CBSE Class 10 Maths (Basic) Question Paper 2025 Solved Code 430-1-2

SECTION A

1. Which of the following is not the criterion for similarity of triangles? [1 Mark]
(A) AAA
(B) SSS
(C) SAS
(D) RHS

Answer: None of the given options is correct; the CBSE marking scheme awarded 1 mark to all candidates who attempted this question.

Teacher's Note:
a) AAA, SSS and SAS are standard similarity criteria for triangles.
b) RHS is normally used for congruence, but the examiner treated the question as ambiguous and awarded marks to everyone.

2. From the figures given below, which of the following is true about the measure of \( \angle P \)? [1 Mark]
(A) \( \angle P = 60^{\circ} \)
(B) \( \angle P = 80^{\circ} \)
(C) \( \angle P = 40^{\circ} \)
(D) The measure of \( \angle P \) cannot be determined

[Figure: Triangle ABC with AB = 3.8 cm, angle A = \(80^{\circ}\), AC = \(3\sqrt{3}\) cm, angle B = \(60^{\circ}\), BC = 6 cm. Triangle PQR (drawn as R at top, P and Q at base) with PR = \(6\sqrt{3}\) cm, RQ = 7.6 cm, PQ = 12 cm.]

Answer: (C) \( \angle P = 40^{\circ} \)

Teacher's Note:
a) Match sides in the same ratio: AB : RQ = AC : RP = BC : QP = 1 : 2.
b) So triangle ABC \(\sim\) triangle RQP, and the third angle \( \angle C = 180^{\circ} - 80^{\circ} - 60^{\circ} = 40^{\circ}\) corresponds to \( \angle P\).

3. If the distance of a tangent to a circle from its centre is 4 cm, then the length of diameter of the circle is : [1 Mark]
(A) 2 cm
(B) 4 cm
(C) 8 cm
(D) 16 cm

Answer: (C) 8 cm

Teacher's Note:
a) The distance of the tangent line from the centre equals the radius, since the tangent is perpendicular to the radius at the point of contact.
b) Diameter = 2 \(\times\) radius = 2 \(\times\) 4 = 8 cm.

4. Which of the following statements is false? [1 Mark]
(A) \( \tan 45^{\circ} = \cot 45^{\circ} \)
(B) \( \sin 90^{\circ} = \tan 45^{\circ} \)
(C) \( \sin 30^{\circ} = \cos 30^{\circ} \)
(D) \( \sin 45^{\circ} = \cos 45^{\circ} \)

Answer: (C) \( \sin 30^{\circ} = \cos 30^{\circ} \)

Teacher's Note:
a) \( \sin 30^{\circ} = \frac{1}{2}\) and \( \cos 30^{\circ} = \frac{\sqrt{3}}{2}\), so they are not equal.
b) Learn the standard trigonometric ratio table for quick checks.

5. The value of \( \left(\cot^{2} A - \dfrac{1}{\sin^{2} A}\right) \) is : [1 Mark]
(A) more than 1
(B) 1
(C) 0
(D) \( -1 \)

Answer: (D) \( -1 \)

Teacher's Note:
a) Use the identity \( 1 + \cot^{2} A = \csc^{2} A \), so \( \cot^{2}A - \csc^{2}A = -1\).
b) Remember \( \csc^{2}A = \frac{1}{\sin^{2}A}\).

6. In the given figure, which of the following angles represents the angle of depression? [1 Mark]
(A) x
(B) y
(C) z
(D) a

[Figure: Observer at top with a horizontal dashed line drawn from the observer's eye. Angle z is between the horizontal line and the line of sight going down. Angle y is below z. A right angle a is at the bottom of the vertical line where it meets the ground. The line of sight reaches an object on the ground, making angle x there.]

Answer: (C) z

Teacher's Note:
a) The angle of depression is always measured from the horizontal line at the observer's eye level down to the line of sight.
b) Do not confuse it with the angle of elevation, which is measured at the object.

7. The perimeter of the shaded region in the given figure is : [1 Mark]
(A) l
(B) l + a
(C) l + 2r
(D) l + 2r + a

[Figure: Circle with centre O. Two radii of length r each are drawn from O to the circle, enclosing angle a at O and an arc of length l between them; the shaded region is bounded by the two radii and the arc l.]

Answer: (C) l + 2r

Teacher's Note:
a) The boundary of the shaded sector consists of the two straight radii and the curved arc.
b) Perimeter of a sector = arc length + 2 \(\times\) radius.

8. The ratio of the area of a quadrant of a circle to the area of the same circle is : [1 Mark]
(A) 1 : 2
(B) 2 : 1
(C) 1 : 4
(D) 4 : 1

Answer: (C) 1 : 4

Teacher's Note:
a) A quadrant is a quarter of the circle, so its area is \( \frac{1}{4}\) of the circle's area.
b) Area of quadrant \( = \frac{1}{4}\pi r^{2}\).

9. For which of the following solids is the lateral / curved surface area and total surface area the same? [1 Mark]
(A) Cube
(B) Cuboid
(C) Hemisphere
(D) Sphere

Answer: (D) Sphere

Teacher's Note:
a) A sphere has no flat base, so its curved surface area equals its total surface area.
b) A hemisphere has a flat circular base, so its total surface area includes an extra term.

10. The class mark of the median class of the following data is :
Class Interval: 10-25 | 25-40 | 40-55 | 55-70 | 70-85 | 85-100
Frequency: 2 | 3 | 7 | 6 | 6 | 6 [1 Mark]
(A) 40
(B) 55
(C) 47·5
(D) 62·5

Answer: (D) 62·5

Teacher's Note:
a) Total frequency = 30, so \( \frac{n}{2} = 15\); the cumulative frequency first exceeds 15 in the class 55-70.
b) Class mark of 55-70 = \( \frac{55+70}{2} = 62.5\).

11. The following distribution shows the number of runs scored by some batsmen in test matches :
Runs Scored: 3000-4000 | 4000-5000 | 5000-6000 | 6000-7000
Number of Batsmen: 5 | 10 | 9 | 8
The lower limit of the modal class is : [1 Mark]
(A) 3000
(B) 4000
(C) 5000
(D) 6000

Answer: (B) 4000

Teacher's Note:
a) The modal class is the class with the highest frequency, which is 10 for 4000-5000.
b) Lower limit of modal class = 4000.

12. A bag contains 3 red, 4 white and 7 green balls. A ball is drawn at random. The probability that the ball drawn is not of red colour is : [1 Mark]
(A) \( \frac{1}{11} \)
(B) \( \frac{3}{14} \)
(C) \( \frac{11}{14} \)
(D) \( \frac{3}{11} \)

Answer: (C) \( \frac{11}{14} \)

Teacher's Note:
a) Total balls = 3 + 4 + 7 = 14; not-red balls = 4 + 7 = 11.
b) P(not red) = \( \frac{11}{14}\).

13. If the HCF of two positive integers a and b is 1, then their LCM is : [1 Mark]
(A) a + b
(B) a
(C) b
(D) ab

Answer: (D) ab

Teacher's Note:
a) Use the relation HCF \(\times\) LCM = a \(\times\) b.
b) Since HCF = 1, LCM = ab.

14. \( \dfrac{\sqrt{3}-3}{\sqrt{3}} \) is : [1 Mark]
(A) a rational number
(B) an irrational number
(C) an integer
(D) a natural number

Answer: (B) an irrational number

Teacher's Note:
a) \( \dfrac{\sqrt3 -3}{\sqrt3} = 1 - \sqrt3 \), which is irrational.
b) A rational number minus an irrational number is always irrational.

15. The discriminant of the quadratic equation \( -x^{2} - 5x + 6 = 0 \) is : [1 Mark]
(A) 1
(B) \( -1 \)
(C) 49
(D) 7

Answer: (C) 49

Teacher's Note:
a) Discriminant \( = b^{2} - 4ac\), here \( a=-1, b=-5, c=6\).
b) \( D = (-5)^{2} - 4(-1)(6) = 25+24 = 49\).

16. The equation \( x + \dfrac{1}{x} = 3 \; (x \neq 0) \) is expressed as a quadratic equation in the form of \( ax^{2}+bx+c=0 \). The value of \( a - b + c \) is : [1 Mark]
(A) 5
(B) 2
(C) 1
(D) \( -1 \)

Answer: (A) 5

Teacher's Note:
a) Multiplying by x gives \( x^{2} - 3x + 1 = 0\), so \( a=1, b=-3, c=1\).
b) \( a - b + c = 1-(-3)+1 = 5\).

17. The distance of a point P(3, - 7) from y-axis is : [1 Mark]
(A) 3
(B) 7
(C) \( -7 \)
(D) \( \sqrt{58} \)

Answer: (A) 3

Teacher's Note:
a) Distance of a point from the y-axis equals the absolute value of its x-coordinate.
b) Here x = 3, so the distance is 3 units.

18. The mid-point of a line segment divides the line segment in the ratio : [1 Mark]
(A) 1 : 2
(B) 2 : 1
(C) 1 : 1
(D) \( \frac{1}{2} : 2 \)

Answer: (C) 1 : 1

Teacher's Note:
a) The mid-point divides a segment into two equal parts.
b) Ratio 1 : 1 gives the mid-point formula \( \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\).

19. Assertion (A) : The value of p for which the system of equations \( 4x + py + 8 = 0 \) and \( 2x + 2y + 2 = 0 \) is consistent is 4.
Reason (R) : The system of equations \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \) is consistent with infinitely many solutions, if \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \). [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Teacher's Note:
a) Check: \( \frac{4}{2}=\frac{p}{2}=\frac{8}{2}\) gives \( p=4\), but this makes the two equations identical, so this value alone should be verified against the reason's condition.
b) The reason statement about consistency with infinitely many solutions is a correct general rule.

20. Assertion (A) : For any two natural numbers a and b, the HCF of a and b is a factor of the LCM of a and b.
Reason (R) : HCF of any two natural numbers divides both the numbers. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:
a) Since HCF \(\times\) LCM = a \(\times\) b, the HCF always divides the LCM.
b) The reason correctly explains why the assertion is true.

SECTION B

21. Two concentric circles are of radii 6 cm and 10 cm. Find the length of the chord of the larger circle which touches the smaller circle. [2 Marks]

Answer:
1. Let O be the common centre; the chord AB of the larger circle touches the smaller circle at C, so OC \(\perp\) AB and OC = 6 cm, OA = 10 cm.
2. \( AC = \sqrt{OA^{2}-OC^{2}} = \sqrt{100-36} = 8\) cm.
3. Since OC bisects AB, \( AB = 2 \times AC = 16\) cm.

Teacher's Note:
a) The perpendicular from the centre to a chord bisects the chord; use this with the Pythagoras theorem.
b) Draw a clear figure showing the two circles and the tangent chord to avoid sign errors.

22. (a) Find the values of A and B \( (0 \leq A \lt 90^{\circ}, 0 \leq B \lt 90^{\circ}) \), if \( \tan (A + B) = 1 \) and \( \tan (A - B) = \dfrac{1}{\sqrt{3}} \). [2 Marks]

Answer:
1. \( \tan(A+B)=1 \Rightarrow A+B = 45^{\circ}\).
2. \( \tan(A-B)=\frac{1}{\sqrt3} \Rightarrow A-B=30^{\circ}\).
3. Solving the two equations: \( A = 37.5^{\circ}, B = 7.5^{\circ}\).

Teacher's Note:
a) Convert each trigonometric equation into a linear angle equation using known standard angles.
b) Add and subtract the two equations to find A and B quickly.

OR

(b) Prove that \( \tan 45^{\circ} = 1 \) geometrically. [2 Marks]

Answer:
1. Consider an isosceles right triangle ABC with the right angle at B and AB = BC = x.
2. By the angle sum property, \( \angle A = \angle C = 45^{\circ}\).
3. \( \tan 45^{\circ} = \dfrac{AB}{BC} = \dfrac{x}{x} = 1\).

Teacher's Note:
a) Draw the isosceles right triangle clearly, marking the right angle and the two \(45^{\circ}\) angles.
b) Using equal legs (AB = BC) is the key step that gives the ratio 1.

23. In the given figure, two concentric circles with centre O and radii 2 cm and 3 cm are shown. Find the perimeter of the shaded region. [2 Marks]

[Figure: Two concentric circles with centre O; inner circle radius 2 cm, outer circle radius 3 cm. Two radii from O to the outer circle make an angle of \(60^{\circ}\) at O; the shaded region lies between the two arcs and the two radial segments.]

Answer:
1. Outer arc length \( = \dfrac{60}{360}\times 2\pi \times 3 = \dfrac{22}{7}\) cm.
2. Inner arc length \( = \dfrac{60}{360}\times 2\pi \times 2 = \dfrac{44}{21}\) cm.
3. Perimeter \( = \dfrac{22}{7}+\dfrac{44}{21}+2 = \dfrac{152}{21}\) cm or 7.24 cm (approx).

Teacher's Note:
a) The perimeter includes both curved arcs plus the two straight radial segments of length (3-2) cm each.
b) Use \( \pi = \frac{22}{7}\) as instructed in the paper.

24. Solve for x and y :
0.1x + 0.3y = 1
0.2x - 0.1y = -0.1 [2 Marks]

Answer:
1. Multiplying both equations by 10: \( x+3y=10\) and \( 2x-y=-1\).
2. From the second equation, \( y=2x+1\); substituting: \( x+3(2x+1)=10 \Rightarrow 7x=7 \Rightarrow x=1\).
3. Then \( y = 2(1)+1 = 3\). So \( x=1, y=3\).

Teacher's Note:
a) Clear decimals first by multiplying throughout by 10 to avoid calculation errors.
b) Verify the answer by substituting back into both original equations.

25. (a) In the given figure, if PQ \( \parallel \) RS, then prove that \( \triangle POQ \sim \triangle SOR \). [2 Marks]

[Figure: Two line segments PQ and RS crossing at point O, with P and R at the top, Q and S at the bottom, and arrowheads on PQ and RS showing they are parallel.]

Answer:
1. Since PQ \( \parallel \) RS, \( \angle P = \angle S\) and \( \angle Q = \angle R\) (alternate interior angles).
2. Therefore \( \triangle POQ \sim \triangle SOR\) by the AA similarity criterion.

Teacher's Note:
a) Identify alternate interior angles carefully using the parallel lines and the common transversal.
b) AA similarity only needs two pairs of equal angles.

OR

(b) In the given figure, \( \triangle OSR \sim \triangle OQP \), \( \angle ROQ = 125^{\circ} \) and \( \angle ORS = 70^{\circ} \). Find the measures of \( \angle OSR \) and \( \angle OQP \). [2 Marks]

[Figure: Lines PQ and SR crossing at O; S and R at the top ends, P and Q at the bottom ends; angle ROQ = \(125^{\circ}\) at O, angle ORS = \(70^{\circ}\) at R.]

Answer:
1. By the exterior angle property of triangle ORS, \( \angle OSR = 125^{\circ} - 70^{\circ} = 55^{\circ}\).
2. Since \( \triangle OSR \sim \triangle OQP\), corresponding angles are equal, so \( \angle OQP = \angle OSR = 55^{\circ}\).

Teacher's Note:
a) Exterior angle of a triangle equals the sum of the two opposite interior angles.
b) Once similarity is given, matching corresponding angles gives the second answer directly.

SECTION C

26. (a) Solve the following system of equations graphically :
x + 3y = 6; 2x - 3y = 12 [3 Marks]

Answer:
1. For x + 3y = 6: points (0, 2) and (6, 0).
2. For 2x - 3y = 12: points (6, 0) and (0, -4).
3. Plotting both lines, they intersect at (6, 0), so \( x = 6, y = 0\).

Teacher's Note:
a) Plot at least two clear points for each line using a suitable scale.
b) The point of intersection of the two lines is the required solution.

OR

(b) x and y are complementary angles such that x : y = 1 : 2. Express the given information as a system of linear equations in two variables and hence solve it. [3 Marks]

Answer:
1. Since x and y are complementary, \( x+y=90^{\circ}\).
2. Since \( x:y=1:2\), \( y=2x\).
3. Substituting: \( x+2x=90 \Rightarrow x=30^{\circ}, y=60^{\circ}\).

Teacher's Note:
a) Complementary angles always add up to \(90^{\circ}\).
b) Convert the ratio into an equation before solving the linear system.

27. Prove that a rectangle circumscribing a circle is a square. [3 Marks]

[Figure: A circle with centre O inscribed in a rectangle ABCD; the circle touches AB at P, BC at Q, CD at R and DA at S.]

Answer:
1. Since tangents drawn from an external point to a circle are equal in length: AP = AS, BP = BQ, CR = CQ, DR = DS.
2. Adding these: AB + CD = BC + AD.
3. Since ABCD is a rectangle, AB = CD and BC = AD, so this gives \( AB = AD\).
4. As adjacent sides of the rectangle are equal, ABCD is a square.

Teacher's Note:
a) The equal-tangent-length property is the key idea used here.
b) A rectangle with two equal adjacent sides is, by definition, a square.

28. Prove the following trigonometric identity :
\( \sqrt{\dfrac{\csc A - 1}{\csc A + 1}} = \sec A - \tan A \) [3 Marks]

Answer:
1. \( \text{LHS} = \sqrt{\dfrac{\csc A - 1}{\csc A + 1}\times\dfrac{\csc A - 1}{\csc A - 1}} = \sqrt{\dfrac{(\csc A - 1)^{2}}{\cot^{2}A}}\).
2. This simplifies to \( \dfrac{\csc A - 1}{\cot A}\).
3. \( = \dfrac{\csc A}{\cot A} - \dfrac{1}{\cot A} = \sec A - \tan A = \text{RHS}\).

Teacher's Note:
a) Multiply numerator and denominator by \( (\csc A -1)\) to create a perfect square under the root.
b) Use \( \csc^{2}A - 1 = \cot^{2}A\) and convert cosec, cot into sec, tan.

29. A lot consists of 200 pens of which 180 are good and the rest are defective. A customer will buy a pen if it is not defective. The shopkeeper draws a pen at random and gives it to the customer. What is the probability that the customer will not buy it? Another lot of 100 pens containing 80 good pens is mixed with the previous lot of 200 pens. The shopkeeper now draws one pen at random from the entire lot and gives it to the customer. What is the probability that the customer will buy the pen? [3 Marks]

Answer:
1. Defective pens in first lot = 200 - 180 = 20, so P(customer will not buy) = \( \dfrac{20}{200} = \dfrac{1}{10}\).
2. After mixing, total pens = 200 + 100 = 300 and total good pens = 180 + 80 = 260.
3. P(customer will buy the pen) = \( \dfrac{260}{300} = \dfrac{13}{15}\).

Teacher's Note:
a) The customer buys the pen only if it is a good pen.
b) Recalculate the total number of pens and good pens after mixing before finding the second probability.

30. (a) Prove that \( \sqrt{3} \) is an irrational number. [3 Marks]

Answer:
1. Suppose \( \sqrt3\) is rational, so \( \sqrt3 = \dfrac{p}{q}\), where p and q are co-prime integers, \( q \neq 0\).
2. Then \( 3q^{2}=p^{2}\), so 3 divides \( p^{2}\), and hence 3 divides p. Let \( p=3m\).
3. Substituting: \( 3q^{2}=9m^{2} \Rightarrow q^{2}=3m^{2}\), so 3 divides q also.
4. This contradicts that p and q are co-prime; hence our assumption is wrong and \( \sqrt3\) is irrational.

Teacher's Note:
a) This is the standard proof by contradiction method for irrationality.
b) The key step is showing that 3 divides both p and q, which contradicts co-primality.

OR

(b) The factor tree of a number x is shown below :
x branches into 2 and y.
y branches into 2 and 210.
210 branches into a and 70.
70 branches into 2 and 35.
35 branches into 5 and b.
Find the values of x, y, a and b. Hence, write the product of the prime factors of the number x so obtained. [3 Marks]

[Figure: Factor tree with x at top, branching to 2 and y; y branches to 2 and 210; 210 branches to a and 70; 70 branches to 2 and 35; 35 branches to 5 and b.]

Answer:
1. \( 35 = 5 \times b \Rightarrow b = 7\).
2. \( 210 = a \times 70 \Rightarrow a = 3\).
3. \( y = 2 \times 210 = 420\), and \( x = 2 \times y = 2 \times 420 = 840\).
4. \( x = 840 = 2^{3} \times 3 \times 5 \times 7\).

Teacher's Note:
a) Work from the bottom of the tree upward, multiplying the branch values.
b) Always express the final answer as a product of prime factors in increasing order.

31. Find a quadratic polynomial, sum and product of whose zeroes are 5 and - 6, respectively. Also, find the zeroes of the polynomial so obtained. [3 Marks]

Answer:
1. Required polynomial: \( x^{2} - (\text{sum})x + (\text{product}) = x^{2} - 5x - 6\).
2. Factorising: \( x^{2}-5x-6 = (x-6)(x+1)\).
3. Zeroes are \( x = 6\) and \( x = -1\).

Teacher's Note:
a) The standard form for a quadratic polynomial from sum and product of zeroes is \( x^{2} - (\text{sum})x + (\text{product})\).
b) Factorise or use the quadratic formula to find the zeroes.

SECTION D

32. State "Basic Proportionality Theorem" and use it to prove the following :
A line through the mid-point of one side of a triangle, parallel to another side, bisects the third side. [5 Marks]

[Figure: Triangle ABC with P as the mid-point of AB and line PQ drawn parallel to BC, meeting AC at Q.]

Answer:
1. Statement (BPT): 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.
2. Given: In triangle ABC, P is the mid-point of AB and PQ \( \parallel \) BC.
3. To prove: Q is the mid-point of AC.
4. Proof: Since PQ \( \parallel \) BC, by BPT, \( \dfrac{AP}{PB} = \dfrac{AQ}{QC}\). Since P is the mid-point, \( AP = PB\), so \( \dfrac{AP}{PB} = 1\).
5. Therefore \( \dfrac{AQ}{QC} = 1 \Rightarrow AQ = QC\), so Q is the mid-point of AC.

Teacher's Note:
a) State the theorem exactly as given in the textbook before applying it.
b) The proof depends only on substituting \( AP=PB\) into the BPT ratio.

33. (a) A toy is in the form of a cone surmounted on a hemisphere. The cone and hemisphere have the same radii. The height of the conical part of the toy is equal to the diameter of its base. If the radius of the conical part is 5 cm, find the volume of the toy. [5 Marks]

[Figure: A toy shaped like a cone sitting on top of a hemisphere, both with radius 5 cm; the cone's height equals its base diameter, 10 cm.]

Answer:
1. Radius \( r = 5\) cm, height of cone \( h = 2r = 10\) cm.
2. Volume of toy = volume of hemisphere + volume of cone = \( \dfrac{2}{3}\pi r^{3} + \dfrac{1}{3}\pi r^{2}h\).
3. \( = \dfrac{2}{3}\times\dfrac{22}{7}\times 125 + \dfrac{1}{3}\times\dfrac{22}{7}\times 25\times 10 = \dfrac{5500}{21}+\dfrac{5500}{21}\).
4. Volume \( = \dfrac{11000}{21} \) cu. cm or 523.81 cu. cm (approx).

Teacher's Note:
a) Total volume of a combined solid is the sum of the volumes of its parts.
b) Substitute the height as twice the radius before calculating.

OR

(b) A cubical block is surmounted by a hemisphere of radius 3·5 cm. What is the smallest possible length of the edge of the cube so that the hemisphere can totally lie on the cube? Find the total surface area of the solid so formed. [5 Marks]

[Figure: A cube with a hemisphere of radius 3.5 cm placed on top, touching all four edges of the top face.]

Answer:
1. The smallest edge of the cube must equal the diameter of the hemisphere: \( a = 2\times3.5 = 7\) cm.
2. Total surface area of the solid = \( 6a^{2} + 2\pi r^{2} - \pi r^{2} = 6a^{2}+\pi r^{2}\).
3. \( = 6\times 7\times7 + \dfrac{22}{7}\times3.5\times3.5 = 294 + 38.5\).
4. Total surface area \( = 332.5\) sq. cm (or \( \dfrac{665}{2}\) sq. cm).

Teacher's Note:
a) The circular base area of the hemisphere is subtracted since it merges with the cube's top face.
b) Always convert the radius to the correct edge length before substituting.

34. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the monthly mean consumption from the data.
Monthly Consumption (in units): 50-100 | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 | 350-400
Number of Consumers: 4 | 5 | 13 | 20 | 14 | 8 | 4 [5 Marks]

Answer:
1. Taking class marks \( x_i\): 75, 125, 175, 225 (assumed mean a = 225), 275, 325, 375 with \( u_i = \dfrac{x_i - a}{50}\): -3, -2, -1, 0, 1, 2, 3.
2. Corresponding \( f_iu_i\): -12, -10, -13, 0, 14, 16, 12; sum \( f_i = 68\), sum \( f_iu_i = 7\).
3. Mean \( = a + \dfrac{\Sigma f_iu_i}{\Sigma f_i}\times h = 225 + \dfrac{7}{68}\times 50\).
4. Mean \( = 230.15\) units.

Teacher's Note:
a) The step-deviation method is fastest when class marks are large numbers.
b) Double check that \( \Sigma f_i = 68\) matches the total number of consumers given.

35. (a) The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers. [5 Marks]

Answer:
1. Let the greater number be x and the smaller number be y.
2. \( x^{2}-y^{2}=180\) and \( y^{2}=8x\).
3. Substituting: \( x^{2}-8x=180 \Rightarrow x^{2}-8x-180=0\).
4. Factorising: \( (x-18)(x+10)=0 \Rightarrow x=18\) (x = -10 is rejected, being negative).
5. \( y^{2}=8\times18=144 \Rightarrow y=12\). The numbers are 18 and 12.

Teacher's Note:
a) Always reject negative roots when the problem specifies positive numbers.
b) Substitute one equation into the other to reduce to a single-variable quadratic.

OR

(b) Find the value(s) of k for which the equation \( 2x^{2}+kx+3=0 \) has real and equal roots. Hence, find the roots of the equations so obtained. [5 Marks]

Answer:
1. For real and equal roots, discriminant \( b^{2}-4ac=0\): \( k^{2}-24=0 \Rightarrow k=\pm2\sqrt6\).
2. For \( k=2\sqrt6\): equation is \( 2x^{2}+2\sqrt6x+3=0\), giving roots \( x=-\sqrt{\dfrac{3}{2}}, -\sqrt{\dfrac{3}{2}}\).
3. For \( k=-2\sqrt6\): equation is \( 2x^{2}-2\sqrt6x+3=0\), giving roots \( x=\sqrt{\dfrac{3}{2}}, \sqrt{\dfrac{3}{2}}\).

Teacher's Note:
a) Equal roots always satisfy \( x=-\dfrac{b}{2a}\).
b) Consider both positive and negative values of k, since squaring removes the sign.

SECTION E

Case Study - 1

36. In a garden, saplings of rose flowers were planted at equal intervals to form a spiral pattern. The spiral is made up of successive semicircles, with centres alternatively at A and B, starting with centre at A, of radii 50 cm, 100 cm, 150 cm, ....... as shown in the figure given below. Spiral 1 has 10 flowers, Spiral 2 has 20 flowers, Spiral 3 has 30 flowers and so on.

[Figure: A photo of a spiral flower pattern, and a diagram of semicircles labelled \(l_1, l_2, l_3, l_4\), with centres alternating between points A and B on a horizontal line, radii increasing outward.]

(i) What is the radius of the 13th spiral? [1 Mark]

Answer: The radii form an AP with first term 50 and common difference 50, so \( a_{13} = 50+(13-1)\times50 = 650\) cm.

Teacher's Note:
a) Use the AP formula \( a_n = a+(n-1)d\).
b) Here \( a=d=50\).

(ii) If the radius of the nth spiral is 500 cm, find the value of n. [1 Mark]

Answer: \( 50+(n-1)\times50=500 \Rightarrow n=10\).

Teacher's Note:
a) Rearranging the AP formula for n gives a simple linear equation.
b) Cross-check by substituting n = 10 back into the formula.

(iii) (a) Find the total number of saplings till the 11th spiral. [2 Marks]

Answer:
1. The number of flowers per spiral forms an AP: 10, 20, 30, ... with \( a=10, d=10\).
2. \( S_{11} = \dfrac{11}{2}[2\times10+(11-1)\times10] = \dfrac{11}{2}\times120 = 660\).

Teacher's Note:
a) Use the AP sum formula \( S_n=\dfrac{n}{2}[2a+(n-1)d]\).
b) Do not confuse the flower-count AP with the radius AP; they are different series.

OR

(b) Till which spiral, will there be a total of 450 saplings? [2 Marks]

Answer:
1. \( 450=\dfrac{n}{2}[2\times10+(n-1)\times10]\).
2. Simplifying: \( n^{2}+n-90=0\).
3. Solving the quadratic: \( n=9\) (rejecting the negative root).

Teacher's Note:
a) Set up the sum formula equal to the given total and simplify to a quadratic in n.
b) Always reject the negative value of n since it represents a spiral number.

Case Study - 2

37. In a society, there is a circular park having two gates. The gates are placed at points A(10, 20) and B(50, 50), as shown in the figure below. Two fountains are installed at points P and Q on AB such that AP = PQ = QB.

[Figure: A circular park with gates at A(10, 20) and B(50, 50); AB is a diameter passing through the centre C; fountains P and Q lie on AB such that AP = PQ = QB.]

(i) Find the coordinates of the centre C. [1 Mark]

Answer: C is the mid-point of AB: \( C = \left(\dfrac{10+50}{2}, \dfrac{20+50}{2}\right) = (30, 35)\).

Teacher's Note:
a) The centre of a circle lies at the midpoint of any diameter.
b) Use the mid-point formula carefully with the correct coordinate pairs.

(ii) Find the radius of the circular park. [1 Mark]

Answer: Radius \( = \sqrt{(30-10)^{2}+(35-20)^{2}} = \sqrt{400+225} = \sqrt{625} = 25\) units.

Teacher's Note:
a) The radius is the distance from the centre to either gate.
b) Use the distance formula \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).

(iii) (a) Find the coordinates of the point P. [2 Marks]

Answer:
1. Since AP = PQ = QB, point P divides AB in the ratio 1 : 2 from A.
2. By the section formula, \( P = \left(\dfrac{1\times50+2\times10}{3}, \dfrac{1\times50+2\times20}{3}\right) = \left(\dfrac{70}{3}, 30\right)\).

Teacher's Note:
a) Identify the correct ratio (1:2) before applying the section formula.
b) Keep the coordinates as fractions for accuracy.

OR

(b) Find the distance of the fountain at Q from gate A. [2 Marks]

Answer:
1. AB \( = 2 \times 25 = 50\) units (diameter).
2. Since AQ covers two of the three equal parts, \( AQ = \dfrac{2}{3}\times AB = \dfrac{2}{3}\times50 = \dfrac{100}{3}\) units.

Teacher's Note:
a) AQ is two-thirds of AB since AP = PQ = QB divides AB into three equal parts.
b) Use the diameter length found from twice the radius.

Case Study - 3

38. An injured bird was found on the roof of a building. The building is 15 m high. A fireman was called to rescue the bird. The fireman used an adjustable ladder to reach the roof. He placed the ladder in such a way that the ladder makes an angle of 60° with the ground in order to reach the roof.

[Figure: A right triangle with the building of height 15 m as the vertical side, the ladder as the hypotenuse making an angle of \(60^{\circ}\) with the ground.]

(i) Find the length of the ladder used by the fireman to reach the roof. [1 Mark]

Answer: \( \sin60^{\circ} = \dfrac{15}{a} \Rightarrow a = \dfrac{30}{\sqrt3} = 10\sqrt3\) m.

Teacher's Note:
a) The ladder is the hypotenuse; the building height is the side opposite the given angle.
b) Rationalise \( \dfrac{30}{\sqrt3}\) to get \( 10\sqrt3\).

(ii) Find the distance of the point on the ground at which the ladder was fixed from the bottom of the building. [1 Mark]

Answer: \( \tan60^{\circ} = \dfrac{15}{x} \Rightarrow x = \dfrac{15}{\sqrt3} = 5\sqrt3\) m.

Teacher's Note:
a) Use tangent since both the height and the base distance are involved (no hypotenuse needed).
b) Simplify \( \dfrac{15}{\sqrt3}\) to \( 5\sqrt3\) by rationalising.

(iii) In order to avoid skidding, the fireman placed the ladder in such a way that the bottom of the ladder touches the base of the wall which is opposite to the building, making an angle of 30° with the ground.

(a) Draw a neat diagram to represent the above situation and hence find the width of the road between the building and the wall. [2 Marks]

[Figure: Right triangle with the top of the building A at height 15 m, base of the building B, and the foot of the ladder C at the base of the opposite wall; angle at C = \(30^{\circ}\); BC is the road width y.]

Answer: \( \tan30^{\circ} = \dfrac{15}{y} \Rightarrow y = 15\sqrt3\) m.

Teacher's Note:
a) The triangle now uses the same building height but a different base angle (30°).
b) Always draw the new right triangle separately to avoid confusing it with part (i) and (ii).

OR

(b) Find the length of the ladder used by the fireman in this case. [2 Marks]

Answer: \( \sin30^{\circ} = \dfrac{15}{l} \Rightarrow l = \dfrac{15}{\frac{1}{2}} = 30\) m.

Teacher's Note:
a) The ladder is again the hypotenuse of the right triangle with the building height as the opposite side.
b) Recall \( \sin30^{\circ}=\dfrac{1}{2}\) to solve quickly.

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