In this chapter, we provide NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 3 Coordinate Geometry for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 3 Coordinate Geometry pdf, free NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 3 Coordinate Geometry book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 3 |

Chapter Name | Coordinate Geometry |

Category | NCERT Exemplar |

## NCERT Exemplar Class 9 Maths Solutions Coordinate Geometry

**Exercise 3.1: Multiple Choice Questions (MCQs)**

**Question 1:**

Point (-3, 5) lies in the**(a)** first quadrant ** (b)** second quadrant**(c)** third quadrant ** (d)** fourth quadrant

Thinking Process**(i)** Firstly, check the sign of each coordinate of a point.**(ii)** If both coordinates x and y has same positive sign i.e., (+, +), then the point lies in first quadrant.**(iii)** If x-coordinate has negative sign and y-coordinate has positive sign i.e., (-, +), then the point lies in second quadrant.**(iv)** If both coordinate x and y has negative sing i.e, (-, -), then the point lies in third quadrant.**(v)** If x-coordinate has positive sign and y-coordinate has negative sign i.e., (+, -), then the point lies in fourth quadrant**Solution:****(b)** In point (-3, 5), x-coordinate is negative and y-coordinate is positive. So, the point lies in the second quadrant.

**Question 2:**

Signs of the abscissa and ordinate of a point in the second quadrant are respectively.**(a)** +, + **(b)**-,- **(c)**-,+ ** (d)** +, –**Solution:****(C)** In second quadrant, X-axis is negative and Y-axis is positive. So, sign of abscissa of a point is negative and sign of ordinate of a point is positive.

**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

Thinking Process**(i)** Firstly, check whether any coordinate of point is zero or not.**(a)** If x-coordinate is zero and y-coordinate is non-zero, then the point lies on Y-axis.**(b)** If y-coordinate is zero and x-coordinate is non-zero, then the point lies on X- axis.**(c)** If x-coordinate and y- coordinate are zero, then the point lies on origin (or on both the axes).**(d)** If none of the coordinates is zero, then the point lies in any one of the four quadrants**Solution:****(c)** In point (0, -7) x-coordinate is zero, so it lies on Y-axis and y-coordinate is negative, so

the point (0, – 7) lies on the Y-axis in the negative direction.

**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**Solution:****(a)** In point (-10, 0) y-coordinate is zero, so it lies on X-axis and its x-coordinate is negative, so the point (-10, 0) lies on the X-axis in the negative direction.

**Question 5:**

Abscissa of all the points on the X-axis is**(a)** 0 ** (b)** 1 ** (c)** 2 ** (d)** any number**Solution:****(d)** Abscissa of all the points on the X-axis is any number because X-axis is a number line which contains many real numbers on it.

**Question 6:**

Ordinate of all points on the X-axis is**(a)** 0 ** (b)** 1 ** (c)** – 1 ** (d)** any’number**Solution:****(a)** Ordinate of all points on the X-axis is zero. Because ordinate (or y-coordinate) of a point is perpendicular distance of this point from the X-axis measured along the Y-axis.

If point lies on X- axis, then the perpendicular distance of point from X- axis will be zero, so ordinate will be zero.

**Question 7:**

The point at which the two coordinate axes meet is called the**(a)** abscissa ** (b)** ordinate ** (c)** origin ** (d)** quadrant**Solution:****(c)** The point at which the two coordinate axes meet is called the origin.

**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**Solution:****(c)** A point both of whose coordinates are negative will lie in III quadrant because, in III quadrant x-coordinate and y-coordinate both are negative.

**Question 9:**

Points (1, -1), (2, – 2), (4, – 5) and (-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**Solution:****(d)** In points (1, -1), (2, -2) and (4, -5) x-coordinate is positive and y-coordinate is negative, So, they all lie in IV quadrant. In point (-3, – 4) x-coordinate is negative and y-coordinate is negative. So, it lies in III quadrant So, given points do not lie in the same quadrant.

**Question 10:**

If 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**Solution:****(c)** If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

**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**Solution:****(c)** In point (-5,2), x-coordinate is negative and y-coordinate is positive, so it lies in II quadrant and in point (2, – 5), x- coordinate is positive and y-coordinate is negative, so it lies in 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 X-axis, then the point P has**(a)** x-coordinate = -5 ** (b)** y-coordinate = 5 only**(c)** y-coordinate = – 5 only ** (d)** y-coordinate = 5 or – 5**Solution:****(d)** We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

**Question 13:**

On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. Which of the following figure is obtained?**(a)** Square ** (b)** Rectangle**(c)** Trapezium ** (d)** Rhombus**Solution:****(b)** Here, point 0 (0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B (3, 4) lies

in 1st quadrant and C (0, 4) lines on positive direction of Y-axis. On joining OA AB, BC

and CO the figure obtained is a rectangle.

**Question 14:**

If P(-l, 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 is/are**(a)** P and ** (b)** Q and R**(c)** Only S ** (d)** P and R**Solution:****(b)** In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant.

Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4),

It is clear from the graph that points R and Q lie in fourth quadrant.

**Question 15:**

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is**(a)**-5 **(b)** 1 ** (c)**-1 ** (d)** -2**Solution:****(b)** We have, points P(- 2, 3) and Q(- 3, 5)

Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = – 2 – (-3) =-2 + 3=1.

**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 points) on the X-axis is/are**(a)** P and R** (b)** R and S ** (c)** Only Q ** (d)** Q and O**Solution:****(d)** We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

**Question 17:**

Abscissa of a point is positive in**(a)** I and II quadrants ** (b)** I and IV quadrants**(c)** I quadrant ** (d)** II quadrant**Solution:****(b)** Abscissa of a point is positive in I and IV quadrants.

**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**Solution:****(d)** The points whose abscissa and ordinate have different signs will be of the form (-x, y) or (x, – y)and these points will lie in II and IV quadrants.

**Question 19:**

In following figure, coordinates of P are**(a)** (-4,2)**(b)** (-2,4)**(c)** (4,-2)**(d)** (2, -4)**Solution:****(b)** Here, given point P lies in II quadrant, so its abscissa will be negative and ordinate wilt be positive. Also, its perpendicular distance from X-axis is 4, so y-coordinate of P is 4 and its perpendicular distance from Y-axis is 2, so x-coordinate is -2. Hence, coordinates of P are (-2, 4).

**Question 20:**

In following figure, the point identified by the coordinates (-5, 3) is**(a)** T ** (b)** R ** (c)** L ** (d)** S**Solution:****(c)** In point (-5, 3), x-coordinate is negative and y-coordinate is positive, so it will lie in II quadrant. Now, we see that perpendicular distance of L from V-axis is 5 and from X-axis is 3. So, the required point is L.

**Question 21:**

The point whose ordinate is 4 and which lies on K-axis is**(a)** (4,0) ** (b)** (0,4) ** (c)** (1,4) ** (d)** (4,2)**Solution:****(b)** Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

**Question 22:**

Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and 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**Solution:****(c)** We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

**Question 23:**

The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is**(a)** (0,5) ** (b)** (5,0) ** (c)** (0,-5) ** (d)**(-5,0)**Solution:****(C)** Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.

Hence, the required point is (0, – 5).

**Question 24:**

The perpendicular distance of the point P(3, 4) from the Y-axis is**(a)** 3 ** (b)** 4 ** (c)** 5 ** (d)** 7**Solution:****(a)** We know that, abscissa or the x-coordinate of a point is its perpendicular distance from the Y-axis. So, perpendicular distance of the point P(3, 4)from Y-axis

= Abscissa = 3

**Exercise 3.2: Very Short Answer Type Questions**

**Question 1:**

Write whether the following statements are true or false? Justify your answer.**(i)** Point (3, 0) lies in the first quadrant.**(ii)** Points (1 -1) and (-1, 1) lie in the same quadrant.**(iii)** The coordinates of a point whose ordinate is – ½ and abscissa is 1 are (-½ ,1).**(iv)** A point lies on Y-axis at a distance of 2 units from the X-axis. Its coordinates are (2, 0).**(v)** (-1,7) is a point in the second quadrant.**Solution:****(i)** False, since the ordinate of the point (3, 0) is zero. So, the point lies on X-axis.**(ii)** False, because in point (1, -1) x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant and in (-1,1), x-coordinate is negative and y-coordinate is positive, so it lies in II quadrant.**(iii)** False, because stating the coordinates of a point abscissa comes first and then ordinate (1,-½).**(iv)** False, because point (2, 0) lies on X-axis whose distance from Y-axis is 2 units.**(v)** True, because in a point (-1, 7) abscissa is negative and ordinate is positive.

**Exercise 3.3: Short Answer Type Questions**

**Question 1:**

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure

Thinking Process**(i)** Firstly, draw the perpendicular lines from the point to the coordinates axes.**(ii)** Further, measure the distance from intersecting points to the origin along their sign.**(iii)** Finally, write the x unit distance and y unit distance in pair.**Solution:**

Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S from X-axis is 1 and from Y-axis is 2, so coordinates of S are (2,1). Point 0 lies on X-axis in negative direction so its y-coordinate will be zero and x-coordinate will be -3. So, coordinates of 0 are (-3 0). Point R lies in III quadrant, so its both coordinates will be negative. Now, its perpendicular distance from X-axis is 3 and from Y-axis is 2, so coordinates of point R are (-2, – 3).

Point T lies in IV quadrant, so its x-coordinate will be positive and y-coordinate will be negative. Now, its. perpendicular distance from X-axis is 2 and from Y-axis is 4, so coordinates of T are (4, -2). Point 0 is the intersection of both axes, so it is the origin and its coordinates are 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)andS(2, 2).**Solution:**

Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, – 3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

**Question 3:**

Plot the points (x, y) given by the following table.**Solution:**

On plotting the given points on the graph, we get the points P(2,4), Q(4,2) R (-3, 0), S (-2, 5), T (3, – 3)and O (0 0)

**Question 4:**

Plot the following points and check whether they are collinear or not**(i)** (X 3), (-X -1), (-2, – 3)

**(ii)** (1,1), (2, – 3), (-X – 2)

(iii) (0,0),(2,2),(5,5)

Thinking Process**(i)** Firstly, plot all three points on a graph paper and join them.**(ii)** If it lives a straight line, then points are collinear otherwise non-collinear.**Solution:****(i)** Plotting the points P (1, 3), Q (-1, -1) and R (-2, – 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear.**(ii)** Plotting the points P (1,1), 0 (2, – 3) and R (-1, -2) on the graph paper and join these three points, we get three lines i.e., the given points do not lie on the same line. So, given points are not collinear.**(iii)** Plotting the points 0 (0, 0), A (2,2)and 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given 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.

Thinking Process**(i)** Firstly, write the giver) coordinates in a point form and check the sign of each coordinate of a point.**(ii)** Signs of the coordinates of a point in first quadrant are (+, +) in the second quadrant (-, +), in the third quadrant and in the fourth quadrant (+,-).**Solution:****(i)** The given point is (- 3, 5). Here, abscissa is negative and ordinate is positive, so it lies in II quadrant.**(ii)** The given point is (-5, – 3). Here, abscissa and ordinate both are negative, so it lies in III quadrant.**(iii)** The given point is (-5, 3). Here, abscissa is negative and ordinate is positive, so it lies in II quadrant.**(iv)** The given point is (3, 5). Here, abscissa and ordinate both are positive, so it lies in I quadrant.

**Question 6:**

In 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?**Solution:**

Given, LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units.**(i)** Coordinate of point P = (3, 2) [since, its perpendicular distance from X-axis is 2] Coordinate of point 0 = (3, -1) [since, its perpendicular distance from X-axis is 1 in negative direction of Y-axis]. Coordinate of point R = (3, 0) [since its lies on X-axis, so its y-coordinate is zero].**(ii)** Abscissa of point L = 3, abscissa of point M = 3

Difference between the abscissa of the points L and M = 3-3 = 0

**Question 7:**

In which quadrant or on which axis each of the following points lie?

(-3, 5), (4,-1), (2,0), (2, 2), (-3,-6)**Solution:****(i)** In point (-3, 5), x-coordinate is negative and y-coordinate is positive, so it lies in

II quadrant.**(ii)** In point (4,-1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant.**(iii)** In point (2,0), x-coordinate is positive and y-coordinate is zero, so it lies on X-axis.**(iv)** In point (2,2), x-coordinate and y-coordinate both are positive, so it lies in I quadrant.**(v)** In point (-3, – 6), x-coordinate and y-coordinate both are negative, so it lies in III quadrant.

**Question 8:**

Which of the following points lies on Y-axis?

A(l, 1), B(1, 0), C(0, 1), D(0, 0), E(0, -1), F(-1, 0), G(0, 5), H(-7, 0) and I(3 ,3).

Thinking Process

The point lies on Y-axis means x-coordinate of point will be zero. Check this condition for every given point and find out the correct point.**Solution:**

We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also, D(0, 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

**Question 9:**

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit.**Solution:**

Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

**Question 10:**

A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates? What will be the coordinates, if it lies on negative direction of Y-axis at a distance of 7 units from X-axis?**Solution:**

Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative direction of X-axis, then its x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

**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 Y-axis.**(iii)** whose abscissa is 5 and which lies on X-axis.**Solution:****(i)** The point which lies on X and Y-axes both is origin whose coordinates are (0, 0).**(ii)** The point whose ordinate is – 4 and which lies on Y-axis, i.e., whose x-coordinate is zero, is (0,-4).**(iii)** The point whose abscissa is 5 and which lies on X-axis, i.e., whose y-coordinate is zero, is (5, 0).

**Question 12:**

Taking 0.5 cm as 1 unit, plot the following points on the graph paper A( 1, 3), 6(-3, -1,), C( 1, -4), D(-2, 3), E(0, -8) and F(1, 0).**Solution:**

Here, in point 4(1, 3) both x and y-coordinates are positive, so it lies in I quadrant. In point 8(-3, -1),both x and y-coordinates are negative, so it lies in III quadrant. In point C(1, -4), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant.

In point D(-2, 3), x-coordinate is negative and y-coordinate is positive, so it lies in II quadrant.

In point E(0,-8) x-coordinate is zero, so it lies on Y-axis and in point F(1,0) y-coordinate is zero, so it lies on X-axis.

On plotting the given points, we get the following graph.

**Exercise 3.4: Long Answer Type Questions**

**Question 1:**

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. Plot these points on a graph paper and hence, find the coordinates of the vertex C.

Thinking Process**(i)** Firstly, plot the given points on a graph and join in order.**(ii)** Now, we extend a line from point D parallel to X-axis and extend an other line from point 8 parallel to Y-axis, which will meet at point C.**(iii)** Further, we measure the distance from point C to the coordinate axis.**Solution:**

The graph obtained by plotting the points A, B and C andD is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal.

So, abscissa of C should be equal to abscissa of B i.e., -2 and ordinate of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-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.**Solution:**

Given, length of a rectangle = 5 units and breadth of a rectangle = 3 units

One vertex is at origin i.e., (0, 0) and one of the other vertices lies in III quadrant. So, the length of the rectangle is 5 units in the negative direction of X-axis and then vertex is A(-5, 0). Also, the breadth of the rectangle is 3 units in the negative,direction of y-axis and then vertex is C(0, -3). The fourth vertex B is (-5, – 3).

**Question 3:**

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square.**Solution:**

In point P( 1, 0), y-coordinate is zero, so it lies on X-axis. In point Q(4, 0), y-coordinate is zero so it lies on X-axis. In point S (1, 3), both coordinates are positive, so it lies in I quadrant. On plotting these points, we get the following graph.

Now, take a point R on the graph such that PQRS is a square. Then, all sides will be equal i.e., PQ = QR= RS = PS. So, abscissa of R should be equal to abscissa of Q i.e., 4 and ordinate of R should be equal to ordinate of S i.e., 3. Hence, the coordinates of R are (4, 3).

**Question 4:**

From the given figure, answer the following questions**(i)** Write the points whose abscissa is 0.**(ii)** Write the points whose ordinate is 0.**(iii)** Write the points whose abscissa is – 5,**Solution:****(i)** We know that, the point whose abscissa is 0 will lie on Y-axis. So, the required points whose abscissa is 0 are A, L and O.**(ii)** We know that, the point whose ordinate is 0 will lie on X-axis. So, the required points, whose ordinate is 0 are G,l and O.**(iii)** Here, abscissa ‘-5’ is negative, which shows that point with abscissa -5 will lie in II and III quadrants. So, the required points whose abscissa is -5, are D and H.**Note:** We know that, origin O is the intersection point of both axes. So, we can consider it on X-axis as well as on Y-axis.

**Question 5:**

Plot the points A (1, – 1) and B (4, 5).**(i)** Draw the 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.**Solution:**

In point A(1, -1), x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant. In point B(4, 5), both coordinates are positive, so it lies in I quadrant. On plotting these point, we get the following graph.**(i)** On joining the points A and B, we get the line segment AB. Now, to find the coordinates of a point on this line segment between A and B draw a perpendicular to X-axis from x = 2 and 3.

[since, x = 2 and 3 lies between A and B] say it intersect line segment AB at P and p’. Now, draw a perpendicular to Y-axis from P and p’, they intersect Y-axis at y = 1 and 3, respectively. Thus, we get points (2,1) and (3, 3) which lie between line segment AB.**(ii)** Extent the line segment AB. Now, draw a perpendicular to X-axis from x = 5, say it intersects extended line segment at Q. Again, draw a perpendicular to Y-axis from Q and it intersects Y-axis at y = 7. Thus, we get the point Q(5,7) which lies outside the line segment AB.

**All Chapter NCERT Exemplar Problems Solutions For Class 9 maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 9**

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