

$\frac{3}{180} = \frac{5}{x}$

यहाँ, $x$ वह दूरी है जो 5 घंटे में तय की जाएगी।

समीकरण को हल करके हम पाएंगे:

$x = \frac{5 \times 180}{3} = 300$

इस प्रकार, 5 घंटे में 300 किलोमीटर की यात्रा की जाएगी।

किसी भी अनुपात को हल करते समय यह सुनिश्चित करना महत्वपूर्ण है कि हम सही तरीके से प्रत्येक मात्राओं के बीच स्थिर अनुपात बनाए रखें।

कभी-कभी हम "क" (constant) को निर्धारित करके अन्य जानकारी निकाल सकते हैं। उदाहरण के लिए, यदि हम जानते हैं कि $y = kx$ और हमें $k$ का मान दिया गया है, तो हम दूसरे मानों को सरलता से निकाल सकते हैं।

ध्यान रखें कि प्रत्यक्ष अनुपात तब ही सही होता है जब दोनों राशियाँ एक दूसरे के साथ सीधे रूप से बढ़ती या घटती हैं।

Following are the car parking charges near a railway station upto:
4 hours Rs. 60
8 hours Rs. 100
12 hours Rs. 140
24 hours Rs. 180

Check if the parking charges are in direct proportion to the parking time.

Solution:

Check if the parking charges are in direct proportion to the parking time

Let the parking time be represented by $t$ (in hours), and the parking charges be represented by $C$ (in Rs.).

We are given the following data:

  • For 4 hours, the charge is Rs. 60.
  • For 8 hours, the charge is Rs. 100.
  • For 12 hours, the charge is Rs. 140.
  • For 24 hours, the charge is Rs. 180.

To check if the charges are in direct proportion to the time, we need to check if the ratio of charge to time remains constant for all given data points.

The ratio of charge to time for 4 hours is:

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

The ratio of charge to time for 8 hours is:

$$\frac{100}{8} = 12.5$$

The ratio of charge to time for 12 hours is:

$$\frac{140}{12} \approx 11.67$$

The ratio of charge to time for 24 hours is:

$$\frac{180}{24} = 7.5$$

Since the ratio of charge to time is not constant, the parking charges are not in direct proportion to the parking time.

A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.

Parts of red pigment 1 4 7 12 20

Parts of base 8 ... ... ... ...

Solution:

Solution:

We are given that the mixture is prepared by mixing 1 part of red pigment with 8 parts of base. This means the ratio of red pigment to base is 1:8.

Let the parts of base be $b$ for each corresponding value of red pigment in the table.

Using the given ratio of 1:8, we can find the parts of base for each corresponding part of red pigment by multiplying the red pigment value by 8.

Parts of Red Pigment Parts of Base
1 8
4 4 * 8 = 32
7 7 * 8 = 56
12 12 * 8 = 96
20 20 * 8 = 160

In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Solution:

Question 1

A mixture of paint is prepared by mixing 1 part of red pigment with 8 parts of base. In the following table, find the parts of base that need to be added.

Parts of Red Pigment Parts of Base
1 8
4 8 × 4 = 32
7 8 × 7 = 56
12 8 × 12 = 96
20 8 × 20 = 160
Question 2

If 1 part of red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?

Let the number of parts of red pigment required be $x$. The ratio of red pigment to base is 1:75, so we can write the equation:

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

Cross-multiplying, we get:

$75x = 1800$

$x = \frac{1800}{75} = 24$

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

A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?

Solution:

Solution:

Let the number of bottles filled in one hour be $x$.

According to the given information, in 6 hours, the machine fills 840 bottles. Therefore, in 1 hour, the number of bottles filled is:

$x = \frac{840}{6} = 140$ bottles per hour.

Now, in 5 hours, the number of bottles filled will be:

Number of bottles filled in 5 hours = $140 \times 5 = 700$ bottles.

A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Solution:

2. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?

Let the actual length of the bacteria be denoted by $L$. The photograph is enlarged 50,000 times, so the relationship between the actual length and the enlarged length can be given as:

$\text{Enlarged Length} = \text{Actual Length} \times \text{Magnification}$

From the given information:

$5 = L \times 50,000$

Solving for $L$:

$L = \frac{5}{50,000}$

$L = 0.0001$ cm

So, the actual length of the bacteria is $0.0001$ cm.

Now, if the photograph is enlarged 20,000 times, the enlarged length $L_{\text{new}}$ will be:

$L_{\text{new}} = 0.0001 \times 20,000$

$L_{\text{new}} = 2$ cm

Thus, if the photograph is enlarged 20,000 times, the enlarged length would be 2 cm.

In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

Solution:

Question: In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?

We are given that:

  • Height of the mast of the model ship = 9 cm
  • Height of the mast of the actual ship = 12 m
  • Length of the actual ship = 28 m

We need to find the length of the model ship.

We can use the concept of similar triangles, where the ratio of corresponding heights of the actual ship and the model ship is the same as the ratio of their corresponding lengths.

Let the length of the model ship be $x$ meters.

The ratio of the height of the mast of the model ship to the actual ship is:

$$ \frac{9 \text{ cm}}{12 \text{ m}} = \frac{x}{28 \text{ m}} $$

Converting 9 cm to meters, we get 0.09 m. So the equation becomes:

$$ \frac{0.09}{12} = \frac{x}{28} $$

Now, cross-multiply to find $x$:

$$ 0.09 \times 28 = 12 \times x $$ $$ 2.52 = 12x $$

Solving for $x$:

$$ x = \frac{2.52}{12} = 0.21 \text{ m} $$

Therefore, the length of the model ship is 0.21 m or 21 cm.

Suppose 2 kg of sugar contains 9 × 10$^6$ crystals. How many sugar crystals are there in (i) 5 kg of sugar? (ii) 1.2 kg of sugar?

Solution:

Given:

2 kg of sugar contains 9 × 10$^6$ crystals.

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

Let the number of sugar crystals in 1 kg of sugar be $ \frac{9 × 10^6}{2} $ = 4.5 × 10$^6$ crystals.

Therefore, the number of sugar crystals in 5 kg of sugar = 5 × 4.5 × 10$^6$ = 22.5 × 10$^6$ crystals.

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

Number of sugar crystals in 1.2 kg of sugar = 1.2 × 4.5 × 10$^6$ = 5.4 × 10$^6$ crystals.

Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?

Solution:

Solution:

We are given that the scale of the map is 1 cm = 18 km. Rashmi drives 72 km on the road. We need to find the distance covered on the map.

Let the distance covered on the map be $x$ cm.

Using the scale, we can set up a proportion:

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

Now, solve for $x$:

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

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

$x = 4$

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

A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time (i) the length of the shadow cast by another pole 10 m 50 cm high (ii) the height of a pole which casts a shadow 5m long.

Solution:

Question:

A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time:

  1. the length of the shadow cast by another pole 10 m 50 cm high
  2. the height of a pole which casts a shadow 5 m long
Solution:

Let the height of the first pole be $h_1 = 5$ m $60$ cm $= 5 + \dfrac{60}{100} = 5.6$ m.

The length of the shadow of the first pole is $l_1 = 3$ m $20$ cm $= 3 + \dfrac{20}{100} = 3.2$ m.

For similar triangles, the ratio of height to shadow length for both poles is constant.

Thus, $\dfrac{h_1}{l_1} = \dfrac{h_2}{l_2}$, where $h_2$ is the height of the second pole and $l_2$ is the length of its shadow.

We are given $h_2 = 10$ m $50$ cm $= 10 + \dfrac{50}{100} = 10.5$ m.

Using the proportion: $$\dfrac{h_1}{l_1} = \dfrac{h_2}{l_2} \Rightarrow \dfrac{5.6}{3.2} = \dfrac{10.5}{l_2}$$

Solving for $l_2$: $$ l_2 = \dfrac{10.5 \times 3.2}{5.6} = 6 \text{ m.} $$

So, the length of the shadow cast by the second pole is 6 m.

Now, for part (ii), we are asked to find the height of a pole which casts a shadow of 5 m. Let the height of the pole be $h_3$ and the shadow length be $l_3 = 5$ m.

Using the same ratio of heights to shadow lengths, we have: $$\dfrac{h_1}{l_1} = \dfrac{h_3}{l_3}$$

Substitute the known values: $$\dfrac{5.6}{3.2} = \dfrac{h_3}{5}$$

Solving for $h_3$: $$ h_3 = \dfrac{5.6 \times 5}{3.2} = 8.75 \text{ m.} $$

So, the height of the pole is 8.75 m.

A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Solution:

Question: A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?

Let the speed of the truck be $S$ km per minute.

Speed $S = \frac{14}{25}$ km per minute.

Now, the truck will travel in 5 hours, which is $5 \times 60 = 300$ minutes.

Distance traveled in 300 minutes = Speed $S \times$ Time

Distance = $\frac{14}{25} \times 300 = 168$ km.