11.3-Inverse Proportion

11.3-Inverse Proportion Important Formulae

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11.3 - Inverse Proportion
  • Inverse proportion means when one quantity increases, the other decreases.
  • Mathematically, if $x$ is inversely proportional to $y$, then $x \propto \frac{1}{y}$ or $x = \frac{k}{y}$, where $k$ is a constant.
  • When two quantities are in inverse proportion, their product remains constant. That is, $x \times y = k$.
  • Example: If 5 workers take 20 days to complete a task, then 10 workers will take 10 days, as work done is inversely proportional to number of workers.

11.3 - Inverse Proportion

Inverse proportion is a concept in mathematics where one quantity increases while the other decreases, such that their product remains constant. This relationship is opposite to direct proportion, where both quantities change in the same manner. In inverse proportion, if one quantity increases, the other must decrease, and vice versa, to maintain the constant product.

In simpler terms, two quantities are said to be in inverse proportion if increasing one leads to a decrease in the other, while the product of the two quantities remains unchanged.

Mathematically, if two quantities, say $x$ and $y$, are inversely proportional, we express this relationship as:

Inverse Proportion Formula: $x \propto \frac{1}{y}$ or equivalently, $x \cdot y = k$ where $k$ is a constant of proportionality.

From this, we can derive the equation for inverse proportion. If $x$ is inversely proportional to $y$, then for any two values $x_1$ and $x_2$ corresponding to $y_1$ and $y_2$, we can write:

$x_1 \cdot y_1 = x_2 \cdot y_2 = k$

Thus, the product of the two quantities remains constant for all pairs of values of $x$ and $y$.

For example, if $x = 5$ and $y = 2$, then their product is $5 \cdot 2 = 10$. If $x$ changes to $10$, the corresponding value of $y$ must be $1$ (because $10 \cdot 1 = 10$), keeping the product constant.

Inverse proportion can also be expressed in terms of ratios. If $x$ and $y$ are inversely proportional, then:

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

This property is important because it shows the relationship between the two variables when their product is constant, and it allows us to solve problems involving inverse proportion by simply setting up and solving equations.

Some real-life examples of inverse proportion include:

  • The time taken to complete a task and the number of workers. If the number of workers increases, the time to finish the task decreases, assuming the total work remains constant.
  • The speed of a vehicle and the time taken to cover a fixed distance. As speed increases, the time taken decreases.

Inverse proportion is widely used in solving problems related to rates, speeds, work, and time, among others. Understanding the concept and its formulae can help solve a variety of word problems in both mathematics and real-world applications.

11.3-इनवर्स प्रपोर्शन

इनवर्स प्रपोर्शन (Inverse Proportion) वह स्थिति होती है जब दो मात्राएँ एक-दूसरे के साथ विपरीत रूप से संबंधित होती हैं। अर्थात, जब एक मात्रा बढ़ती है, तो दूसरी मात्रा घटती है, और जब एक मात्रा घटती है, तो दूसरी मात्रा बढ़ती है।

अगर दो मात्राएँ $x$ और $y$ इनवर्सली प्रपोर्शनल हैं, तो उनका गुणनफल स्थिर रहेगा। इसका मतलब है कि:

समीकरण:

$x \times y = k$

यहाँ, $k$ एक स्थिरांक है। इसका मतलब है कि अगर $x$ बढ़ता है, तो $y$ घटता है, और यदि $x$ घटता है, तो $y$ बढ़ता है।

इसे इस प्रकार भी व्यक्त किया जा सकता है:

$x \propto \dfrac{1}{y}$

यहाँ, "$\propto$" का अर्थ है "प्रपोर्शनल" (related)। इस संबंध में, जब $x$ और $y$ के मान बदलते हैं, तो उनका गुणनफल हमेशा $k$ के बराबर रहता है।

इनवर्स प्रपोर्शन के उदाहरण

मान लीजिए एक कार 3 घंटे में 120 किलोमीटर यात्रा करती है। अब अगर कार की गति को दोगुना कर दिया जाए, तो यात्रा का समय आधा हो जाएगा। इसका मतलब है कि दूरी और समय इनवर्स प्रपोर्शनल हैं।

उदाहरण के रूप में, अगर $x$ और $y$ दो इनवर्स प्रपोर्शनल मात्राएँ हैं, तो उनके बीच का संबंध इस प्रकार होगा:

$x_1 \times y_1 = x_2 \times y_2 = k$

यहाँ, $x_1$, $y_1$ और $x_2$, $y_2$ अलग-अलग स्थिति के लिए मान हैं, और $k$ एक स्थिरांक है।

इनवर्स प्रपोर्शन का उपयोग

इनवर्स प्रपोर्शन का उपयोग विभिन्न वास्तविक जीवन स्थितियों में किया जाता है। कुछ सामान्य उदाहरण इस प्रकार हैं:

  • गति और समय के बीच संबंध।
  • किसी टैंक में पानी भरने का समय और पाइप का व्यास।
  • किसी वस्तु के मूल्य और उसकी आपूर्ति के बीच संबंध।
समीकरण को हल करने की प्रक्रिया

यदि किसी समस्या में इनवर्स प्रपोर्शन के आधार पर समीकरण दिया गया हो, तो उसे हल करने की प्रक्रिया निम्नलिखित है:

  1. प्रारंभिक स्थितियों के आधार पर $x$ और $y$ के मानों को पहचानें।
  2. यह सुनिश्चित करें कि उनका गुणनफल स्थिरांक $k$ के बराबर है।
  3. समस्या में दिए गए किसी नए मान से संबंधित समीकरण तैयार करें और उसे हल करें।

उदाहरण के लिए, यदि $x_1 = 4$ और $y_1 = 5$ हों, तो इनका गुणनफल होगा:

$4 \times 5 = k = 20$

अब यदि $x_2 = 8$ हो, तो $y_2$ का मान निम्नलिखित होगा:

$8 \times y_2 = 20$

अतः, $y_2 = \dfrac{20}{8} = 2.5$

इस प्रकार, इनवर्स प्रपोर्शन में, जब एक मात्रा बदलती है, तो दूसरी मात्रा उसी अनुपात में बदलती है ताकि उनका गुणनफल समान रहे।

1. Which of the following are in inverse proportion?
(i)  The number of workers on a job and the time to complete the job. 

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

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

(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:

Which of the following are in inverse proportion?

(i) The number of workers on a job and the time to complete the job.
This is in inverse proportion because as the number of workers increases, the time to complete the job decreases.

(ii) The time taken for a journey and the distance travelled in a uniform speed.
This is not in inverse proportion because time and distance are directly related when speed is constant.

(iii) Area of cultivated land and the crop harvested.
This is not in inverse proportion because a larger area of cultivated land generally leads to more crop harvested.

(iv) The time taken for a fixed journey and the speed of the vehicle.
This is in inverse proportion because as the speed increases, the time taken for the journey decreases.

(v) The population of a country and the area of land per person.
This is in inverse proportion because as the population increases, the area of land per person decreases.

In a Television game show, the prize money of Rs. 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Number of winners 1 2 4 5 8 10 20
Prize for each winner (in Rs.) 1,00,000 50,000 ... ... ... ... ...

Solution:

Number of winners and Prize for each winner
Number of winners Prize for each winner (in Rs.)
1 1,00,000
2 50,000
4 25,000
5 20,000
8 12,500
10 10,000
20 5,000

The prize for each winner is inversely proportional to the number of winners. As the number of winners increases, the prize for each winner decreases.

Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.

Number of spokes 4 6 8 10 12
Angle between a pair of consecutive spokes 90° 60° ... ... ...

Solution:

Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consecutive spokes are equal. Help him by completing the following table.


Number of spokes: 4, 6, 8, 10, 12
Angle between a pair of consecutive spokes: 90°, 60°, 45°, 36°, 30°

* Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion? Yes, the number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion. As the number of spokes increases, the angle between consecutive spokes decreases, and vice versa. * (ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes. The angle between consecutive spokes is calculated using the formula: $$\text{Angle} = \frac{360^\circ}{\text{Number of spokes}}$$ For 15 spokes: $$\text{Angle} = \frac{360^\circ}{15} = 24^\circ$$ * (iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°? Using the formula: $$\text{Number of spokes} = \frac{360^\circ}{\text{Angle}}$$ For an angle of 40°: $$\text{Number of spokes} = \frac{360^\circ}{40^\circ} = 9$$ So, 9 spokes would be needed.

4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?

Solution:

4. If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?

Let the total number of sweets in the box be $S$.

When the box is divided among 24 children, each child gets 5 sweets. So, we can write:

$S = 24 \times 5 = 120$

Now, if the number of children is reduced by 4, the new number of children is $24 - 4 = 20$. Let the number of sweets each child gets be $x$. Then, we can write:

$S = 20 \times x$

Substitute the value of $S$ from the first equation:

$120 = 20 \times x$

Solving for $x$:

$x = \frac{120}{20} = 6$

Therefore, each child will get 6 sweets.

A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?

Solution:

4. A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?

Let the amount of food required to feed 1 animal for 1 day be $F$. Then, the total amount of food required for 20 animals for 6 days is:

Total food = $20 \times 6 \times F = 120F$

Now, if 10 more animals are added, the total number of animals becomes $20 + 10 = 30$. Let the number of days the food will last be $D$. Then, the total amount of food required for 30 animals for $D$ days is:

Total food = $30 \times D \times F$

Since the amount of food remains the same, we equate the two expressions for total food:

$120F = 30 \times D \times F$

Canceling $F$ from both sides:

$120 = 30D$

Solving for $D$:

$D = \frac{120}{30} = 4$

Thus, the food will last for 4 days if there were 10 more animals in the cattle.

A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?

Solution:

Question:

A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?

Solution:

Let the number of days required for 4 persons to complete the job be $x$ days.

According to the given information, the total work done is constant. The work done is the product of the number of persons and the number of days they work. Therefore, we have:

For 3 persons: $3 \times 4 = 12$ person-days of work

For 4 persons: $4 \times x$ person-days of work

Since the total work is the same, we set the two expressions equal to each other:

$3 \times 4 = 4 \times x$

$12 = 4x$

Solving for $x$, we get:

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

Therefore, 4 persons will take $3$ days to complete the job.

A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Solution:

Math Problem

A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?

Number of bottles in the batch = $25 \times 12 = 300$ bottles.

Now, number of boxes when packed with 20 bottles in each box = $\frac{300}{20} = 15$ boxes.

A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?

Solution:

Question:

A factory requires 42 machines to produce a given number of articles in 63 days. How many machines would be required to produce the same number of articles in 54 days?

Solution:

Let the number of machines required to produce the articles in 54 days be $x$.

Using the concept of inverse variation, we have:

Number of machines $\propto$ $\frac{1}{\text{Number of days}}$.

This implies that:

42 machines $\times$ 63 days = $x$ machines $\times$ 54 days

Therefore,

$42 \times 63 = x \times 54$

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

$x = \frac{2646}{54}$

$x = 49$

Hence, the number of machines required is $49$.

A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?

Solution:

Given:

The car takes 2 hours to reach the destination at a speed of 60 km/h.

Formula:

Time = Distance / Speed

Step 1: Find the distance traveled by the car.

Distance = Speed × Time

Distance = 60 km/h × 2 hours = 120 km

Step 2: Calculate the time taken when the car travels at 80 km/h.

Time = Distance / Speed

Time = 120 km / 80 km/h = 1.5 hours

Answer:

The car will take 1.5 hours to reach the destination when traveling at 80 km/h.

Two persons could fit new windows in a house in 3 days. (i) One of the persons fell ill before the work started. How long would the job take now? (ii) How many persons would be needed to fit the windows in one day?

Solution:

Question: Two persons could fit new windows in a house in 3 days.

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

Let the rate of work of one person be $r$ windows per day.

For 2 persons working together, their combined rate is $2r$ windows per day.

According to the problem, the job takes 3 days for 2 persons to complete the task. Hence, the total amount of work is:

Total work = $2r \times 3 = 6r$ windows.

Now, if one person fell ill, only one person is left to do the work. The rate of work of 1 person is $r$ windows per day.

To complete the same $6r$ windows, the time taken will be:

Time taken = $ \frac{6r}{r} = 6$ days.

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

Let the number of persons required be $n$. The rate of work of $n$ persons is $nr$ windows per day. To complete the task in one day, we need:

Total work = $nr \times 1 = 6r$ windows (since the total work is $6r$ as calculated earlier).

Hence, $nr = 6r$.

Solving for $n$, we get:

$n = 6$ persons.

A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?

Solution:

4. A school has 8 periods a day each of 45 minutes duration. How long would each period be, if the school has 9 periods a day, assuming the number of school hours to be the same?

Given:

  • Number of periods initially = 8
  • Duration of each period initially = 45 minutes
  • Number of periods after change = 9

Total time for the school day initially = $8 \times 45 = 360$ minutes.

Since the number of school hours remains the same, the total time for the school day with 9 periods will still be 360 minutes. Let the duration of each new period be $x$ minutes.

Therefore, the total time for the school day with 9 periods is $9 \times x = 360$ minutes.

Now, solve for $x$:

$9x = 360$

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

$x = 40$

Thus, each period will be 40 minutes long.