Madhya Pradesh Board Class 10 Mathematics — NCERT Solutions

Concept: Prime factorisation — divide the number successively by the smallest prime factor until the quotient is 1.

(i) 140
$$140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7$$
$$\boxed{140 = 2^2 \times 5 \times 7}$$

(ii) 156
$$156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$$
$$\boxed{156 = 2^2 \times 3 \times 13}$$

(iii) 3825
$$3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3 \times 3 \times 5 \times 85 = 3 \times 3 \times 5 \times 5 \times 17$$
$$\boxed{3825 = 3^2 \times 5^2 \times 17}$$

(iv) 5005
$$5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$$
$$\boxed{5005 = 5 \times 7 \times 11 \times 13}$$

(v) 7429
$$7429 = 17 \times 437 = 17 \times 19 \times 23$$
$$\boxed{7429 = 17 \times 19 \times 23}$$

Concept: Express each number as a product of prime factors. HCF = product of the smallest powers of common prime factors. LCM = product of the greatest powers of all prime factors. Verification: LCM × HCF = product of the two numbers.

(i) 26 and 91

Prime factorisations:
$$26 = 2 \times 13, \quad 91 = 7 \times 13$$

Common prime factor: $13$
$$\text{HCF}(26, 91) = 13$$
$$\text{LCM}(26, 91) = 2 \times 7 \times 13 = 182$$

Verification:
$$\text{LCM} \times \text{HCF} = 182 \times 13 = 2366$$
$$26 \times 91 = 2366 \quad \checkmark$$

(ii) 510 and 92

Prime factorisations:
$$510 = 2 \times 255 = 2 \times 3 \times 85 = 2 \times 3 \times 5 \times 17$$
$$92 = 2 \times 46 = 2 \times 2 \times 23 = 2^2 \times 23$$

Common prime factor: $2$ (smallest power $= 2^1$)
$$\text{HCF}(510, 92) = 2$$
$$\text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$$

Verification:
$$\text{LCM} \times \text{HCF} = 23460 \times 2 = 46920$$
$$510 \times 92 = 46920 \quad \checkmark$$

(iii) 336 and 54

Prime factorisations:
$$336 = 2 \times 168 = 2^4 \times 3 \times 7$$
$$54 = 2 \times 27 = 2 \times 3^3$$

Common prime factors: $2$ and $3$ (smallest powers $2^1$ and $3^1$)
$$\text{HCF}(336, 54) = 2^1 \times 3^1 = 6$$
$$\text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$$

Verification:
$$\text{LCM} \times \text{HCF} = 3024 \times 6 = 18144$$
$$336 \times 54 = 18144 \quad \checkmark$$

The number of zeroes of $p(x)$ equals the number of times the graph of $y = p(x)$ intersects (or touches) the $x$-axis.

(i) The graph does not intersect the $x$-axis at all.
$$\text{Number of zeroes} = 0$$

(ii) The graph intersects the $x$-axis at exactly one point.
$$\text{Number of zeroes} = 1$$

(iii) The graph intersects the $x$-axis at exactly three points.
$$\text{Number of zeroes} = 3$$

(iv) The graph intersects the $x$-axis at exactly two points.
$$\text{Number of zeroes} = 2$$

(v) The graph intersects the $x$-axis at exactly four points.
$$\text{Number of zeroes} = 4$$

(vi) The graph intersects the $x$-axis at exactly three points.
$$\text{Number of zeroes} = 3$$

Concept: The zeroes of a quadratic polynomial $ax^2+bx+c$ are found by factorisation (or formula). If $\alpha$ and $\beta$ are the zeroes, then:
$$\alpha+\beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}$$

(i) $p(x) = x^2 - 2x - 8$

Splitting the middle term:
$$x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = x(x-4)+2(x-4) = (x-4)(x+2)$$

Zeroes: $x - 4 = 0 \Rightarrow x = 4$ and $x + 2 = 0 \Rightarrow x = -2$

So $\alpha = 4,\ \beta = -2$.

Verification: Here $a=1,\ b=-2,\ c=-8$.
$$\alpha+\beta = 4+(-2) = 2 = -\frac{(-2)}{1} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = 4\times(-2) = -8 = \frac{-8}{1} = \frac{c}{a} \checkmark$$

(ii) $p(s) = 4s^2 - 4s + 1$

Splitting the middle term:
$$4s^2 - 4s + 1 = 4s^2 - 2s - 2s + 1 = 2s(2s-1)-1(2s-1) = (2s-1)(2s-1)$$

Zeroes: $2s - 1 = 0 \Rightarrow s = \dfrac{1}{2}$ (repeated)

So $\alpha = \beta = \dfrac{1}{2}$.

Verification: Here $a=4,\ b=-4,\ c=1$.
$$\alpha+\beta = \frac{1}{2}+\frac{1}{2} = 1 = -\frac{(-4)}{4} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4} = \frac{1}{4} = \frac{c}{a} \checkmark$$

(iii) $p(x) = 6x^2 - 7x - 3$

(Rewriting: $6x^2 - 3 - 7x = 6x^2 - 7x - 3$)

Splitting the middle term (product $= 6\times(-3)=-18$; factors $-9$ and $+2$):
$$6x^2 - 9x + 2x - 3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)$$

Zeroes: $3x+1=0 \Rightarrow x = -\dfrac{1}{3}$ and $2x-3=0 \Rightarrow x = \dfrac{3}{2}$

So $\alpha = -\dfrac{1}{3},\ \beta = \dfrac{3}{2}$.

Verification: Here $a=6,\ b=-7,\ c=-3$.
$$\alpha+\beta = -\frac{1}{3}+\frac{3}{2} = \frac{-2+9}{6} = \frac{7}{6} = -\frac{(-7)}{6} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = \left(-\frac{1}{3}\right)\times\frac{3}{2} = -\frac{1}{2} = \frac{-3}{6} = \frac{c}{a} \checkmark$$

(iv) $p(u) = 4u^2 + 8u$

Factorising:
$$4u^2 + 8u = 4u(u + 2)$$

Zeroes: $4u = 0 \Rightarrow u = 0$ and $u+2=0 \Rightarrow u = -2$

So $\alpha = 0,\ \beta = -2$.

Verification: Here $a=4,\ b=8,\ c=0$.
$$\alpha+\beta = 0+(-2) = -2 = -\frac{8}{4} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = 0\times(-2) = 0 = \frac{0}{4} = \frac{c}{a} \checkmark$$

(v) $p(t) = t^2 - 15$

Factorising:
$$t^2 - 15 = \left(t-\sqrt{15}\right)\left(t+\sqrt{15}\right)$$

Zeroes: $t = \sqrt{15}$ and $t = -\sqrt{15}$

So $\alpha = \sqrt{15},\ \beta = -\sqrt{15}$.

Verification: Here $a=1,\ b=0,\ c=-15$.
$$\alpha+\beta = \sqrt{15}+(-\sqrt{15}) = 0 = -\frac{0}{1} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = \sqrt{15}\times(-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{c}{a} \checkmark$$

(vi) $p(x) = 3x^2 - x - 4$

Splitting the middle term (product $= 3\times(-4)=-12$; factors $-4$ and $+3$):
$$3x^2 - 4x + 3x - 4 = x(3x-4)+1(3x-4) = (x+1)(3x-4)$$

Zeroes: $x+1=0 \Rightarrow x=-1$ and $3x-4=0 \Rightarrow x=\dfrac{4}{3}$

So $\alpha = -1,\ \beta = \dfrac{4}{3}$.

Verification: Here $a=3,\ b=-1,\ c=-4$.
$$\alpha+\beta = -1+\frac{4}{3} = \frac{-3+4}{3} = \frac{1}{3} = -\frac{(-1)}{3} = -\frac{b}{a} \checkmark$$
$$\alpha\beta = (-1)\times\frac{4}{3} = -\frac{4}{3} = \frac{-4}{3} = \frac{c}{a} \checkmark$$

Given: Total students = 10; Number of girls is 4 more than number of boys.

Let number of boys = $x$ and number of girls = $y$.

Equations formed:
$$x + y = 10 \quad \text{...(1)}$$
$$y - x = 4 \quad \text{...(2)}$$

Solutions for Equation (1): $x + y = 10$

Solutions for Equation (2): $y = x + 4$

Plot these points and draw both lines on the same graph.

The two lines intersect at the point $(3, 7)$.

$$\therefore x = 3, \quad y = 7$$

Verification: $3 + 7 = 10$ ✓ and $7 - 3 = 4$ ✓

Answer: Number of boys = 3 and number of girls = 7.

Let cost of one pencil = ₹ $x$ and cost of one pen = ₹ $y$.

Equations formed:
$$5x + 7y = 50 \quad \text{...(1)}$$
$$7x + 5y = 46 \quad \text{...(2)}$$

Solutions for Equation (1): $y = \dfrac{50 - 5x}{7}$

Solutions for Equation (2): $y = \dfrac{46 - 7x}{5}$

Plot these points and draw both lines on the same graph.

The two lines intersect at the point $(3, 5)$.

$$\therefore x = 3, \quad y = 5$$

Verification: $5(3) + 7(5) = 15 + 35 = 50$ ✓ and $7(3) + 5(5) = 21 + 25 = 46$ ✓

Answer: Cost of one pencil = ₹ 3 and cost of one pen = ₹ 5.

Given: $(x + 1)^2 = 2(x - 3)$

Simplifying LHS and RHS:
$$x^2 + 2x + 1 = 2x - 6$$
$$x^2 + 2x + 1 - 2x + 6 = 0$$
$$x^2 + 7 = 0$$

This is of the form $ax^2 + bx + c = 0$ where $a = 1,\ b = 0,\ c = 7$ and $a \neq 0$.

Conclusion: The given equation is a quadratic equation.

(i) Taxi fare situation:

Given: Fare for 1st km = ₹15, each additional km = ₹8.

Fare after 1st km = ₹15
Fare after 2nd km = ₹15 + ₹8 = ₹23
Fare after 3rd km = ₹23 + ₹8 = ₹31
Fare after 4th km = ₹31 + ₹8 = ₹39

The list is: 15, 23, 31, 39, …

Differences: $23 - 15 = 8$, $31 - 23 = 8$, $39 - 31 = 8$

Since each successive difference is the same (= 8), this list forms an AP with $a = 15$ and $d = 8$.

(ii) Air in cylinder:

Let initial amount of air = $V$.

After 1st pump: $V - \dfrac{V}{4} = \dfrac{3V}{4}$

After 2nd pump: $\dfrac{3V}{4} - \dfrac{1}{4}\cdot\dfrac{3V}{4} = \dfrac{3V}{4}\times\dfrac{3}{4} = \dfrac{9V}{16}$

After 3rd pump: $\dfrac{9V}{16}\times\dfrac{3}{4} = \dfrac{27V}{64}$

The list is: $V,\ \dfrac{3V}{4},\ \dfrac{9V}{16},\ \dfrac{27V}{64},\ldots$

Difference: $\dfrac{3V}{4} - V = -\dfrac{V}{4}$

$\dfrac{9V}{16} - \dfrac{3V}{4} = \dfrac{9V}{16} - \dfrac{12V}{16} = -\dfrac{3V}{16}$

Since $-\dfrac{V}{4} \neq -\dfrac{3V}{16}$, the differences are not equal.

This list does NOT form an AP.

(iii) Cost of digging a well:

Cost for 1st metre = ₹150
Cost for 2nd metre = ₹150 + ₹50 = ₹200
Cost for 3rd metre = ₹200 + ₹50 = ₹250
Cost for 4th metre = ₹250 + ₹50 = ₹300

The list is: 150, 200, 250, 300, …

Differences: $200-150=50$, $250-200=50$, $300-250=50$

Since each successive difference is the same (= 50), this list forms an AP with $a = 150$ and $d = 50$.

(iv) Compound interest:

Amount after 1st year = $10000\left(1+\dfrac{8}{100}\right)^1 = 10800$

Amount after 2nd year = $10000\left(1+\dfrac{8}{100}\right)^2 = 11664$

Amount after 3rd year = $10000\left(1+\dfrac{8}{100}\right)^3 = 12597.12$

Differences: $11664 - 10800 = 864$; $12597.12 - 11664 = 933.12$

Since the differences are not equal, this list does NOT form an AP.

(i) All circles are similar.
Reason: All circles have the same shape; they differ only in size (radius), so they are similar but not necessarily congruent.

(ii) All squares are similar.
Reason: All squares have all angles equal to 90° and all sides in the same ratio (1:1 for any two squares scaled appropriately), so they are always similar.

(iii) All equilateral triangles are similar.
Reason: In every equilateral triangle each angle is 60°, so all equilateral triangles have equal corresponding angles and are therefore similar by AAA criterion.

(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.

Given: Points $(2,3)$ and $(4,1)$.

Formula: Distance $= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Working:
$$d = \sqrt{(4-2)^2+(1-3)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$

Answer: The distance is $2\sqrt{2}$ units.

Given: △ABC right-angled at B, AB = 24 cm, BC = 7 cm.

Step 1: Find the hypotenuse AC.

By Pythagoras theorem:
$$AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625$$
$$AC = 25 \text{ cm}$$

(i) For angle A:

• Side opposite to A = BC = 7 cm
• Side adjacent to A = AB = 24 cm
• Hypotenuse = AC = 25 cm

$$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$$

$$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$$

(ii) For angle C:

• Side opposite to C = AB = 24 cm
• Side adjacent to C = BC = 7 cm
• Hypotenuse = AC = 25 cm

$$\sin C = \frac{AB}{AC} = \frac{24}{25}$$

$$\cos C = \frac{BC}{AC} = \frac{7}{25}$$

Let the rope be the hypotenuse of a right triangle. The height of the pole is the side opposite the $30^\circ$ angle.

Using $\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}}$,
$$\sin 30^\circ = \frac{h}{20}$$
$$\frac{1}{2} = \frac{h}{20}$$
$$h = 20 \times \frac{1}{2} = 10$$
So, the height of the pole is 10 m.

A circle can have infinitely many tangents.

Reason: At every point on the circumference of a circle, a unique tangent can be drawn. Since a circle has infinitely many points on it, infinitely many tangents can be drawn to a circle.

Given: Radius $r = 6$ cm, angle of sector $\theta = 60°$

Formula: Area of sector $= \dfrac{\theta}{360} \times \pi r^2$

Solution:
$$\text{Area of sector} = \frac{60}{360} \times \frac{22}{7} \times 6 \times 6$$
$$= \frac{1}{6} \times \frac{22}{7} \times 36$$
$$= \frac{22 \times 36}{6 \times 7} = \frac{792}{42} = \frac{132}{7}$$
$$= 18\frac{6}{7} \text{ cm}^2$$

Answer: Area of the sector $= \dfrac{132}{7}$ cm² $= 18\dfrac{6}{7}$ cm²

Each cube has volume $64\text{ cm}^3$, so its edge is $\sqrt[3]{64}=4$ cm. Two such cubes joined end to end form a cuboid of dimensions $8\text{ cm} \times 4\text{ cm} \times 4\text{ cm}$.\n\nSurface area of cuboid $=2(lb+bh+hl)$\n\n$=2(8\times 4+4\times 4+8\times 4)$\n\n$=2(32+16+32)=2\times 80=160\text{ cm}^2$.\n\nBut the two cubes are joined along one face, so that joined face is not outside. The resulting cuboid’s actual dimensions are still $8\times 4\times 4$, and the surface area is\n\n$2(8\times4+4\times4+8\times4)=160\text{ cm}^2$.\n\nIf the textbook’s intended result is the surface area after joining two cubes end to end, it is $160\text{ cm}^2$. (The printed chapter example for two cubes of volume 64 cm³ does not appear here, so if your book/teacher uses a different convention, follow that.)

Given: Frequency distribution of number of plants in 20 houses.

Method Used: Direct Method (since the values of $x_i$ are small and easy to compute).

Formula: $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$

Step 1: Find the midpoint $x_i$ of each class.

Step 2: Calculate the mean.
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1$$

Answer: The mean number of plants per house is 8.1.

Reason for choosing Direct Method: The class midpoints $x_i$ are small integers, making direct multiplication simple without needing an assumed mean.

(i) 1

(ii) 0; such an event is called an impossible event.

(iii) 1; such an event is called a sure event or certain event.

(iv) 1

(v) greater than or equal to 0 and less than or equal to 1.