Madhya Pradesh Board Class 9 Mathematics — NCERT Solutions

Given: The number zero (0).

Concept: A number is rational if it can be expressed as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

Working:
Yes, zero is a rational number. We can write:
$$0 = \frac{0}{1} = \frac{0}{2} = \frac{0}{3} = \frac{0}{-1} \text{ etc.}$$
In each case, $p = 0$ (an integer) and $q \neq 0$ (an integer).

Conclusion: Zero is indeed a rational number and can be written in the form $\frac{p}{q}$ in infinitely many ways.

Given: Two rational numbers 3 and 4.

Concept: To find $n$ rational numbers between two numbers $a$ and $b$, multiply numerator and denominator to create a gap. Here we need 6 rational numbers, so we write:
$$3 = \frac{3 \times 7}{7} = \frac{21}{7} \quad \text{and} \quad 4 = \frac{4 \times 7}{7} = \frac{28}{7}$$

Working:
The rational numbers between $\frac{21}{7}$ and $\frac{28}{7}$ are:
$$\frac{22}{7},\ \frac{23}{7},\ \frac{24}{7},\ \frac{25}{7},\ \frac{26}{7},\ \frac{27}{7}$$

Answer: Six rational numbers between 3 and 4 are:
$$\frac{22}{7},\ \frac{23}{7},\ \frac{24}{7},\ \frac{25}{7},\ \frac{26}{7},\ \frac{27}{7}$$
(Note: There are infinitely many such rational numbers; this is one possible set.)

Concept: A polynomial in one variable is an expression of the form $a_n x^n + a_{n-1}x^{n-1} + \cdots + a_0$ where all exponents of the variable are whole numbers (non-negative integers).

(i) $4x^{2} - 3x + 7$
All exponents of $x$ are whole numbers (2, 1, 0). It is a polynomial in one variable $x$.

(ii) $y^{2} + \sqrt{2}$
All exponents of $y$ are whole numbers (2, 0). It is a polynomial in one variable $y$.

(iii) $3\sqrt{t} + t\sqrt{2}$
$3\sqrt{t} = 3t^{1/2}$. The exponent $\frac{1}{2}$ is not a whole number. Hence it is not a polynomial.

(iv) $y + \frac{2}{y}$
$\frac{2}{y} = 2y^{-1}$. The exponent $-1$ is not a whole number. Hence it is not a polynomial.

(v) $x^{10} + y^3 + t^{50}$
This expression contains three variables $x$, $y$, and $t$. Hence it is not a polynomial in one variable (it is a polynomial in three variables).

Concept: The coefficient of $x^2$ is the number multiplied with $x^2$ in the expression.

(i) $2 + x^{2} + x$
The term containing $x^2$ is $1 \cdot x^2$.
Coefficient of $x^2$ = $\mathbf{1}$

(ii) $2 - x^{2} + x^{3}$
The term containing $x^2$ is $-1 \cdot x^2$.
Coefficient of $x^2$ = $\mathbf{-1}$

(iii) $\frac{\pi}{2} x^2 + x$
The term containing $x^2$ is $\frac{\pi}{2} x^2$.
Coefficient of $x^2$ = $\dfrac{\boldsymbol{\pi}}{\mathbf{2}}$

(iv) $\sqrt{2} x - 1$
There is no $x^2$ term in this expression.
Coefficient of $x^2$ = $\mathbf{0}$

Given: A table lamp is placed on a study table.

Concept Used: To describe the position of any object in a plane, we need two reference lines (perpendicular to each other) and the distances of the object from those lines.

Solution:

Step 1: Consider the study table as a plane surface.

Step 2: Choose any two adjacent edges of the table as reference lines. Let one edge along the length of the table be the reference line in the horizontal direction, and one edge along the width be the reference line in the vertical direction.

Step 3: Measure the perpendicular distance of the lamp from the horizontal reference edge. Suppose this distance is $x$ cm.

Step 4: Measure the perpendicular distance of the lamp from the vertical reference edge. Suppose this distance is $y$ cm.

Step 5: The position of the table lamp can now be described to another person by saying: "The lamp is placed at a distance of $x$ cm from one edge (along the length) and $y$ cm from the adjacent edge (along the width) of the table."

Conclusion: Thus, using two perpendicular reference lines (the two adjacent edges of the table), the position of the lamp can be uniquely described by the ordered pair $(x, y)$.

Given:

• Two main roads cross at the centre of the city along North-South and East-West directions.
• All other streets are parallel to these roads and 200 m apart.
• There are 5 streets in each direction.
• Scale: 1 cm = 200 m.
• Convention: Cross-street $(m, n)$ means the crossing of the $m$-th North-South street and the $n$-th East-West street.

Model of the City:
Draw 5 vertical lines (representing North-South streets) and 5 horizontal lines (representing East-West streets), each 1 cm apart on the notebook. This gives a grid of $5 \times 5 = 25$ cross-streets.

Part (i): Cross-streets referred to as (4, 3)

The cross-street $(4, 3)$ is the point where the 4th North-South street meets the 3rd East-West street.

In the grid, there is exactly one 4th North-South street and exactly one 3rd East-West street. They can meet at only one point.

$$\boxed{\text{Only } 1 \text{ cross-street can be referred to as } (4, 3).}$$

Part (ii): Cross-streets referred to as (3, 4)

The cross-street $(3, 4)$ is the point where the 3rd North-South street meets the 4th East-West street.

Similarly, there is exactly one 3rd North-South street and exactly one 4th East-West street. They meet at only one point.

$$\boxed{\text{Only } 1 \text{ cross-street can be referred to as } (3, 4).}$$

Note: $(4, 3)$ and $(3, 4)$ refer to different cross-streets because the first number denotes the North-South street and the second denotes the East-West street. Interchanging the numbers gives a different location.

Given: Cost of a notebook = ₹x, Cost of a pen = ₹y.

Statement: The cost of a notebook is twice the cost of a pen.

Forming the equation:
$$x = 2y$$

This can be written in standard form as:
$$x - 2y = 0$$

or equivalently $1 \cdot x + (-2) \cdot y + 0 = 0$.

Answer: The required linear equation in two variables is $x = 2y$ (or $x - 2y = 0$).

(i) False.
Through a single (one) point, infinitely many lines can pass. Euclid's postulate states that a unique line passes through two distinct points, not one. Hence the statement is false.

(ii) False.
By Euclid's first postulate, one and only one line can pass through two distinct points. Hence the statement is false.

(iii) True.
By Euclid's second postulate, a terminated line (line segment) can be produced indefinitely on both sides, giving an infinite straight line. Hence the statement is true.

(iv) True.
If two circles are equal, they can be superimposed on each other such that they coincide completely. Therefore their centres coincide and their radii must be equal. Hence the statement is true.

(v) True.
Given: $AB = PQ$ and $PQ = XY$.
By Euclid's first axiom — Things which are equal to the same thing are equal to one another — we conclude $AB = XY$. Hence the statement is true.

Given: Lines AB and CD intersect at O. ∠AOC + ∠BOE = 70° and ∠BOD = 40°.

Step 1: Find ∠AOC.

Since AB is a straight line, ∠AOC and ∠BOC are supplementary... but more directly, ∠AOC and ∠BOD are vertically opposite angles (formed by intersecting lines AB and CD).

$$\angle AOC = \angle BOD = 40^\circ \quad (\text{Vertically opposite angles})$$

Step 2: Find ∠BOE.

We are given:
$$\angle AOC + \angle BOE = 70^\circ$$
$$40^\circ + \angle BOE = 70^\circ$$
$$\angle BOE = 70^\circ - 40^\circ = 30^\circ$$

Step 3: Find ∠COE.

Since AB is a straight line, the angles on one side of AB along line CD sum to 180°.

Ray OE lies between OB and OC (from the figure). The angles ∠BOE, ∠COE together with consideration of the straight line:

$$\angle BOC = 180^\circ - \angle AOC = 180^\circ - 40^\circ = 140^\circ \quad (\text{Linear pair, since AOB is a line})$$

Now, ∠COE = ∠BOC − ∠BOE:
$$\angle COE = 140^\circ - 30^\circ = 110^\circ$$

Step 4: Find reflex ∠COE.
$$\text{Reflex } \angle COE = 360^\circ - 110^\circ = 250^\circ$$

Answers: $\angle BOE = 30^\circ$ and reflex $\angle COE = 250^\circ$.

Given: In quadrilateral ACBD, AC = AD and AB bisects ∠A, i.e., ∠CAB = ∠DAB.

To prove: △ABC ≅ △ABD

Proof:

Consider △ABC and △ABD.

$$AC = AD \quad \text{(Given)}$$

$$\angle CAB = \angle DAB \quad \text{(AB bisects } \angle A\text{)}$$

$$AB = AB \quad \text{(Common side)}$$

Therefore, by SAS congruence rule:
$$\Delta ABC \cong \Delta ABD$$

About BC and BD:
Since △ABC ≅ △ABD, by CPCT:
$$BC = BD$$
So BC and BD are equal, i.e., B is equidistant from C and D.

Given: ABCD is a parallelogram in which diagonal AC = diagonal BD.

To prove: ABCD is a rectangle.

Proof:

In $\triangle ABC$ and $\triangle DCB$:

• $AB = DC$ (opposite sides of a parallelogram)
• $BC = BC$ (common)
• $AC = DB$ (given, diagonals are equal)

By SSS congruence rule:
$$\triangle ABC \cong \triangle DCB$$

Therefore, $\angle ABC = \angle DCB$ (CPCT)

Since $AB \parallel DC$ and $BC$ is a transversal:
$$\angle ABC + \angle DCB = 180^\circ \quad \text{(co-interior angles)}$$

$$\Rightarrow 2\angle ABC = 180^\circ$$

$$\Rightarrow \angle ABC = 90^\circ$$

Since ABCD is a parallelogram with one angle $= 90^\circ$, ABCD is a rectangle. $\blacksquare$

Given: Two congruent circles with centres O and O′ and equal radii r. AB and CD are equal chords of the two circles respectively, i.e., AB = CD.

To Prove: ∠AOB = ∠CO′D

Proof:

In △AOB and △CO′D:

$$OA = O'C = r \quad \text{(radii of congruent circles)}$$

$$OB = O'D = r \quad \text{(radii of congruent circles)}$$

$$AB = CD \quad \text{(given)}$$

By SSS congruence criterion:
$$\triangle AOB \cong \triangle CO'D$$

Therefore, by CPCT:
$$\angle AOB = \angle CO'D$$

Hence proved. Equal chords of congruent circles subtend equal angles at their centres.

Given: Diameter = 10.5 cm, so radius $r = \dfrac{10.5}{2} = 5.25$ cm; slant height $l = 10$ cm.

Formula: Curved Surface Area of cone $= \pi r l$

Calculation:
$$\text{CSA} = \frac{22}{7} \times 5.25 \times 10$$
$$= \frac{22}{7} \times 52.5$$
$$= 22 \times 7.5$$
$$= 165 \text{ cm}^2$$

Answer: The curved surface area of the cone is $\mathbf{165 \text{ cm}^2}$.

Given: Data on causes of illness and death among women aged 15–44 worldwide.

(i) Graphical Representation (Bar Graph):

We draw a bar graph with the causes on the X-axis and the female fatality rate (%) on the Y-axis.

• Take a suitable scale on the Y-axis, e.g., 1 unit = 5%.
• Draw bars of equal width for each cause with heights proportional to the fatality rate:

All bars are drawn with equal width and gaps between them. The bar for "Reproductive health conditions" is the tallest.

(ii) Major cause:

From the bar graph (and the table), Reproductive health conditions is the major cause of women's ill health and death worldwide, with the highest fatality rate of 31.8%.

(iii) Two factors responsible:

Two major factors that play a role in reproductive health conditions being the leading cause are:

1. Lack of proper medical facilities and healthcare during pregnancy and childbirth, especially in developing countries.
2. Poor nutrition and lack of awareness about reproductive health among women, leading to complications.