VMC VIQ Scholarship Test
ApplyRegister for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.
Linear Equations in One Variable Class 8 Questions And Answers provided here. These NCERT Solutions are prepared by subject matter expert of Careers360 considering latest syllabus and pattern of CBSE 2023-24. this chapter created a strong foundation for algebra unit. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable is comprehensively covering each problem related to this particular topic.
In this chapter of NCERT Syllabus for Class 8 Maths, there will be an expression consisting of variables as well as numbers and you have to find the value of that variable. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable are covering solutions to word problems using the linear equations of one variable. This chapter has a total of 6 exercises consisting of a total of 65 questions in all. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable has the solution to all 65 questions to boost your overall preparation level.
Linear equations with a linear expression on one side:
Transpose the number to the side with all the numbers, maintaining the sign.
Solve (add/subtract) the equation on both sides to simplify and find the variable's value.
Linear equations with variables on both sides:
Transpose both the number and the variable to one side, maintaining the sign of the number.
Solve (add/subtract) the equation on both sides to simplify and find the variable's value.
Linear equations with a number in the denominator and variables on both sides:
Find the LCM (Least Common Multiple) of the denominators on both sides.
Multiply both sides by the LCM to simplify the equation.
Solve it like a linear equation with variables on both sides to find the variable's value.
Linear equations that are reducible to the linear form:
If the equation is of the form (x + a / x + b) = c / d, cross-multiply the numerator and denominator to simplify it to a linear form, such as (x + a) * d = c * (x + b).
Solve this linear equation with variables on both sides to find the variable's value.
Free download NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable for CBSE Exam.
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable - Excercise: 2.1
Q1 Solve the following equations x – 2 = 7
Answer: 1. Transposing -2 to the RHS, we get
x = 7 + 2 = 9
Q2 Solve the following equation y + 3 = 10
Answer: Transposing 3 to the RHS, we get
y = 10 - 3 = 7
y = 7
Q3 Solve the following equation 6 = z + 2
Answer: Transposing 2 to the LHS, we get
6 - 2 = z => z= 4
Thus z = 4
Q 7 Solve the following equations.
Answer: 7.
Multiplying both sides by 3, we get
Now, dividing both sides by 2, we get
Q9 Solve the following equation 7x – 9 = 16
Answer: Transposing 9 to the RHS, we get
7x = 16 + 9 = 25
Now, dividing both sides by :
Q10 Solve the following equation 14y – 8 = 13
Answer: Transposing -8 to the RHS, we get
14y = 13 + 8 = 21
Now dividing both sides by 14:
Q11 Solve the following equation 17 + 6p = 9
Answer: At first, transposing 17 to the RHS:
6p = 9 - 17 = -8
Now, dividing both sides by 6, we get
Q12 Solve the following equations.
Answer: 12.
Transposing 1 to the RHS, we get
Now multiplying both sides by 3, it becomes
Class 8 maths chapter 2 question answer - Exercise: 2.2
Q1 If you subtract from a number and multiply the result by , you get . What is the number?
Answer: Assume the number to be x.
Thus according to the question,
Multiplying both sides by 2, we get
Now transposing to the RHS, we get
Thus the number was x = 3/4
Answer: Let the breadth of the pool be x m.
According to the question, the length of the pool = 2x + 2 m.
Perimeter of recatangle = 2(l + b) = 154 m .
i.e., 2(2x + 2 + x) = 154
2(3x + 2) = 154
Dividing both sides by 2, we get
3x + 2 = 77
Now transposing 2 to the RHS, we get
3x = 77 - 2 = 75
Dividing both sides by 3, we get
Thus breadth of pool = 25 m
and lenght of the pool = 2 25 + 2 = 52 m
Answer: In isosceles triangles, we have 2 sides of equal length.
Given that its perimeter is cm.
Let's assume the length of the equal side is x cm.
Also,
Transposing to the RHS side it becomes,
Now dividing both sides by 2, we get
Hence, the length of the equal sides of the isosceles triangle is cm.
Q4 Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Answer: Assume one number to be x. Then the other number will be x + 15.
Now it is given that sum of the two numbers is 95.
Thus quation becomes x + x + 15 = 95
or 2x +15 = 95
Transposing 15 to the RHS, it becomes
2x = 95 - 15 = 80
Dividng both sides by 2, we get
Thus two numbers are 40 and 55
Q5 Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Answer: Given that numbers are in the ratio of 5:3, so we can assume numbers to be 5x and 3x.
Also, the difference between these numbers is 18, so the equation becomes
5x - 3x = 18
2x = 18
Dividing both sides by 2, we get
x = 9
Hence the two numbers are 5 9 = 45 and 3 9 = 27
Q6 Three consecutive integers add up to 51. What are these integers?
Answer: Let the consecutive integers be x, x+1, x+2.
Sum of these integers is given to be 51.
Thus equation becomes : x + x + 1 + x + 2 = 51
3x + 3 = 51
Transposing 3 to the RHS,
3x = 48
Now dividing both sides by 3, we get
Hence the consecutive integers are 16, 17, 18.
Q7 The sum of three consecutive multiples of 8 is 888. Find multiples.
Answer: Let x be the multiple of 8.
Then the three consecutive integers (multiple of 8) are x, x+8, x+16.
Given their sum is 888,
thus the equation becomes: x + x + 8 + x + 16 = 888
or 3x + 24 = 888
Transposing 24 to the RHS
3x = 888 - 24 = 864
Now dividing both sides by 3, we get
Thus the required consecutive integers are 288, 296, 304.
Answer: Let the consecutive integers be x, x+1, x+2.
Now according to the question equation becomes,
2x + 3(x+1) + 4(x+2) = 74
or 2x + 3x + 3 + 4x + 8 = 74
or 9x + 11 = 74
Transposing 11 to the RHS we get,
9x = 74 - 11 = 63
Divinding both sides by 9.
Therefore required consecutive integers are 7, 8, 9.
Answer: Let the current age of Rahul and Haroon be 5x and 7x respectively(Since there age ratio is given as 5:7).
Four years later their ages become 5x + 4 and 7x + 4 respectively.
According to the question,
5x + 4 +7x + 4 = 56
or 12x + 8 = 56
Transposing 8 to the RHS we get,
12x = 48
Dividing both sides by 12:
Thus current age of Rahul = 5 4 = 20
Haroon = 7 4 = 28
Answer: Let us assume the number of boys and the number of girls is 7x and 5x.
According to the given data in the question,
7x = 5x + 8
Transposing 5x to the LHS we get,
7x - 5x = 8
2x = 8
Dividing both sides by 2
x = 4
Hence number of boys = 7 4 = 28
and number of girls = 5 4 = 20
So total number of students = 28 + 20 = 48.
Answer: Let the age of Baichung be x years.
Then according to question age of Baichung's father = x + 29 years
and age of Baichung's grandfather = x + 29 + 26 years
The sum of their ages is given 135 years. According to that, x + x + 29 + x + 29 + 26 = 135
3x + 84 = 135
Transposing 81 on the RHS we get,
3x = 135 - 84 = 51
Dividing both sides by 3
Thus, the age of Baichung = 17 years
The age of Baichung's father = 17 + 29 = 46 years
The age of Baichung's grandfather = 46 + 26 = 72 years
Q12 Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Answer: Let us assume Ravi's present age to be x years.
According to the question,
x + 15 = 4x
Transposing x to the RHS
15 = 3x
Now dividing both sides by 3, we get
Hence Ravi's current age is 5 years.
Answer: Let the rational number be x.
According to the question,
Transposing to the RHS:
Multiplication by 2 in both sides, we get
Dividing both sides by 5, we get
Answer: Since the ratio of currency notes is 2:3:5. Therefore,
Let the number of currency notes of Rs.100, Rs.50 and Rs.10 be 2x, 3x, 5x respectively.
Hence according to the question equation becomes,
100 2x + 50 3x + 10 5x = 400000
or 200x + 150x + 50x = 400000
or 400x = 400000
Dividing both sides by 400 we get,
x = 1000 .
No. of Rs.100 notes = 2 1000 = 2000 notes
No. of Rs.50 notes = 3 1000 = 3000 notes
No. of Rs.10 notes = 5 1000 = 5000 notes
Answer: Let the no. of Rs.1 coin be y.
And the no. of Rs.5 coin be x.
Thus according to question no. of Rs.2 coin will be 3x.
Also, the total no. of coins = 160
This implies : y + x + 3x = 160
or y + 4x = 160
Transposing 4x to the RHS
y = 160 - 4x = No. of Rs.1 coin.
Now, it is given that total amount is Rs.300.
i.e., 1(160-4x) + 2(3x) + 5(x) = 300
or 160 - 4x + 6x + 5x = 300
or 160 + 7x = 300
Transposing 160 to the RHS :
7x = 300 - 160 = 140
Dividing both sides by 7:
x = 20
Thus number of Rs.1 coin = 160 - 4x = 160 - 80 = 80
number of Rs.2 coin = 3x = 60
number of Rs.5 coin = x = 20
Answer: Let the no. of winners be x.
Since total no. of participants is 63, thus no. of participants who does not win = 63 - x
According to the question, equation becomes:
x(100) + (63-x)(25) = 3000
or 100x + 1575 - 25x = 3000
or 75x + 1575 = 3000
Transposing 1575 to the RHS :
75x = 3000 - 1575
or 75x = 1425
Dividing both sides by 75, we get
x = 19
Therefore no. of winners = 19
Class 8 maths chapter 2 ncert solutions - Exercise: 2.3
Q1 Solve the following equations and check your results 3x = 2x + 18
Answer: We have
3x = 2x + 18
Subtracting 2x from both sides
3x - 2x = 2x - 2x + 18
or x = 18
Check :- Put x = 18 in both LHS and the RHS.
LHS = 3x = 3(18) = 54
RHS = 2x + 18 = 2(18) + 18 = 36 + 18 = 54
Thus, LHS = RHS
Q2 Solve the following equation and check your results 5t – 3 = 3t – 5
Answer: We have
5t - 3 = 3t - 5
Transposing 3t to the LHS and -3 to the RHS, we get:
5t - 3t = -5 + 3
or 2t = -2
Dividing both sides by 2 :
t = -1
Check :- Put t = -1 in the LHS we have,
5t - 3 = 5(-1) - 3 = -5 -3 = -8
Similarly put t = -1 in the RHS:
3t - 5 = 3(-1) - 5 = -3 - 5 = -8
Hence, LHS = RHS
Q3 Solve the following equation and check your result 5x + 9 = 5 + 3x
Answer: We have
5x + 9 = 5 + 3x
Transposing 3x to the LHS and 9 to the RHS, we get:
5x - 3x = 5 - 9
or 2x = -4
Dividing both sides by 2 :
x = -2
Check :- Put x = - 2 in both LHS and RHS
LHS :- 5x + 9 = 5(-2) + 9 = -10 + 9 = -1
RHS :- 5 + 3x = 5 + 3(-2) = 5 - 6 = -1
Hence, LHS = RHS
Q4 Solve the following equation and check your result. 4z + 3 = 6 + 2z
Answer: We have
4z + 3 = 6 +2z
Transposing 2z to the LHS and 3 to the RHS, we get:
4z - 2z = 6 - 3
or 2z = 3
DIviding both sides by 2,
Check :- Put in LHS as well as RHS, we have
LHS :- 4z + 3
RHS :- 6 + 2z
Thus, LHS = RHS
Q5 Solve the following equation and check your result 2x – 1 = 14 – x
Answer: We have
2x - 1 = 14 - x
Transposing -x to the LHS and -1 to the RHS
2x + x = 14 + 1
or 3x = 15
Dividing both sides by 3, we get
x = 5
Check :- Put x = 5 in both the LHS and the RHS
LHS :- 2x - 1 = 2(5) - 1 = 10 - 1 = 9
RHS :- 14 - x = 14 - 5 = 9
Hence, LHS = RHS
Q6 Solve the following equation and check your result 8x + 4 = 3 (x – 1) + 7
Answer: We have
8x + 4 = 3 (x - 1) + 7
or 8x + 4 = 3x - 3 + 7 = 3x + 4
Transposing 3x to the LHS and 4 to the RHS, we get
8x - 3x = 4 - 4 = 0
5x = 0
Dividing both sides by 5:
x = 0
Check :- Putting x = 0 in both LHS and RHS:
LHS :- 8x + 4 = 8(0) + 4 = 4
RHS :- 3(x-1) + 7 = 3(-1) + 7 = -3 + 7 = 4
Hence, LHS = RHS
Q7 Solve the following equations and check your results.
Answer: We have
Transposing to the left-hand side:
or
or
Multiplying both sides by 5
x = 40
Check :- Put x = 40 in both the LHS and the RHS.
LHS :- x = 40
RHS :-
Thus, LHS = RHS
Q8 Solve the following equation and check your result.
Answer: We have
Transposing to the LHS and 1 to the RHS:
or
or
or x = 10
Check :- Put x = 10 in both the LHS and the RHS
LHS :
RHS :
Thus, LHS = RHS
Q9 Solve the following equations and check your results.
Answer: We have
Transposing -y to the LHS and to the RHS, we get
or 3y = 7
Dividing both sides by 3:
i
Check :- Put in both the LHS and the RHS:
LHS :
RHS :
Hence, LHS = RHS
Q10 Solve the following equation and check your result.
Answer: We have
Transposing 5m to the LHS
or
Dividing both sides by -2, we get
Check :- Put in the LHS and the RHS:
LHS :-
RHS :-
Hence, LHS = RHS
Class 8 linear equations in one variable NCERT solutions - Exercise: 2.4
Answer: Let us assume that the number Amina thought was x.
As per the question,
or
Transposing 3x to the LHS and -20 to the RHS, we get
8x - 3x = 20
or 5x = 20
Dividing both sides by 5.
x = 4
Therefore the number Amina thought was 4.
Answer: Let the No.1 = x
Then the second no. No.2 = 5x.
Now the question states that if we add 21 to both numbers one number becomes twice of the other.
i.e., 2(x + 21) = 5x + 21
or 2x + 42 = 5x + 21
Transposing 2x to the RHS and 21 to the LHS:
21 = 5x - 2x = 3x
or x = 7
Thus the two numbers are 7 and 35 . (Since No.2 = 5x)
Answer: Given that sum of the two digits is 9.
Let us assume the digit of units place be x.
Then the digit of tens place will be 9-x.
Thus the two digit number is 10(9-x) + x
Now if we reverse the digits, the number becomes 10x + (9-x).
As per the question,
10x + (9-x) = 10(9-x) + x + 27
or 9x + 9 = 90 - 10x + x + 27
or 9x + 9 = 117 - 9x
Transposing -9x to the LHS and 9 to the RHS:
9x + 9x = 117 - 9
or 18x = 108
x = 6
Thus two digit number is 36.
Answer: Let us assume that the units place of a two digit number is x.
Then as per the question, tens place of the number is 3x.
Thus the two digit number is 10(3x) + x.
If we interchange the digits the number becomes : 10x + 3x
According to the question,
10(3x) + x + 10x + 3x = 88
or 30x + 14x = 88
or 44x = 88
Dividing both sides by 44, we get
x = 2
Hence the two digit number is 10(6) + 2 = 62
Note that the two-digit number can also be found if we reverse the digits. So one more possible number is 26.
Answer: Let us assume that the Shobo's present age be x.
So, Shobo's mother's present age = 6x.
Shobo's age five years from now will be = x + 5.
It is given that Shobo’s age five years from now will be one-third of his mother’s present age.
So,
or x + 5 = 2x
Transposing x to the RHS,
5 = 2x - x = x
Hence Shobo's present age is x = 5 years
and Shobo's mother's present age is 6x = 30 years
Answer: Let the length and breadth of the plot be 11x and 4x respectively.
(Since they are in the ratio11:4)
We know that the fencing will be done on the boundary of the plot, so we need to calculate its perimeter.
Perimeter of rectangle = 2(lenght + breadth)
Total cost to fence the plot = Rs.75000 = Rs.100(Perimeter of plot)
Thus the equation becomes,
100 [2(11x + 4x)] = 75000
or 200(15x) = 75000
or 3000x = 75000
Dividing both sides by 3000:
x = 25
Hence length of plot is 11x = 275 m
and breadth of the plot is 4x = 100m
Answer: Let the total shirt material that Hasan bought be x metre.
and the total trouser material is y metre.
As per the question,
Thus trouser material
Now we will make another equation by using cost of the materials.
Profit for shirt is 12% i.e., selling price of shirt is
and selling price for trouser is
Hence equation becomes:
or 56x + 66x = 36600
or 122x = 36600
Dividing both sides by 122:
x = 300
Thus trouser material that Hasan bought was = 200 meter
Answer: Let us assume the number of deer in the herd to be x.
Half of a herd of deer are grazing in the field i.e., .
Now half are remaining.
Three-fourths of the remaining are playing nearby, implies :
The remaining number of deer will be :
Given that rest is 9, so
Hence the number of deer in the herd = 72
Answer: Let the age of granddaughter is x years.
So the age of grandfather is x + 54.
Also, grandfather is ten times older than his granddaughter.
i.e., x + 54 = 10x
Transposing x to the RHS :
54 = 10x - x
or 54 = 9x
or x = 6
Therefore age of granddaughter = 6 years
and the age of grandfather = 6 + 54 = 60 years
Answer: Let the age of Aman's son is x years.
So the age of Aman will be 3x years.
Ten years ago, the age of Aman's son = x - 10 and Aman's age = 3x - 10.
So according to question,
5(x - 10) = 3x - 10
or 5x - 50 = 3x - 10
Transposing 3x to the LHS and -50 to the RHS:
5x - 3x = -10 + 50 = 40
or 2x = 40
or x = 20
So Aman's age = 3x = 60 years and
Aman's son's age = x = 20 years.
NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Excercise: 2.5
Q1 Solve the following linear equations.
Answer: The detailed solution of the above-written question is here,
We have
or
or
or
Multiplying both sides by 6:
Thus
Q2 Solve the following linear equations.
Answer: We have
Multiplying both sides 12.
(We multiplied both sides by 12 because it is the lowest common factor.)
6n - 9n + 10n = 252
or 7n = 252
or n = 36
Q3 Solve the following linear equations.
Answer: We have
Multiplying both sides by 6, we get
6x + 42 - 16x = 17 - 15x
or -10x +42 = 17 - 15x
Transposing -15x to the LHS and 42 to the RHS :
-10x + 15x = 17 - 42
or 5x = -25
or x = -5
Q4 Solve the following linear equations.
Answer: We have
Multiplying both sides by 15 . (because the LCM of denomenator is 15 )
5x - 25 = 3x - 9
Transposing 3x to the LHS and -25 to the RHS.
5x - 3x = -9 + 25
or 2x = 16
x = 8
Q5 Solve the following linear equations.
Answer: We have
Multiplying both sides by 12, we get
9t - 6 - ( 8t + 12 ) = 8 - 12t
or 9t - 6 - 8t - 12 = 8 - 12t
or t - 18 = 8 - 12t
Transposing -18 to the RHS and -12t to the LHS:
12t + t = 8 + 18 = 26
or 13t = 26
Thus, t = 2
Q6 Solve the following linear equation.
Answer: We have
Multiplying both sides by 6 (as it is LCM of the denominator.)
6m - 3(m - 1) = 6 - 2(m - 2)
or 6m - 3m + 3 = 6 - 2m + 4
or 3m + 3 = 10 - 2m
Transposing -2m to the LHS and 3 to the RHS:
3m + 2m = 10 - 3
or 5m = 7
Q7 Simplify and solve the following linear equation. 3(t – 3) = 5(2t + 1)
Answer: Let us open the brackets :
LHS : 3(t - 3) = 3t - 9
RHS : 5(2t + 1) = 10t + 5
The equation is 3t - 9 = 10t + 5
or -9 = 10t - 3t + 5
or -9 - 5 = 7t
or -14 = 7t
Therefore t = -2
Q8 Simplify and solve the following linear equation . 15(y – 4) –2(y – 9) + 5(y + 6) = 0
Answer: Let us open the brackets
Equation becomes: 15y - 60 - 2y + 18 + 5y + 30 = 0
or 18y - 12 = 0
or 18y = 12
Q9 Simplify and solve the following linear equation. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Answer: Let us open the brackets.
LHS: 3(5z - 7) - 2(9z - 11)
= 15z - 21 - 18z + 22
= -3z + 1
RHS: 4(8z - 13) - 17
= 32z - 52 - 17
= 32z - 69
So the equation becomes : -3z + 1 = 32z - 69
or 69 + 1 = 32z + 3z
or 70 = 35z
z = 2
Q10 Simplify and solve the following linear equations. 0.25(4f – 3) = 0.05(10f – 9)
Answer: We have
0.25(4f – 3) = 0.05(10f – 9)
Let us open the brackets:
f - 0.75 = 0.5f - 0.45
or f - 0.5f = - 0.45 + 0.75
or 0.5f = 0.30
Dividing both sides by 0.5, we get
f = 0.6
NCERT linear equations in one variable class 8 questions and answers - Exercise: 2.6
Q1 Solve the following equation.
Answer: We observe that the equation is not a linear equation since LHS is not linear.
So we multiply 3x to both the sides to make it linear.
We get, 8x - 3 = 2 3x
or 8x - 3 = 6x
or 8x - 6x = 3
or 2x = 3
Q2 Solve the following equation.
Answer: We will convert the given equation into a linear equation by multiplying
(7 - 6x) to both sides.
Equation becomes: 9x = 15(7 - 6x)
or 9x = 105 - 90x
or 99x = 105
Q3 Solve the following equation.
Answer: We have
Cross-multiplication gives:
9z = 4z + 60
or 9z - 4z = 60
or 5z = 60
or z = 12
Q4 Solve the following equation.
Answer:
We have
By cross-multiplication we get:
5(3y + 4) = -2(2 - 6y)
or 15y + 20 = -4 + 12y
or 15y - 12y = -4 - 20
or 3y = -24
y = -8
Q5 Solve the following equation.
Answer: We have
Cross-multiplication gives,
3(7y + 4) = -4(y + 2)
or 21y + 12 = -4y - 8
or 21y + 4y = -8 - 12
or 25y = -20
Answer: Let the present ages of Hari and Harry be 5x and 7x respectively.
(Since there are in the ratio of 5:7)
Four years from age of Hari will be = 5x + 4
and Harry's age will be = 7x + 4.
According to question four years from now, the ratio of their ages will be 3:4.
So the equation becomes :
Cross-multiplication gives:
4(5x + 4) = 3(7x + 4)
or 20x + 16 = 21x + 12
or x = 4
Hence present age of Hari = 5x = 20 and
present age of Harry = 7x = 28
Answer: Let the numerator of the rational number be x.
Then denominator will be x + 8.
Further, according to question,
or
Cross-multiplication gives:
2(x + 17) = 3(x + 7)
or 2x + 34 = 3x + 21
or x = 13
Hence the rational number is .
Chapter -1 | |
Chapter -2 | Linear Equations in One Variable |
Chapter-3 | |
Chapter-4 | |
Chapter-5 | |
Chapter-6 | |
Chapter-7 | |
Chapter-8 | |
Chapter-9 | |
Chapter-10 | |
Chapter-11 | |
Chapter-12 | |
Chapter-13 | |
Chapter-14 | |
Chapter-15 | |
Chapter-16 |
Step-by-Step Solutions: Detailed step-by-step solutions of maths chapter 2 class 8 to each problem, ensuring clarity and understanding for students.
Variety of Problems: A diverse set of problems covering different aspects of linear equations in one variable. Practice class 8 maths ch 2 question answer to command the concepts.
Concept Clarity: A focus on explaining the fundamental concepts and principles related to linear equations in a clear and concise manner provided in this ch 2 maths class 8.
How to use NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable
Keep working hard and happy learning!
Multiplication and division of algebraic equations, identities, factorization, and the method of solving linear equations in one variable are the important topics of this chapter.
CBSE class 8 maths is base and foundation for the upcoming classes. Most of the topics are related to previous classes. It is basic maths and very easy.
No, CBSE doesn’t provide NCERT solutions for any class or subject.
Here you will get the detailed NCERT solutions for class 8 by clicking on the link.
Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.
NCERT textbook is the best book for CBSE class 8 maths. You don't need to buy any supplementary book. All you need to do is a rigorous practice of NCERT problems.
Admit Card Date:04 October,2024 - 29 November,2024
Admit Card Date:04 October,2024 - 29 November,2024
Application Date:07 October,2024 - 22 November,2024
Application Correction Date:08 October,2024 - 27 November,2024
Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
Accepted by more than 11,000 universities in over 150 countries worldwide
Register now for PTE & Unlock 20% OFF : Use promo code: 'C360SPL20'. Valid till 30th NOV'24! Trusted by 3,500+ universities globally
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE