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

1. 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:

1) For any two positive integers, the product of their HCF and LCM equals the product of the numbers.
2) When the HCF is 1, the numbers are co-prime.
3) Therefore, their LCM is simply their product, \( ab \).

2. The number \( 3 + \sqrt{2} \) 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:

1) The sum of a rational number and an irrational number is always irrational.
2) Here, 3 is a rational number and \( \sqrt{2} \) is an irrational number.
3) Thus, their sum is strictly irrational.

3. The discriminant of the quadratic equation \( x^2 - 3x - 2 = 0 \) is : [1 Mark]
(a) \( 1 \)
(b) \( 17 \)
(c) \( \sqrt{17} \)
(d) \( -\sqrt{17} \)
Answer: (b) 17

Teacher's Note:

1) The discriminant formula for a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( D = b^2 - 4ac \).
2) Substitute \( a = 1 \), \( b = -3 \), and \( c = -2 \) into the formula.
3) We get \( D = (-3)^2 - 4(1)(-2) = 9 + 8 = 17 \).

4. 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:

1) Multiply the entire equation by \( x \) to get the standard quadratic form: \( x^2 - 3x + 1 = 0 \).
2) Identify the coefficients: \( a = 1 \), \( b = -3 \), and \( c = 1 \).
3) Evaluate the expression: \( a - b + c = 1 - (-3) + 1 = 1 + 3 + 1 = 5 \).

5. For a point \( (3, -5) \), the value of \( (\text{abscissa} - \text{ordinate}) \) is : [1 Mark]
(a) \( -8 \)
(b) \( -2 \)
(c) \( 2 \)
(d) \( 8 \)
Answer: (d) 8

Teacher's Note:

1) The abscissa of a point is its \( x \)-coordinate, which is \( 3 \).
2) The ordinate of a point is its \( y \)-coordinate, which is \( -5 \).
3) Subtract the ordinate from the abscissa: \( 3 - (-5) = 3 + 5 = 8 \).

6. 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:

1) By definition, a mid-point is equidistant from both endpoints of a line segment.
2) This means it splits the segment into two equal halves.
3) Thus, the division ratio is always \( 1 : 1 \).

7. 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. (Note: One mark given to all attempting students)

Teacher's Note:

1) AAA, SSS, and SAS are standard similarity criteria for triangles.
2) RHS is a congruence criterion, but right-angled triangles can also be similar by AA (a special case of AAA). Due to ambiguity in standard options, grace marks are typically awarded.

8. 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
Answer: (c) \( \angle P = 40^{\circ} \)

Teacher's Note:

1) Find the missing angle in the first triangle using the angle sum property: \( \angle C = 180^{\circ} - (80^{\circ} + 60^{\circ}) = 40^{\circ} \).
2) Check the ratio of corresponding sides of both triangles to establish similarity.
3) Corresponding angles in similar triangles are equal, leading to \( \angle P = 40^{\circ} \).

9. In the given figure, PA is a tangent to a circle with centre O. If \( \text{OP} = 10\text{ cm} \), then the length of AP is : [1 Mark]
(a) \( 10\sqrt{3}\text{ cm} \)
(b) \( 20\text{ cm} \)
(c) \( 5\text{ cm} \)
(d) \( 5\sqrt{3}\text{ cm} \)
Answer: (d) \( 5\sqrt{3}\text{ cm} \)

Teacher's Note:

1) The radius through the point of contact is perpendicular to the tangent, forming a right-angled triangle OAP at A.
2) Using trigonometric ratios in right triangle OAP, \( \cos(30^{\circ}) = \frac{\text{AP}}{\text{OP}} \).
3) Substitute values: \( \frac{\sqrt{3}}{2} = \frac{\text{AP}}{10} \), which gives \( \text{AP} = 5\sqrt{3}\text{ cm} \).

10. 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:

1) We know that \( \sin 30^{\circ} = \frac{1}{2} \) and \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \). Since they are unequal, statement (c) is false.
2) All other statements evaluate to true trigonometric identities.

11. The value of \( \left(\tan^2 A - \frac{1}{\cos^2 A}\right) \) is : [1 Mark]
(a) more than 1
(b) 1
(c) 0
(d) -1
Answer: (d) -1

Teacher's Note:

1) Recall the reciprocal identity: \( \frac{1}{\cos^2 A} = \sec^2 A \).
2) The expression transforms into \( \tan^2 A - \sec^2 A \).
3) Using the fundamental trigonometric identity \( 1 + \tan^2 A = \sec^2 A \), we get \( \tan^2 A - \sec^2 A = -1 \).

12. In the given figure, which of the following angles represents the angle of depression ? [1 Mark]
(a) x
(b) y
(c) z
(d) a
Answer: (c) z

Teacher's Note:

1) The angle of depression is formed by the line of sight and the horizontal line when looking downward from an observer.
2) In the given diagram, angle z lies between the horizontal line and the downward line of sight.

13. The perimeter of the shaded region in the given figure is : [1 Mark]
(a) \( l \)
(b) \( l + a \)
(c) \( l + 2r \)
(d) \( l + 2r + a \)
Answer: (c) \( l + 2r \)

Teacher's Note:

1) The perimeter of a region is the total length of its outer boundary.
2) The boundary consists of the curved arc of length \( l \) and two radii \( r \) meeting at the center.
3) Thus, total perimeter = \( l + r + r = l + 2r \).

14. 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:

1) A quadrant represents one-fourth of a full circle.
2) Therefore, the area of a quadrant is \( \frac{1}{4}\pi r^2 \), while the circle's area is \( \pi r^2 \).
3) The ratio simplifies to \( 1 : 4 \).

15. 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:

1) A sphere has only one continuous curved surface without any flat bases or tops.
2) Consequently, its curved surface area and total surface area are identical (\( 4\pi r^2 \)).

16. 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:

1) Total frequency \( N = 30 \), so \( \frac{N}{2} = 15 \).
2) Cumulative frequencies are 2, 5, 12, 18, 24, 30. The median class is 55-70.
3) Class mark = \( \frac{55 + 70}{2} = 62.5 \).

17. The following distribution shows the number of runs scored by some batsmen in test matches : [1 Mark]
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 :
(a) \( 3000 \)
(b) \( 4000 \)
(c) \( 5000 \)
(d) \( 6000 \)
Answer: (b) 4000

Teacher's Note:

1) The modal class is the class interval with the highest frequency.
2) Here, the maximum frequency is 10, corresponding to the interval 4000-5000.
3) The lower limit of this modal class is 4000.

18. In a random experiment of throwing a die, which of the following is a sure event ? [1 Mark]
(a) Getting a number between 1 and 6
(b) Getting an odd number \( < 7 \)
(c) Getting an even number \( < 7 \)
(d) Getting a natural number \( < 7 \)
Answer: (d) Getting a natural number < 7

Teacher's Note:

1) A standard die has faces numbered 1, 2, 3, 4, 5, and 6.
2) All these outcomes are natural numbers strictly less than 7.
3) Thus, this event is certain to occur on every roll.

19. 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. [1 Mark]
Reason (R) : HCF of any two natural numbers divides both the numbers.
Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Teacher's Note:

1) HCF is always a factor of LCM for any pair of natural numbers.
2) The given reason correctly states the divisibility property of the HCF.

20. 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. [1 Mark]
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 \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Answer: (d) Assertion (A) is false, but Reason (R) is true.

Teacher's Note:

1) Consistency covers both unique and infinitely many solutions.
2) Testing the ratios reveals that the system is consistent for values other than 4 as well, making the assertion incorrect.

21. Solve the following system of equations for x and y : [2 Marks]
\( \frac{x}{2} + \frac{2y}{3} = -1 \) and \( x - \frac{y}{3} = 3 \)

Multiply the first equation by 6 to clear denominators: \( 3x + 4y = -6 \)...(1)

Multiply the second equation by 3 to clear denominators: \( 3x - y = 9 \)...(2)

Subtract equation (2) from equation (1): \( 5y = -15 \implies y = -3 \).

Substitute \( y = -3 \) into equation (2): \( 3x - (-3) = 9 \implies 3x + 3 = 9 \implies 3x = 6 \implies x = 2 \).

Thus, the solution is \( x = 2 \) and \( y = -3 \).

Teacher's Note:

1) Clear fractions first to simplify algebraic manipulation.
2) Use elimination method by matching coefficients of x.
3) Verify answers by plugging them back into both original equations.

22. (a) In the given figure, if \( PQ \parallel RS \), then prove that \( \Delta POQ \sim \Delta SOR \). [2 Marks]
OR
(b) In the given figure, \( \Delta OSR \sim \Delta OQP \), \( \angle ROQ = 125^{\circ} \) and \( \angle ORS = 70^{\circ} \). Find the measures of \( \angle OSR \) and \( \angle OQP \).

(a) Given \( PQ \parallel RS \), alternate interior angles are equal, so \( \angle P = \angle S \) and \( \angle Q = \angle R \). By AA similarity criterion, \( \Delta POQ \sim \Delta SOR \).

OR

(b) Using the exterior angle property in triangle ORQ, \( \angle OSR = 125^{\circ} - 70^{\circ} = 55^{\circ} \). Since \( \Delta OSR \sim \Delta OQP \), corresponding angles are equal, so \( \angle OQP = \angle OSR = 55^{\circ} \).

Teacher's Note:

1) Identify parallel lines to establish alternate interior angles.
2) Apply triangle properties such as the exterior angle theorem accurately.

23. 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]

Let the radii be \( r = 6\text{ cm} \) and \( R = 10\text{ cm} \). The radius is perpendicular to the tangent at the point of contact. Using the Pythagorean theorem in the right-angled triangle formed, half the chord length \( BC \) is given by \( BC = \sqrt{R^2 - r^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\text{ cm} \).

Thus, the total length of the chord is \( 2 \times 8 = 16\text{ cm} \).

Teacher's Note:

1) Draw a clear diagram showing the perpendicular from the center to the chord.
2) Use the property that the perpendicular from the center bisects the chord.

24. (a) Find the values of A and B (\( 0^{\circ} \le A < 90^{\circ} \), \( 0^{\circ} \le B < 90^{\circ} \)), if \( \tan(A + B) = 1 \) and \( \tan(A - B) = \frac{1}{\sqrt{3}} \). [2 Marks]
OR
(b) Prove that \( \tan 45^{\circ} = 1 \) geometrically.

(a) Given \( \tan(A + B) = 1 \implies A + B = 45^{\circ} \). Also, \( \tan(A - B) = \frac{1}{\sqrt{3}} \implies A - B = 30^{\circ} \). Adding both equations gives \( 2A = 75^{\circ} \implies A = 37.5^{\circ} \), and subtracting gives \( 2B = 15^{\circ} \implies B = 7.5^{\circ} \).

OR

(b) Consider an isosceles right-angled triangle ABC with equal perpendicular sides of length \( x \). The acute angles are \( 45^{\circ} \) each. By definition, \( \tan 45^{\circ} = \frac{\text{Perpendicular}}{\text{Base}} = \frac{x}{x} = 1 \).

Teacher's Note:

1) Relate trigonometric values to standard angle table entries.
2) Solve linear simultaneous equations carefully.

25. A chord of a circle of diameter 20 cm subtends an angle of \( 60^{\circ} \) at the centre of the circle. Find the area of the corresponding minor segment of the circle. (Use \( \pi = 3.14 \) and \( \sqrt{3} = 1.73 \)) [2 Marks]

The radius of the circle is \( r = \frac{20}{2} = 10\text{ cm} \). Since the central angle is \( 60^{\circ} \), the sector is an equilateral triangle combined with a segment. Area of the minor segment = Area of sector - Area of triangle.

\( \text{Area} = \left(\frac{60^{\circ}}{360^{\circ}} \times 3.14 \times 10^2\right) - \left(\frac{\sqrt{3}}{4} \times 10^2\right) = \left(\frac{1}{6} \times 3.14 \times 100\right) - \left(\frac{1.73}{4} \times 100\right) \)

\( \text{Area} = 52.33 - 43.25 = 9.08\text{ sq. cm} \) (or \( \frac{109}{12}\text{ sq. cm} \)).

Teacher's Note:

1) Convert diameter to radius before starting calculations.
2) Subtract the triangle's area from the sector's area to isolate the segment.

26. (a) Prove that \( \sqrt{3} \) is an irrational number. [3 Marks]
OR
(b) The factor tree of a number x is shown below : Find the values of x, y, a and b. Hence, write the product of the prime factors of the number x so obtained.

(a) Assume \( \sqrt{3} \) is rational, such that \( \sqrt{3} = \frac{p}{q} \) where p and q are co-prime integers and \( q \neq 0 \). Squaring both sides gives \( 3q^2 = p^2 \), implying 3 divides \( p^2 \) and hence divides p. Let \( p = 3m \), substituting yields \( 3q^2 = 9m^2 \implies q^2 = 3m^2 \), meaning 3 divides q as well. This contradicts our assumption that p and q are co-prime. Therefore, \( \sqrt{3} \) is irrational.

OR

(b) Working upwards from the bottom of the factor tree: \( b = 7 \), \( a = 3 \), \( y = 210 \times 2 = 420 \), and \( x = 420 \times 2 = 840 \). The prime factorization of \( 840 \) is \( 2^3 \times 3 \times 5 \times 7 \).

Teacher's Note:

1) Use proof by contradiction for irrationality proofs.
2) Work step-by-step from bottom to top when solving factor trees.

27. Find a quadratic polynomial whose sum and product of zeroes are 0 and -9, respectively. Also, find the zeroes of the polynomial so obtained. [3 Marks]

A quadratic polynomial is given by \( x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \).

Substituting the given values: \( P(x) = x^2 - 0x + (-9) = x^2 - 9 \).

To find the zeroes, set \( P(x) = 0 \implies x^2 - 9 = 0 \implies (x - 3)(x + 3) = 0 \).

Thus, the zeroes are \( x = 3 \) and \( x = -3 \).

Teacher's Note:

1) Remember the standard template for constructing a polynomial from its zeroes.
2) Factorize using the difference of squares identity.

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

(a) Plot the lines for both equations. For \( x + 3y = 6 \), intercepts are at \( (6, 0) \) and \( (0, 2) \). For \( 2x - 3y = 12 \), intercepts are at \( (6, 0) \) and \( (0, -4) \). The lines intersect at the point \( (6, 0) \), giving the solution \( x = 6 \) and \( y = 0 \).

OR

(b) Since x and y are complementary, \( x + y = 90^{\circ} \). Given the ratio \( x : y = 1 : 2 \), we get \( 2x = y \) or \( 2x - y = 0 \). Substituting \( y = 2x \) into the first equation: \( x + 2x = 90^{\circ} \implies 3x = 90^{\circ} \implies x = 30^{\circ} \). Thus, \( y = 60^{\circ} \).

Teacher's Note:

1) Label graph axes clearly and mark intersection points accurately.
2) Translate word problems into linear equations correctly.

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

Let ABCD be a rectangle circumscribing a circle. The lengths of tangents drawn from an external point to a circle are equal, so \( AP = AS \), \( BP = BQ \), \( CR = CQ \), and \( DR = DS \). Adding these equations gives \( (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \), which simplifies to \( AB + CD = AD + BC \text{ or } 2AB = 2BC \implies AB = BC \). Since adjacent sides of the rectangle are equal, ABCD is a square.

Teacher's Note:

1) Use the tangent property from an external point.
2) Combine segment lengths carefully to prove equal adjacent sides.

30. Prove that : [3 Marks]
\( \frac{1 + \cot^2 A}{1 + \tan^2 A} = \left(\frac{1 - \cot A}{1 - \tan A}\right)^2 \)

Starting with the Left-Hand Side (LHS):

\( \text{LHS} = \frac{1 + \frac{\cos^2 A}{\sin^2 A}}{1 + \frac{\sin^2 A}{\cos^2 A}} = \frac{\frac{\sin^2 A + \cos^2 A}{\sin^2 A}}{\frac{\cos^2 A + \sin^2 A}{\cos^2 A}} = \frac{\frac{1}{\sin^2 A}}{\frac{1}{\cos^2 A}} = \frac{\cos^2 A}{\sin^2 A} = \cot^2 A \).

Now simplify the Right-Hand Side (RHS):

\( \text{RHS} = \left(\frac{1 - \frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}\right)^2 = \left(\frac{\frac{\sin A - \cos A}{\sin A}}{\frac{\cos A - \sin A}{\cos A}}\right)^2 = \left(\frac{-\cos A}{\sin A}\right)^2 = \cot^2 A \).

Since \( \text{LHS} = \text{RHS} \), the identity is proven.

Teacher's Note:

1) Convert all trigonometric ratios to sine and cosine for easier simplification.
2) Handle negative signs carefully during fraction inversion.

31. 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]

Total pens in first lot = 200, Defective pens = \( 200 - 180 = 20 \).

Probability that the customer will not buy the pen (i.e., it is defective) = \( \frac{20}{200} = \frac{1}{10} \).

After mixing the new lot, total pens = \( 200 + 100 = 300 \), and total good pens = \( 180 + 80 = 260 \).

Probability that the customer will buy the pen (i.e., it is good) = \( \frac{260}{300} = \frac{13}{15} \).

Teacher's Note:

1) Carefully distinguish between good and defective items at each stage.
2) Update total sample space and favorable outcomes after mixing new lots.

32. (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]
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.

(a) Let the greater number be x and the smaller number be y. According to the problem, \( x^2 - y^2 = 180 \) and \( y^2 = 8x \). Substituting the second equation into the first gives \( x^2 - 8x - 180 = 0 \). Factorizing yields \( (x - 18)(x + 10) = 0 \), giving \( x = 18 \) (since numbers are positive, \( x = -10 \) is rejected). Thus, \( y^2 = 8(18) = 144 \implies y = 12 \). The numbers are 18 and 12.

OR

(b) For real and equal roots, the discriminant must be zero: \( b^2 - 4ac = 0 \implies k^2 - 4(2)(3) = 0 \implies k^2 - 24 = 0 \implies k = \pm 2\sqrt{6} \). Substituting k back gives equations \( 2x^2 \pm 2\sqrt{6}x + 3 = 0 \), which simplify to \( (\sqrt{2}x \pm \sqrt{3})^2 = 0 \), yielding repeated roots at \( x = \pm \frac{\sqrt{3}}{\sqrt{2}} \).

Teacher's Note:

1) Set up quadratic equations carefully using substitution.
2) Check the condition for equal roots using the discriminant \( D = 0 \).

33. State "Basic Proportionality Theorem" and use it to prove the following : In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that \( \frac{AO}{BO} = \frac{CO}{DO} \) as shown in the given figure. Prove that ABCD is a trapezium. [5 Marks]

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.

Proof: Given \( \frac{AO}{BO} = \frac{CO}{DO} \), we can rewrite this as \( \frac{AO}{CO} = \frac{BO}{DO} \). Construct a line \( OE \parallel AB \) meeting AD at E. By Basic Proportionality Theorem in triangle DAB, \( \frac{DE}{AE} = \frac{DO}{BO} \). Combining ratios shows that \( \frac{DE}{AE} = \frac{CO}{AO} \). By the converse of BPT in triangle ADC, \( OE \parallel DC \). Since \( OE \parallel AB \) and \( OE \parallel DC \), we get \( AB \parallel DC \). Hence, ABCD is a trapezium.

Teacher's Note:

1) State the theorem precisely before applying it.
2) Use a construction line parallel to one of the bases to connect triangle properties.

34. (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]
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.

(a) Given radius \( r = 5\text{ cm} \). Height of the cone \( h = 2r = 10\text{ cm} \). Volume of the toy = Volume of hemisphere + Volume of cone = \( \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h \).

\( \text{Volume} = \frac{1}{3}\pi r^2 (2r + h) = \frac{1}{3} \times \frac{22}{7} \times 5^2 \times (10 + 10) = \frac{1}{3} \times \frac{22}{7} \times 25 \times 20 = \frac{11000}{21}\text{ cu. cm} \) (or \( 523.81\text{ cu. cm} \)).

OR

(b) The smallest edge of the cube must equal the diameter of the hemisphere, so edge \( a = 2 \times 3.5 = 7\text{ cm} \). Total surface area of the solid = Surface area of cube - Base area of hemisphere + Curved surface area of hemisphere = \( 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2 \).

\( \text{Total Area} = 6(7)^2 + \frac{22}{7}(3.5)^2 = 294 + 38.5 = 332.5\text{ sq. cm} \).

Teacher's Note:

1) Combine formula expressions before substituting numerical values to avoid calculation errors.
2) Account for overlapping surface areas in composite solid problems.

35. The following data gives the information on the observed lifetime (in hours) of 200 electrical components : [5 Marks]
Lifetime (in hours): 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120
Number of components: 10 | 35 | 50 | 60 | 30 | 15
Find the mean lifetime (in hours) of the electrical components.

Using the assumed mean method: Let assumed mean \( a = 50 \) and class size \( h = 20 \).

Class marks (\( x_i \)): 10, 30, 50, 70, 90, 110.

Frequencies (\( f_i \)): 10, 35, 50, 60, 30, 15 (Total \( \Sigma f_i = 200 \)).

Step-deviation values (\( u_i = \frac{x_i - a}{h} \)): -2, -1, 0, 1, 2, 3.

Products (\( f_i u_i \)): -20, -35, 0, 60, 60, 45 (Total \( \Sigma f_i u_i = 110 \)).

\( \text{Mean} = a + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h = 50 + \left(\frac{110}{200}\right) \times 20 = 50 + 11 = 61\text{ hours} \).

Teacher's Note:

1) Construct a frequency distribution table with clear columns for \( x_i \) and \( u_i \).
2) Apply the step-deviation formula correctly to simplify arithmetic.

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 \( 60^{\circ} \) with the ground in order to reach the roof.

36. Based on the above information, answer the following questions :
(i) Find the length of the ladder used by the fireman to reach the roof. [1 Mark]
(ii) Find the distance of the point on the ground at which the ladder was fixed from the bottom of the building. [1 Mark]
(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 \( 30^{\circ} \) 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]
OR
(b) Find the length of the ladder used by the fireman in this case. [2 Marks]

(i) Let ladder length be a. Using \( \sin 60^{\circ} = \frac{15}{a} \implies \frac{\sqrt{3}}{2} = \frac{15}{a} \implies a = \frac{30}{\sqrt{3}} = 10\sqrt{3}\text{ m} \).

(ii) Let distance be x. Using \( \tan 60^{\circ} = \frac{15}{x} \implies \sqrt{3} = \frac{15}{x} \implies x = \frac{15}{\sqrt{3}} = 5\sqrt{3}\text{ m} \).

(iii)(a) Let road width be y. Using \( \tan 30^{\circ} = \frac{15}{y} \implies \frac{1}{\sqrt{3}} = \frac{15}{y} \implies y = 15\sqrt{3}\text{ m} \).

OR

(iii)(b) Let ladder length be l. Using \( \sin 30^{\circ} = \frac{15}{l} \implies \frac{1}{2} = \frac{15}{l} \implies l = 30\text{ m} \).

Teacher's Note:

1) Choose appropriate trigonometric ratios based on the sides given and required.
2) Rationalize denominators where necessary in radical answers.

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.

37. Based on the above information, answer the following questions :
(i) What is the radius of the \( 13^{\text{th}} \) spiral ? [1 Mark]
(ii) If the radius of the \( n^{\text{th}} \) spiral is 500 cm, find the value of n. [1 Mark]
(iii) (a) Find the total number of saplings till the \( 11^{\text{th}} \) spiral. [2 Marks]
OR
(b) Till which spiral, will there be a total of 450 saplings ? [2 Marks]

(i) The radii form an AP: 50, 100, 150, ... with \( a = 50 \) and \( d = 50 \). The radius of the \( 13^{\text{th}} \) spiral is \( a_{13} = 50 + (13 - 1)50 = 50 + 600 = 650\text{ cm} \).

(ii) Given \( a_n = 500 \implies 50 + (n - 1)50 = 500 \implies 50n = 500 \implies n = 10 \).

(iii)(a) The number of flowers forms an AP: 10, 20, 30, ... with \( a = 10 \) and \( d = 10 \). Total saplings till 11 spirals: \( S_{11} = \frac{11}{2}[2(10) + (11 - 1)10] = \frac{11}{2}[20 + 100] = \frac{11}{2} \times 120 = 660 \).

OR

(iii)(b) Given \( S_n = 450 \implies \frac{n}{2}[2(10) + (n - 1)10] = 450 \implies n(n + 1) = 90 \implies n^2 + n - 90 = 0 \implies (n - 9)(n + 10) = 0 \implies n = 9 \) (rejecting negative value).

Teacher's Note:

1) Identify arithmetic progression patterns for radii and quantities.
2) Use standard AP formulas for \( n^{\text{th}} \) term and sum of n terms.

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.

38. Based on the above information, answer the following questions :
(i) Find the coordinates of the centre C. [1 Mark]
(ii) Find the radius of the circular park. [1 Mark]
(iii) (a) Find the coordinates of the point P. [2 Marks]
OR
(b) Find the distance of the fountain at Q from gate A. [2 Marks]

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

(ii) Radius is the distance from C(30, 35) to A(10, 20): \( R = \sqrt{(30 - 10)^2 + (35 - 20)^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25\text{ units} \).

(iii)(a) Since \( AP : PB = 1 : 2 \), using the section formula, coordinates of P are \( \left(\frac{1(50) + 2(10)}{3}, \frac{1(50) + 2(20)}{3}\right) = \left(\frac{70}{3}, 30\right) \).

OR

(iii)(b) Total diameter length \( AB = 2 \times 25 = 50 \). Since \( AQ = \frac{2}{3}AB \), distance \( AQ = \frac{2}{3} \times 50 = \frac{100}{3}\text{ units} \).

Teacher's Note:

1) Use the mid-point formula for circle centers given diameter endpoints.
2) Apply the section formula accurately for internal division points.

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