NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.4) Exercise 2.4

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CBSE NCERT Solutions for Class 8 Maths Chapter 2 – Exercise 2.4

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Important Topics under NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.4) Exercise 2.4

Class 8 Maths Chapter 2 Exercise 2.4, includes questions based on some applications of Linear Equations in One Variable. This exercise primarily covers linear equations with variables on both sides. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4, greatly reduces students’ burden as they attempt to solve these questions. The problems presented in this exercise have clear solutions in the Extramarks provided NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4, are written by knowledgeable subject experts with consideration for students’ level of comprehension in Class 8. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4, are written using the most recent curriculum that the CBSE has approved for use in Class 8. As a result, students no longer need to search for reliable solutions elsewhere or worry about the syllabus. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4, are a one-stop destination for Class 8 students to find thorough, trustworthy solutions to all of the exercises’ questions.

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Importance of Linear Equations in One Variable

Linear Equations in One Variable are one of the most important concepts in Mathematics. Not only in Mathematics, linear equations are used in many other fields of study, such as Physics, Chemistry, Economics, Biology, etc. They are useful for various types of problems in daily life. Linear equations are used to solve a wide range of problems, such as problems on ages, numbers, area, perimeter, cost, etc. Students need to have the best study resources available to understand these concepts. Due to the fact that Extramarks provides a 360-degree solution for all of their study needs, students do not need to worry about looking for the appropriate resources. Students learn a lot of new concepts about Linear Equations in this chapter. Some of these are initially difficult to grasp. But students can get better at these with consistent practice. The 3D animations provided by Extramarks help students understand each topic better. Through the use of the chapter-wise worksheets and practice questions provided by Extramarks, they can understand all the concepts required by the curriculum. Students should usually check the CBSE syllabus to be aware of the topics they need to study. They should read each chapter of the textbook in detail to understand the concepts of linear equations and their applications. They should attempt to answer every question in the exercises of the textbook. Students can practice the exercises with the help of Extramarks’ NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4.

Access NCERT solutions for Math Chapter 2 – Linear Equations in One

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Exercise 2.4 NCERT Solutions Class 8 Maths Chapter 2

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NCERT Solutions Class 8 Math Chapter 2 All Exercises

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Solutions for Linear Equations in One Variable NCERT Questions

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NCERT Solutions for Class 8

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Q.1

Amina thinks of a number and subtracts 52 from it. She multiplies the result by 8. The result now obtained is3 times the same number she thought of. What is the number?

Ans

Let the number be x.
According to the given question,

8(x52)=3x

8x − 20 = 3x
Transposing 3x to L.H.S and −20 to R.H.S, we obtain
8x − 3x = 20
5x = 20
On dividing both sides by 5, we get
x = 4
Hence, the number is 4.

Q.2 A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers? What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Ans

Let the numbers be x and 5x.

According to the question, the equation becomes 21+5x=2(x+21)21+5x=2x+42Transposing 2x to L.H.S and 21 to R.H.S, we obtain5x2x=42213x=21Dividing both sides by 3, we obtainx=75x=5×7=35Hence, the numbers are 7 and 35 respectively.

Q.3 Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Ans

Let the digits at tens place and ones place be x and 9 − x respectively.
Therefore, original number = 10x + (9 − x) = 9x + 9
On interchanging the digits, the digits at ones place and tens place will be x and 9 − x respectively.

Therefore, new number after interchanging the digits = 10(9 − x) + x
= 90 − 10x + x
= 90 − 9x

According to the given question,
New number = Original number + 27
90 − 9x = 9x + 9 + 27
90 − 9x = 9x + 36

Transposing 9x to R.H.S and 36 to L.H.S, we obtain
90 − 36 = 18x
54 = 18x

Dividing both sides by 18, we obtain
3 = x and 9 − x = 6

Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.

Therefore, the two-digit number is 9x + 9
= 9 × 3 + 9
= 36

Q.4 One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Ans

Let the digits at tens place and ones place be ‘x and ‘3x.the original number=10x+3x=13xOn interchanging the digits, the digits at ones place will be ‘x’and tens place will be ‘3x.New number=10×3x+x=30x+x=31xAccording to the given question,the equationbecomes 8813x+ 31x= 8844x= 88

Dividing both sides by 44, we obtain44x44=8844x= 2, original number=13x=13×2=26Hence, the twodigit number may be 26 or 62.

Q.5 Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Ans

Let Shobo’s age be x years.
His mother’s age will be 6x years.
According to the given question, the equation becomes
:

x+5=6x3

Multiplying both sides by 3

(x+5)×3=6x3×33x+15=6xTransposing 3x to R.H.S,we get15=6x3x15=3xDividing both sides by 3, we get153=3x3x=56x=6×5=30

Therefore, the present ages of Shobo and his mother will be 5 years and 30 years.

Q.6 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate ₹100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?

Ans

Let the ratio between the length and breadth of the rectangular plot be x.

Hence, the length and breadth of the rectangular plot will be 11x m and 4x m respectively.

Perimeter of the plot = 2(Length + Breadth)

=[2(11x+4x)]m=30x

It is given that the cost of fencing the plot at the rate of ₹ 100 per metre is ₹ 75, 000.

100 × Perimeter = 75000

100 × 30x = 75000

3000x = 75000

Dividing both sides by 3000, we obtain

3000x 3000 = 75000 3000 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakeaadaWcaaqaaiaabodacaaIWaGaaGimaiaaicdacaWG4baabaGaaG4maiaaicdacaaIWaGaaGimaaaacqGH9aqpcaqGGaWaaSaaaeaacaqG3aGaaeynaiaaicdacaaIWaGaaGimaaqaaiaaiodacaaIWaGaaGimaiaaicdaaaaaaa@482E@

x = 25

Therefore, Length = 11x m = (11 × 25) m = 275 m

and Breadth = 4x m = (4 × 25) m = 100 m

Hence, the dimensions of the plot are 275 m and 100 m.

Q.7 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,600. How much trouser material did he buy?

Ans

Let the trouser and shirt material bought by Hasan be 2x and 3x .
The selling price of trouser material is

=(90+90×12100)=100.80

The selling price of Shirt material is

=( 50+ 50×10 100 ) = 55 MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiGacaGaaeqabaWaaqaafaaakqaabeqaaiabg2da9iaaysW7tCvAUfKttLearyGu1bxzLbIrVjxyKLwyUbacgaGae8hyaa2aaeWaaeaacaaI1aGaaGimaiabgUcaRmaalaaabaGaaGynaiaaicdacqGHxdaTcaaIXaGaaGimaaqaaiaaigdacaaIWaGaaGimaaaaaiaawIcacaGLPaaaaeaacqGH9aqpcaaMe8Uae8hyaaMae8hiaaIaaGynaiaaiwdaaaaa@5607@

Total sale =36,600

100.80×(2x)+55×(3x)=36,600201.60x+165x= 36,600366.60x=36600Dividing both sides by 366.60, we obtain366.60x366.60=36600366.60x=99.836x=100Trouser material=2xm =(2×100)m =200 m

Q.8 Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Ans

Let the number of deer be x.

Number of deer grazing in the field = x / 2

Remaining deer= x – (x / 2)

Therefore, number of deer playing nearby is

=34×(xx2)=34×x2=3x8

The rest 9 are drinking water from the pond, so the equation becomes:

x(x2+3x8)=9x(4x+3x8)=9x7x8=9

8x7x8=9x8=9

Mutiplying both sides by 8,we getx8×8=9×8x=72

Therefore, the total number of deer in the herd is 72.

Q.9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Ans

Let the granddaughter’s age be x years.

Therefore, grandfather’s age will be 10x years.

According to the question, the equation becomes:

10x = x + 54

Transposing x to L.H.S, we obtain

10xx = 54

9x = 54

x = 6

Granddaughter’s age = x years = 6 years

Grandfather’s age = 10x years

= (10 × 6) years

= 60 years

Therefore, the grandfather’s age is 60 years and granddaughter’s age is 6 years.

Q.10 Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Ans

Let Aman’s son’s age be x years.
Therefore, Aman’s age will be 3x years.
Ten years ago, their age was (x − 10) years and (3x − 10) years.
According to the question, the equation is:

3x10=5(x10)3x10=5x50On transposing 5x to L.H.S​ and 10 to R.H.S,we get3x5x=10502x=402x=40

On dividing both the sides by 2,we get2x2=402x=20

Therefore,
Aman’s son’s age is 20 years and
Aman’s age = 3 × 20 years = 60 years.

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FAQs (Frequently Asked Questions)

1. Where are the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4 available for Class 8 students to access?

Students can find the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4 on the Extramarks website and mobile app. The NCERT solutions are available online in PDF format for students to download. This makes the solutions available offline so that students can use them whenever and wherever they like without being connected to the internet.

2. Are the concepts under the NCERT Class 8 Maths Chapter 2 Exercise 2.4 clarified by the NCERT solutions?

For students in Class 8, NCERT solutions are crucial for exam preparation. In order to prepare for the annual exams, students can use learning modules like the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4.  Students can develop the ability to respond to any question of a similar nature that appears in the exam by becoming more familiar with the steps and procedure.

3. When preparing for the board exams, how should students use the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.4?

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