There is a saying in Mathematics that the area where algebra and geometry intersect is called coordinate geometry. In coordinate geometry, every point has a story, and the graph will show the visual representation. In Chapter 3 of the NCERT exemplar class 9 maths solutions, students will learn about coordinate geometry and its real-life applications. Furthermore, they will learn about the Cartesian system, the origin, quadrants, and plotting points on graph paper using given coordinates. After finishing the NCERT textbook exercises, when students need an extra resource for practice, these class 9 NCERT exemplars become handy.
This Story also Contains
- NCERT Exemplar Solutions Class 9 Maths Chapter 3 Important Topics:
- NCERT Class 9 Exemplar Solutions for other Subjects:
- NCERT Exemplar Class 9 Maths Solutions Chapter-Wise
- Importance of NCERT Exemplar Class 9 Maths Solutions Chapter 3
- NCERT Solutions for class 9 Mathematics: Chapter-wise
- NCERT Solution Subject Wise
- NCERT Notes Subject Wise
- NCERT Books and NCERT Syllabus
The NCERT exemplar maths solutions for class 9 of coordinate geometry will build a strong base of this chapter early so that students have no problem dealing with this chapter in higher classes. Careers360 teachers have prepared these solutions, explaining each step alongside the relevant formulas so that the learning process becomes easier. Students can also check the NCERT solutions for class 9 for more information.
Exercise: 3.1 Total Questions: 24 Page Numbers: 25-27 |
Question:1
Point (–3, 5) lies in the
(A) first quadrant
(B) second quadrant
(C) third quadrant
(D) fourth quadrant
Answer: [B]
Solution:

We know that in
First quadrant : (+, +)
Second quadrant : (–, +)
Third quadrant : (–, –)
Fourth quadrant : (+, –)
Thus, the value of x is –3 and y is 5
So, that point (–3, 5) lies in the second quadrant.
Therefore, option (B) is correct.
Question:2
Signs of the abscissa and ordinate of a point in the second quadrant are respectively
(A) +, +
(B) –, –
(C) –, +
(D) +, –
Answer: [C] –, +
Solution:

We know that the x-coordinate is called the abscissa, and the y-coordinate is also called the ordinate.
Thus, in the second quadrant abscissa is negative, and the ordinate is positive.
So that, the in the second quadrant, signs of the abscissa and ordinate is ( –, +)
Therefore, option (C) is correct.
Question:3
Point (0, –7) lies
(A) on the x –axis
(B) in the second quadrant
(C) on the y-axis
(D) in the fourth quadrant
Answer: [C] on the y-axis
Solution:
We know that
Coordinate of the x-axis is (x, 0), i.e., the value of the y coordinate is zero
Coordinate of the y-axis is (0, y), i.e., the value of the x coordinate is zero
So the point $(0, -7)$ is on the y-axis because here the value of x is 0 and the value of y is 7.
Therefore, option (C) is correct
Question:4
Point $(-10, 0)$ lies
(A) on the negative direction of the x-axis
(B) on the negative direction of the y-axis
(C) in the third quadrant
(D) in the fourth quadrant
Answer: [A] on the negative direction of the x-axis
Solution:
Here, the x-coordinate and y-coordinate are $-10$ and 0, respectively.
We know that coordinates on the negative direction of the x-axis are in the form $(-x, 0)$
And the coordinates on the negative direction of the y-axis is in the form $(0, -y)$
Hence, point $(-10, 0)$ lies on the negative direction of the x-axis.
Therefore, option (A) is correct
Question:5
Abscissa of all the points on the x-axis is
(A) 0
(B) 1
(C) 2
(D) any number
Answer: [D] Any number
Solution.
We know that the coordinate of any point on the x-axis is (x, 0),
where x can take any value.
So, the abscissa of any point on the x-axis is any number.
Therefore, option (D) is correct
Question:6
Ordinate of all points on the x-axis is
(A) 0
(B) 1
(C) -1
(D) any number
Answer: [A]
Solution.
We know that the coordinate of every point on the x-axis is (x, 0)
i.e., (7, 0), (1, 0), etc.
So the y-coordinate on the x-axis is always zero.
Hence, the ordinate of every point on the x-axis is 0.
Therefore, option (A) is correct
Question:7
The point at which the two coordinate axes meet is called the
(A) abscissa
(B) ordinate
(C) origin
(D) quadrant
Answer: [C] origin
Solution.

We know that the coordinate axis x and y divide the plane into four parts called quadrants and the point of intersection of the axis is called the origin.
Coordinate of the origin are (0, 0).
Abscissa is the x-axis (horizontal) coordinate
Ordinate is the y-axis (vertical) coordinate
Therefore, option (C) is correct
Question:8
A point both of whose coordinates are negative will lie in
(A) I quadrant
(B) II quadrant
(C) III quadrant
(D) IV quadrant
Answer: (C) III quadrant
Solution.
We know that the coordinates of a point are of the form (+, +) in the first quadrant.
(-, +) in the second quadrant
(-,-) in the third quadrant and
(+, -) in the fourth quadrant.
Where + denotes a positive real number and - denotes a negative real number. So that we can say that two coordinates that are negative lie in the third quadrant.
Therefore, option (C) is correct. Question:9
Points $(1,-1),(2,-2),(4,-5),(-3,-4)$
(A) lie in II quadrant
(B) lie in III quadrant
(C) lie in IV quadrant
(D) do not lie in the same quadrant
Answer:
Answer: [D] Do not lie in the same quadrant
Solution.

We know that coordinates of a point are of the form (+, +) in the first quadrant
(–, +) in the second quadrant
(–,–) in the third quadrant and
(+, –) in the fourth quadrant.
Where + denotes a positive real number and - denotes a negative real number.
So, points $(1, -1), (2, -2)$ and $(4, -5)$ all lie in the IV quadrant but $(-3, -4)$ lies in IIIrd quadrant.
So, we can say that the given points do not lie in the same quadrant.
Therefore, option (D) is correct
Question:10
If the y coordinate of a point is zero, then this point always lies
(A) In I quadrant
(B) In II quadrant
(C) on x - axis
(D) on y-axis
Answer: (C) on x - axis

We know that coordinate of x axis are (x, 0)
i.e., the y coordinate is zero on the x axis.
So, if the y-coordinate of a point is zero, then this point always lies on the x-axis.
Therefore, option (C) is correct
Question:11
The points $(-5, 2)$ and $(2, - 5)$ lie in the
(A) same quadrant
(B) II and III quadrants, respectively
(C) II and IV quadrants, respectively
(D) IV and II quadrants, respectively
Answer: (C) II and IV quadrants, respectively
We know that
A point in the first quadrant is (+, +)
A point in the second quadrant is (–, +)
A point in the third quadrant is (–, –)
A point in the fourth quadrant is (+, –)

Here $(-5, 2)$ and $(2, -5)$ both are lie in the different quadrants.
Point $(-5, 2)$ lies in the II quadrant and point $(2, -5)$is lies in the IV quadrant.
Question:12
If the perpendicular distance of a point P from the x-axis is 5 units and the foot of the perpendicular lies on the negative direction of the x-axis, then the point P has
(A) x coordinate = -5
(B) y coordinate = 5 only
(C) y coordinate = -5only
(D) y coordinate = 5 or -5
Answer: (D) y coordinate = 5 or -5
Solution.
We know that the perpendicular distance of any point from the x-axis gives the y-coordinate of that point.
i.e., the x coordinate is always perpendicular to the y-axis.

(A) Here x-coordinate is -5. It lies on the negative direction of the x-axis, so it is incorrect
(B) Here y-coordinate is 5. It lies on the positive direction of the y-axis and is perpendicular to the x-axis, so it is incorrect
(C) Here y-coordinate is -5. It lies on the negative direction of the y-axis and is perpendicular to the x-axis.
So it is incorrect
(D) Here y-coordinate is 5 or -5, i.e., the perpendicular distance can be in the II quadrant or III quadrant. So that we can say the point P has y-coordinate 5 or -5
Therefore, option (D) is correct
Question:13
On plotting the points O (0, 0), A (3, 0), B (3, 4), C (0, 4) and joining OA, AB, BC and CO, which of the following figures is obtained?
(A) Square
(B) Rectangle
(C)Trapezium
(D) Rhombus
Answer: (B) Rectangle
Solution.

(C)Rectangle: We know that in the rectangle, opposite sides are equal in measurement and are parallel to each other.
So here OA = BC = 3 units and OC = AB = 4 units
Hence, it is a rectangle.
Question:14
If $P (- 1, 1), Q (3, - 4), R(1, -1), S(-2, -3)$ and $T (- 4, 4)$ are plotted on the graph paper, then the point(s) in the fourth quadrant are:
(A) P and T
(B) Q and R
(C) Only S
(D) P and R
Answer: (B) Q and R

(A) Here point $P(-1, 1)$ and $T(-4, 4)$ lie in II quadrant.
(B) Here point $Q(3, -4)$ and $R(1, -1)$ lie in IV quadrant.
(C) Here point $S(-2, -3)$ lies in III quadrant
(D) Here point $P(-1, 1)$ and $R(1, -1)$ both lie in different quadrants
P lies in II quadrant and R lies in IV quadrant.
Therefore, option (B) is correct
Question:16
If P (5, 1), Q (8, 0), R (0, 4), S (0, 5) and O (0, 0) are plotted on the graph paper, then the point(s) on the x-axis are:
(A) P and R
(B) R and S
(C) Only Q
(D) Q and O
Answer: (D) Q and O
Plotting the given points, we have:

(A) Points P and R lie on a different axis. So it is incorrect.
(B) Points R and S lie on the y-axis because, on the y-axis, the x coordinate is always zero. So it is incorrect.
(C) Point Q (8, 0) lies on the x-axis because on the x-axis, the y coordinate is always zero. So it is incorrect.
(D) Points Q and O have a y-coordinate as 0, so they are plotted on the x-axis. So this is correct.
Therefore, option (D) is correct
Question:17
Abscissa of a point is positive in
(A) I and II quadrants
(B) I and IV quadrants
(C) I quadrant only
(D) II quadrant only
Answer: (B) I and IV quadrants
Solution.

The abscissa is the x-axis (horizontal) coordinate
(A) We know that abscissa is positive and negative in I quadrant and II quadrants respectively, so this option is incorrect.
(B) Abscissa of a point is positive in 1 quadrant and IV quadrants. Because in I quadrant we have (+, +) and in II quadrant we have (+, –).
So, the abscissa is positive in both quadrants.
Hence this option is correct.
(C) We know that in the I and IV quadrant, the abscissa is positive.
So this option is incorrect.
(D) In II quadrant, abscissa is negative.
So this option is incorrect.
Therefore, option (B) is correct. Question:18
The points whose abscissa and ordinate have different signs will lie in
(A) I and II quadrants
(B) II and III quadrants
(C) I and III quadrants
(D) II and IV quadrants
Answer: (D) II and IV quadrants
Solution.
The abscissa is the x-axis (horizontal) coordinate
The ordinate is the y-axis (vertical) coordinate
We know that
The sign of coordinates in the first quadrant is (+, +)
The sign of coordinates in the second quadrant is (–, +)
The sign of coordinates in the third quadrant is (–, –)
The sign of coordinates in the fourth quadrant is (+, –)
So that we can say that in II and III quadrant abscissa and ordinate have different signs.
Therefore, option (D) is correct
Question:19
In Fig. 3.1, coordinates of P are

(A) $(- 4, 2)$
(B) $(-2, 4)$
(C) $(4, - 2)$
(D) $(2, - 4)$
Answer: (B) (-2, 4)
Solution.

From the figure, we can see that:
Here x coordinate is -2, and the y coordinate is 4
So that coordinate of point P(-2, 4).
Therefore, option (B) is correct
Question:21
The point whose ordinate is 4 and which lies on the y-axis is
(A) (4, 0)
(B) (0, 4)
(C) (1, 4)
(D) (4, 2)
Answer: (B) (0, 4)
Solution.
The ordinate is the y-axis (vertical) coordinate
(A) (4, 0) is lying on the x-axis because here abscissa is 4 and the ordinate is 0.
(B) (0, 4) is lying on the y-axis because here abscissa is 0, and the ordinate is 4
(C) (1, 4) lies in I quadrant because here abscissa is 1 and ordinate is 4
(D) (4, 2) lies in I quadrant because here abscissa is 4 and ordinate is 2
Therefore, option (B) is correct
Question:22
Which of the points P(0. 3), Q(1, 0), R(0, -1), S(-5, 0), T(1, 2) do not lie on the x-axis?
(A) P and R only
(B) Q and S only
(C) P, R and T
(D) Q, S and T
Answer: (C) P, R and T
Solution.
We know that points on the x-axis have coordinates (x, 0).
Points on the y-axis have coordinates (0, y). So, that we can say Q(1, 0), S(-5, 0) lies on x-axis
and P(0, 3), R(0, -1) lies on the y-axis and T(1, 2) lies in the I quadrant.
Hence, P, R and T do not lie on the x-axis.
Therefore, option (C) is correct.
Question:23
The point which lies on the y-axis at a distance of 5 units in the negative direction of the y-axis is
(A) (0, 5)
(B) (5, 0)
(C) (0,-5)
(D) (-5,0)
Answer: (C) $(0, -5)$
Solution.
We know that the point lies on the y-axis, so its x-coordinate is zero.
Also, it is a distance of 5 units in the negative direction of the y-axis.
So, the y-coordinate is negative with an ordinate of 5.
(A) (0, 5) lies in the positive direction in the y-axis.
(B) (5, 0) lies on the positive direction in the x-axis.
(C) $(0, -5)$lies in the negative direction in the y-axis.
(D) $(-5, 0)$ lies in the negative direction in the x-axis.
Therefore, option (C) is correct
Question:24
The perpendicular distance of the point P (3, 4) from the y-axis is
(A) 3
(B) 4
(C) 5
(D) 7
Answer: [A] 3
Solution.
Plotting the point (3, 4) on the graph:
We know that the perpendicular distance from the y-axis is the x-coordinate (abscissa).
So here 3 is the perpendicular distance from the y-axis, and 4 is the perpendicular distance from the x-axis.
Hence, 3 is the correct answer.
Therefore, option (A) is correct
Exercise: 3.3 Total Questions: 12 Page Numbers: 29-31 |
Question:1
Write the coordinates of each of the points P, Q, R, S, T and O from the Figure.

Answer:
Coordinates of any point are in the form (x, y).
From the given graph, we can see:
Coordinate of point P=(1, 1)
Coordinate of point Q=(-3, 0)
Coordinate of point R = (-2, -3)
Coordinate of point S = (2, 1)
Coordinate of point T = (4, -2)
Coordinate of point O = (0, 0)
Question:2
Plot the following points and write the name of the figure obtained by joining them in order: $P (-3,2), Q(-7,-3), R(6,-3), S(2,2)$
Answer:
Solution.
The given points
$P(-3,2), Q(-7,-3),R(6,-3),S(2,2)$
are plotted as follows:

From the figure we can see that PS is parallel to QR. Distance between them is fixed, i.e., 5 units.
Also, PQ and RS are non-parallel.
Hence, the obtained figure is a trapezium.
Question:4(i)
Plot the following points and check whether they are collinear or not : (1, 3), (– 1, – 1), (– 2, – 3)
Answer:
Plotting the given points:
(1, 3), (– 1, – 1), (– 2, – 3)

Hence, these points lie on a straight line.
These points are collinear
Question:4(ii)
Plot the following points and check whether they are collinear or not :
(1, 1), (2, – 3), (– 1, – 2)
Answer: Non-Collinear
The given points can be plotted as follows:
(1, 1), (2, – 3), (– 1, – 2)

The given points do not lie on the same straight line.
Hence, they are non-collinear
Question:4(iii)
Plot the following points and check whether they are collinear or not :
(0, 0), (2, 2), (5, 5)
Answer: Collinear
Plotting the given points:
(0, 0), (2, 2), (5, 5)

Hence, these points lie on a straight line.
These points are collinear.
Question:5
Without plotting, the points indicate the quadrant in which they will lie if
(i) ordinate is 5 and abscissa is -3
(ii) abscissa is -5 and ordinate is -3
(iii) abscissa is -5 and ordinate is 3
(iv) ordinate is 5 and abscissa is 3
Answer:
(i) II quadrant
(ii) III quadrant
(iii) II quadrant
(iv) I quadrant
Solution.
We know that sign in I quadrant = (+, +)
In II quadrant = (-, +)
In III quadrant = (-,-)
In IV quadrant = (+,-)
Abscissa is the x-axis (horizontal) coordinate
Ordinate is the y-axis (vertical) coordinate
(i) Hence abscissa is -3, and the ordinate is 5.
So, it lies in II quadrant.
(ii) Here x coordinate is -5, and the y-coordinate is -3
So both are negative.
Its lies in III quadrant.
(iii) Abscissa is -5 and ordinate is 3
Here x coordinate is -5, and the y-coordinate is 3
So it lies in II quadrant.
(iv) Ordinate is 5 and abscissa is 3
Here x coordinate is 3, and the y-coordinate is 5
Both are positive.
So it lies in I quadrant
Question:6
In the figure, LM is a line parallel to the y-axis at a distance of 3 units.
(i) What are the coordinates of the points P, R and Q?
(ii) What is the difference between the abscissa of the points L and M?

Answer:
(i) P = (3, 2)
R = (3, 0)
Q = (3, -1)
(ii) 0
Solution.
We have,

All the points lie on a line where the x coordinate is fixed, i.e., 3.
(i) Coordinate of the points P
Distance from x-axis: 2 units
So coordinates are (3, 2)
Coordinate of the points R
Distance from x-axis: 0 unit
So coordinates are (3, 0)
Coordinate of the points Q
Distance from x-axis: $-1 unit$
So coordinates are (3, -1)
(ii) Abscissa is the x-axis (horizontal) coordinate
All the points lie on a line where the x coordinate is fixed, i.e., 3.
So the abscissa of L and M is the same, i.e., 3
Hence difference = 3 - 3 = 0
Question:7
In which quadrant or on which axis does each of the following points lie? (– 3, 5), (4, – 1), (2, 0), (2, 2), (– 3, – 6)
Answer:
(-3,5): 2nd Quadrant
(4, - 1): 4th Quadrant
(2, 0): x-axis
(2, 2): 1st Quadrant
(- 3, - 6): 3rd Quadrant
Solution.
Plotting the given points,(-3,5),(4,-1),(2,0),(2,2),(-3,-6)

Question:8
Which of the following points lie on the y-axis?
A (1, 1), B (1, 0), C (0, 1), D (0, 0), E (0, - 1), F (- 1, 0), G (0, 5), H (- 7, 0), I (3, 3).
Answer:
C(0, 1), D (0, 0), E(0, -1), G(0, 5)
Solution.
We know that for points that lie on the y-axis, the x coordinate should be zero. The abscissa is the x-axis (horizontal) coordinate, which should be zero.
Question:11
Find the coordinates of the point
(i) which lies on x and y axes both.
(ii) whose ordinate is -4 and which lies on the y-axis.
(iii) whose abscissa is 5 and which lies on the x-axis.
Answer:
(i) (0, 0)
(ii) (0, -4)
(iii) (5, 0)
Solution.
(i) We know that the point which lies on the x-axis and y-axis both is the origin whose coordinates are (0, 0)
(ii) We know that the point that lies on the y-axis must have its x-coordinate as zero.
Here, it is given that a point lies on the y-axis with ordinate -4
The ordinate is the y-axis (vertical) coordinate
So y-coordinate = -4
Hence coordinates are (0, -4)
(iii) We know that the point that lies on the x-axis must have its y-coordinate as zero.
Here, it is given that the point lies on the x-axis with abscissa 5
The abscissa is the x-axis (horizontal) coordinate
So x-coordinate = 5
Hence coordinates are (5, 0)
The points are shown as follows:

Question:12
Taking 0.5 cm as 1 unit, plot the following points on the graph paper:A (1,3), B(-3,-1), C(1,-4) ,D(-2,3) ,E(0,-8), F(1,0)
Answer:
$Scale 0.5 cm = 1 units$

Exercise: 3.4 Total Questions: 5 Page Numbers: 32 |
Question:1
Points A (5, 3), B (- 2, 3), and D (5, -4) are three vertices of a square ABCD. Plot these points on a graph paper and find the coordinates of the vertex C.
Answer: $(-2, -4)$
Solution.
Points $A (5, 3), B (- 2, 3)$ and $D (5, - 4)$ are three vertices of a square ABCD
If we plot them, we get:

Length of AD = Length of BC
Length of AB = Length of CD
From point D, we have made a parallel line to AB, towards the third quadrant.
Similarly, from point B, we have made a parallel line to AD, towards the third quadrant.
These lines intersect at the point $C (-2, -4).$
Hence, the square ABCD is formed, and C is $(-2, -4).$
Question:2
Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units, respectively, one vertex at the origin, the longer side lies on the x-axis, and one of the vertices lies in the third quadrant.
Answer:
Answer:
Coordinates of A (-5, 0)
Coordinates of B (-5, -3)
Coordinates of C (0, -3)
Coordinates of O (0, 0)
Solution.
Given rectangle has one vertex at the origin. Let the vertex be O.
∴ The coordinates of O are O (0, 0).
Let the rectangle be OABC.
Length = 5 units
Breadth = 3 units.
Now as the longer side is on the x-axis, we get that length lies on x-axis and breadth lies on the y axis.
So,
The coordinates of A are A (-5, 0)
The coordinates of C will be (0, -3).
Given that one of the vertex is in the third quadrant.
So, the rectangle lies in the III quadrant.
The coordinates of B will be (-5, -3).

Question:3
Plot the points P (1, 0), Q (4, 0) and S (1, 3). Find the coordinates of point R such that PQRS is a square.
Answer: (4, 3)
Given
P (1, 0): Lies on the x-axis at a distance of 1 unit from the y-axis
Q (4, 0): Lies on the x-axis at a distance of 1 unit from the y-axis
Length of PQ = 4 -1 = 3 units
Let the coordinates of point R be (x, y)
Now, as PQRS is a square (PQ = QR = RS = SP)
Consider,
PQ = RS
3 = x - 1
So, x = 4
Similarly,
PQ = QR
3 = y - 0
So, y = 3
Hence, R is (4, 3)
The graph is as follows:

Question:4
From the Fig. 3.8, answer the following:
(i) Write the points whose abscissa is 0.
(ii) Write the points whose ordinate is 0.
(iii) Write the points whose abscissa is -5.

Answer:
(i) A (0, 3); L (0, -4)
(ii) G (5, 0); I (-2, 0)
(iii) D (-5, 1); H (-5, -3)
Abscissa is the x-axis (horizontal) coordinate
Ordinate is the y-axis (vertical) coordinate
(i) Abscissa is given as 0
So, the x-coordinate must be zero.
Hence points are A (0, 3); L (0, -4)
(ii) Ordinate is given as 0
So, the y-coordinate must be zero.
Hence points are G (5, 0); I (-2, 0)
(iii) Abscissa is given as -5
So, the x-coordinate must be -5.
Hence, points are D (-5, 1); H (-5, -3)
Question:5
Plot the points A (1, -1) and B (4, 5)
(i) Draw a line segment joining these points. Write the coordinates of a point on this line segment between the points A and B.
(ii) Extend this line segment and write the coordinates of a point on this line which lies outside the line segment AB.
Answer:
(i) (3, 3)
(ii) (5, 7)
Solution.
(i) Coordinates of a point on this line segment between the points A and B:C (3, 3), D (2.5, 2)

(ii) Coordinates of a point on this line which lies outside the line segment AB.E (5, 7), F (4.5, 6)

NCERT Exemplar Solutions Class 9 Maths Chapter 3 Important Topics:
NCERT exemplar Class 9 Maths solutions chapter 3 on Coordinate geometry deals with the following topics:
◊ Cartesian system: Any point in the plane is expressed.
◊ Coordinates of a point: An ordered pair to represent any point on the plane (X, Y).
◊ NCERT exemplar Class 9 Maths solutions chapter 3 includes physical interpretation of coordinates: Length of abscissa and ordinate.
◊ The possibility of coordinate: Different possibilities of coordinates in four parts of the plane.
◊ Horizontal axis and vertical axis: X-axis as the horizontal axis and Y-axis as the vertical axis.
◊ Importance of origin: The intersection point of two axes will be the reference point.
NCERT Class 9 Exemplar Solutions for other Subjects:
The following links will lead students to the solutions of other subjects' Exemplar solutions.
Importance of NCERT Exemplar Class 9 Maths Solutions Chapter 3
These Class 9 Maths NCERT exemplar chapter 3 solutions will help the students grasp the basics of coordinate geometry and turn out to be extremely useful in higher Classes or competitive exams. Here are some of the important features of these solutions.
- Students will have a clear idea of how to draw coordinates on a graph paper when the coordinates are given.
- The concept of the Cartesian system and Quadrants will also become much clearer.
- Students will learn about shortcuts and alternative methods of solving these problems after checking these solutions.
NCERT Solution Subject Wise
Students can check the links below for the NCERT textbook solutions of other subjects.
NCERT Notes Subject Wise
NCERT notes are a useful tool in the learning process. The following links are beneficial for that cause.
NCERT Books and NCERT Syllabus
At the start of the preparation process, students should check the latest CBSE syllabus for changes and updates. They can use the links below to check the latest syllabus. Also, there are some reference books that students can use for further assistance.