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📝 Section – A (Questions: 1 to 24) (1 mark each) [24]

Choose the correct option (Questions: 1 to 6)

1. Two sticks are shown in the following figure. One white and other is of black. From the lengths given in the figure, the length of white stick will be _____ cm. [1]The figure shows the total length of the two sticks combined is $22$ cm, and the length of the black stick is $5$ cm. The white stick is the remainder of the total length.Length of white stick $= 22 \text{ cm} – 5 \text{ cm} = 17 \text{ cm}$.(The options provided in the source (A) 11, (B) 13.5, (C) 8.5, (D) 6 seem to be incorrect based on the visual information given in the source image1111. Assuming the length of the white stick is the difference between $22 \text{ cm}$ and $5 \text{ cm}$):Answer: $17$ cm. (Since $17$ is not an option, there may be an error in the question or the options. Assuming the question intended to ask for $22-5=17$ based on the diagram.)
2. For the quadratic equation $x^{2}-2x+1=0,$ value of $x+\frac{1}{x}=\_\_\_\_\_.$ [1]The quadratic equation is $x^{2}-2x+1=0$.This can be factored as $(x-1)^2 = 0$.The root is $x=1$.Therefore, $x+\frac{1}{x} = 1 + \frac{1}{1} = 1 + 1 = 2$.Answer: (C) 2 ²²²²
3. For an A.P., if $d=-4$, $a_{7}=4$ then its first term $a = \_\_\_\_\_.$ [1]The formula for the $n^{th}$ term of an A.P. is $a_n = a + (n-1)d$.$a_7 = a + (7-1)d$$4 = a + 6(-4)$$4 = a – 24$$a = 4 + 24 = 28$.Answer: (D) 28 ³³³³
4. The distance between the points $(2,-1)$ and $(-1,-5)$ is _____ units. [1]Using the distance formula $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:$D = \sqrt{(-1-2)^2 + (-5-(-1))^2}$$D = \sqrt{(-3)^2 + (-5+1)^2}$$D = \sqrt{9 + (-4)^2}$$D = \sqrt{9 + 16} = \sqrt{25} = 5$.Answer: (B) 5 ^444444444
5. If $\sin^{2}\theta=\frac{1}{2}$ then value of $\tan^{2}\theta=\_\_\_\_\_.$ [1]If $\sin^{2}\theta=\frac{1}{2}$, then $\sin\theta = \frac{1}{\sqrt{2}}$. This means $\theta = 45^\circ$.Therefore, $\tan^{2}\theta = \tan^{2}45^{\circ} = (1)^2 = 1$.Alternatively, from $\sin^{2}\theta=\frac{1}{2}$, we have $\cos^{2}\theta = 1 – \sin^{2}\theta = 1 – \frac{1}{2} = \frac{1}{2}$.$\tan^{2}\theta = \frac{\sin^{2}\theta}{\cos^{2}\theta} = \frac{1/2}{1/2} = 1$.Answer: (D) 1 ⁵⁵⁵⁵
6. For some data, if mean and median are $21$ and $23$ respectively then mode _____. [1]Using the empirical relationship for unimodal frequency distributions:Mode $\approx 3 \times$ Median $- 2 \times$ MeanMode $\approx 3(23) – 2(21)$Mode $\approx 69 – 42$Mode $\approx 27$.Answer: (A) 27 ⁶

Fill in the blanks with correct option (Questions: 7 to 12)

7. If $1080=2^{x}\times3^{y}\times5$ then $x – y = \_\_\_\_\_.$ (2, 0, 3) [1]Find the prime factorization of $1080$:$1080 = 108 \times 10 = (2^2 \times 3^3) \times (2 \times 5) = 2^3 \times 3^3 \times 5^1$.Comparing this with $2^{x}\times3^{y}\times5$:$x=3$ and $y=3$.$x-y = 3-3 = 0$.Answer: 0 ⁷
8. If $a$ and $b$ are the zeroes of the polynomial $P(x)=x^{2}-2x+5$ then $a \times b = \_\_\_\_\_.$ (3, 4, 5) [1]For a quadratic polynomial $Ax^2+Bx+C$, the product of the zeroes is $\frac{C}{A}$.Here, $A=1, B=-2, C=5$.Product $a \times b = \frac{5}{1} = 5$.Answer: 5 ⁸
9. A balanced dice is tossed once. Then the total number of possible outcomes are _____. (6, 12, 36) [1]The possible outcomes when a single die is tossed are $\{1, 2, 3, 4, 5, 6\}$.The total number of possible outcomes is 6.Answer: 6 ⁹⁹⁹⁹
10. $\sin 30^{\circ}=\_\_\_\_\_.$ ($\frac{1}{2},\frac{1}{\sqrt{2}},\frac{\sqrt{3}}{2}$) [1]Answer: $\frac{1}{2}$ ¹⁰
11. _____ tangents can be drawn from the point lying in the interior of the circle. (2, 1, 0) [1]A point lying in the interior of the circle cannot have any tangent drawn from it to the circle. Answer: 0 11
12. For a given data 2, 6, 4, 5, 0, 3, 1, 3, 2, 3, mode = _____. (2, 3, 4) [1]The mode is the value that appears most frequently in a data set.The number 3 appears three times, which is more than any other number.Answer: 3 ¹²

State whether the following statements are true or false (Questions: 13 to 16)

13. $\sqrt{2}$ is an irrational number. [1]Answer: True ¹³
14. HCF of 12, 15 and 21 is 1. [1]Prime factorizations: $12 = 2^2 \times 3$, $15 = 3 \times 5$, $21 = 3 \times 7$.The Highest Common Factor (HCF) is 3.Answer: False ¹⁴
15. $\sqrt{3}x+5$ is a linear polynomial. [1]Yes, a linear polynomial is of the form $ax+b$, where $a \ne 0$. Here $a=\sqrt{3}$ and $b=5$.Answer: True ¹⁵
16. The sum of probabilities of Event E and ‘Event not E’ is 1. [1]$P(E) + P(\overline{E}) = 1$.Answer: True ¹⁶

Answer the following in one sentence or one word or number (Questions: 17 to 20)

17. $1, 1, 1, 2, 2, 2, 3, 3, 3, \ldots$ is an Arithmetic Progression or not? [1]An A.P. has a constant difference ($d$) between consecutive terms.$1-1 = 0$$1-1 = 0$$2-1 = 1$Since the difference is not constant, it is not an A.P.Answer: Not an Arithmetic Progression (or No) ¹⁷
18. How many tangents can a circle have? [1]A circle can have infinitely many tangents.Answer: Infinitely many ¹⁸
19. If $P(A)=0.65$ then find $P(\overline{A})$. [1]$P(\overline{A}) = 1 – P(A) = 1 – 0.65 = 0.35$.Answer: 0.35 ¹⁹
20. For the following frequency distribution find the modal class. [1]The modal class is the class interval with the highest frequency.The highest frequency is $8$, which corresponds to the class interval $3-5$.Answer: 3-5 ^20202020

Match the pairs (Questions: 21 to 24)

A	B	Match
21) Curved surface area of a cylinder	(a) $\frac{1}{3}\pi r^{2}h$	(c) $2\pi rh$ 21
22) Volume of a cone	(b) $2\pi r^{2}$	(a) $\frac{1}{3}\pi r^{2}h$ 22

A	B	Match
23) The circumference of a circle with radius $r$	(a) $\frac{\pi r\theta}{180}$	(b) $2 \pi r$ 23
24) The area of a minor sector of a circle of an angle $\theta$	(b) $2 \pi r$	(c) $\frac{\pi r^{2}\theta}{360}$ 24

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⚙️ Section – B (Any 9) (Questions: 25 to 37) (2 marks each) [18]

25. Find the roots of the quadratic equation $x^{2}-x-20=0$ [2]By factoring:We need two numbers that multiply to $-20$ and add up to $-1$. These are $-5$ and $4$.$x^{2}-5x+4x-20=0$$x(x-5) + 4(x-5) = 0$$(x-5)(x+4) = 0$The roots are $x=5$ and $x=-4$. 25
26. Find a quadratic polynomial, the sum and product of whose zeroes are $-3$ and $2$ respectively. [2]A quadratic polynomial can be written as $k[x^2 – (\text{Sum of zeroes})x + (\text{Product of zeroes})]$.Let Sum of zeroes ($S$) $=-3$ and Product of zeroes ($P$) $=2$.The polynomial is $k[x^2 – (-3)x + 2] = k(x^2 + 3x + 2)$.Taking $k=1$, the polynomial is $x^2 + 3x + 2$. 26
27. If one root of quadratic polynomial $6x^{2}+37x-(P-2)$ is inverse of the other root, then find the value of P. [2]For a quadratic polynomial $Ax^2+Bx+C$, if one root is $\alpha$ and the other is $\frac{1}{\alpha}$, then the product of the roots is $\alpha \times \frac{1}{\alpha} = 1$.The product of the roots is also given by $\frac{C}{A}$.Here, $A=6$ and $C=-(P-2) = 2-P$.$\frac{C}{A} = 1$$\frac{2-P}{6} = 1$$2-P = 6$$P = 2-6 = -4$. ²⁷
28. Find $20^{th}$ term of an A.P.: $2, 7, 12, \ldots$ [2]First term $a=2$.Common difference $d = 7 – 2 = 5$.The $n^{th}$ term formula is $a_n = a + (n-1)d$.For $n=20$:$a_{20} = 2 + (20-1)5$$a_{20} = 2 + (19)5$$a_{20} = 2 + 95 = 97$. ²⁸
29. Find the sum of all integers from 51 to 100. [2]The series is an A.P.: $51, 52, \ldots, 100$.First term $a=51$.Last term $l=100$.Common difference $d=1$.Number of terms $n = 100 – 51 + 1 = 50$.The sum formula is $S_n = \frac{n}{2}(a+l)$.$S_{50} = \frac{50}{2}(51+100)$$S_{50} = 25(151)$$S_{50} = 3775$. ²⁹
30. Find the coordinates of the point which divides the line segment joining the points $(4,-3)$ and $(8, 5)$ in the ratio $3:1$ internally. [2]Using the section formula for internal division:$P(x,y) = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right)$Here, $(x_1, y_1) = (4, -3)$, $(x_2, y_2) = (8, 5)$, and $m:n = 3:1$.$x = \frac{3(8) + 1(4)}{3+1} = \frac{24+4}{4} = \frac{28}{4} = 7$$y = \frac{3(5) + 1(-3)}{3+1} = \frac{15-3}{4} = \frac{12}{4} = 3$The coordinates of the point are $(7, 3)$. 30
31. A circle with centre P, whose diameter is XY. The coordinates of X and Y are $(3,-10)$ and $(1, 4)$. Find the coordinates of P. [2]The center P is the midpoint of the diameter XY.Midpoint formula: $P(x,y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$Here, $(x_1, y_1) = (3, -10)$ and $(x_2, y_2) = (1, 4)$.$x = \frac{3+1}{2} = \frac{4}{2} = 2$$y = \frac{-10+4}{2} = \frac{-6}{2} = -3$The coordinates of P are $(2, -3)$. 31
32. Prove that $\cos^{2}\theta-\sin^{2}\theta=2\cos^{2}\theta-1$ [2]Start with the Left-Hand Side (LHS):LHS $= \cos^{2}\theta-\sin^{2}\theta$Using the identity $\sin^{2}\theta + \cos^{2}\theta = 1$, we have $\sin^{2}\theta = 1 – \cos^{2}\theta$.LHS $= \cos^{2}\theta – (1 – \cos^{2}\theta)$LHS $= \cos^{2}\theta – 1 + \cos^{2}\theta$LHS $= 2\cos^{2}\theta – 1$LHS $=$ RHS. Hence Proved. 32
33. Find the value: $4\cot^{2}45^{\circ}-\sec^{2}60^{\circ}+\sin^{2}60^{\circ}+\cos^{2}90^{\circ}.$ [2]Substitute the standard values:$\cot 45^{\circ} = 1$$\sec 60^{\circ} = 2$$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$$\cos 90^{\circ} = 0$Value $= 4(1)^2 – (2)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + (0)^2$Value $= 4(1) – 4 + \frac{3}{4} + 0$Value $= 4 – 4 + \frac{3}{4} = \frac{3}{4}$. 33
34. The angle of elevation of the top of a tower from a point on the ground, which is $30 \text{ m}$ away from the foot of the tower, is $30^{\circ}$. Find the height of the tower. [2]Let the height of the tower be $h$ and the distance from the foot of the tower to the observation point be $d=30 \text{ m}$. The angle of elevation is $\theta=30^{\circ}$.In the right-angled triangle, we have $\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{h}{d}$. $\tan 30^{\circ} = \frac{h}{30}$$\frac{1}{\sqrt{3}} = \frac{h}{30}$$h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m}$.The height of the tower is $10\sqrt{3} \text{ m}$ (or approximately $17.32 \text{ m}$). 34
35. Find the total surface area of a cube with edge $6 \text{ cm}$. [2]The formula for the Total Surface Area (TSA) of a cube with edge $a$ is $\text{TSA} = 6a^2$.Here, $a=6 \text{ cm}$.$\text{TSA} = 6 \times (6 \text{ cm})^2 = 6 \times 36 \text{ cm}^2 = 216 \text{ cm}^2$. ³⁵
36. The height and the diameter of a base of a cone are $6 \text{ cm}$ and $5 \text{ cm}$ respectively. Find the slant height of the cone. [2]Height $h=6 \text{ cm}$.Diameter $D=5 \text{ cm}$, so radius $r = \frac{D}{2} = \frac{5}{2} = 2.5 \text{ cm}$.The slant height $l$ is found using the formula $l = \sqrt{r^2 + h^2}$. $l = \sqrt{(2.5)^2 + (6)^2}$$l = \sqrt{6.25 + 36}$$l = \sqrt{42.25}$$l = 6.5 \text{ cm}$.The slant height of the cone is $6.5 \text{ cm}$. 36
37. If for some frequency distribution $l=40$, $f_{1}=7$, $f_{0}=3$, $f_{2}=6$ and $h=15$. Then find the mode. [2]The formula for the Mode is:Mode $= l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h$Mode $= 40 + \left( \frac{7 – 3}{2(7) – 3 – 6} \right) \times 15$Mode $= 40 + \left( \frac{4}{14 – 9} \right) \times 15$Mode $= 40 + \left( \frac{4}{5} \right) \times 15$Mode $= 40 + 4 \times 3$Mode $= 40 + 12 = 52$. 37373737

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➕ Section – C (Any 6) (Questions: 38 to 46) (3 marks each) [18]

38. Alok has some Pigeons and Cows. The total number of their eyes is 120 and total number of their legs is 180. How many Pigeons and Cows the Alok has? [3]Let $P$ be the number of Pigeons and $C$ be the number of Cows.
    • Eyes Equation: Both Pigeons and Cows have 2 eyes each.$2P + 2C = 120$$P + C = 60$ (Equation 1)
    • Legs Equation: Pigeons have 2 legs, and Cows have 4 legs.$2P + 4C = 180$ (Equation 2)From Equation 1, $P = 60 – C$. Substitute this into Equation 2:$2(60 – C) + 4C = 180$$120 – 2C + 4C = 180$$120 + 2C = 180$$2C = 180 – 120$$2C = 60$$C = 30$Substitute $C=30$ back into Equation 1:$P + 30 = 60$$P = 30$Alok has 30 Pigeons and 30 Cows. 38
39. Solve the linear pair of equations in two variables $x+y=5$, $2x-3y=4$ by elimination method. [3](1) $x + y = 5$(2) $2x – 3y = 4$Multiply Equation (1) by 3:(3) $3(x + y) = 3(5) \implies 3x + 3y = 15$Add Equation (3) and Equation (2):$(3x + 3y) + (2x – 3y) = 15 + 4$$5x = 19$$x = \frac{19}{5}$Substitute the value of $x$ into Equation (1):$\frac{19}{5} + y = 5$$y = 5 – \frac{19}{5} = \frac{25}{5} – \frac{19}{5} = \frac{6}{5}$The solution is $x = \frac{19}{5}$ and $y = \frac{6}{5}$. 39
40. If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first 20 terms. [3]The sum of $n$ terms of an A.P. is $S_n = \frac{n}{2}(2a + (n-1)d)$.Given $S_7 = 49$:$S_7 = \frac{7}{2}(2a + 6d) = 49$$7(a + 3d) = 49 \implies a + 3d = 7$ (Equation 1)Given $S_{17} = 289$:$S_{17} = \frac{17}{2}(2a + 16d) = 289$$17(a + 8d) = 289 \implies a + 8d = \frac{289}{17} = 17$ (Equation 2)Subtract Equation 1 from Equation 2:$(a + 8d) – (a + 3d) = 17 – 7$$5d = 10 \implies d = 2$Substitute $d=2$ into Equation 1:$a + 3(2) = 7 \implies a + 6 = 7 \implies a = 1$Now, find the sum of the first 20 terms ($S_{20}$):$S_{20} = \frac{20}{2}(2a + (20-1)d)$$S_{20} = 10(2(1) + 19(2))$$S_{20} = 10(2 + 38) = 10(40) = 400$.The sum of the first 20 terms is 400. 40
41. Find the coordinates of the points which divide the line segment joining $A(-2,2)$ and $B(2,8)$ into four equal parts. [3]Let the points be $P_1, P_2, P_3$. These points divide $AB$ into 4 equal parts, so:
    • $P_1$ divides $AB$ in the ratio $1:3$.
    • $P_2$ divides $AB$ in the ratio $2:2$, meaning $P_2$ is the midpoint of $AB$.
    • $P_3$ divides $AB$ in the ratio $3:1$.1. Point $P_2$ (Midpoint, 2:2 or 1:1 ratio):$P_2 = \left( \frac{-2 + 2}{2}, \frac{2 + 8}{2} \right) = \left( \frac{0}{2}, \frac{10}{2} \right) = (0, 5)$2. Point $P_1$ (1:3 ratio):$x = \frac{1(2) + 3(-2)}{1+3} = \frac{2-6}{4} = \frac{-4}{4} = -1$$y = \frac{1(8) + 3(2)}{1+3} = \frac{8+6}{4} = \frac{14}{4} = \frac{7}{2} = 3.5$$P_1 = (-1, 3.5)$3. Point $P_3$ (3:1 ratio):$x = \frac{3(2) + 1(-2)}{3+1} = \frac{6-2}{4} = \frac{4}{4} = 1$$y = \frac{3(8) + 1(2)}{3+1} = \frac{24+2}{4} = \frac{26}{4} = \frac{13}{2} = 6.5$$P_3 = (1, 6.5)$The coordinates of the points are $(-1, 3.5)$, $(0, 5)$, and $(1, 6.5)$. 41
42. Show that the points $(1, 7)$, $(4, 2)$, $(-1,-1)$ and $(-4, 4)$ are the vertices of a square. [3]Let the points be $A(1, 7)$, $B(4, 2)$, $C(-1, -1)$, and $D(-4, 4)$.A quadrilateral is a square if all four sides are equal and the diagonals are equal.Using the distance formula $D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
    • Sides:$AB = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9+25} = \sqrt{34}$$BC = \sqrt{(-1-4)^2 + (-1-2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}$$CD = \sqrt{(-4-(-1))^2 + (4-(-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9+25} = \sqrt{34}$$DA = \sqrt{(1-(-4))^2 + (7-4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25+9} = \sqrt{34}$Since $AB=BC=CD=DA$, the quadrilateral is either a rhombus or a square.
    • Diagonals:$AC = \sqrt{(-1-1)^2 + (-1-7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4+64} = \sqrt{68}$$BD = \sqrt{(-4-4)^2 + (4-2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64+4} = \sqrt{68}$Since $AC=BD$, the diagonals are equal.Since all sides are equal and the diagonals are equal, the points $(1, 7)$, $(4, 2)$, $(-1,-1)$, and $(-4, 4)$ are the vertices of a square. 42
43. Prove that: The tangent at any point of a circle is perpendicular to the radius through the point of contact. [3]Statement: Let a circle have center $O$ and $XY$ be a tangent to the circle at point $P$. We need to prove that $OP \perp XY$. Proof:
    1. Take any point $Q$ on the tangent $XY$, other than $P$.
    2. Join $OQ$. Since $Q$ is a point on the tangent other than the point of contact $P$, $Q$ must lie outside the circle. (If $Q$ were inside, $XY$ would be a secant, not a tangent).
    3. Therefore, the line segment $OQ$ is longer than the radius $OP$. $OQ > OP$.
    4. Since this is true for every point $Q$ (except $P$) on $XY$, the shortest distance from the center $O$ to the line $XY$ is the segment $OP$.
    5. The shortest distance from a point to a line is the perpendicular distance.
    6. Hence, $OP$ is perpendicular to $XY$. $OP \perp XY$. ⁴³
44. Two concentric circles are of radii $5 \text{ cm}$ and $3 \text{ cm}$. Find the length of the chord of the larger circle which touches the smaller circle. [3]Let $O$ be the center of the concentric circles.Radius of the larger circle ($R$) $= 5 \text{ cm}$.Radius of the smaller circle ($r$) $= 3 \text{ cm}$.Let $AB$ be the chord of the larger circle that touches the smaller circle at point $P$.The radius of the smaller circle $OP$ is perpendicular to the tangent $AB$ at the point of contact $P$ (as proven in Q43).$OP = 3 \text{ cm}$, $OB = 5 \text{ cm}$. $OP \perp AB$.In the right-angled $\triangle OPB$, by Pythagoras theorem:$OP^2 + PB^2 = OB^2$$3^2 + PB^2 = 5^2$$9 + PB^2 = 25$$PB^2 = 25 – 9 = 16$$PB = 4 \text{ cm}$The perpendicular from the center to a chord of the larger circle bisects the chord, so $AB = 2 \times PB$.Length of chord $AB = 2 \times 4 \text{ cm} = 8 \text{ cm}$.The length of the chord is $8 \text{ cm}$. 44444444
45. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: [3]Determine the modal lifetimes of the components.The formula for the Mode is: Mode $= l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h$

Lifetime (in hours)	Frequency (f)
0-20	10
20-40	35
40-60	52 ($f_0$)
60-80	61 ($f_1$)
80-100	38 ($f_2$)
100-120	29

* **Modal Class:** 60-80 (has the highest frequency, 61)
* Lower limit of modal class ($l$) $= 60$
* Frequency of modal class ($f_1$) $= 61$
* Frequency of preceding class ($f_0$) $= 52$
* Frequency of succeeding class ($f_2$) $= 38$
* Class size ($h$) $= 80 - 60 = 20$
Substitute the values into the formula:
Mode $= 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20$
Mode $= 60 + \left( \frac{9}{122 - 90} \right) \times 20$
Mode $= 60 + \left( \frac{9}{32} \right) \times 20$
Mode $= 60 + \frac{180}{32} = 60 + 5.625$
Mode $= 65.625$
The modal lifetime of the components is $65.625$ hours. [cite: 224, 227, 228]

46. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be – [3]Total number of marbles = $5 (\text{Red}) + 8 (\text{White}) + 4 (\text{Green}) = 17$.The probability of an event $E$ is $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.(i) red?Number of red marbles = 5.$P(\text{red}) = \frac{5}{17}$ 45(ii) white?Number of white marbles = 8.$P(\text{white}) = \frac{8}{17}$ 46(iii) not green?Number of marbles that are not green = $5 (\text{Red}) + 8 (\text{White}) = 13$.Alternatively, $P(\text{not green}) = 1 – P(\text{green})$. $P(\text{green}) = \frac{4}{17}$.$P(\text{not green}) = \frac{13}{17}$ 47

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➗ Section – D (Any 5) (Questions: 47 to 54) (4 marks each) [20]

47. Prove that: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. [4]This is the statement of the Basic Proportionality Theorem (BPT) or Thales Theorem.Statement: In $\triangle ABC$, let $DE$ be a line segment such that $DE || BC$, and $DE$ intersects $AB$ at $D$ and $AC$ at $E$. We need to prove that $\frac{AD}{DB} = \frac{AE}{EC}$. Construction:
    • Join $BE$ and $CD$.
    • Draw $EM \perp AB$ and $DN \perp AC$.Proof:
    1. The area of $\triangle ADE$, $\text{Area}(ADE) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EM$.
    2. $\text{Area}(DBE) = \frac{1}{2} \times DB \times EM$ (as $EM$ is the height for base $DB$ of $\triangle DBE$).
    3. Ratio $\frac{\text{Area}(ADE)}{\text{Area}(DBE)} = \frac{\frac{1}{2} \times AD \times EM}{\frac{1}{2} \times DB \times EM} = \frac{AD}{DB}$ (Equation 1)
    4. Similarly, $\text{Area}(ADE) = \frac{1}{2} \times AE \times DN$.
    5. $\text{Area}(ECD) = \frac{1}{2} \times EC \times DN$.
    6. Ratio $\frac{\text{Area}(ADE)}{\text{Area}(ECD)} = \frac{\frac{1}{2} \times AE \times DN}{\frac{1}{2} \times EC \times DN} = \frac{AE}{EC}$ (Equation 2)
    7. Since $\triangle DBE$ and $\triangle ECD$ are on the same base $DE$ and between the same parallel lines $DE$ and $BC$, their areas are equal: $\text{Area}(DBE) = \text{Area}(ECD)$.
    8. From (1), (2), and (7): $\frac{AD}{DB} = \frac{AE}{EC}$. Hence Proved. ⁴⁸
48. $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle PQR$. For each of the following cases, state whether $EF || QR$. [4]By the converse of the Basic Proportionality Theorem, $EF || QR$ if and only if $\frac{PE}{EQ} = \frac{PF}{FR}$.(i) $PE=3.9 \text{ cm}$, $EQ=3 \text{ cm}$, $PF=3.6 \text{ cm}$ and $FR=2.4 \text{ cm}$.$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$$\frac{PF}{FR} = \frac{3.6}{2.4} = \frac{36}{24} = \frac{3}{2} = 1.5$Since $1.3 \ne 1.5$, $\frac{PE}{EQ} \ne \frac{PF}{FR}$.Conclusion: $EF$ is not parallel to $QR$. 49494949(ii) $PE=4 \text{ cm}$, $QE=4.5 \text{ cm}$, $PF=8 \text{ cm}$ and $RF=9 \text{ cm}$.$\frac{PE}{QE} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9}$$\frac{PF}{RF} = \frac{8}{9}$Since $\frac{8}{9} = \frac{8}{9}$, $\frac{PE}{QE} = \frac{PF}{RF}$.Conclusion: $EF$ is parallel to $QR$. 50
49. Find two consecutive positive integers, sum of whose squares is 365. [4]Let the two consecutive positive integers be $x$ and $x+1$.The sum of their squares is 365:$x^2 + (x+1)^2 = 365$$x^2 + (x^2 + 2x + 1) = 365$$2x^2 + 2x + 1 = 365$$2x^2 + 2x – 364 = 0$Divide by 2:$x^2 + x – 182 = 0$Factor the quadratic equation: We need two numbers that multiply to $-182$ and add up to $1$. These are $14$ and $-13$.$x^2 + 14x – 13x – 182 = 0$$x(x + 14) – 13(x + 14) = 0$$(x-13)(x+14) = 0$The possible values for $x$ are $x=13$ or $x=-14$.Since the integers must be positive, $x=13$.The two consecutive positive integers are $x=13$ and $x+1=14$.(Check: $13^2 + 14^2 = 169 + 196 = 365$). 51
50. Ramkali saved $\text{₹}5$ in the first week of a year and then increased her weekly savings by $\text{₹}1.75$. If in the $n^{th}$ week, her weekly savings become $\text{₹}20.75$, find $n$. [4]Ramkali’s weekly savings form an Arithmetic Progression (A.P.):
    • First term ($a$) $= 5$ (Savings in 1st week)
    • Common difference ($d$) $= 1.75$ (Increase per week)
    • $n^{th}$ term ($a_n$) $= 20.75$ (Savings in $n^{th}$ week)The formula for the $n^{th}$ term is $a_n = a + (n-1)d$.$20.75 = 5 + (n-1)1.75$$20.75 – 5 = (n-1)1.75$$15.75 = (n-1)1.75$$n-1 = \frac{15.75}{1.75}$$n-1 = \frac{1575}{175} = 9$$n = 9 + 1 = 10$The weekly savings become $\text{₹}20.75$ in the $\mathbf{10^{th}}$ week. 52
51. The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers. [4]We will use the Assumed Mean Method (or Step-Deviation Method) to find the mean ($\bar{x}$).

Class Interval (Ci​)	Frequency (fi​)	Mid-point (xi​)	Deviation di​=xi​−A (A=40)	fi​di​
15-25	6	20	-20	-120
25-35	11	30	-10	-110
35-45	7	40 ($A$)	0	0
45-55	4	50	10	40
55-65	4	60	20	80
65-75	2	70	30	60
75-85	1	80	40	40
Total	$\sum f_i = 35$			$\sum f_i d_i = -10$

Assume Mean ($A$) $= 40$.
The formula for the Mean ($\bar{x}$) using the Assumed Mean Method is:
$\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$
$\bar{x} = 40 + \frac{-10}{35}$
$\bar{x} = 40 - \frac{2}{7}$
$\bar{x} \approx 40 - 0.2857$
$\bar{x} \approx 39.7143$
The mean percentage of female teachers is approximately $\mathbf{39.71\%}$. [cite: 284, 285, 287, 288]

52. 30 women were examined by AIIMS doctors and the number of heart beats per minute was recorded… [4]

Number of Heart Beats per minute (Ci​)	Number of Women (fi​)	Cumulative Frequency (cf)
65-68	2	2
68-71	4	6
71-74	3	9
74-77	8	17 $\rightarrow$ Median Class
77-80	7	24 $\rightarrow$ Modal Class
80-83	4	28
83-86	2	30
Total	$\mathbf{N = 30}$

**(i) How many women have heart beat in range of 68-77.** [1]
Women in $68-71$ range $= 4$
Women in $71-74$ range $= 3$
Women in $74-77$ range $= 8$
Total women $= 4 + 3 + 8 = 15$.
**Answer:** 15 [cite: 302, 303]

**(ii) What is the median class of heart beats per minute for these women?** [1]
Total number of observations $N=30$.
We need $\frac{N}{2} = \frac{30}{2} = 15$.
The cumulative frequency ($cf$) just greater than or equal to 15 is 17, which corresponds to the class interval 74-77.
**Answer:** 74-77 [cite: 304]

**(iii) Find the mode for the heart beat per minute for these women.** [2]
The modal class is $77-80$ (highest frequency $f_1=7$).
* Lower limit of modal class ($l$) $= 77$
* Frequency of modal class ($f_1$) $= 7$
* Frequency of preceding class ($f_0$) $= 8$ (Error here, the highest frequency is 8, the modal class is 74-77)
Let's re-evaluate the modal class: The highest frequency is 8, so the Modal Class is $\mathbf{74-77}$.
* Lower limit of modal class ($l$) $= 74$
* Frequency of modal class ($f_1$) $= 8$
* Frequency of preceding class ($f_0$) $= 3$
* Frequency of succeeding class ($f_2$) $= 7$
* Class size ($h$) $= 3$
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Mode $= 74 + \left( \frac{8 - 3}{2(8) - 3 - 7} \right) \times 3$
Mode $= 74 + \left( \frac{5}{16 - 10} \right) \times 3$
Mode $= 74 + \left( \frac{5}{6} \right) \times 3$
Mode $= 74 + \frac{15}{6} = 74 + 2.5 = 76.5$.
**Answer:** 76.5 beats per minute [cite: 305, 306]

53. A die is thrown once. Find the probability of getting: [4]Total possible outcomes when a die is thrown: $\{1, 2, 3, 4, 5, 6\}$. Total outcomes $= 6$.(i) a prime numberPrime numbers in the set are $\{2, 3, 5\}$. Number of favorable outcomes $= 3$.$P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$ 53(ii) a number lying between 2 and 6Numbers between 2 and 6 are $\{3, 4, 5\}$. Number of favorable outcomes $= 3$.$P(\text{between 2 and 6}) = \frac{3}{6} = \frac{1}{2}$ 54(iii) an odd numberOdd numbers in the set are $\{1, 3, 5\}$. Number of favorable outcomes $= 3$.$P(\text{odd}) = \frac{3}{6} = \frac{1}{2}$ 55(iv) 7The number 7 is not in the set of outcomes. Number of favorable outcomes $= 0$.$P(7) = \frac{0}{6} = 0$ 56
54. One card is drawn from a well-shuffled deck of 52 cards. Find the Probability of getting: [4]Total number of cards $= 52$.(i) a king of red colourThere are two red suits (Hearts and Diamonds), and one King in each.Favorable outcomes $= 2$ (King of Hearts, King of Diamonds).$P(\text{red king}) = \frac{2}{52} = \frac{1}{26}$ 57(ii) the jack of heartsThere is only one Jack of Hearts.Favorable outcomes $= 1$.$P(\text{jack of hearts}) = \frac{1}{52}$ 58(iii) a spadeThere are 13 spades in the deck.Favorable outcomes $= 13$.$P(\text{a spade}) = \frac{13}{52} = \frac{1}{4}$ 59(iv) a red face cardFace cards are Jack, Queen, King (3 per suit).There are 2 red suits (Hearts, Diamonds).Favorable outcomes $= 3 \times 2 = 6$.$P(\text{red face card}) = \frac{6}{52} = \frac{3}{26}$ 60

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