Our ncert solutions for Exercise 2.4 Class 8 maths 2. Linear Equations in One Variable - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.
Exercise 2.4 Class 8 maths 2. Linear Equations in One Variable - ncert solutions - Toppers Study
Topper Study classes prepares ncert solutions on practical base problems and comes out with the best result that helps the students and teachers as well as tutors and so many ecademic coaching classes that they need in practical life. Our ncert solutions for Exercise 2.4 Class 8 maths 2. Linear Equations in One Variable - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.
Exercise 2.4 Class 8 maths 2. Linear Equations in One Variable - ncert solutions - Toppers Study
CBSE board students who preparing for class 8 ncert solutions maths and Mathematics solved exercise chapter 2. Linear Equations in One Variable available and this helps in upcoming exams 2024-2025.
You can Find Mathematics solution Class 8 Chapter 2. Linear Equations in One Variable
- All Chapter review quick revision notes for chapter 2. Linear Equations in One Variable Class 8
- NCERT Solutions And Textual questions Answers Class 8 Mathematics
- Extra NCERT Book questions Answers Class 8 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.
NCERT Solutions do not only help you to cover your syllabus but also will give to textual support in exams 2024-2025 to complete exercise 2.4 maths class 8 chapter 2 in english medium. So revise and practice these all cbse study materials like class 8 maths chapter 2.4 in english ncert book. Also ensure to repractice all syllabus within time or before board exams for ncert class 8 maths ex 2.4 in english.
See all solutions for class 8 maths chapter 2 exercise 2 in english medium solved questions with answers.
Exercise 2.4 class 8 Mathematics Chapter 2. Linear Equations in One Variable
Sure! The following topics will be covered in this article
- Exercise 2.4 Class 8 Maths 2. Linear Equations In One Variable - Ncert Solutions - Toppers Study
- Class 8 Ncert Solutions
- Solution Chapter 2. Linear Equations In One Variable Class 8
- Solutions Class 8
- Chapter 2. Linear Equations In One Variable Exercise 2.4 Class 8
Notice: Undefined offset: 5 in /home/atpeduca/public_html/toppersstudy.com/view-home.php on line 123
2. Linear Equations in One Variable
| Exercise 2.4 |
Exercise 2.4 Class 8 maths 2. Linear Equations in One Variable - ncert solutions - Toppers Study
Q1. Amna thinks of a number and subtract 5/2 from it. She multiplies result from 8 the result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let, the number be x
So, A.T.Q the equation will be: (x - 5/2 ) 8 = 3x
⇒ 8x - 40/2 = 3x
⇒ 8x – 3x = 20
⇒ 5x = 20
⇒ x = 20/5 = 4
So, the number is x = 4.
Q2. A positive number is 5 times another number. If 21 are added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution:
Let, the first number is x
And the second number is 5x
2(x + 21) = 5x + 21
⇒ 2x + 42 = 5x + 21
⇒ 5x – 2x = 42 – 21
⇒ 3x = 21
⇒ x = 21/7 =7
So, the both numbers will be x = 7 × 1 = 7 and 5x = 5 × 7 = 35 respectively.
Q3. 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?
Solution:
Let, the tens digit of number is = x
The once digit of number is = y
So, the sum of digits x + y = 9---------------- (1)
So, the number will be (10 × x) + y
= 10x + y ----------------- (2)
When we under change the digits then the numbers will be: 10y + x ------------ (3)
By equation (1) and (2)
(10x + y) – (10y + x) = 27
⇒ 10x + y – 10y – x = 27
⇒ 9x – 9y = 27
⇒ 9(x – y) = 27
⇒ x – y = 27/3 = 3
So, x – y = 3 -------------- (4)
Adding equation (3) and (4)
x + y = 9
+ x – y = 3 .
2x = 12
⇒ x = 12/2 = 6
Putting x = 6 in equation (1)
6 + y = 9
⇒ y = 9 – 6 = 3
At last the number = 10x + y = 10 × 6 + 3 = 63
Q4. 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?
Solution:
Let, the first digit be x
So, the second digit will be 3x
So, the number will be 10 × x + 3x = 13x
Interchanging the digits 10 × 3x + x = 31x
A.T.Q the equation will be 13x + 31x = 88
⇒ 44x = 88
⇒ x = 44/88 = 2
So, the original number is 13x = 13 × 2 = 26
Q5. 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?
Solution:
Let the age of Shobo be x
So, the age of her mother is 6x
ATQ, the equation will be: 3(x + 5) = 6x
⇒ 3x + 15 = 6x
⇒ 6x – 3x = 15
⇒ 3x = 15
⇒ x = 15/3 = 5
so the age of shobo x = 5
and the age of shobo mother 6x = 6 × 5 = 30
Q6. 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 meter it will cost the village panchayat 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Let, the length and the breadth of the plot is 11x and 4x respectively.
So, the perimeter of plot = 2(l+b)
= 2(11x+4x) = 22x + 8x= 30x
Perimeter of plot = the total cost of plot/per meter cost
So, the length of the plot 11x = 11×25 = 275m
The breadth of the plot 4x = 4×25 = 100m
Q7. Hasan buys two kinds of cloths materials for school uniforms, shirt materials that cost him R.s 50 per meter and trouser material that costs him R.s 90 per meter. For every 3 meters of the shirt material he buys 2 meters of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is ` 36,600. How much trouser material did he buy?
Solution:
Q9. Grandfather is ten times older than his granddaughter.
He is also 54 years older than her. Find their present ages.
Solution:
Let, the age of granddaughter be x and the age of grandfather 10x
According to the question the equation will be:
⇒ 10x – x = 54
⇒ 9x = 54
so, the age of grandfather is 10x = 10 × 6 = 60
and the age of granddaughter x = 6
Q10. 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.
Solution:
Let, the age of Aman’s son xv and the age of Aman be 3x
According to the question the equation will be
5(x-10) = 3x-10
⇒ 5x – 50 = 3x - 10
⇒ 5x – 3x = -10 + 50
⇒ 2x = 40
So, the age of Aman's son x = 20
and the age of Aman 3x = 3 × 20 = 60
Other Pages of this Chapter: 2. Linear Equations in One Variable
Select Your CBSE Classes
Important Study materials for classes 06, 07, 08,09,10, 11 and 12. Like CBSE Notes, Notes for Science, Notes for maths, Notes for Social Science, Notes for Accountancy, Notes for Economics, Notes for political Science, Noes for History, Notes For Bussiness Study, Physical Educations, Sample Papers, Test Papers, Mock Test Papers, Support Materials and Books.
Mathematics Class - 11th
NCERT Maths book for CBSE Students.
books
Study Materials List:
Solutions ⇒ Class 8th ⇒ Mathematics
Topper's Study
New Books