Ch 7: Proportional Reasoning – 1

Observing Similarity in Change, Ratios, and Simplest Form

Chapter 7: Proportional Reasoning – 1

Welcome to the world of proportional reasoning! This powerful mathematical concept helps us understand relationships, make predictions, and ensure fairness and consistency in everything from recipes and maps to engineering and art.

Page 1 of 5: Observing Similarity, Ratios, and Simplest Form

Concept Introduction

Have you ever tried to resize a photograph on your computer? If you pull it from a corner, it gets bigger or smaller but looks perfectly normal. But if you pull it from just one side, the image gets stretched or squashed, and the person in it might look strangely tall and thin, or short and wide! This simple act holds the key to understanding proportionality.

Things that look "right" when resized are proportional. This means every part of the object has changed by the same multiplicative factor. The stretched image is not proportional because its width and height changed by different factors. This idea of scaling things correctly is the foundation of ratios and proportions, a concept we use every day without even realizing it.

{{FORMULA: expr=a : b :: c : d | symbols=a,b,c,d:terms of the ratios, '::':is proportional to}}

Definitions & Formulas

Here are the fundamental terms we will use throughout this chapter. Understanding them clearly is the first step to mastering proportional reasoning.

The Logic of Proportionality

How do we mathematically prove that two shapes, like the tiger images, are proportional? We don't just guess! We follow a logical process of comparing their ratios. The core idea is that proportional ratios are identical when simplified.

Here is the step-by-step method to check if two ratios, say a : b and c : d, are in proportion.

1. Identify the Ratios: Write down the two ratios you want to compare. For example, let's compare the ratios of width to height for two images: 60 : 40 and 90 : 60.

2. Find the HCF for the First Ratio: Find the Highest Common Factor (HCF) of the terms a and b. For 60 : 40, the factors of 60 are (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60) and for 40 are (1, 2, 4, 5, 8, 10, 20, 40).

HCF(60, 40) = 20

3. Simplify the First Ratio: Divide both terms of the first ratio by their HCF. This reduces the ratio to its simplest form.

(60 ÷ 20) : (40 ÷ 20)  →  3 : 2

4. Find the HCF for the Second Ratio: Now, repeat the process for the second ratio, c : d. For 90 : 60, the HCF of 90 and 60 is 30.

HCF(90, 60) = 30

5. Simplify the Second Ratio: Divide both terms of the second ratio by their HCF.

(90 ÷ 30) : (60 ÷ 30)  →  3 : 2

6. Compare the Simplest Forms: Look at the simplified forms of both ratios. If they are exactly the same, the original ratios are in proportion.

3 : 2 is the same as 3 : 2

Therefore, we can state that 60 : 40 :: 90 : 60. The images are proportional!

{{VISUAL: diagram: Four images of a tiger. Images A, C, and D are rectangles of different sizes but look proportional. Image B is stretched vertically. Image E is a square and looks compressed horizontally.}}

Solved Examples

Let's walk through some problems, starting from easy and moving to more complex ones.

Example 1: Simplifying a Ratio (Easy)

Given: A class has 24 boys and 18 girls.

To Find: The ratio of boys to girls in its simplest form.

Solution:

1. First, write the ratio of boys to girls.

24 : 18

2. Next, find the Highest Common Factor (HCF) of the two terms, 24 and 18. The factors of 24 are {1, 2, 3, 4, 6, 8, 12, 24} and the factors of 18 are {1, 2, 3, 6, 9, 18}. The HCF is 6.

HCF(24, 18) = 6

3. Divide both terms of the ratio by their HCF to get the simplest form.

(24 ÷ 6) : (18 ÷ 6)

4. Calculate the final simplified ratio.

4 : 3

Final Answer: The ratio of boys to girls in its simplest form is 4 : 3.

Example 2: Checking for Proportion (Medium)

Given: Two ratios, 21 : 6 and 35 : 10.

To Find: Are these ratios proportional?

Solution:

1. Take the first ratio, 21 : 6, and find the HCF of its terms. The HCF of 21 and 6 is 3.

HCF(21, 6) = 3

2. Simplify the first ratio by dividing both terms by the HCF.

(21 ÷ 3) : (6 ÷ 3)  →  7 : 2

3. Now, take the second ratio, 35 : 10, and find the HCF of its terms. The HCF of 35 and 10 is 5.

HCF(35, 10) = 5

4. Simplify the second ratio by dividing both terms by its HCF.

(35 ÷ 5) : (10 ÷ 5)  →  7 : 2

5. Compare the simplest forms of both ratios. Since both simplify to 7 : 2, they are equal.

7 : 2 = 7 : 2

Final Answer: Yes, the ratios are proportional. We can write 21 : 6 :: 35 : 10.

{{KEY: type=concept | title=The Multiplicative Factor Rule | text=Two figures or quantities are proportional only if you can get from one to the other by MULTIPLYING (or dividing) all corresponding parts by the SAME number. Adding or subtracting the same number breaks the proportion and causes distortion.}}

Example 3: Proportionality in a Real-World Scenario (Hard)

Given: A baker, Priya, uses 4 cups of flour for every 3 cups of sugar in her cake recipe. Her friend, Rahul, uses 10 cups of flour for every 7 cups of sugar.

To Find: Whose cake recipe is sweeter (i.e., has a higher ratio of sugar to flour)?

Solution:

1. To compare sweetness, we should compare the ratio of sugar to flour for each recipe.
2. Write down Priya's ratio of sugar to flour.

Priya's Ratio = 3 : 4

3. This ratio is already in its simplest form, as the HCF of 3 and 4 is 1.
4. Write down Rahul's ratio of sugar to flour.

Rahul's Ratio = 7 : 10

5. This ratio is also in its simplest form, as the HCF of 7 and 10 is 1.
6. To compare 3 : 4 and 7 : 10, we need a common baseline. Let's find out how much sugar each person uses for the same amount of flour. The LCM (Lowest Common Multiple) of the flour amounts (4 and 10) is 20.
7. Scale Priya's ratio so that the flour part becomes 20. To get from 4 to 20, we multiply by 5. We must do the same to the sugar part.

(3 × 5) : (4 × 5)  →  15 : 20

8. Scale Rahul's ratio so that the flour part becomes 20. To get from 10 to 20, we multiply by 2.

(7 × 2) : (10 × 2)  →  14 : 20

9. Now compare. For 20 cups of flour, Priya uses 15 cups of sugar, while Rahul uses 14. Since 15 > 14, Priya's recipe uses more sugar for the same amount of flour.

Final Answer: Priya's cake recipe is sweeter.

Example 4: Ratios with Different Units (Tricky)

Given: An advertisement claims a car travels 5 kilometers on 200 milliliters of petrol. Another car claims to travel 12 kilometers on 0.5 liters of petrol.

To Find: Are their fuel efficiencies proportional?

Solution:

1. The first step is to make the units consistent. We cannot compare milliliters (mL) to liters (L) directly. Let's convert everything to milliliters. We know that 1 liter = 1000 milliliters.

0.5 L = 0.5 × 1000 mL = 500 mL

2. Now, write the ratio of distance (km) to fuel (mL) for the first car.

Car 1 Ratio = 5 : 200

3. Simplify this ratio. The HCF of 5 and 200 is 5.

(5 ÷ 5) : (200 ÷ 5)  →  1 : 40

This means Car 1 travels 1 km for every 40 mL of petrol.

4. Write the ratio of distance (km) to fuel (mL) for the second car, using our converted unit.

Car 2 Ratio = 12 : 500

5. Simplify this ratio. The HCF of 12 and 500 is 4.

(12 ÷ 4) : (500 ÷ 4)  →  3 : 125

6. Compare the two simplified ratios: 1 : 40 and 3 : 125. They are clearly not the same.

Final Answer: No, the fuel efficiencies of the two cars are not proportional.

Tips & Tricks

Use these techniques to solve problems faster and with more confidence.

Common Mistakes

Many students make these simple errors. Be careful to avoid them!

Brain-Teaser Questions

Time to challenge yourself! Think carefully before answering.

1. The ratio of blue, green, and red marbles in a bag is 12 : 30 : 18. What is this ratio in its simplest form?

   💡 Answer: Find the HCF of all three numbers: 12, 30, and 18. The HCF is 6. Divide all three terms by 6: (12÷6) : (30÷6) : (18÷6) gives the simplified ratio 2 : 5 : 3.

2. The ratio of the length to the width of a rectangular field is 5 : 3. If the perimeter of the field is 160 meters, what is its area?

   💡 Answer: Let the length be 5x and the width be 3x. Perimeter = 2 × (length + width). So, 160 = 2 × (5x + 3x), which means 160 = 2 × (8x) or 160 = 16x. Solving for x, we get x = 10. The length is 5 × 10 = 50 m and the width is 3 × 10 = 30 m. The area is length × width = 50 × 30 = 1500 m².

3. Two friends, A and B, have money in the ratio 4 : 5. If A gets ₹20 more, the new ratio of their money becomes 6 : 5. How much money did B have originally?

   💡 Answer: Let their initial money be 4x and 5x. After A gets ₹20, A has 4x + 20. B's money remains 5x. The new ratio is (4x + 20) : 5x = 6 : 5. This means (4x + 20) / 5x = 6 / 5. Multiplying both sides by 5x gives 4x + 20 = 6x. Solving this, 2x = 20, so x = 10. B's original money was 5x, which is 5 × 10 = ₹50.

Mini Cheatsheet

Here's a quick summary of today's lesson. Screenshot this for last-minute revision!

Problem Solving with Proportional Reasoning — Part 1

Page 2: Problem Solving with Proportional Reasoning — Part 1

Welcome to the next step in your journey with ratios! We've learned what it means for two ratios to be proportional — they represent the same relationship or value. Now, we'll learn a powerful technique to solve real-world problems using this idea.

Imagine you have a recipe for a chocolate cake that serves 4 people and requires 2 cups of flour. What if 12 guests are coming over? You can't just guess how much flour to add. You need the exact amount to keep the cake tasting perfect. This is where proportional reasoning comes in. By keeping the ratio of flour : people constant, you can calculate precisely what you need. This lesson focuses on finding the 'magic number' that links one ratio to another.

{{FORMULA: expr=a : b :: c : d | symbols=a,b,c,d: terms of the proportion, :: : "is proportional to"}}

Definitions & Formulas

Understanding the language of proportions is the first step to mastering them. Here are the key terms we'll use.

The Logic of "Factor of Change"

The core idea behind solving proportions is that to get from one ratio to an equivalent one, you must multiply (or divide) both parts of the ratio by the exact same number. This number is what we call the factor of change. Let's break down the method.

1. Set up the Proportion First, arrange your known ratio and the new ratio with its unknown part. Let's say we have a : b :: c : ?, where the question mark is the value we need to find, d.

2. Identify Corresponding Terms Look at the parts of the ratios that are in the same position. In a : b :: c : d, a and c are the first terms in each ratio, so they correspond. Similarly, b and d are the second terms and correspond to each other.

3. Calculate the Factor of Change (k) To find the factor that changes a into c, we simply divide c by a.

   k = c ÷ a
   

   This k is our factor of change.

{{VISUAL: diagram: Two ratios, a:b and c:d, are shown side-by-side. An arrow points from 'a' to 'c' labelled "× k", and another arrow points from 'b' to 'd' also labelled "× k", illustrating the constant factor of change.}}

4. Apply the Factor to the Other Term For the ratios to remain proportional, the same multiplicative change must happen to the second terms. So, we multiply b by the same factor k to find our missing value d.

   d = b × k
   

5. State the Final Proportion Once you've found d, you can write the complete, balanced proportion.

   a : b :: c : d
   

This method works for scaling up (when k > 1), scaling down (when k < 1), and even when the factor is a fraction!

{{KEY: type=concept | title=Multiplicative, Not Additive | text=Proportional relationships are always based on multiplication or division, never addition or subtraction. As shown in the NCERT example about Neelima's age, adding 9 years to both her and her mother's age changed the ratio. To keep a ratio the same, you must multiply by the same factor.}}

Solved Examples

Let's apply this method to some problems, starting easy and getting more challenging.

Example 1: Lemonade Recipe (Easy)

Given: Kesang uses 10 spoons of sugar for 6 glasses of lemonade. The ratio is 6 glasses : 10 spoons. She now wants to make 18 glasses.

To Find: The number of spoons of sugar needed for 18 glasses of lemonade to maintain the same sweetness.

Solution:

1. Set up the proportion with the unknown value (x). The relationship is glasses : sugar.

   6 : 10 :: 18 : x
   

2. Identify the corresponding known terms: the number of glasses, which are 6 and 18.

3. Calculate the factor of change (k) from the first term of the first ratio to the first term of the second ratio.

   k = 18 ÷ 6 = 3
   

   The number of glasses has been multiplied by 3.

4. Apply this same factor to the second term (spoons of sugar) to find x.

   x = 10 × 3 = 30
   

Final Answer:

Kesang should add 30 spoons of sugar. The proportion is 6 : 10 :: 18 : 30.

Example 2: Scaling Down a Photo (Medium)

Given: A large poster has a width of 150 cm and a height of 100 cm. The ratio is 150 width : 100 height. You want to print a smaller version with a width of 30 cm.

To Find: The height of the smaller version to keep the photo proportional.

Solution:

1. Set up the proportion. The relationship is width : height. Let the new height be h.

   150 : 100 :: 30 : h
   

2. Identify the corresponding known terms: the widths, which are 150 and 30.

3. Calculate the factor of change (k) from the old width to the new width. Notice we are scaling down.

   k = 30 ÷ 150
   

4. To make the division easier, simplify the fraction.

   k = 30/150 = 3/15 = 1/5
   

   The factor of change is 1/5. This means the new photo is 1/5th the size of the original.

5. Apply this same factor to the original height (100 cm) to find the new height h.

   h = 100 × (1/5) = 100 ÷ 5 = 20
   

Final Answer:

The height of the smaller photo should be 20 cm. The proportion is 150 : 100 :: 30 : 20.

Example 3: Finding a Missing Term with a Fractional Factor (Hard)

Given: The ratio 14 : 21 is proportional to 6 : x.

To Find: The value of the missing term x.

Solution:

1. Set up the proportion as given.

   14 : 21 :: 6 : x
   

2. Identify the corresponding known terms: the first terms, 14 and 6.

3. Calculate the factor of change (k) that transforms 14 into 6.

   k = 6 ÷ 14
   

4. Simplify this fraction to its lowest terms.

   k = 6/14 = 3/7
   

   This means we need to multiply the terms of the first ratio by 3/7 to get the second ratio.

5. Apply this fractional factor to the second term (21) to find x.

   x = 21 × (3/7)
   

6. Perform the multiplication. You can divide 21 by 7 first.

   x = (21 ÷ 7) × 3 = 3 × 3 = 9
   

Final Answer:

The missing number is 9. The proportion is 14 : 21 :: 6 : 9.

Example 4: Construction Site Cement (Tricky)

Given: A construction plan requires 4 bags of cement for every 10 metres of wall built. A team has already built a 25-metre wall.

To Find: How many bags of cement should they have used to be on track with the plan?

Solution:

1. First, establish the base ratio from the plan. The relationship is cement bags : wall length.

   Ratio = 4 : 10
   

2. Now, set up the proportion. The new wall length is 25 metres. We need to find the corresponding number of cement bags (c).

   4 : 10 :: c : 25
   

   Be careful! Notice the positions. 10 and 25 are the corresponding terms (wall length).

3. Find the factor of change (k) that transforms 10 metres into 25 metres.

   k = 25 ÷ 10
   

4. Calculate k. It's better to work with fractions or decimals.

   k = 25/10 = 5/2 = 2.5
   

5. Apply this same factor to the first term of the original ratio (bags of cement, 4) to find c.

   c = 4 × k = 4 × 2.5
   

6. Calculate the final value of c.

   c = 10
   

Final Answer:

The team should have used 10 bags of cement for a 25-metre wall.

Tips & Tricks

Use these shortcuts to solve problems faster and more accurately.

Common Mistakes

Many students make similar errors when they start. Here’s what to watch out for.

Brain-Teaser Questions

Test your understanding with these tricky problems!

1. A recipe for a smoothie calls for 3 bananas and 2 apples to serve 2 people. How many apples do you need to make enough smoothie for 5 people?

   💡 Answer: The ratio is people : apples which is 2 : 2 or 1 : 1. To serve 5 people, you need 5 apples. The number of bananas would be 5 × (3/2) = 7.5 bananas! Proportions can lead to fractional answers.

2. The ratio of the width to the height of a school's blackboard is 5 : 3. A student draws a proportional rectangle in their notebook. If the height of their drawing is 12 cm, what is its perimeter?

   💡 Answer: The proportion is 5 : 3 :: width : 12. The factor of change from height 3 to 12 is 12 ÷ 3 = 4. So, the width of the drawing is 5 × 4 = 20 cm. The perimeter is 2 × (width + height) = 2 × (20 + 12) = 2 × 32 = 64 cm.

3. In a school library, the ratio of science books to math books is 4 : 5. The ratio of math books to history books is 2 : 3. If there are 80 science books, how many history books are there?

   💡 Answer: This is a two-step problem. First, find the number of math books: 4 : 5 :: 80 : math. Factor is 80 ÷ 4 = 20. So, math books = 5 × 20 = 100. Now use the second ratio: 2 : 3 :: 100 : history. Factor is 100 ÷ 2 = 50. So, history books = 3 × 50 = 150. There are 150 history books.

Mini Cheatsheet

Here's a quick summary of everything on this page. Screenshot this for your notes!

Problem Solving with Proportional Reasoning — Part 2

{{FORMULA: expr=a : b :: c : d | symbols=a:First Term, b:Second Term, c:Third Term, d:Fourth Term}}

Problem Solving with Proportional Reasoning — Part 2

Welcome back! In our last session, we explored what it means for two ratios to be proportional by simplifying them. Now, we'll learn a powerful, systematic method to solve problems where one value is missing. This technique is a cornerstone of mathematics and is used everywhere, from calculating ingredients in a kitchen to planning large construction projects.

This method, known in classical mathematics as the Rule of Three (or Trairasika in ancient Indian mathematics), is a simple yet effective way to find the fourth, unknown value in a proportional relationship when you know the other three. It’s like having a map where one location is missing, and you use the other three known points to pinpoint its exact position.

Imagine you're a graphic designer scaling an image. The original image is 400 pixels wide and 600 pixels tall. You need to resize it to be 1000 pixels wide for a website banner, but you must maintain the original aspect ratio to avoid distortion. How tall should the new image be? This is a classic proportion problem: 400 : 600 :: 1000 : ?. The Rule of Three gives us a direct way to find that missing height and keep the image looking perfect.

Definitions & Formulas

When we write a proportion as a : b :: c : d, the terms have special names. Understanding these helps in applying the core formula of this section. The fundamental rule is that for any true proportion, the product of the "outer" terms equals the product of the "inner" terms.

The Logic Behind Cross-Multiplication

Where does the rule "Product of Extremes = Product of Means" come from? It's not magic! It's a logical consequence of what it means for two ratios to be equal. Let's derive it step-by-step.

1. We start with the statement of proportionality. This means the two ratios are equivalent.

   a : b :: c : d

2. Writing this statement in fraction form makes the relationship clearer.

   a/b = c/d
   

3. Our goal is to remove the denominators to get a simpler, linear equation. We can do this by multiplying both sides of the equation by b.

   (a/b) × b = (c/d) × b
   

4. This simplifies the left side, leaving a by itself.

   a = (c × b) / d
   

5. Now, to eliminate the remaining denominator d, we multiply both sides of the equation by d.

   a × d = ((c × b) / d) × d
   

6. This cancels out the d on the right side, giving us the final, powerful result.

   a × d = b × c
   

   This proves that the product of the extremes (a and d) is indeed equal to the product of the means (b and c). This technique is often called cross-multiplication because, in the fraction form a/b = c/d, we are multiplying the numbers diagonally across the equals sign.

Solved Examples

Let's put this theory into practice. We'll start with a simple problem and gradually increase the difficulty.

Example 1: Finding the Missing Number (Easy)

Given: The proportion 5 : 8 :: 15 : x, where x is the unknown fourth term.

To Find: The value of x.

Solution:

1. Identify the terms. Here, a=5, b=8, c=15, and d=x. These are the extremes (5, x) and the means (8, 15).

2. Apply the rule: Product of Extremes = Product of Means.

   5 × x = 8 × 15
   

3. Calculate the product of the means.

   5 × x = 120
   

4. To find x, divide both sides by 5.

```
x = 120 ÷ 5
```

5. Perform the division to get the final answer.

```
x = 24
```

Final Answer:

The value of x is 24. So, the complete proportion is 5 : 8 :: 15 : 24.

Example 2: The Cost of Books (Medium)

Given: A stack of 6 identical notebooks costs ₹150.

To Find: The cost of 15 such notebooks.

Solution:

1. First, set up the proportion. We must keep the order consistent. Let's use the format (number of notebooks) : (cost). Let the unknown cost be x.

   Notebooks₁ : Cost₁ :: Notebooks₂ : Cost₂
   

2. Substitute the given values into this structure.

   6 : 150 :: 15 : x
   

3. Apply the cross-multiplication rule (a × d = b × c).

   6 × x = 150 × 15
   

4. Calculate the product on the right side.

   6 × x = 2250
   

5. Isolate x by dividing both sides by 6.

   x = 2250 ÷ 6
   

6. Calculate the final value of x.

   x = 375
   

Final Answer:

The cost of 15 notebooks is ₹375.

Example 3: The Recipe Problem (Hard)

Given: A recipe for 16 cupcakes requires 400 grams of flour. You only have 350 grams of flour.

To Find: The maximum number of cupcakes you can make while maintaining the recipe's proportions.

Solution:

1. Set up the proportion. We will compare the ratio of flour to cupcakes. Let x be the number of cupcakes we can make.

   Flour₁ : Cupcakes₁ :: Flour₂ : Cupcakes₂
   

2. Plug in the known values.

   400 : 16 :: 350 : x
   

3. Use the cross-multiplication formula.

   400 × x = 16 × 350
   

4. Calculate the product of the means.

   400 × x = 5600
   

5. Solve for x by dividing by 400.

   x = 5600 ÷ 400
   

6. Simplify the division. You can cancel the two zeros from both numbers.

   x = 56 ÷ 4
   

7. Calculate the final result.

   x = 14
   

Final Answer:

With 350 grams of flour, you can make a maximum of 14 cupcakes.

{{KEY: type=concept | title=Maintaining Consistency is Crucial | text=When setting up a proportion from a word problem, the order of quantities in both ratios must be identical. If the first ratio is A : B, the second ratio must also be A : B (e.g., km : hours :: km : hours), not B : A (hours : km).}}

Example 4: The Map Scale Challenge (Tricky)

Given: A map has a scale where 2 cm on the map represents an actual distance of 5 km. The distance between two cities on this map is 11.4 cm.

To Find: The actual distance between the two cities in kilometers.

Solution:

1. Establish the proportional relationship. The ratio of map distance to actual distance is constant. Let x be the unknown actual distance.

   Map Distance₁ : Actual Distance₁ :: Map Distance₂ : Actual Distance₂
   

2. Substitute the given numbers into the proportion.

   2 cm : 5 km :: 11.4 cm : x km
   

3. Apply the rule: Product of Extremes = Product of Means.

   2 × x = 5 × 11.4
   

4. Calculate the product on the right side.

   2 × x = 57
   

5. Isolate x by dividing both sides by 2.

   x = 57 ÷ 2
   

6. Perform the division to find the actual distance.

   x = 28.5
   

Final Answer:

The actual distance between the two cities is 28.5 km.

Tips & Tricks

Cross-multiplication is fantastic, but here are a few other ways to think about these problems that can sometimes be faster.

Common Mistakes to Avoid

Proportional reasoning is straightforward, but small setup errors can lead to big mistakes. Here are some common pitfalls.

Brain-Teaser Questions

Ready for a challenge? These problems require careful setup and more than one step.

1. A 5-meter tall streetlamp casts a shadow that is 3 meters long. At the exact same time, a nearby building casts a shadow that is 12 meters long. What is the height of the building?

   💡 Answer: The proportion is height₁ : shadow₁ :: height₂ : shadow₂. 5 : 3 :: x : 12 5 × 12 = 3 × x → 60 = 3x → x = 20. The building is 20 meters tall.

2. A recipe for a fruit punch that serves 4 people requires 2 parts apple juice, 1 part grape juice, and 1 part soda water. If you need to make enough punch for 10 people, how much grape juice will you need if the total volume for 4 people is 800 mL?

   💡 Answer: First, find the amount of grape juice for 4 people. The total parts are 2+1+1=4. So grape juice is 1/4 of the total volume. Grape juice for 4 people = (1/4) × 800 mL = 200 mL. Now set up the proportion: people₁ : juice₁ :: people₂ : juice₂. 4 : 200 :: 10 : x 4 × x = 200 × 10 → 4x = 2000 → x = 500. You will need 500 mL of grape juice.

3. On a city blueprint, a scale of 2 cm : 5 m is used. A rectangular park is drawn on the blueprint with dimensions 6 cm by 10 cm. If fencing costs ₹250 per meter, what is the total cost to fence the actual park?

   💡 Answer: This is a multi-step problem. Step 1: Find the actual length. 2 cm : 5 m :: 10 cm : L → 2L = 50 → L = 25 m. Step 2: Find the actual width. 2 cm : 5 m :: 6 cm : W → 2W = 30 → W = 15 m. Step 3: Calculate the actual perimeter. Perimeter = 2 × (L + W) = 2 × (25 + 15) = 2 × 40 = 80 m. Step 4: Calculate the total cost. Cost = Perimeter × Cost per meter = 80 m × ₹250/m = ₹20,000. The total cost is ₹20,000.

Mini Cheatsheet

Here’s a quick summary of the key ideas from this page. Screenshot this for your last-minute revision!

Sharing, but Not Equally!

Page 4 of 5: Sharing, but Not Equally!

Concept Introduction

Imagine you and your friend decide to bake and sell cookies. You do most of the baking (3 hours), while your friend handles the sales (1 hour). At the end of the day, you make a profit of ₹400. Would it be fair to split the money equally? Probably not. Since you put in more work, it seems fair that you get a larger share of the profit.

This is where ratios come into play. Instead of a simple 1:1 equal split, you might decide to share the profit in the ratio of the hours you worked, which is 3:1. This means for every 1 part your friend gets, you get 3 parts. Proportional sharing is a powerful mathematical tool that helps us divide a whole quantity into unequal parts based on a specific, fair relationship or ratio. It's a fundamental concept used everywhere from business partnerships and recipes to mixing chemicals and creating art.

{{FORMULA: expr=Share = (Part of ratio / Sum of ratio parts) × Total Quantity | symbols=Part of ratio:The number for a specific share (e.g., m), Sum of ratio parts:The total of the numbers in the ratio (e.g., m + n), Total Quantity:The whole amount to be divided (e.g., x)}}

Definitions & Formulas

This method allows us to find the exact value of each share when a total quantity x is divided in the ratio m : n.

The Logic: How It Works

Dividing a quantity by a ratio is a logical, step-by-step process. Let's break down how to divide a quantity x in the ratio m : n.

1. Understand the Total Parts A ratio of m : n means the whole is conceptually split into m parts for the first share and n parts for the second share. This gives a total of m + n equal parts.

   Total Parts = m + n
   

2. Find the Value of a Single Part If the total quantity x is made up of m + n equal parts, we can find the value of just one of these parts by dividing the total quantity by the total number of parts.

   Value of One Part = x ÷ (m + n)
   

   {{VISUAL: diagram: A bar representing a total quantity 'x' is divided into (m+n) equal blocks. The first 'm' blocks are colored blue and the next 'n' blocks are colored yellow, visually showing the m:n split.}}

3. Calculate the First Share The first share corresponds to m parts of the ratio. To find its value, we multiply the number of parts (m) by the value of a single part we found in the previous step.

   First Share = m × (x / (m + n))
   

4. Calculate the Second Share Similarly, the second share corresponds to n parts of the ratio. We multiply n by the value of a single part.

   Second Share = n × (x / (m + n))
   

5. Verification As a final check, the sum of the calculated shares must equal the original total quantity x.

   (First Share) + (Second Share) = x
   

Solved Examples

Example 1: Dividing Money (Easy)

Given: Total amount = ₹4,500, Ratio = 2 : 3

To Find: The value of each of the two parts.

Solution:

1. First, find the sum of the ratio parts. This tells us how many total "groups" we are dividing the money into.

   Sum of ratio parts = 2 + 3 = 5
   

2. Next, find the value of one single part by dividing the total amount by the sum of the parts.

   Value of one part = ₹4500 ÷ 5 = ₹900
   

3. Now, calculate the first share, which corresponds to 2 parts of the ratio.

   First Share = 2 × ₹900 = ₹1800
   

4. Calculate the second share, which corresponds to 3 parts of the ratio.

   Second Share = 3 × ₹900 = ₹2700
   

Final Answer: The two parts are ₹1,800 and ₹2,700.

Example 2: Business Profit Sharing (Medium)

Given: Prashanti's investment = ₹75,000, Bhuvan's investment = ₹25,000, Total profit = ₹4,000. Profit is shared in the ratio of investment.

To Find: Each person's share of the profit.

Solution:

1. First, determine the ratio of their investments. We can simplify this ratio to its lowest terms.

   Ratio = 75000 : 25000
   

2. Divide both sides by 25,000 to simplify.

   Simplified Ratio = 3 : 1
   

3. Find the sum of the ratio parts.

   Sum of ratio parts = 3 + 1 = 4
   

4. Calculate the value of one part of the profit.

   Value of one part = ₹4000 ÷ 4 = ₹1000
   

5. Calculate Prashanti's share (3 parts).

   Prashanti's share = 3 × ₹1000 = ₹3000
   

6. Calculate Bhuvan's share (1 part).

   Bhuvan's share = 1 × ₹1000 = ₹1000
   

Final Answer: Prashanti's share is ₹3,000 and Bhuvan's share is ₹1,000.

Example 3: Mixture Problem (Hard)

Given: Original mixture of 40 kg contains sand and cement in the ratio 3 : 1. The new desired ratio of sand to cement is 5 : 2.

To Find: How much cement should be added to the mixture.

Solution:

1. First, find the initial quantities of sand and cement in the 40 kg mixture using the ratio 3 : 1. The sum of parts is 3 + 1 = 4.

   Weight of Sand = (3 / 4) × 40 kg = 30 kg
   

   Weight of Cement = (1 / 4) × 40 kg = 10 kg
   

2. In the new mixture, the amount of sand remains the same (30 kg). The new ratio is 5 : 2 (Sand : Cement). Let the new quantity of cement be C.

   New Ratio = (Weight of Sand) : (New Weight of Cement) = 5 : 2
   

3. We can set up a proportion to find the required amount of cement for 30 kg of sand.

   5 / 2 = 30 / C
   

4. Solve for C. We can see that 30 is 6 × 5. So, C must be 6 × 2.

   C = (30 × 2) / 5 = 12 kg
   

5. The new mixture needs 12 kg of cement. The original mixture already has 10 kg. Calculate the difference to find how much to add.

   Cement to be added = 12 kg - 10 kg = 2 kg
   

Final Answer: 2 kg of cement should be added.

{{KEY: type=concept | title=Isolate the Constant Quantity | text=In mixture problems where one component is added, first calculate the initial quantities of both components. The component that is not being changed is your constant. Use this constant amount to calculate the required amount of the other component in the new ratio.}}

Example 4: Working with Differences (Tricky)

Given: A sum of money is divided between Rohan and Sania in the ratio 5 : 9. Sania receives ₹1,200 more than Rohan.

To Find: The total sum of money.

Solution:

1. Let the common multiplier for the ratio be k. Rohan's share = 5k Sania's share = 9k

2. The problem states that the difference between their shares is ₹1,200.

   Sania's share - Rohan's share = ₹1200
   

3. Substitute the expressions from step 1 into this equation.

   9k - 5k = 1200
   

4. Solve for k.

   4k = 1200
   

   k = 1200 ÷ 4 = 300
   

5. Now that we know k, we can find the value of each share. Rohan's share = 5k = 5 × 300 = ₹1500 Sania's share = 9k = 9 × 300 = ₹2700

6. The total sum is the sum of their individual shares.

   Total Sum = ₹1500 + ₹2700 = ₹4200
   

Final Answer: The total sum of money is ₹4,200.

Tips & Tricks

Common Mistakes to Avoid

Brain-Teaser Questions

1. A sum of ₹13,600 is to be divided among three friends, Anjali, Bimal, and Charu, in the ratio ½ : ⅓ : ¼. How much money does Bimal receive?

   💡 Answer: First, convert the fractional ratio to a whole number ratio by finding the LCM of the denominators (2, 3, 4), which is 12. Multiply each fraction by 12: (½ × 12) : (⅓ × 12) : (¼ × 12) → 6 : 4 : 3. Now, divide ₹13,600 in the ratio 6:4:3. Sum of parts = 6+4+3=13. Value of one part = 13600 ÷ 13 = 1046.15... wait, let me recheck my math. Ah, the LCM method is correct, let's assume the question had a cleaner number. Let's re-do. Sum of parts = 13. Let's make the total ₹13,000 for a clean example. Total = 13,000. Value of one part = 13000 / 13 = 1000. Bimal's share is 4 parts. So, 4 × 1000 = ₹4000. The logic is to first convert fractional ratios to integer ratios. Let me redo the original number. Sum of parts is 6+4+3=13. Value of one part is 13600 ÷ 13. This doesn't divide cleanly. Let me assume a typo in the question and use a number like ₹11,700 for a clean calculation. Total = ₹11,700. 11700 ÷ 13 = 900. Bimal's share (4 parts) = 4 × 900 = ₹3600. Okay, let me create a better question. Let's use the same logic but a clean number. "A sum of ₹2,600 is divided..." Yes, this works.

   New Question 1: A sum of ₹2,600 is to be divided among Anjali, Bimal, and Charu, in the ratio ½ : ⅓ : ¼. How much money does Bimal receive?

   💡 Answer: First, convert the fractional ratio into a whole number ratio. Find the LCM of the denominators (2, 3, 4), which is 12. Multiply each part of the ratio by the LCM: (12 × ½) : (12 × ⅓) : (12 × ¼), which simplifies to 6 : 4 : 3. Now, divide ₹2,600 in this new ratio. The sum of parts is 6 + 4 + 3 = 13. The value of one part is ₹2600 ÷ 13 = ₹200. Bimal's share corresponds to 4 parts, so his share is 4 × ₹200 = ₹800.

2. The salaries of three employees, A, B, and C, are in the ratio 2 : 3 : 5. If their salaries are increased by 15%, 10%, and 20% respectively, what will be the new ratio of their salaries?

   💡 Answer: Let their initial salaries be 2k, 3k, and 5k. A's new salary = 2k + (15% of 2k) = 2k + 0.15 × 2k = 2.3k. B's new salary = 3k + (10% of 3k) = 3k + 0.10 × 3k = 3.3k. C's new salary = 5k + (20% of 5k) = 5k + 0.20 × 5k = 6.0k. The new ratio is 2.3k : 3.3k : 6.0k. We can cancel k and multiply by 10 to remove decimals. The new ratio is 23 : 33 : 60.

3. In a bag, there are coins of 25 paise, 10 paise, and 5 paise in the ratio 1 : 2 : 3. If their total value is ₹30, find the number of 5 paise coins.

   💡 Answer: Let the number of coins be 1k (25p), 2k (10p), and 3k (5p). Now, convert the number of coins to their value in rupees. Value of 25p coins = k × 0.25 = 0.25k. Value of 10p coins = 2k × 0.10 = 0.20k. Value of 5p coins = 3k × 0.05 = 0.15k. The total value is ₹30. So, 0.25k + 0.20k + 0.15k = 30. This gives 0.60k = 30. Solving for k, we get k = 30 / 0.60 = 50. The number of 5 paise coins is 3k, so 3 × 50 = 150.

Mini Cheatsheet

Unit Conversions & Summary

Unit Conversions in Proportional Reasoning

Imagine you're following a recipe from an international cookbook. It asks for 2 pounds of flour, but your kitchen scale only measures in kilograms. Or, you're planning a trip abroad and see a road sign that says "100 miles to the next city," but you're used to thinking in kilometers. In these moments, you're facing a common challenge: different systems of measurement.

Unit conversion is the process of changing a quantity from one unit of measurement to another, without changing its actual value. This skill is crucial in proportional reasoning because ratios and proportions only work correctly when the quantities are compared in the same units. If you're comparing the cost of two plots of land, you can't compare one price in 'per square foot' and another in 'per acre' directly. You must first convert them to a common unit to make a fair comparison.

{{FORMULA: expr=F = (9/5) × C + 32 | symbols=F:Temperature in Fahrenheit, C:Temperature in Celsius}}

Definitions & Formulas

The following table lists some of the most common unit conversions you will encounter. These are standard values used globally in science, trade, and daily life.

Logic: Deriving the Temperature Formulas

The two temperature conversion formulas are not independent; one can be derived from the other using simple algebra. Let's see how to get the Celsius formula from the Fahrenheit formula.

1. Start with the formula for Fahrenheit.

   F = (9/5) × C + 32
   

2. Our goal is to isolate C. First, subtract 32 from both sides of the equation to move the constant term.

   F – 32 = (9/5) × C
   

3. Now, C is being multiplied by the fraction 9/5. To undo this, we can multiply both sides by the reciprocal of this fraction, which is 5/9.

   (5/9) × (F – 32) = (5/9) × (9/5) × C
   

4. Simplify the right side of the equation. The fractions 5/9 and 9/5 cancel each other out, leaving just C.

   (5/9) × (F – 32) = 1 × C
   

5. This gives us the final formula for converting Fahrenheit to Celsius.

   C = (5/9) × (F – 32)
   

Solved Examples

Example 1: Room Area Conversion (Easy)

Given: A rectangular room has an area of 20 square metres.

To Find: The area of the room in square feet.

Solution:

1. Identify the relevant conversion factor from our table.

   1 square metre = 10.764 square feet
   

2. This is a direct proportion. To find the area in square feet for 20 square metres, we multiply the number of square metres by the conversion factor.

   Area in sq ft = 20 × 10.764
   

3. Calculate the final product.

   Area in sq ft = 215.28
   

Final Answer: The area of the room is 215.28 square feet.

Example 2: Farming Manure Calculation (Medium)

Given: A farmer plans to grow tomatoes on a plot of size 200 ft by 500 ft. The recommended amount of manure is 10 tonnes per acre.

To Find: How much manure should the farmer buy?

Solution:

1. First, calculate the total area of the farmer's plot in square feet.

   Area = Length × Width = 200 ft × 500 ft = 100,000 sq feet
   

2. Next, we need to convert this area from square feet to acres to use the given ratio. We know the conversion factor.

   1 acre = 43,560 sq feet
   

3. To find the number of acres, we divide the total square footage by the number of square feet in one acre.

   Area in acres = 100,000 ÷ 43,560 ≈ 2.295 acres
   

4. The recommendation is 10 tonnes of manure per acre. We can set up a proportion or simply multiply the area in acres by this rate.

   Manure needed = 2.295 acres × 10 tonnes/acre
   

5. Calculate the final amount of manure.

   Manure needed ≈ 22.95 tonnes
   

Final Answer: The farmer should buy approximately 23 tonnes of manure.

Example 3: Temperature Conversion (Hard)

Given: The weather forecast for a city in the USA is 86°F.

To Find: The temperature in degrees Celsius.

Solution:

1. Select the correct formula for converting Fahrenheit to Celsius.

   C = (5/9) × (F – 32)
   

2. Substitute the given Fahrenheit value (F = 86) into the formula.

   C = (5/9) × (86 – 32)
   

3. Follow the order of operations (BODMAS/PEMDAS). First, solve the expression inside the parentheses.

   C = (5/9) × (54)
   

4. Now, perform the multiplication. You can simplify 54 ÷ 9 first.

   C = 5 × (54 ÷ 9) = 5 × 6
   

5. Calculate the final temperature in Celsius.

   C = 30
   

Final Answer: The temperature is 30°C.

Example 4: Cost of Land (Tricky)

Given: One acre of land costs ₹15,00,000.

To Find: The cost of 2,400 square feet of the same land.

Solution:

1. This problem requires us to find the price per square foot first. We are given the price per acre. We need the conversion factor.

   1 acre = 43,560 sq feet
   

2. Calculate the cost per square foot by dividing the total cost of one acre by the number of square feet in an acre.

   Cost per sq ft = ₹15,00,000 ÷ 43,560 sq ft
   

   Cost per sq ft ≈ ₹34.435
   

3. Now that we have the rate per square foot, we can find the cost of 2,400 square feet by multiplying.

   Total cost = 2,400 sq ft × ₹34.435 / sq ft
   

4. Calculate the final cost.

   Total cost ≈ ₹82,644
   

Final Answer: The cost of 2,400 square feet of land is approximately ₹82,644.

{{KEY: type=concept | title=The Unity Fraction Method | text=To convert units, multiply by a fraction that equals 1. For example, since 1 m = 3.281 ft, the fractions (1 m / 3.281 ft) and (3.281 ft / 1 m) both equal 1. Choose the fraction that cancels the unit you want to eliminate.}}

Tips & Tricks

Common Mistakes

Key Chapter Takeaways

• Ratios in the form of a : b indicate that for every ‘a’ units of the first quantity, there are ‘b’ units of the second quantity.
• Two ratios a : b and c : d are proportional (written a : b :: c : d) if their terms change by the same factor, i.e., if a × d = b × c.
• If a total quantity x is divided into two parts in the ratio m : n, then the first part is m × (x / (m + n)) and the second part is n × (x / (m + n)).

Brain-Teaser Questions

1. A car is travelling at a constant speed of 100 kilometers per hour. What is its speed in feet per second?

   💡 Answer: First, convert km/hr to m/s. 100 km/hr = 100 × (1000 m / 3600 s) ≈ 27.78 m/s. Now convert metres to feet: 27.78 m/s × 3.281 ft/m ≈ 91.16 ft/s. The car travels at about 91 feet every second!

2. The density of a substance is its mass per unit volume. The density of aluminum is 2700 kg/m³. If you have a solid block of aluminum with a volume of 5000 cubic centimetres (cc), what is its mass in kilograms?

   💡 Answer: First, convert the volume from cm³ to m³. We know 1 m = 100 cm, so 1 m³ = (100)³ cm³ = 1,000,000 cm³. Volume in m³ = 5000 cm³ ÷ 1,000,000 cm³/m³ = 0.005 m³. Mass = Density × Volume = 2700 kg/m³ × 0.005 m³ = 13.5 kg.

3. A baker from Canada is using an American recipe that says, "Bake at 400°F". The baker's oven only shows Celsius, with markings at every 10°C (e.g., 190°C, 200°C, 210°C). What is the closest and safest temperature setting the baker should use?

   💡 Answer: Convert 400°F to Celsius: C = (5/9) × (400 – 32) = (5/9) × 368 ≈ 204.4°C. The closest markings are 200°C and 210°C. To be safe, it's often better to bake at a slightly lower temperature to avoid burning. Therefore, setting the oven to 200°C would be the most sensible choice.

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