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8th class > Linear Equations in One Variable > Enhanced Curriculum Support

Enhanced Curriculum Support

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:

1.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.

2.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.

3.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.

4.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 Questions

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.

SecA

1.Find the value of x in the equation: 7x+3 = 5x+15.

Sol

1.Solution: 7x+3 = 5x+15

7x - 5x = 15 -3

2x = 12

x = 6

SecB

1.A father is three times as old as his son. Five years ago, he was four times as old as his son. Find their present ages.

2.Ramesh bought 3 pens and 4 notebooks for ₹100. If a notebook costs ₹5 more than a pen, find the cost of each pen and notebook.

3.The sum of three consecutive integers is 51. Find the integers.

Sol

1.Solution:

Let x be the present age of the son.

Let 3x be the present age of the father (since the father is three times as old as his son).

Five years ago, the son's age was x−5.

Five years ago, the father's age was 3x−5.

According to question: 3x−5 = 4(x−5) ⟹ 3x-5 = 4x - 20 ⟹ -5 + 20 = 4x -3x ⟹ x = 15

Thus, son's present age is 15 years while father's present age is 3 × 15 = 45 years.

2.Solution:

Let the cost of a pen be ₹ x.

Then, the cost of a notebook will be ₹ x+5 (since a notebook costs ₹5 more than a pen).

The cost for 3 pens is 3×x.

The cost for 4 notebooks is 4×(x+5).

According to the problem, the total cost for pens and notebooks is ₹100: 3x + 4(x+5) = 100

⟹ 3x + 4x + 20 = 100 ⟹ 7x = 80 ⟹ x = 807 = ₹ 11.43

Thus, the cost of each pen is ₹11.43 while cost of a notebook is ₹ x+5 = 11.43 + 5 = ₹ 16.43

3.Solution:

Let the three consecutive integers be x, x + 1 and x + 2.

Thus, x + (x+1) + (x+2) = 51 ⟹ 3x + 3 = 51 ⟹ 3x = 48 ⟹ x = 16.

Thus, the consecutive integers are: 16, 17 (16 + 1) and 18 (16 + 2).

SecC

1.Solve the following equations:

(a) 2x34 = x+12

(b) 5x+73 = 2x12

(c) 3x+25 - x43 = 1

2.The denominator of a fraction is 4 more than its numerator. If 1 is subtracted from both the numerator and the denominator, the fraction becomes 12. Find the fraction.

Sol

1.Solution:

(a) 2x34 = x+1222x3 = 4x+14x6 = 4x+4

4x4x6 = 4 ⟹ -6 ≠ 4

Thus, there are no solutions.

(b) 5x+73 = 2x1225x+7 = 32x110x+14 = 6x3

10x6x = 143 ⟹ 4x = -17 ⟹ x = 174

(c) 3x+25 - x43 = 1

The common denominator for 5 and 3 is 15.

33x+215 - 5x415 = 1

33x+25x415 = 1

9x+65x+2015 = 1

4x+26 = 15 ⟹ 4x = -11 ⟹ x = 114

2.Solution:

Let the numerator be x.

Then, the denominator will be x+4.

The fraction is xx+4.

Therefore, x1x+41 = 12

2(x-1) = 1((x+4) -1)

2x - 2 = x + 3

2x - x = 3 + 2

x = 5

Thus, the fraction is : xx+4 = 55+4 = 59

Value Based Questions

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.

Problem 1

  1. An energy-saving campaign suggests reducing electricity usage by 10% per month. If a household currently uses 500 kWh per month, how much will they use after 3 months of consistent reduction? Why is energy conservation crucial for the environment?

Sol

Solution:

Current electricity usage per month: 500 kWh

Reduction rate: 10% per month

Calculation for Each Month:

First Month:

Reduction: 10% of 500 kWh

Reduction = 0.1 × 500 = 50 kWh

Electricity used in the first month:

Electricity used = 500 − 50 = 450 kWh

Second Month:

Starting with 450 kWh from the first month, Reduction: 10% of 450 kWh

Reduction = 0.1 × 450 = 45 kWh Electricity used in the second month:

Electricity used = 450 − 45 = 405 kWh

Third Month:

Starting with 405 kWh from the second month,

Reduction: 10% of 405 kWh

Reduction = 0.1 × 405 = 40.5 kWh Electricity used in the third month:

Electricity used = 405 − 40.5 = 364.5 kWh

Total Electricity Used After 3 Months

After 3 months of consistent reduction, the household will use: 364.5 kWh.

Energy conservation not only reduces electricity usage and costs but also plays a vital role in protecting the environment by mitigating climate change, preserving natural resources, and promoting sustainable development. These benefits highlight the importance of adopting energy-efficient practices at both individual and collective levels.

Problem 2

2.A student aims to read 5 books every month to improve their knowledge. If they have read 30 books so far, how many months will it take to read a total of 75 books? Why is continuous learning important in personal and professional growth?

Sol

Solution:

Reading goal per month: 5 books

Books read so far: 30 books

Total books to read: 75 books

Let x represent the number of months required to reach the total of 75 books.

The number of additional books needed to reach 75 books: 75 − 30 = 45 books

Since the student reads 5 books per month, to find out how many months x, it will take to read 45 books at a rate of 5 books per month:

5 x = 45

x = 455 = 9

Thus, it will take the student 9 months to read a total of 75 books.

Continuous learning is essential for individuals seeking personal growth, professional development, and overall success in a rapidly changing world. It enables individuals to adapt, innovate, and thrive in their careers while fulfilling their intellectual curiosity and contributing positively to society.

Problem 3

3.A dietitian recommends consuming 30 grams of fiber per day. If a person gets 𝑥 grams of fiber from fruits and needs an additional 18 grams from vegetables, how much fiber do they get from fruits if the total intake is 30 grams? Discuss the importance of a balanced diet.

Sol

Solution:

Total fiber intake recommended: 30 grams per day

Fiber obtained from fruits: 𝑥 grams Additional fiber needed from vegetables: 18 grams

The total fiber intake is the sum of fiber from fruits and vegetables: 𝑥 + 18

According to the dietitian's recommendation, the total fiber intake should be 30 grams:

𝑥 + 18 = 30

𝑥 = 30 − 18

𝑥 = 12

Therefore, the person gets 12 grams of fiber from fruits.

A balanced diet is essential for achieving and maintaining good health throughout life. It provides the necessary nutrients for physical and mental well-being, reduces the risk of chronic diseases, supports digestive health, and fosters long-term sustainability in dietary habits. By making informed food choices and prioritizing nutrient-rich foods, individuals can optimize their health and quality of life.

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.

Q1

The formula for the perimeter of a rectangle is P = 2l + 2w, where 𝑙 is the length and w is the width. If the perimeter of a rectangle is 30 units and the length is 4 units longer than the width, find the dimensions of the rectangle. Explain your solution process.

Sol

Solution:

The formula for the perimeter of a rectangle is P = 2l + 2w.

The perimeter 𝑃 is 30 units.

The length 𝑙 is 4 units longer than the width 𝑤.

P = 2l + 2w

30 = 2(w+4) + 2w

30 = 2w + 8 + 2w

30 = 4w + 8

22 = 4w

𝑤 = 224

w = 5.5

Find the length: l = w + 4

l = 5.5 + 4 = 9.5

Verification;

P = 2l + 2w

P = 2(9.5) + 2(5.5)

P = 19+11

P = 30

Q2

A car rental company charges Rs.30 per day plus Rs.0.25 per km driven. If a customer spends Rs.75 for renting a car for one day, how many kilometres did they drive? Explain how you arrived at your answer.

Sol

Solution:

Daily rental charge: Rs. 30

Cost per km driven: Rs. 0.25

Total amount spent: Rs. 75

Let's denote: x as the number of kms driven.

The total cost can be represented as: 30 + 0.25x = 75

0.25x = 75 − 30

0.25x = 45

𝑥 = 450.25 = 180

So, the customer drove 180 kms.

Verification of the solution:

Cost for kms driven: 0.25 × 180 = 45

Total cost: 30 + 45 = Rs. 75

The total matches the given information, confirming that x = 180 kms is correct.

Q3

An amusement park charges an entrance fee of Rs.20 per person plus Rs.5 per ride. If a group of friends spends Rs.150 in total for entrance and rides, how many rides did they take? Justify your solution process.

Sol

Solution:

Entrance fee per person: Rs. 20

Cost per ride: Rs. 5

Total amount spent: Rs. 150

Let 𝑥 represent the number of rides taken by the group.

The total cost for entrance and rides can be represented as: 20 + 5x = 150

5x = 150 − 20

5x = 130

x = 1305

x = 26

Verification of the solution: Plug the value of 𝑥 back into the original context to verify:

Entrance fee: 20

Cost per ride: 5 × 26 = 130

Total cost: 20 + 130 = 150

The total matches the given information, confirming that x = 26 rides is correct.

Interpretation: The equation 20 + 5x = 150 accurately represents the relationship between the number of rides taken (𝑥) and the total amount spent by the group. Solving this equation provides a clear answer to how many rides the group enjoyed based on the provided pricing structure. Therefore, the group of friends took 26 rides at the amusement park.

NCERT Exemplar Solutions

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.

Questions

1.A home-owner is installing a fence around the square garden. The garden has a perimeter of 6480 cm. Write and solve the equation to find the garden’s dimensions.

2.Distance between two stations A and B is 690 km. Two cars start simultaneously from A and B towards each other, and the distance between them after 6 hours is 30 km. If the speed of one car is less than the other by 10 km/hr, find the speed of each car.

3.A steamer goes downstream from one point to another in 7 hours. It covers the same distance upstream in 8 hours. If the speed of stream be 2 km/hr, find the speed of the steamer in still water and the distance between the ports.

4.The present age of father is four times the age of his son. After 10 years, age of father will become three times the age of his son. Find their present ages.

5.Solve: x2 + x4 + x5+ 10000 = x

Sol

Solution:

1.Fence around square garden = Perimeter of square garden = s + s + s + s = 6480 cm

4s = 6480 cm

s = 64804 = 1620 cm

Thus, side of garden = 1620 cm

2.Solution:

Let speed of faster car = x km/hr

Then speed of other = (x – 10) km/hr

Let 1st one start from A and other from B.

M and N be their position after 6 hours.

AM = 6x,

BN = 6(x –10)

According to condition,

6x + 6x – 60 + 30 = 690

12x = 690 + 30

12x = 720

x = 60 km/hr

Speed of other car = 60 km/hr.

3.Solution:

Let speed of steamer in still water = x km/hr

Speed of stream = 2 km/hr

Speed downstream = (x + 2) km/hr

Speed upstream = (x – 2) km/hr

Distance covered in 7 hours while downstream = 7(x + 2)

Distance covered in 8 hours while upstream = 8(x – 2)

According to the condition,

7(x + 2) = 8(x – 2)

7x + 14 = 8x – 16

x = 30 km/hr

Total Distance = 7(x + 2) km

= 7(30 + 2) km

= 7 × 32 km = 224 km

Total Distance between ports is 224 km

4.Solution:

Let the present age of son be x years

Thus, the present age of father = 4x years

After 10 years:

Age of son = (x + 10) years

Age of father = (4x + 10) years

According to the given condition,

4x + 10 = 3(x + 10)

4x + 10 = 3x + 30

4x – 3x = 30 – 10

x = 20

Therefore, Present age of son = 20 years and present age of father = 4x = 4 × 20 = 80 years

5.Solution: x2 + x4 + x5+ 10000 = x

x2 + x4 + x5 - x = 10000

10x+5x+4x20x20 = 10000

x20 = 10000

x = 200000

State whether statements are True or False:

Questions

(a)Three consecutive even numbers whose sum is 156 are 51, 52 and 53.

(b) x = –12 is the solution of the linear equation 5x –3(2x + 1) = 21 + x

Sol

Solution:

(a) False

(b) True

Choose the correct option for the below given questions

Questions

1.If 3x – 4 (64 – x) = 10, then the value of x is:

(a) –266

(b) 133

(c) 66.5

(d) 38

2.If x = a, then which of the following is not always true for an integer k:

(a) kx = ak

(b) xk = ak

(c) x – k = a – k

(d) x + k = a + k

Sol

Solution:

  1. (d)

  2. (b)

Fill in the blanks:

Questions

(a) Fifteen added to thrice a whole number gives 93. The number is ?.

(b) If 13 - x = 23 then x is ?.

Sol

Solution:

(a) 26

(b) 1

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.

Q1

Anima left one-half of her property to her daughter, one-third to her son and donated the rest to an educational institute. If the donation was worth Rs. 1,00,000. Based on the above situation, answer the following questions:

(i) Write the linear equation formed in the above situation.

(ii) How much money did Anima have?

(OR)

How much money educational institute have?

(iii) How much money did Anima‘s son and daughter have?

Sol

Solution:

Anima left:

One-half of her property to her daughter

One-third of her property to her son

The remaining amount was donated to an educational institute, valued at Rs. 100,000.

(i) Write the linear equation formed in the above situation:

Let 𝑃 denote the total amount of money (in Rs.) that Anima had.

The equation based on the distribution is: 12𝑃 + 13 𝑃 + 100,000 = P

This equation represents the total amount of money Anima had, considering she divided her property between her daughter, son, and the donation.

(ii) How much money did Anima have?

To find out how much money Anima had, solve the equation: 12𝑃 + 13 𝑃 + 100,000 = P

36𝑃 + 26 𝑃 + 100,000 = P

56 𝑃 + 100,000 = P

Subtract 56 𝑃 from both sides to isolate terms with P on one side:

100,000 = P − 56P

100,000 = 16P

P = Rs. 600,000

(iii) Now that we know the total amount Anima had, let's calculate how much her son and daughter received:

Amount left for distribution after donation:

Amount left = P − 100,000 = 600,000 − 100,000 = Rs. 500,000

Amount for daughter = 12 × 500,000 = Rs. 250,000

Amount for son = 13 × 500,000 = Rs. 166,666.67

Q2

A home-owner is installing a fence around the square garden. The garden has a perimeter of 7840 cm. Write and solve the equation to find the garden‘s dimensions.

(a) Find out the side of garden.

(b) Why garden is important for us?

Sol

(a) The perimeter P of a square with side length s is given by:

P = 4s

We are given the perimeter of the garden as 7840 cm. So we can write the equation:

4s = 7840

Divide both sides of the equation by 4 to find the side length s:

s = 78404 = 1960 cm

So, the side length of the garden is 1960 cm.

(b) Overall, gardens contribute to the environment, support biodiversity, enhance physical and mental health, provide fresh produce, and offer aesthetic and recreational benefits.