# Extra Curriculum Support

Complete above steps to enable this content

This is a comprehensive educational resource designed to provide students with the tools and guidance necessary to excel. This support system is structured to cater to various aspects of learning, ensuring that students are well-prepared for academic challenges and practical applications of mathematical concepts. Some are the key benefits are mentioned below:

**Comprehensive Learning:**This holistic approach helps students gain a thorough understanding of the subject. Practical Application: The resources encourage students to apply mathematical concepts to real-life scenarios, enhancing their practical understanding and problem-solving skills.**Critical Thinking and Reasoning:**Value-Based and HOTS questions promote critical thinking and reasoning abilities. These skills are crucial for students to tackle complex problems and make informed decisions.**Exam Preparedness:**Sample Question Papers and NCERT Exemplar Solutions provide ample practice for exams. They help students familiarize themselves with the exam format and types of questions, reducing exam anxiety.**Ethical and Moral Development:**Value-Based Questions integrate ethical and moral lessons into the learning process, helping in the overall development of students' character and social responsibility. By incorporating these diverse elements, Enhanced Curriculum Support aims to provide a robust and well-rounded knowledge, preparing students for both academic success and real-world challenges.

**Sample Question Paper:** These are designed to mimic actual exam papers, providing students with a practice platform to gauge their understanding and readiness. They cover a wide range of topics and question types that students might encounter. Regular practice with these papers helps in boosting confidence and improving exam performance.

**Quick Points:**

Practice for real exam scenarios.

Includes various types of questions.

Helps in time management.

Identifies areas of improvement.

**1.Convert the following decimals into fractions and simplify if possible.**

(a) 0.75 (b) 0.125

**Solution :**

(a) Write 0.75 as a fraction: 0.75=

Simplify the fraction by finding the greatest common divisor (GCD) of 75 and 100, which is 25:

(b) Write 0.125 as a fraction: 0.125=

Simplify the fraction by finding the greatest common divisor (GCD) of 125 and 1000, which is 125:

**2. Raju has solved **

**Solution :**

Both Raju and Sameer have solved the same amount of the exercise.

**3.Convert the fraction **

**Solution :**

The fraction

**4.Solve the following problem involving fractions: A recipe requires **

**Solution :**

Flour needed for one batch =

You will need a total of 2.25 cups of flour for 3 batches of the recipe.

Complete above steps to enable this content

## Value Based Question

**Value-Based Questions:** They integrate moral and ethical values into the learning process, encouraging students to think beyond just academic knowledge. These questions aim to develop a student's character and social responsibility through mathematics. They connect mathematical concepts with everyday life and moral lessons.

**Quick Points:**

Promotes critical thinking.

Encourages ethical reasoning.

Relates mathematics to real-life scenarios.

Enhances decision-making skills.

**1.Helping at the Community Garden.**

**Anita and her friends are working at a community garden. They need to plant 72 flower bulbs in equal rows. They want each row to have the same number of bulbs and ensure that every row has more than 1 but fewer than 10 bulbs. What are the possible numbers of bulbs in each row, and why is it important to organize tasks efficiently in group projects?**

**Solution:**

To find the possible numbers of bulbs in each row, we need to find the factors of 72 that are between 2 and 9.

Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

Possible factors between 2 and 9: 2, 3, 4, 6, 8, 9

So, Anita and her friends can plant the bulbs in rows of 2, 3, 4, 6, 8, or 9 bulbs.

**2.Sharing Resources.**

**Ravi and his three friends want to buy a book that costs ₹185.20. They decide to share the cost equally. How much will each person pay? Also, explain why sharing costs fairly is an important value in friendships and teamwork.**

**Solution:**

To find out how much each person will pay, we need to divide the total cost by the number of friends.

Total cost of the book: ₹185.20

Number of friends: 4

Amount each person will pay = Total cost / Number of friends

= ₹185.20 / 4 = ₹46.30

Each person will pay ₹46.30.

Complete above steps to enable this content

## HOTS

**HOTS (Higher Order Thinking Skills):** They require students to apply, analyze, synthesize, and evaluate information rather than just recall facts. These questions are designed to challenge students and stimulate intellectual growth. Engaging with HOTS questions helps students to develop a deeper understanding and prepares them for complex problem-solving.

**Quick Points:**

Develops advanced problem-solving skills.

Encourages deep understanding.

Fosters creativity and innovation.

Enhances analytical abilities.

**1.A picture hall has seats for 820 persons. At a recent film show, one usher guessed it was 34 full, another that it was 23 full. The ticket office reported 648 sales. Which usher (first or second) made the better guess?**

**Solution:**

Total number of seats = 820

Number of ticket sold = 648

For first usher = 34 × 648 = 3 × 162 = 486

For second usher = 23 × 648 = 2 × 216 = 432

Since 432 < 486

Hence, the first usher guessed better.

**2.Reemu reads 15th pages of a book. If she reads further 40 pages, she would have read 710th page of the book. How many pages are left to be read?**

**Solution:**

Let the total number of pages be x.

Number of pages read by Reemu = 15x

If she reads 40 more pages,

Total number of pages read by her = 15x + 40

Complete above steps to enable this content

## NCERT Exemplar

**NCERT Exemplar Solutions:** They provide detailed answers and explanations to problems in NCERT textbooks, aiding students in understanding complex concepts. These solutions serve as a valuable resource for clarifying doubts and reinforcing learning. They are essential for thorough exam preparation and achieving academic excellence.

**Quick Points:**

Comprehensive solutions for NCERT problems.

Clarifies difficult concepts.

Useful for exam preparation.

Provides step-by-step explanations.

**1.If **

(a) 9 (b) 16 (c) 18 (d) 32

**Solution: Correct answer is .**

**2.The next number of the pattern 60, 30, 15,is.**

(a) 10 (b) 5 (c)

**Solution: Correct answer is .**

**3.The decimal expression for 8 rupees 8 paise (in Rupees) is.**

(a) 8.8 (b) 8.08 (c) 8.008 (d) 88.0

**Solution: Correct answer is .**

**4.Each side of a regular hexagon is 3.5cm long. The perimeter of the given polygon is.**

(a) 17.5cm (b) 21cm (c) 18.3cm (d) 20cm

**Solution: Correct answer is .**

__{.m-red}5.Fraction which is reciprocal of**.**

**Solution: /.**

**6.Product of a proper and improper fraction is __ the improper fraction.**

**Solution: than.**

**7.The two non-zero fractions whose product is 1, are called the __ of each other.**

**Solution: .**

__{.m-red}8.5 rupees 5 paise = _**.**

**Solution: .**

**9.45mm = ___ m.**

**Solution: .**

**10.2.4×1000 =**.__

**Solution: .**

**11.Ramu finishes **

**Solution:**

The part of the work finished by Ramu in 1 hour =

So, the part of the work finished by Ramu in 2

Ramu will finish

**12.Kavita had a piece of rope of length 9.5 m. She needed some small pieces of rope of length 1.9 m each. How many pieces of the required length will she get out of this rope?**

**Solution:**

The length of the rope = 9.5m The length of a small piece of rope = 1.9m Number of small pieces = 9.5 m ÷ 1.9m

=

=

So, she will get 5 small pieces of rope.

**13.Three boys earned a total of 235.50. What was the average amount earned per boy?**

**Solution:**

Three boys earned = 235.50 The average amount earned per boy =

14.Observe the 3 products given in Example 32 and now give the answers of the following questions.

(i) Does interchanging the fractions in the example,

(ii) Is the value of the fraction in the product greater or less than the value of either fraction?

__{.m-green}Solution:

(i) By interchanging

This meansthat interchanging the fractions does not affect the answer.

(ii) By observing the 3 products given in the solution of Example 32, we come to know that the value of the fractions in the products are as follows.

(a) The product of two fractions whose value is less than 1 i.e. the proper fractions is less than each of the fractions that are multiplied.

(b) The product of a proper and an improper fraction is less than the improper fractions and greater than the proper fraction.

(c) The product of two improper fractions is greater than each of the two fractions.

**14.A rule for finding the approximate length of diagonal of a square is to multiply the length of a side of the square by 1.414.Find the length of the diagonal when :**

**(a) The length of a side of the square is 8.3 cm.** **(b) The length of a side of the square is exactly 7.875 cm**

**Solution:**

(a) The length of a side of the square is 8.3 cm Multiply the side length by 1.414:

d=8.3×1.414

Calculate the result: d≈8.3×1.414=11.7462

So, the length of the diagonal is approximately 11.7462 cm.

(b) The length of a side of the square is exactly 7.875 cm Multiply the side length by 1.414:

d=7.875×1.414

Calculate the result:d≈7.875×1.414=11.13375

So, the length of the diagonal is approximately 11.13375 cm

Complete above steps to enable this content

## Case Based Questions

**Case-Based Question:** They present real-life situations requiring students to apply their mathematical knowledge to solve problems, promoting practical understanding. These questions enhance the ability to connect theoretical knowledge with practical applications. They are instrumental in developing problem-solving skills relevant to real-world scenarios.

**Quick Points:**

Real-life application of concepts.

Encourages analytical thinking.

Enhances comprehension of practical problems.

Promotes interdisciplinary learning.

**1.Sana's class is organizing a bake sale to raise money for a school trip. They decided to bake cookies and cakes to sell. They recorded their sales and costs for each type of baked good.**

Sales Record:

Cookies:

Price per cookie: ₹15.75 Number of cookies sold: 40 Cakes:

Price per cake: ₹120.50 Number of cakes sold: 15 Ingredients Cost:

Cookies:

Flour: 2.5 kg at ₹20 per kg Sugar: 1.5 kg at ₹30 per kg Chocolate: 0.75 kg at ₹50 per kg Cakes:

Flour: 5 kg at ₹20 per kg Sugar: 3 kg at ₹30 per kg Eggs: 2 dozen at ₹60 per dozen Questions:

**1.What is the total revenue earned from selling cookies and cakes?**

**2.What is the total cost of ingredients for cookies and cakes?**

**3.How much profit did Sana's class make from the bake sale?**

**4.If the class decides to reduce the price of each cake by 10%, how much would each cake cost?**

**5.If they sell 20 more cookies at the original price, how much additional revenue will they generate?**

**1.Solution:**

Cookies Revenue:

Price per cookie: ₹15.75

Number of cookies sold: 40

Revenue from cookies = 15.75 × 40 = ₹630

Cakes Revenue:

Price per cake: ₹120.50

Number of cakes sold: 15

Revenue from cakes = 120.50 × 15 = ₹1,807.50

Total revenue = ₹630 + ₹1,807.50 = ₹2,437.50

**2.Solution:**

Cookies Ingredients Cost:

Flour: 2.5 kg at ₹20 per kg = 2.5 × 20 = ₹50

Sugar: 1.5 kg at ₹30 per kg = 1.5 × 30 = ₹45

Chocolate: 0.75 kg at ₹50 per kg = 0.75 × 50 = ₹37.50

Total cost for cookies = ₹50 + ₹45 + ₹37.50 = ₹132.50

Cakes Ingredients Cost:

Flour: 5 kg at ₹20 per kg = 5 × 20 = ₹100

Sugar: 3 kg at ₹30 per kg = 3 × 30 = ₹90

Eggs: 2 dozen at ₹60 per dozen = 2 × 60 = ₹120

Total cost for cakes = ₹100 + ₹90 + ₹120 = ₹310

Total cost of ingredients = ₹132.50 + ₹310 = ₹442.50

**3.Solution:**

Total revenue = ₹2,437.50

Total cost of ingredients = ₹442.50

Profit = ₹2,437.50 - ₹442.50 = ₹1,995

**4.Solution:**

Original price per cake = ₹120.50

Reduction = 10% of ₹120.50 = 0.10 × 120.50 = ₹12.05

New price per cake = ₹120.50 - ₹12.05 = ₹108.45

**.Solution:**

Price per cookie = ₹15.75

Number of additional cookies sold = 20

Additional revenue = 20 × 15.75 = ₹315