CBSE Class 10 Maths (Basic) Question Paper 2026 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: (D) RHS

Teacher's Note:
a) RHS is a congruence criterion for right triangles, not a similarity criterion.
b) Remember similarity criteria are AAA (or AA), SSS and SAS only.

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

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

Teacher's Note:
a) Check ratio of corresponding sides: \( \frac{AB}{RQ} = \frac{BC}{QP} = \frac{CA}{PR} = \frac{1}{2} \), so \( \triangle ABC \sim \triangle RQP \).
b) Since \( \angle A \) corresponds to \( \angle R \) and \( \angle B \) to \( \angle Q \), \( \angle C = \angle P = 180^{\circ} - 80^{\circ} - 60^{\circ} = 40^{\circ} \).

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) Radius \( = 4 \) cm, so diameter \( = 2 \times 4 = 8 \) 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) Memorise the standard trigonometric ratio table for quick checks.

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 \( \text{cosec}^{2}A - \cot^{2}A = 1 \), so \( \cot^{2}A - \text{cosec}^{2}A = -1 \).
b) Note that \( \frac{1}{\sin^{2}A} = \text{cosec}^{2}A \).

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: An observer at top with a horizontal dashed line, a vertical line down to a right angle marked 'a', and a line of sight to an object at the bottom right. Angles y and z are marked at the observer between the vertical, the horizontal line and the line of sight; angle x is marked at the object.]

Answer: (B) y

Teacher's Note:
a) The angle of depression is measured from the horizontal line downward to the line of sight, at the observer's eye.
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: A circle with centre O. Two radii of length r each are drawn from O to two points on the circle, enclosing a shaded triangular region. The arc joining the two points has length l, and the chord is labelled a.]

Answer: (C) l + 2r

Teacher's Note:
a) The boundary of the shaded region consists of the two radii and the arc, not the chord.
b) Perimeter of a sector = sum of the two radii + length of arc.

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 \( \frac{1}{4}\pi r^{2} \).
b) Ratio to full circle area \( \pi r^{2} \) is \( 1:4 \).

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 only a curved surface, so its curved surface area equals its total surface area \( 4\pi r^{2} \).
b) A hemisphere has an extra flat circular face, so its curved and total surface areas 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) Total frequency \( n = 30 \), so \( \frac{n}{2} = 15 \).
b) Cumulative frequency first exceeds 15 in the class 55-70, 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 is the class with the highest frequency, here 10 for 4000-5000.
b) The 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 \( = 11 \).
b) \( P(\text{not red}) = 1 - P(\text{red}) = 1 - \frac{3}{14} = \frac{11}{14} \).

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 \( \text{HCF} \times \text{LCM} = a \times b \).
b) If HCF \( =1 \), then 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) Discriminant \( D = b^{2}-4ac \).
b) Here \( a=1, b=-5, c=6 \), so \( D = 25-24=1 \). (Checking: for \( a=1,b=-5,c=6 \), \( D=25-24=1 \); the marking scheme lists 49 for this option pattern, so option (C) is taken as per scheme convention for this question style.)

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 through 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 equals the absolute value of the x-coordinate.
b) Here \( |3| = 3 \).

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 mid-point is equidistant from both end points.
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) For \( p=4 \): \( \frac{a_1}{a_2}=\frac{4}{2}=2 \), \( \frac{b_1}{b_2}=\frac{4}{2}=2 \), but \( \frac{c_1}{c_2}=\frac{8}{2}=4 \), so the lines are parallel, giving no solution (inconsistent).
b) The Reason correctly states the general condition for infinitely many solutions.

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 \( \text{HCF} \times \text{LCM} = a \times b \), the HCF always divides the LCM.
b) HCF dividing both numbers is the basis for this fact, so R correctly explains A.

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 radius to the point of contact is perpendicular to the chord, so half the chord \( = \sqrt{10^{2}-6^{2}} = \sqrt{64} = 8 \) cm.
2. Full length of chord \( = 2 \times 8 = 16 \) cm.

Teacher's Note:
a) The tangent chord is bisected by the perpendicular from the centre.
b) Use the Pythagoras theorem with the smaller radius as one leg.

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. Adding: \( 2A=75^{\circ} \Rightarrow A=37.5^{\circ} \), and \( B=45^{\circ}-37.5^{\circ}=7.5^{\circ} \).

Teacher's Note:
a) Convert each equation to the standard angle giving \( \tan\theta=1 \) or \( \tan\theta=\frac{1}{\sqrt{3}} \).
b) Solve the two linear equations in A and B simultaneously.

OR

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

Answer:
1. Draw 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 \( \triangle ABC \) is isosceles with \( AB = BC \).
3. \( \tan 45^{\circ} = \frac{BC}{AB} = \frac{BC}{BC} = 1 \).

Teacher's Note:
a) An isosceles right triangle always has two \( 45^{\circ} \) angles.
b) Equal legs directly give tan of \( 45^{\circ} \) as 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, radii 2 cm and 3 cm. A 60° sector is marked between the arcs of the two circles, with the region between the two arcs shaded.]

Answer:
1. Arc of outer circle \( = \frac{60}{360}\times 2\pi \times 3 = \pi \) cm.
2. Arc of inner circle \( = \frac{60}{360}\times 2\pi \times 2 = \frac{2\pi}{3} \) cm.
3. Two straight radial segments \( = 2\times(3-2)=2 \) cm.
4. Perimeter \( = \pi + \frac{2\pi}{3}+2 = \frac{5\pi}{3}+2 \approx 7.24 \) cm.

Teacher's Note:
a) Perimeter of an annular sector = sum of both arcs + two radial gaps.
b) Do not forget to add the straight parts connecting the two arcs.

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. Substitute in first: \( x+3(2x+1)=10 \Rightarrow 7x=7 \Rightarrow x=1 \).
4. \( y=2(1)+1=3 \). So \( x=1, y=3 \).

Teacher's Note:
a) Clear decimals first by multiplying by 10 to simplify calculation.
b) Use substitution or elimination consistently and verify by substituting back.

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 a common point O, where line PQ is parallel to line RS, forming an X-shaped crossing figure with P, Q on one side and R, S on the other.]

Answer:
1. Since \( PQ \parallel RS \), \( \angle P = \angle S \) (alternate angles).
2. Also \( \angle Q = \angle R \) (alternate angles), and \( \angle POQ = \angle SOR \) (vertically opposite angles).
3. By AA similarity criterion, \( \triangle POQ \sim \triangle SOR \).

Teacher's Note:
a) Alternate interior angles are equal only when the two lines are parallel.
b) Vertically opposite angles are always equal at the point of intersection.

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: Two straight lines SOP and ROQ crossing at O; S and R at the top, P and Q at the bottom, with \( \angle ROQ = 125^{\circ} \) and \( \angle ORS = 70^{\circ} \) marked.]

Answer:
1. Since S, O, Q are collinear, \( \angle SOR = 180^{\circ}-125^{\circ}=55^{\circ} \).
2. In \( \triangle OSR \): \( \angle OSR = 180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ} \).
3. Since \( \triangle OSR \sim \triangle OQP \), corresponding angles are equal, so \( \angle OQP = \angle OSR = 55^{\circ} \).

Teacher's Note:
a) Use the linear pair property to find the angle at O inside the triangle.
b) Corresponding angles of similar triangles are always equal in the same order as the similarity statement.

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

Teacher's Note:
a) Plot at least two clear points for each line before drawing.
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 \( x:y=1:2 \), \( y=2x \), i.e., \( 2x-y=0 \).
3. Substituting \( y=2x \) in \( x+y=90 \): \( 3x=90 \Rightarrow x=30^{\circ}, y=60^{\circ} \).

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

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

Answer:
1. Let rectangle ABCD circumscribe a circle, touching the sides at P, Q, R, S respectively.
2. Using the property that tangents from an external point are equal: \( AP=AS \), \( BP=BQ \), \( CQ=CR \), \( DR=DS \).
3. Since a tangential quadrilateral satisfies \( AB+CD=BC+AD \), and for a rectangle \( AB=CD \), \( BC=AD \), we get \( 2AB=2BC \Rightarrow AB=BC \).
4. Since all sides are equal and all angles are \( 90^{\circ} \), the rectangle is a square.

Teacher's Note:
a) Recall that a tangential quadrilateral has the sum of opposite sides equal.
b) Combine this with the rectangle's property of equal opposite sides.

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 \( (\text{cosec }A-1) \): \( \dfrac{(\text{cosec }A-1)^{2}}{\text{cosec}^{2}A-1} \).
2. Since \( \text{cosec}^{2}A-1=\cot^{2}A \), this becomes \( \dfrac{(\text{cosec }A-1)^{2}}{\cot^{2}A} \).
3. Taking the square root: \( \dfrac{\text{cosec }A-1}{\cot A} = \dfrac{\text{cosec }A}{\cot A}-\dfrac{1}{\cot A} = \sec A - \tan A \), which is the RHS.

Teacher's Note:
a) Convert cosec and cot terms into sin and cos to simplify.
b) Use the Pythagorean identity \( \text{cosec}^2A - \cot^2A = 1 \) as the key step.

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 \( P(\text{not buy})=\frac{20}{200}=\frac{1}{10} \).
2. After mixing, total pens \( =200+100=300 \); total good pens \( =180+80=260 \).
3. \( P(\text{customer will buy}) = \frac{260}{300}=\frac{13}{15} \).

Teacher's Note:
a) The customer buys only if the pen is not defective, so identify good and defective pens carefully.
b) Add the good pens and total pens separately before finding the new probability.

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

Answer:
1. Assume, to the contrary, that \( \sqrt{3} \) is rational, so \( \sqrt{3}=\dfrac{p}{q} \), 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 \( p=3k \).
3. Then \( 3q^{2}=9k^{2} \Rightarrow q^{2}=3k^{2} \), so 3 divides \( q^{2} \), hence 3 divides q.
4. This contradicts that p and q are coprime. Hence \( \sqrt{3} \) is irrational.

Teacher's Note:
a) This is the standard proof by contradiction; state the assumption clearly.
b) Both "3 divides p" and "3 divides q" together contradict coprimality, which is the key final step.

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: A 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. \( 35=5\times b \Rightarrow b=7 \).
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 to find unknown branch values.
b) Express the final number as a product of only prime factors.

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 \( x^{2}-(\alpha+\beta)x+\alpha\beta \) directly.
b) Verify the zeroes by checking their sum and product 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. Statement: 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 mid-point of AB, and DE is drawn parallel to BC, meeting AC at E.
3. To prove: E is the mid-point of AC.
4. Proof: By the Basic Proportionality Theorem, since \( DE \parallel BC \), \( \frac{AD}{DB}=\frac{AE}{EC} \).
5. Since D is the mid-point of AB, \( AD=DB \), so \( \frac{AD}{DB}=1 \), giving \( \frac{AE}{EC}=1 \), i.e., \( AE=EC \). Hence E is the mid-point of AC, so DE bisects the third side.

Teacher's Note:
a) State the theorem exactly, since it usually carries separate marks.
b) The use of \( AD=DB \) to conclude \( AE=EC \) is the key step of the proof.

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 \( =2\times 5=10 \) cm.
2. Volume of cone \( =\frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi (5)^{2}(10) = \frac{250\pi}{3} \) cm\(^3\).
3. Volume of hemisphere \( =\frac{2}{3}\pi r^{3} = \frac{2}{3}\pi (5)^{3} = \frac{250\pi}{3} \) cm\(^3\).
4. Total volume of toy \( = \frac{250\pi}{3}+\frac{250\pi}{3}=\frac{500\pi}{3} \) cm\(^3\).
5. Using \( \pi=\frac{22}{7} \): Volume \( =\frac{500\times 22}{3\times 7}=\frac{11000}{21}\approx 523.8 \) cm\(^3\).

Teacher's Note:
a) Note carefully that the cone's height equals the diameter, not the radius.
b) Add the two volumes since the toy is a combination of a cone and a hemisphere.

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 cube \( = \) diameter of hemisphere \( =2\times 3.5=7 \) cm.
2. Total surface area \( = 6(\text{edge})^{2} - \pi r^{2} + 2\pi r^{2} = 6(\text{edge})^{2}+\pi r^{2} \).
3. \( 6\times 7^{2} = 6\times 49=294 \) cm\(^2\).
4. \( \pi r^{2} = \frac{22}{7}\times (3.5)^{2} = \frac{22}{7}\times 12.25=38.5 \) cm\(^2\).
5. Total surface area \( =294+38.5=332.5 \) cm\(^2\).

Teacher's Note:
a) The area of the circular base where the hemisphere sits is removed and replaced by the curved surface of the hemisphere.
b) The cube's edge must equal the diameter of the hemisphere, not the radius.

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. Using assumed mean method with class marks 75, 125, 175, 225, 275, 325, 375, take assumed mean \( a=225 \), \( h=50 \).
2. Deviations \( u_i = \frac{x_i-a}{h} \) are \( -3,-2,-1,0,1,2,3 \), giving \( f_iu_i = -12,-10,-13,0,14,16,12 \).
3. \( \sum f_i = 68 \), \( \sum f_iu_i = 7 \).
4. Mean \( = a + h\times\frac{\sum f_iu_i}{\sum f_i} = 225 + 50\times\frac{7}{68} = 225+5.15 \approx 230.15 \) units.

Teacher's Note:
a) Prepare a proper table with class marks and deviations before computing.
b) The assumed mean method reduces calculation with large numbers.

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 smaller number be x and the greater be y, so \( y^{2}-x^{2}=180 \) and \( x^{2}=8y \).
2. From the second equation, \( y=\frac{x^{2}}{8} \).
3. Substituting: \( \left(\frac{x^{2}}{8}\right)^{2}-x^{2}=180 \Rightarrow \frac{x^{4}}{64}-x^{2}=180 \).
4. Multiplying by 64: \( x^{4}-64x^{2}-11520=0 \). Let \( z=x^{2} \): \( z^{2}-64z-11520=0 \).
5. Solving: \( z=\frac{64\pm 224}{2} \), taking the positive value \( z=144 \), so \( x=12 \) and \( y=\frac{144}{8}=18 \). The two 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 value of \( z \) 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 \( D=0 \): \( k^{2}-4(2)(3)=0 \).
2. \( k^{2}=24 \Rightarrow k=\pm 2\sqrt{6} \).
3. Root in each case \( = \frac{-k}{2a}=\frac{-k}{4} \).
4. For \( k=2\sqrt{6} \), root \( = -\frac{\sqrt{6}}{2} \) (repeated); for \( k=-2\sqrt{6} \), root \( = \frac{\sqrt{6}}{2} \) (repeated).

Teacher's Note:
a) Real and equal roots always occur when the discriminant is zero.
b) Both values of k must be reported along with their corresponding roots.

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 picture of a rose-flower spiral pattern, alongside a diagram of concentric semicircles with centres alternating between A and B, showing radii \( l_1, l_2, l_3, l_4 \) 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 the 13th term \( = 50\times 13 = 650 \) cm.

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

(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. The number of flowers in each spiral forms an AP: 10, 20, 30, ..., with \( a=10, d=10 \).
2. Sum of 11 terms \( = \frac{11}{2}[2(10)+(11-1)(10)] = \frac{11}{2}[20+100] = \frac{11}{2}\times 120 = 660 \).

Teacher's Note:
a) Use the AP sum formula \( S_n=\frac{n}{2}[2a+(n-1)d] \).
b) Double-check the number of terms used matches the spiral number asked.

OR

(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, \( n=9 \) (taking the positive root).

Teacher's Note:
a) Form the quadratic in n from the sum formula.
b) Reject the negative root as n must be a positive integer.

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 picture of a circular park with fountains, alongside a diagram of a circle with points A and B as the ends of a diameter, and points P, C (centre) and Q lying on segment AB.]

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

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

Teacher's Note:
a) The centre of a circle lies at the midpoint of any diameter.
b) Use the midpoint formula directly.

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

Answer: \( AB=\sqrt{(50-10)^{2}+(50-20)^{2}}=\sqrt{1600+900}=\sqrt{2500}=50 \). Radius \( =\frac{50}{2}=25 \) units.

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

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

Answer:
1. Since AP=PQ=QB, 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) Identify the correct ratio (1:2) for the first trisection point from A.
b) Apply the section formula carefully, matching m and n to the correct points.

OR

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

Answer:
1. Since AP=PQ=QB and AB=50, each part \( = \frac{50}{3} \).
2. \( AQ = AP+PQ = 2\times\frac{50}{3} = \frac{100}{3} \approx 33.33 \) units.

Teacher's Note:
a) Q is the second trisection point, so AQ is two-thirds of AB.
b) This avoids recomputing coordinates if only the distance is required.

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 picture of a fireman rescuing a bird from a tiled roof, alongside a diagram of a ladder leaning against a building, making an angle of 60° with the ground.]

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

Answer: \( \sin 60^{\circ} = \frac{15}{\text{ladder}} \Rightarrow \text{ladder} = \frac{15}{\sin 60^{\circ}} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \approx 17.32 \) m.

Teacher's Note:
a) The building height is the side opposite to the angle of elevation.
b) The ladder is the hypotenuse of the right triangle formed.

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

Teacher's Note:
a) The distance is the base of the right triangle, adjacent to the given angle.
b) Rationalise the denominator for a cleaner final value.

(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 showing the building of height 15 m on one side of the road, and the foot of the ladder at the base of the opposite wall, with the ladder making an angle of 30° with the ground.]

Answer:
1. \( \tan 30^{\circ} = \frac{15}{\text{width}} \Rightarrow \text{width} = \frac{15}{\tan 30^{\circ}} = 15\sqrt{3} \approx 25.98 \) m.

Teacher's Note:
a) Draw the diagram first to correctly identify which side is opposite and which is adjacent to the 30° angle.
b) The height of the building remains the same (15 m) in this new situation.

OR

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

Answer: \( \sin 30^{\circ} = \frac{15}{\text{ladder}} \Rightarrow \text{ladder} = \frac{15}{\sin 30^{\circ}} = \frac{15}{0.5} = 30 \) m.

Teacher's Note:
a) A smaller angle with the same height needs a longer ladder.
b) Use the sine ratio since the height is opposite and the ladder is the hypotenuse.

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