CBSE Class 10 Maths (Basic) Question Paper 2026 Solved Code 430-1-3

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: (D) RHS

Teacher's Note:
a) AAA, SSS and SAS are used to prove similarity of triangles.
b) RHS is a criterion used only for congruence of right triangles, not similarity.

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 \( \angle A = 80^{\circ} \), \( AB = 3.8 \) cm, \( AC = 3\sqrt{3} \) cm, \( \angle B = 60^{\circ} \), \( BC = 6 \) cm; Triangle PQR (vertices P, R, Q) with \( PR = 6\sqrt{3} \) cm, \( RQ = 7.6 \) cm, \( PQ = 12 \) cm]

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

Teacher's Note:
a) Check ratios of corresponding sides: \( \frac{AB}{RQ} = \frac{AC}{RP} = \frac{BC}{PQ} = \frac{1}{2} \), so \( \triangle ABC \sim \triangle RQP \).
b) Since \( C \) corresponds to \( P \), \( \angle P = \angle C = 180^{\circ} - 80^{\circ} - 60^{\circ} = 40^{\circ} \).

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 perpendicular distance from the centre to a tangent equals the radius.
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) \( \sin 30^{\circ} = \frac{1}{2} \) while \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \), so they are not equal.
b) Remember the standard trigonometric ratio table for quick checking.

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

Answer: (D) \( -1 \)

Teacher's Note:
a) Use the identity \( \text{cosec}^{2}A - \cot^{2}A = 1 \).
b) So \( \cot^{2}A - \text{cosec}^{2}A = -1 \), and \( \frac{1}{\sin^{2}A} = \text{cosec}^{2}A \).

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 the top of a vertical line, a horizontal dashed line drawn from the observer, and a line of sight drawn from the observer down to an object; angle z lies between the horizontal line and the line of sight, angle y lies between the line of sight and the vertical line, angle a is the right angle at the foot of the vertical line, angle x is at the object between the horizontal ground line and the line of sight]

Answer: (C) z

Teacher's Note:
a) The angle of depression is always measured from the horizontal line at the observer's eye down to the line of sight.
b) Do not confuse it with the angle of elevation (x), 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: A circle with centre O; two radii of length r each are drawn from O to two points on the circle; the straight chord joining these two points is labelled a; the arc of the sector between the two points is labelled l]

Answer: (C) l + 2r

Teacher's Note:
a) The shaded sector is bounded by two radii and its own arc, not the chord.
b) Perimeter of a sector = sum of the two radii + arc length = 2r + l.

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 a circle, so its area is \( \frac{1}{4} \) of the circle's area.
b) Ratio = 1 : 4.

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 faces, so its curved surface area itself is its total surface area, \( 4\pi r^{2} \).
b) A hemisphere has a flat circular base in addition, so its CSA and TSA differ.

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

Answer: (D) 62·5

Teacher's Note:
a) \( N = 30 \), so \( \frac{N}{2} = 15 \); cumulative frequencies are 2, 5, 12, 18, 24, 30.
b) The median class is 55-70 (first cf \(\geq\) 15), so class mark = \( \frac{55+70}{2} = 62.5 \).

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 has the highest frequency, here 10, for class 4000-5000.
b) Lower limit of this class is 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; non-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 HCF \(\times\) LCM = a \(\times\) b.
b) Since HCF = 1, LCM = ab.

14. \( \frac{\sqrt{3}-3}{\sqrt{3}} \) 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) Simplify: \( \frac{\sqrt{3}-3}{\sqrt{3}} = 1 - \sqrt{3} \).
b) Since \( \sqrt{3} \) is irrational, \( 1-\sqrt{3} \) is also 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) Here \( a = -1, b = -5, c = 6 \).
b) \( D = b^{2}-4ac = 25 - 4(-1)(6) = 25+24 = 49 \).

16. The equation \( x + \frac{1}{x} = 3 \ (x \neq 0) \) 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 from y-axis is simply the absolute value of the x-coordinate.
b) Here x = 3, so distance = 3.

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) A midpoint divides a segment into two equal parts.
b) Hence the ratio is always 1 : 1.

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 \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). [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) Here \( \frac{a_1}{a_2}=2, \frac{b_1}{b_2}=\frac{p}{2} \); for p = 4, both ratios equal 2, but \( \frac{c_1}{c_2}=4 \neq 2 \), so the lines are parallel and the system has no solution.
b) The reason statement about consistency with infinite solutions is a correct general fact.

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, HCF always divides LCM.
b) Recall HCF \(\times\) LCM = product of the two numbers as the key formula.

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. The tangent from the centre is perpendicular to the chord at its point of contact, so the perpendicular distance from centre to chord = radius of smaller circle = 6 cm.
2. Half chord = \( \sqrt{10^{2}-6^{2}} = \sqrt{64} = 8 \) cm, so full chord = 16 cm.

Teacher's Note:
a) The chord, radius of smaller circle, and half the chord form a right triangle with the radius of the larger circle as hypotenuse.
b) Always double the half-chord to get the final answer.

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 \( \tan(A-B) = \frac{1}{\sqrt{3}} \). [2 Marks]

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

Teacher's Note:
a) Convert the trigonometric equations into simple linear equations in A and B first.
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. Construct a right triangle ABC, right angled at B, with \( \angle A = 45^{\circ} \).
2. Since \( \angle A = 45^{\circ} \) and \( \angle B = 90^{\circ} \), \( \angle C = 45^{\circ} \), so the triangle is isosceles with AB = BC.
3. \( \tan 45^{\circ} = \frac{BC}{AB} = \frac{AB}{AB} = 1 \).

Teacher's Note:
a) A 45-45-90 triangle always has its two legs equal.
b) Draw a clean labelled diagram to earn full marks.

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 radius 2 cm and outer radius 3 cm, sector angle 60° at O, with the region between the two arcs (a ring-shaped sector) shaded/hatched]

Answer:
1. Outer arc length = \( \frac{60}{360} \times 2\pi(3) = \frac{22}{7} \) cm.
2. Inner arc length = \( \frac{60}{360} \times 2\pi(2) = \frac{44}{21} \) cm.
3. Two straight parts = \( 2 \times (3-2) = 2 \) cm.
4. Perimeter = \( \frac{22}{7} + \frac{44}{21} + 2 = \frac{152}{21} \approx 7.24 \) cm.

Teacher's Note:
a) The shaded ring-sector's perimeter has two arcs (inner and outer) plus two straight radial segments.
b) Do not use the chord length instead of the arc length.

24. Solve for x and y :
0·1x + 0·3y = 1
0·2x - 0·1y = - 0·1 [2 Marks]

Answer:
1. Multiply both equations by 10: \( x+3y=10 \) and \( 2x-y=-1 \).
2. From the second equation, \( y = 2x+1 \).
3. Substituting: \( x+3(2x+1)=10 \Rightarrow 7x=7 \Rightarrow x=1 \), so \( y=3 \).

Teacher's Note:
a) Clearing decimals first makes the equations easier to solve.
b) Always verify the solution 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 with P, Q on the left and R, S on the right, both meeting a common point O such that P, O, S are collinear and Q, O, R are collinear, forming an X shape, with PQ parallel to RS]

Answer:
1. Since \( PQ \parallel RS \) and PS is a transversal, \( \angle OPQ = \angle OSR \) (alternate angles).
2. Since \( PQ \parallel RS \) and QR is a transversal, \( \angle OQP = \angle ORS \) (alternate angles).
3. By AA similarity, \( \triangle POQ \sim \triangle SOR \).

Teacher's Note:
a) Look for alternate angles whenever parallel lines and a transversal are given.
b) Only two pairs of equal angles are needed for the AA similarity criterion.

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 SQ and RP crossing at O, forming an X shape with S and R at the top and P and Q at the bottom, \( \angle ROQ = 125^{\circ} \) and \( \angle ORS = 70^{\circ} \) marked]

Answer:
1. Since SOQ is a straight line, \( \angle SOR = 180^{\circ} - 125^{\circ} = 55^{\circ} \).
2. In \( \triangle OSR \), \( \angle OSR = 180^{\circ} - 55^{\circ} - 70^{\circ} = 55^{\circ} \).
3. Since \( \triangle OSR \sim \triangle OQP \), \( \angle OQP = \angle OSR = 55^{\circ} \).

Teacher's Note:
a) Use the linear pair property first to find the vertically related angle at O.
b) Match corresponding vertices of similar triangles carefully before equating angles.

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: when x = 0, y = 2; when x = 6, y = 0.
2. For 2x - 3y = 12: when x = 0, y = -4; when x = 6, y = 0.
3. Plot both lines; they intersect at (6, 0), which is the solution: x = 6, y = 0.

Teacher's Note:
a) Take at least two points for each line and plot on the same graph.
b) The point of intersection of the two lines gives the 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, \( x+y=90 \).
2. Since \( x:y=1:2 \), \( y=2x \), i.e., \( 2x-y=0 \).
3. Solving: \( x+2x=90 \Rightarrow x=30 \), so \( y=60 \).

Teacher's Note:
a) Complementary angles always add up to 90°.
b) Convert the given ratio into a linear equation before solving.

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

Answer:
1. Let rectangle ABCD circumscribe a circle touching sides AB, BC, CD, DA at P, Q, R, S respectively.
2. Tangent lengths from each vertex are equal, so using the property that opposite sides of a quadrilateral circumscribing a circle are equal in sum: \( AB+CD = BC+AD \).
3. Since ABCD is a rectangle, \( AB=CD \) and \( BC=AD \), so \( 2AB=2BC \), giving \( AB=BC \).

Teacher's Note:
a) Use the theorem that sum of opposite sides of a circumscribing quadrilateral are equal.
b) A rectangle with all sides equal is by definition a square.

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

Answer:
1. Multiply numerator and denominator inside the root by \( (\text{cosec}\,A - 1) \): LHS \( = \sqrt{\dfrac{(\text{cosec}\,A-1)^{2}}{\text{cosec}^{2}A-1}} \).
2. Since \( \text{cosec}^{2}A-1=\cot^{2}A \), LHS \( = \dfrac{\text{cosec}\,A-1}{\cot A} \).
3. This equals \( \dfrac{\frac{1}{\sin A}-1}{\frac{\cos A}{\sin A}} = \dfrac{1-\sin A}{\cos A} = \sec A - \tan A = \) RHS.

Teacher's Note:
a) Multiplying inside the root by the conjugate is the key first step.
b) Use \( \text{cosec}^{2}A - 1 = \cot^{2}A \) to simplify the square root.

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

Teacher's Note:
a) "Not buy" means the pen drawn is defective.
b) Recalculate the total number of pens and good pens carefully after mixing the two lots.

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

Answer:
1. Assume, to the contrary, that \( \sqrt{3} = \frac{p}{q} \), where p, q are coprime integers, \( q \neq 0 \).
2. Squaring, \( 3q^{2}=p^{2} \), so 3 divides \( p^{2} \), hence 3 divides p. Let \( p=3m \).
3. Then \( 3q^{2}=9m^{2} \Rightarrow q^{2}=3m^{2} \), so 3 divides q also, contradicting that p, q are coprime. Hence \( \sqrt{3} \) is irrational.

Teacher's Note:
a) This is a standard proof by contradiction; always start by assuming the opposite.
b) The key step is showing 3 divides both p and q, which contradicts coprimality.

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 diagram - x splits into 2 and y; y splits into 2 and 210; 210 splits into a and 70; 70 splits into 2 and 35; 35 splits into 5 and b]

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

Teacher's Note:
a) Work from the bottom of the tree upward to find each unknown.
b) The final prime factorisation should include all prime branches of the tree.

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) Use the formula \( x^{2}-(\text{sum of zeroes})x+(\text{product of zeroes}) \).
b) Verify by checking that the sum and product of the factorised zeroes match the given values.

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]

Answer:
1. Basic Proportionality Theorem: 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 \( \triangle ABC \), D is the midpoint of AB, and DE \( \parallel \) BC meets AC at E.
3. To prove: E is the midpoint of AC.
4. By BPT, since DE \( \parallel \) BC, \( \frac{AD}{DB} = \frac{AE}{EC} \).
5. Since D is the midpoint of AB, \( AD=DB \), so \( \frac{AD}{DB}=1 \), giving \( \frac{AE}{EC}=1 \), i.e., \( AE=EC \). Hence DE bisects AC.

Teacher's Note:
a) The statement of BPT must be written exactly and clearly for full marks.
b) Applying BPT directly with D as midpoint is the shortest way to reach the conclusion.

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]

Answer:
1. Radius \( r=5 \) cm; height of cone \( h = \) diameter \( = 10 \) cm.
2. Volume of cone \( = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi(25)(10) = \frac{250\pi}{3} \) cm\(^3\).
3. Volume of hemisphere \( = \frac{2}{3}\pi r^{3} = \frac{2}{3}\pi(125) = \frac{250\pi}{3} \) cm\(^3\).
4. Total volume of toy \( = \frac{250\pi}{3}+\frac{250\pi}{3} = \frac{500\pi}{3} \approx 523.81 \) cm\(^3\).

Teacher's Note:
a) Since both solids share the same radius, add their individual volume formulas directly.
b) Always convert height in terms of the given radius before substituting.

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]

Answer:
1. Smallest edge of the cube = diameter of hemisphere = \( 2 \times 3.5 = 7 \) cm.
2. Total surface area = 6(edge)\(^2\) - \( \pi r^{2} \) (base of hemisphere removed) + \( 2\pi r^{2} \) (curved surface of hemisphere) = \( 6(\text{edge})^{2} + \pi r^{2} \).
3. \( = 6(49) + \frac{22}{7}(12.25) = 294 + 38.5 = 332.5 \) cm\(^2\).

Teacher's Note:
a) The circular base area of the hemisphere is subtracted from the cube's surface, and its curved surface is added.
b) The net change to the cube's surface area is simply adding \( \pi r^{2} \).

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 assumed mean \( a=225 \) (midpoint of 200-250) and class width \( h=50 \), compute \( u_i = \frac{x_i-a}{h} \) for each class.
2. \( f_i u_i \) values: -12, -10, -13, 0, 14, 16, 12; sum \( \Sigma f_i u_i = 7 \).
3. Mean \( = a + h \times \frac{\Sigma f_i u_i}{\Sigma f_i} = 225 + 50 \times \frac{7}{68} = 225 + 5.15 \approx 230.15 \) units.

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

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 smaller number = x, greater number = y. Given \( x^{2}=8y \) and \( y^{2}-x^{2}=180 \).
2. From the first equation, \( y=\frac{x^{2}}{8} \). Substitute into the second: \( \frac{x^{4}}{64}-x^{2}-180=0 \).
3. Let \( z=x^{2} \): \( z^{2}-64z-11520=0 \), which gives \( z=\frac{64\pm224}{2} \), so \( z=144 \) (rejecting negative value).
4. \( x^{2}=144 \Rightarrow x=12 \), and \( y=\frac{144}{8}=18 \). The numbers are 12 and 18.

Teacher's Note:
a) Substitute one variable in terms of the other to reduce to a single quadratic.
b) Reject the negative root for \( z \) since \( x^2 \) cannot be negative.

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 \( D=0 \): \( k^{2}-4(2)(3)=0 \Rightarrow k^{2}=24 \Rightarrow k=\pm 2\sqrt{6} \).
2. For equal roots, \( x=\frac{-k}{2a}=\frac{-k}{4} \).
3. For \( k=2\sqrt{6} \), \( x=-\frac{\sqrt{6}}{2} \) (repeated root); for \( k=-2\sqrt{6} \), \( x=\frac{\sqrt{6}}{2} \) (repeated root).

Teacher's Note:
a) Setting the discriminant equal to zero is the key condition for equal roots.
b) There are two possible values of k, each giving a different repeated root.

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 rose flower spiral pattern; and a diagram showing successive semicircles with alternating centres A and B, radii labelled \( l_1, l_2, l_3, l_4 \) 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 \( r_{13} = 50 \times 13 = 650 \) cm.

Teacher's Note:
a) Use \( a_n = a+(n-1)d \) with a = 50, d = 50.
b) Since a = d = 50 here, \( r_n = 50n \) directly.

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

Answer: \( 50n=500 \Rightarrow n=10 \).

Teacher's Note:
a) Substitute the given radius into the general term formula.
b) Solve the simple linear equation for n.

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

Answer:
1. Saplings per spiral form an AP: 10, 20, 30, ... with a = 10, d = 10.
2. \( S_{11} = \frac{11}{2}[2(10)+(11-1)(10)] = \frac{11}{2}(20+100) = \frac{11}{2}(120) = 660 \).

Teacher's Note:
a) Use the sum of n terms formula of an AP: \( S_n = \frac{n}{2}[2a+(n-1)d] \).
b) Substitute n = 11 carefully to avoid calculation errors.

OR

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

Answer:
1. \( S_n = \frac{n}{2}[2(10)+(n-1)(10)] = 5n(n+1) \).
2. Setting \( 5n(n+1)=450 \Rightarrow n(n+1)=90 \Rightarrow n^{2}+n-90=0 \).
3. Solving, \( n=9 \) (rejecting the negative root).

Teacher's Note:
a) Form a quadratic equation in n using the sum formula.
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 photo of a circular park with fountains; and a diagram of a circle with a diameter line from A to B passing through centre C, with points P and Q marked on the line between A and B such that AP = PQ = QB]

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

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

Teacher's Note:
a) Since AB is a diameter, its midpoint gives the centre of the circle.
b) Use the midpoint formula \( \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \).

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

Answer: \( AB = \sqrt{(50-10)^{2}+(50-20)^{2}} = \sqrt{1600+900} = 50 \); radius \( = \frac{50}{2} = 25 \) units.

Teacher's Note:
a) Use the distance formula to find the length of the diameter AB.
b) Radius is always half the diameter.

(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. Using the section formula: \( P = \left(\frac{1(50)+2(10)}{3}, \frac{1(50)+2(20)}{3}\right) = \left(\frac{70}{3}, 30\right) \).

Teacher's Note:
a) AP:PB = 1:2 because AP is one-third of AB while PB is two-thirds.
b) Apply the section formula carefully with m = 1, n = 2.

OR

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

Answer:
1. Since AP = PQ = QB = \( \frac{AB}{3} = \frac{50}{3} \), distance AQ = AP + PQ = \( 2 \times \frac{50}{3} = \frac{100}{3} \) units.

Teacher's Note:
a) AQ covers two of the three equal segments of AB.
b) Total length AB divided by 3 gives the length of each equal part.

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 photo of a fireman on a ladder against a rooftop with a bird; and a diagram showing a ladder leaning against a vertical building making a 60° angle with the ground]

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

Answer: Ladder length \( = \frac{15}{\sin 60^{\circ}} = \frac{15}{\frac{\sqrt{3}}{2}} = 10\sqrt{3} \approx 17.32 \) m.

Teacher's Note:
a) The ladder acts as the hypotenuse of a right triangle with the building height as the opposite side.
b) Use \( \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \).

(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: Distance \( = \frac{15}{\tan 60^{\circ}} = \frac{15}{\sqrt{3}} = 5\sqrt{3} \approx 8.66 \) m.

Teacher's Note:
a) Use \( \tan\theta = \frac{\text{opposite}}{\text{adjacent}} \) with the building height as opposite side.
b) Rationalise \( \frac{15}{\sqrt{3}} \) to get \( 5\sqrt{3} \).

(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: A right triangle with the building of height 15 m on one side, the ladder as hypotenuse making a 30° angle with the ground at the base of the opposite wall, and the width of the road as the horizontal base]

Answer: Width of road \( = \frac{15}{\tan 30^{\circ}} = 15\sqrt{3} \approx 25.98 \) m.

Teacher's Note:
a) The height of the building stays the same (15 m) as it is the same roof.
b) Use \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \) to find the horizontal width.

OR

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

Answer: Ladder length \( = \frac{15}{\sin 30^{\circ}} = \frac{15}{\frac{1}{2}} = 30 \) m.

Teacher's Note:
a) The building height remains the opposite side of the right triangle.
b) With a smaller angle, the same height needs a longer ladder.

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