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

SECTION A

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

Answer : (D) ab. We know HCF × LCM = a × b. Since HCF = 1, LCM = ab.

Teacher's Note:
Always remember the identity: Product of two numbers = HCF × LCM.
When HCF = 1, the numbers are called co-prime, and their LCM is just their product.

2. [1 mark] The number 3 + √2 is :
(A) a rational number
(B) an irrational number
(C) an integer
(D) a natural number

Answer : (B) an irrational number, since a rational number (3) added to an irrational number (√2) always gives an irrational number.

Teacher's Note:
Sum or difference of a rational and an irrational number is always irrational.
Do not confuse this with the product/sum of two irrationals, which can sometimes be rational.

3. [1 mark] The discriminant of the quadratic equation x² − 3x − 2 = 0 is :
(A) 1
(B) 17
(C) √17
(D) −√17

Answer : (B) 17. Here a = 1, b = −3, c = −2, so D = b² − 4ac = 9 − 4(1)(−2) = 9 + 8 = 17.

Teacher's Note:
Discriminant formula: D = b² − 4ac.
Watch the sign carefully when c is negative, since −4ac becomes positive.

4. [1 mark] The equation x + 1/x = 3 (x ≠ 0) is expressed as a quadratic equation in the form of ax² + bx + c = 0. The value of a − b + c is :
(A) 5
(B) 2
(C) 1
(D) −1

Answer : (A) 5. Multiplying both sides by x: x² + 1 = 3x, so x² − 3x + 1 = 0. Here a = 1, b = −3, c = 1. So a − b + c = 1 − (−3) + 1 = 5.

Teacher's Note:
Clear the fraction first by multiplying through by x, then rearrange to standard form.
Be careful with signs while substituting into a − b + c.

5. [1 mark] For a point (3, − 5), the value of (abscissa − ordinate) is :
(A) − 8
(B) − 2
(C) 2
(D) 8

Answer : (D) 8. Abscissa (x-coordinate) = 3, ordinate (y-coordinate) = −5. So abscissa − ordinate = 3 − (−5) = 8.

Teacher's Note:
Abscissa means the x-coordinate; ordinate means the y-coordinate.
Subtracting a negative number is the same as adding the positive value.

6. [1 mark] The mid-point of a line segment divides the line segment in the ratio :
(A) 1 : 2
(B) 2 : 1
(C) 1 : 1
(D) 1/2 : 2

Answer : (C) 1 : 1, because the mid-point is equidistant from both endpoints.

Teacher's Note:
Mid-point formula is a special case of the section formula with ratio 1:1.
x = (x₁+x₂)/2, y = (y₁+y₂)/2.

7. [1 mark] Which of the following is not the criterion for similarity of triangles ?
(A) AAA
(B) SSS
(C) SAS
(D) RHS

Answer : (D) RHS. RHS (Right angle-Hypotenuse-Side) is a criterion for congruence of right triangles, not for similarity.

Teacher's Note:
Similarity criteria are AAA, SSS and SAS only.
Congruence criteria (SSS, SAS, ASA, RHS) are a separate list - don't mix the two.

8. [1 mark] From the figures given below, which of the following is true about the measure of ∠P ?
In triangle ABC: AB = 3.8 cm, ∠A = 80°, AC = 3√3 cm, ∠B = 60°, BC = 6 cm.
In triangle PQR: PR = 6√3 cm, QR = 7.6 cm, PQ = 12 cm.
(A) ∠P = 60°
(B) ∠P = 80°
(C) ∠P = 40°
(D) The measure of ∠P cannot be determined

Answer : (C) ∠P = 40°.

First, compare the sides of the two triangles. In ΔABC, sides are AB = 3.8, BC = 6, CA = 3√3 ≈ 5.2.
In ΔPQR, sides are QR = 7.6, PQ = 12, RP = 6√3 ≈ 10.4.
Each side of ΔPQR is exactly double the corresponding side of ΔABC (AB↔QR, BC↔PQ, CA↔RP), so ΔABC ~ ΔRQP by SSS similarity.
This correspondence gives A ↔ R, B ↔ Q, C ↔ P.
So ∠P = ∠C = 180° − ∠A − ∠B = 180° − 80° − 60° = 40°.

Teacher's Note:
To match up vertices between two triangles, pair up sides in the same ratio and see which vertices they share.
Once the correspondence is found, corresponding angles are automatically equal.

9. [1 mark] In the given figure, PA is a tangent to a circle with centre O. If OP = 10 cm and ∠OPA = 30°, then the length of AP is :
(A) 10√3 cm
(B) 20 cm
(C) 5 cm
(D) 5√3 cm

Answer : (D) 5√3 cm.

Since PA is tangent to the circle at A, OA ⊥ PA, so ∠OAP = 90°.
In right triangle OAP, ∠P = 30°, hypotenuse OP = 10 cm.
AP = OP × cos 30° = 10 × (√3/2) = 5√3 cm.

Teacher's Note:
Radius drawn to the point of contact is always perpendicular to the tangent.
Use this right angle to apply trigonometric ratios directly.

10. [1 mark] Which of the following statements is false ?
(A) tan 45° = cot 45°
(B) sin 90° = tan 45°
(C) sin 30° = cos 30°
(D) sin 45° = cos 45°

Answer : (C) sin 30° = cos 30° is false, since sin 30° = 1/2 but cos 30° = √3/2, and these are not equal.

Teacher's Note:
Keep the standard trigonometric ratio table for 0°, 30°, 45°, 60°, 90° memorised.
Only at 45° do sine and cosine become equal.

11. [1 mark] The value of (tan² A − 1/cos² A) is :
(A) more than 1
(B) 1
(C) 0
(D) − 1

Answer : (D) −1. Since 1/cos²A = sec²A, the expression becomes tan²A − sec²A = −(sec²A − tan²A) = −1, using the identity sec²A − tan²A = 1.

Teacher's Note:
Remember the identity sec²A − tan²A = 1.
Rewriting 1/cos²A as sec²A is the key first step here.

12. [1 mark] In the given figure, which of the following angles represents the angle of depression ? (An observer stands with a horizontal dashed line drawn from the eye; the line of sight goes down to an object; angle z is between the horizontal line and the line of sight at the observer; angle y is below z at the observer; angle a is the right angle at the foot of the vertical; angle x is at the object between the line of sight and the ground level.)
(A) x
(B) y
(C) z
(D) a

Answer : (C) z. The angle of depression is always measured between the horizontal line at the observer's eye and the line of sight going down to the object, which is angle z.

Teacher's Note:
Angle of depression is always at the observer's eye level, between the horizontal and the line of sight.
Angle of elevation, by contrast, is measured at the object looking up.

13. [1 mark] The perimeter of the shaded region in the given figure is : (A circle with centre O; the shaded region is bounded by two radii r, r drawn to the ends of an arc, with arc length l and chord length a marked below the arc.)
(A) l
(B) l + a
(C) l + 2r
(D) l + 2r + a

Answer : (C) l + 2r. The shaded region is a sector, so its boundary consists of the two straight radii (length r each) and the curved arc (length l).

Teacher's Note:
Perimeter of a sector = sum of its two radii + arc length.
The chord (a) is not part of the sector's boundary, so it is not included here.

14. [1 mark] The ratio of the area of a quadrant of a circle to the area of the same circle is :
(A) 1 : 2
(B) 2 : 1
(C) 1 : 4
(D) 4 : 1

Answer : (C) 1 : 4, since a quadrant is exactly one-fourth of the full circle (90° out of 360°).

Teacher's Note:
Area of quadrant = (90/360) × πr² = (1/4) × πr².
So its ratio to the full circle's area is always 1:4, regardless of the radius.

15. [1 mark] For which of the following solids is the lateral/curved surface area and total surface area the same ?
(A) Cube
(B) Cuboid
(C) Hemisphere
(D) Sphere

Answer : (D) Sphere. A sphere has only a curved surface and no flat base, so its curved surface area equals its total surface area (both = 4πr²).

Teacher's Note:
A hemisphere has an extra flat circular base, so its total surface area (3πr²) differs from its curved surface area (2πr²).
Only the sphere has no flat face at all.

16. [1 mark] The class mark of the median class of the following data is :

(A) 40
(B) 55
(C) 47.5
(D) 62.5

Answer : (D) 62.5.

Total frequency N = 2+3+7+6+6+6 = 30, so N/2 = 15.
Cumulative frequencies are 2, 5, 12, 18, 24, 30. The class where cf first exceeds 15 is 55–70, so this is the median class.
Class mark = (55+70)/2 = 62.5.

Teacher's Note:
First build the cumulative frequency column to locate the median class.
Class mark = (upper limit + lower limit)/2 of that class.

17. [1 mark] The following distribution shows the number of runs scored by some batsmen in test matches :

The lower limit of the modal class is :
(A) 3000
(B) 4000
(C) 5000
(D) 6000

Answer : (B) 4000. The modal class is the class with the highest frequency, which is 4000–5000 (frequency 10), so its lower limit is 4000.

Teacher's Note:
Modal class = class interval with the maximum frequency.
No calculation is needed for this - just spot the highest frequency.

18. [1 mark] In a random experiment of throwing a die, which of the following is a sure event ?
(A) Getting a number between 1 and 6
(B) Getting an odd number less than 7
(C) Getting an even number less than 7
(D) Getting a natural number less than 7

Answer : (D) Getting a natural number less than 7, because every possible outcome of a die (1, 2, 3, 4, 5, 6) is a natural number less than 7, so this event always happens.

Teacher's Note:
A sure event is one that includes every possible outcome of the experiment.
Options (B) and (C) fail since only odd or only even numbers appear on some throws, not all.

19. [1 mark] 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.
(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). Since HCF divides both numbers a and b, and LCM is always a multiple of both a and b, HCF necessarily divides LCM as well.

Teacher's Note:
Recall Product of numbers = HCF × LCM, which shows HCF always divides LCM exactly.
This is a standard true fact worth remembering directly.

20. [1 mark] 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 a₁x + b₁y = c₁ and a₂x + b₂y = c₂ is consistent with infinitely many solutions, if a₁/a₂ = b₁/b₂ = c₁/c₂.
(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.

Here a₁ = 4, b₁ = p, c₁ = 8, a₂ = 2, b₂ = 2, c₂ = 2, so a₁/a₂ = 2 and c₁/c₂ = 4.
For infinitely many solutions we would need p/2 = 2 = 4, which is impossible, so the lines can never be coincident.
The system has a unique solution whenever a₁/a₂ ≠ b₁/b₂, i.e. whenever p ≠ 4; at exactly p = 4 the lines become parallel (a₁/a₂ = b₁/b₂ but c₁/c₂ is different), which makes the system inconsistent, not consistent.
So Assertion (A) is false, while Reason (R) states a correct general fact.

Teacher's Note:
Always check all three ratios a₁/a₂, b₁/b₂, c₁/c₂ carefully before concluding consistency.
A common mistake is to just check when two lines "match" without confirming the third ratio for coincident lines.

SECTION B

21. [2 marks] Solve the following system of equations for x and y :
x/2 + 2y/3 = −1 and x − y/3 = 3

Answer :

First, clear the fractions.
Multiplying the first equation by 6: 3x + 4y = −6 ... (i)
Multiplying the second equation by 3: 3x − y = 9 ... (ii)

Subtracting (ii) from (i): (3x + 4y) − (3x − y) = −6 − 9
5y = −15
y = −3

Substituting y = −3 in (ii): 3x − (−3) = 9
3x + 3 = 9
3x = 6
x = 2

So, x = 2 and y = −3.

Teacher's Note:
Clear all fractions first by multiplying by the LCM of the denominators.
Elimination method works well once both equations have the same coefficient for one variable.

22. [2 marks] (a) In the given figure, if PQ ∥ RS, then prove that ΔPOQ ~ ΔSOR. (Two lines PS and QR cross at point O, forming an hourglass shape with P and Q on one side and R, S on the other; PQ and RS are the slanted top segments.)

Answer (a) :

In ΔPOQ and ΔSOR :
∠POQ = ∠SOR (vertically opposite angles)
∠OPQ = ∠OSR (alternate angles, since PQ ∥ RS and PS is the transversal)

Therefore, ΔPOQ ~ ΔSOR (by AA similarity criterion).

OR

(b) In the given figure, ΔOSR ~ ΔOQP, ∠ROQ = 125° and ∠ORS = 70°. Find the measures of ∠OSR and ∠OQP.

Answer (b) :

Since S, O, Q lie on one straight line, ∠SOR and ∠ROQ form a linear pair.
∠SOR = 180° − ∠ROQ = 180° − 125° = 55°

In ΔOSR, the angles add up to 180°:
∠SOR + ∠ORS + ∠OSR = 180°
55° + 70° + ∠OSR = 180°
∠OSR = 55°

Since ΔOSR ~ ΔOQP, corresponding angles are equal, so ∠OQP = ∠OSR = 55°.

So, ∠OSR = 55° and ∠OQP = 55°.

Teacher's Note:
When two straight lines cross, vertically opposite angles are equal and adjacent angles on a line add to 180°.
In similar triangles, always match corresponding vertices in the same order as given (O↔O, S↔Q, R↔P here).

23. [2 marks] 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.

Answer :

Let O be the common centre, and let chord AB of the larger circle touch the smaller circle at point M.
Since AB is tangent to the smaller circle at M, OM ⊥ AB, and OM = 6 cm (radius of smaller circle), OA = 10 cm (radius of larger circle).

In right triangle OMA, by Pythagoras theorem:
AM² = OA² − OM² = 10² − 6² = 100 − 36 = 64
AM = 8 cm

Since the perpendicular from the centre bisects the chord, AB = 2 × AM = 2 × 8 = 16 cm.

So, the required chord length is 16 cm.

Teacher's Note:
The perpendicular from the centre of a circle to a chord always bisects the chord.
This question combines that property with the Pythagoras theorem in the right triangle formed by the two radii.

24. [2 marks] (a) Find the values of A and B (0 ≤ A < 90°, 0 ≤ B < 90°), if tan (A + B) = 1 and tan (A − B) = 1/√3.

Answer (a) :

Since tan (A + B) = 1 = tan 45°, we get A + B = 45° ... (i)
Since tan (A − B) = 1/√3 = tan 30°, we get A − B = 30° ... (ii)

Adding (i) and (ii): 2A = 75°, so A = 37.5°
Subtracting (ii) from (i): 2B = 15°, so B = 7.5°

So, A = 37.5° and B = 7.5°.

OR

(b) Prove geometrically that tan 45° = 1.

Answer (b) :

Consider a right triangle ABC, right-angled at B, with ∠A = 45°.
Since the angles of a triangle add up to 180°: ∠A + ∠B + ∠C = 180°, so 45° + 90° + ∠C = 180°, giving ∠C = 45°.

Since ∠A = ∠C = 45°, the sides opposite to these equal angles are equal, so BC = AB.

Now, tan A = (side opposite A)/(side adjacent to A) = BC/AB = 1, since BC = AB.

Hence, tan 45° = 1.

Teacher's Note:
Recognising standard values like tan 45° = 1 and tan 30° = 1/√3 lets you convert directly into angle equations.
For the geometric proof, use the fact that equal angles in a triangle are opposite equal sides.

25. [2 marks] A chord of a circle of diameter 20 cm subtends an angle of 60° at the centre of the circle. Find the area of the corresponding minor segment of the circle. (Use π = 3.14 and √3 = 1.73)

Answer :

Radius, r = 20/2 = 10 cm, and the angle at the centre, θ = 60°.

Area of sector = (θ/360°) × πr² = (60/360) × 3.14 × 10² = (1/6) × 314 = 52.33 cm² (approx.)

Since the angle is 60° and both radii are equal, triangle OAB is equilateral with side = 10 cm.
Area of triangle OAB = (√3/4) × (side)² = (1.73/4) × 100 = 43.25 cm²

Area of minor segment = Area of sector − Area of triangle = 52.33 − 43.25 = 9.08 cm² (approx.)

So, the area of the minor segment is approximately 9.08 cm².

Teacher's Note:
Minor segment area = sector area − triangle area, always.
When the central angle is exactly 60°, the triangle formed by the two radii and chord is equilateral - use this shortcut.

SECTION C

26. [3 marks] (a) Prove that √3 is an irrational number.

Answer (a) :

Let us assume, to the contrary, that √3 is a rational number.
Then we can write √3 = p/q, where p and q are coprime integers (no common factor other than 1) and q ≠ 0.

Squaring both sides: 3 = p²/q², so p² = 3q².
This means 3 divides p², and therefore 3 divides p (since 3 is prime).
Let p = 3m for some integer m. Substituting back: (3m)² = 3q², so 9m² = 3q², i.e. q² = 3m².
This means 3 divides q², and therefore 3 divides q as well.

So both p and q are divisible by 3, which contradicts our assumption that p and q have no common factor.
This contradiction shows that our assumption was wrong. Hence, √3 is an irrational number.

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.

Answer (b) :

Starting from the bottom of the tree: 35 = 5 × b, so b = 7.
Next, 70 = 2 × 35, which checks out (2 × 35 = 70).
Next, 210 = a × 70, so a = 210/70 = 3.
Next, y = 2 × 210 = 420.
Finally, x = 2 × y = 2 × 420 = 840.

So, x = 840, y = 420, a = 3, b = 7.

Writing x as a product of primes: 840 = 2 × 2 × 2 × 3 × 5 × 7 = 2³ × 3 × 5 × 7.

Teacher's Note:
For the irrationality proof, always start "let us assume the opposite" and end by pointing out the contradiction clearly.
For a factor tree, work from the bottom (smallest known factors) upward to find each unknown one step at a time.

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

Answer :

A quadratic polynomial with given sum and product of zeroes is written as:
x² − (sum of zeroes)x + (product of zeroes)
= x² − (0)x + (−9)
= x² − 9

To find the zeroes, set the polynomial equal to zero:
x² − 9 = 0
x² = 9
x = ± 3

So, the required polynomial is x² − 9, and its zeroes are 3 and −3.

Teacher's Note:
Quadratic polynomial formula: x² − (sum)x + (product).
Here the "sum = 0" term simply disappears, leaving a difference-of-squares polynomial that factorises easily.

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

Answer (a) :

For x + 3y = 6: when x = 0, y = 2; when y = 0, x = 6. So points are (0, 2) and (6, 0).
For 2x − 3y = 12: when x = 0, y = −4; when y = 0, x = 6. So points are (0, −4) and (6, 0).

Plotting both lines on graph paper, they intersect at the common point (6, 0).

So, the solution is x = 6, y = 0.

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.

Answer (b) :

Since x and y are complementary angles, their sum is 90°:
x + y = 90 ... (i)

Since x : y = 1 : 2, we get y = 2x, i.e.
2x − y = 0 ... (ii)

Substituting y = 2x into equation (i): x + 2x = 90
3x = 90
x = 30

Then y = 2 × 30 = 60.

So, the two linear equations are x + y = 90 and 2x − y = 0, and the solution is x = 30°, y = 60°.

Teacher's Note:
For the graphical method, always find at least two clean points per line and check both lines meet the same point.
Complementary angles always add to 90°, while supplementary angles add to 180° - do not mix these up.

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

Answer :

Let ABCD be a rectangle circumscribing a circle, touching the sides AB, BC, CD and DA at points P, Q, R and S respectively.

Since the lengths of tangents drawn from an external point to a circle are equal:
AP = AS (tangents from A)
BP = BQ (tangents from B)
CQ = CR (tangents from C)
DR = DS (tangents from D)

Adding AP, BP, CR and DR on one side and their equals on the other:
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC

But ABCD is a rectangle, so its opposite sides are equal: AB = CD and AD = BC.
Substituting: AB + AB = AD + AD, i.e. 2AB = 2AD, so AB = AD.

Since ABCD is a rectangle (all angles 90°) with adjacent sides AB and AD equal, all four sides of ABCD are equal.
Hence, ABCD is a square.

Teacher's Note:
Key property: tangents drawn from the same external point to a circle are always equal in length.
A rectangle becomes a square exactly when its adjacent sides are also equal.

30. [3 marks] Prove that :
(1 + cot² A)/(1 + tan² A) = ((1 − cot A)/(1 − tan A))²

Answer :

First, simplify the left-hand side (LHS).
LHS = (1 + cot²A)/(1 + tan²A) = cosec²A/sec²A (using the identities 1+cot²A=cosec²A and 1+tan²A=sec²A)
= (1/sin²A) / (1/cos²A) = cos²A/sin²A = cot²A

Now simplify the right-hand side (RHS).
1 − cot A = 1 − cosA/sinA = (sinA − cosA)/sinA
1 − tan A = 1 − sinA/cosA = (cosA − sinA)/cosA = −(sinA − cosA)/cosA

So, (1 − cot A)/(1 − tan A) = [(sinA − cosA)/sinA] ÷ [−(sinA − cosA)/cosA] = −cosA/sinA = −cot A

Squaring: RHS = (−cot A)² = cot²A

Since LHS = cot²A = RHS, the identity is proved.

Teacher's Note:
Convert everything to sinA and cosA when an identity looks complicated - this often reveals common factors.
Remember 1+tan²A=sec²A and 1+cot²A=cosec²A, two of the most frequently used Pythagorean identities.

31. [3 marks] 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 ?

Answer :

In the first lot of 200 pens, 180 are good, so the number of defective pens = 200 − 180 = 20.
The customer will not buy the pen only if it is defective.
P(customer will not buy it) = (number of defective pens)/(total pens) = 20/200 = 1/10

Now, the second lot has 100 pens with 80 good pens, so it has 100 − 80 = 20 defective pens.
When mixed with the first lot: total pens = 200 + 100 = 300
Total good pens = 180 + 80 = 260

The customer will buy the pen only if it is good.
P(customer will buy the pen) = 260/300 = 13/15

So, the required probabilities are 1/10 and 13/15.

Teacher's Note:
Probability = (favourable outcomes)/(total outcomes); always recompute the total when lots are combined.
"Will not buy" and "will buy" are complementary conditions based on whether the pen is defective or good.

SECTION D

32. [5 marks] (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.

Answer (a) :

Let the greater number be y and the smaller number be x.

Given: difference of squares is 180, so y² − x² = 180 ... (i)
Given: square of smaller number is 8 times the greater number, so x² = 8y ... (ii)

Substituting (ii) into (i): y² − 8y = 180
y² − 8y − 180 = 0

Using the quadratic formula: y = [8 ± √(64 + 720)]/2 = [8 ± √784]/2 = [8 ± 28]/2

So y = 36/2 = 18 or y = −20/2 = −10 (rejected, since the number must be positive).

So y = 18. Substituting into (ii): x² = 8 × 18 = 144, so x = 12.

Check: 18² − 12² = 324 − 144 = 180. Correct.

So, the two numbers are 12 and 18.

OR

(b) Find the value(s) of k for which the equation 2x² + kx + 3 = 0 has real and equal roots. Hence, find the roots of the equations so obtained.

Answer (b) :

For real and equal roots, the discriminant must be zero: D = b² − 4ac = 0
Here a = 2, b = k, c = 3.
k² − 4(2)(3) = 0
k² = 24
k = ± 2√6

For equal roots, each root = −b/2a = −k/4.

When k = 2√6, the root is −2√6/4 = −√6/2 (repeated twice).
When k = −2√6, the root is 2√6/4 = √6/2 (repeated twice).

So, k = 2√6 gives roots −√6/2, −√6/2, and k = −2√6 gives roots √6/2, √6/2.

Teacher's Note:
"Real and equal roots" always means Discriminant = 0 - set up and solve this condition first.
Once D=0, both roots automatically equal −b/2a, so no separate quadratic formula step is needed after finding k.

33. [5 marks] State the "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 AO/BO = CO/DO, as shown in the given figure. Prove that ABCD is a trapezium.

Answer :

Basic Proportionality Theorem (Thales' 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.

Proof: Draw a line EO through O, parallel to AB, meeting AD at E.

In ΔABD, since EO ∥ AB, by the Basic Proportionality Theorem:
DE/EA = DO/OB, i.e. AE/ED = BO/DO ... (i)

Now, from the given condition AO/BO = CO/DO, cross-multiplying gives:
AO × DO = BO × CO
Dividing both sides by CO × DO: AO/CO = BO/DO ... (ii)

From (i) and (ii): AE/ED = BO/DO = AO/OC

So, in ΔACD, we have E on AD and O on AC with AE/ED = AO/OC.
By the converse of the Basic Proportionality Theorem, EO ∥ DC.

But by construction, EO ∥ AB. Since EO is parallel to both AB and DC, we conclude AB ∥ DC.

Since ABCD has one pair of parallel sides (AB ∥ DC), ABCD is a trapezium. Hence proved.

Teacher's Note:
The converse of the Basic Proportionality Theorem is just as important as the theorem itself for "prove trapezium" style questions.
The key trick is drawing an extra parallel line through the intersection point to create the two triangles needed.

34. [5 marks] (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.

Answer (a) :

Radius, r = 5 cm. Height of cone = diameter of base = 2 × 5 = 10 cm.

Volume of cone = (1/3)πr²h = (1/3) × π × 5² × 10 = (1/3) × π × 250 = 250π/3 cm³

Volume of hemisphere = (2/3)πr³ = (2/3) × π × 5³ = (2/3) × π × 125 = 250π/3 cm³

Total volume of toy = volume of cone + volume of hemisphere = 250π/3 + 250π/3 = 500π/3 cm³

Using π = 22/7: Volume = (500 × 22)/(3 × 7) = 11000/21 ≈ 523.81 cm³

So, the volume of the toy is 500π/3 cm³, which is approximately 523.81 cm³.

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.

Answer (b) :

For the hemisphere (radius 3.5 cm) to lie entirely on the top face of the cube, the edge of the cube must be at least equal to the diameter of the hemisphere.
Smallest edge = 2 × 3.5 = 7 cm

Total surface area of the solid = (surface area of cube) − (area of circular base of hemisphere removed) + (curved surface area of hemisphere)
= 6(edge)² − πr² + 2πr²
= 6(edge)² + πr²

= 6 × 7² + π × (3.5)²
= 6 × 49 + (22/7) × 12.25
= 294 + 38.5
= 332.5 cm²

So, the smallest edge of the cube is 7 cm, and the total surface area of the solid is 332.5 cm².

Teacher's Note:
When a hemisphere sits on a cube/cuboid, the flat circular base of the hemisphere disappears into the cube's surface - subtract it once and add the curved surface area once.
The smallest possible edge always equals the diameter of the hemisphere so that it fits without overhanging.

35. [5 marks] The following data gives the information on the observed lifetime (in hours) of 200 electrical components :

Find the mean lifetime (in hours) of the electrical components.

Answer :

First, find the class mark (x_i) of each class interval, and then f_ix_i.

Sum of frequencies, Σf_i = 10+35+50+60+30+15 = 200
Sum of f_ix_i, Σf_ix_i = 100+1050+2500+4200+2700+1650 = 12200

Mean = Σf_ix_i/Σf_i = 12200/200 = 61

So, the mean lifetime of the electrical components is 61 hours.

Teacher's Note:
Direct method for mean: Mean = Σf_ix_i/Σf_i, using the class mark (midpoint) of each interval.
Keep the table organised in columns to avoid arithmetic slips while adding up large sums.

SECTION E

Case Study – 1

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.

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

Answer :

The building's height (15 m) is opposite to the 60° angle, and the ladder is the hypotenuse.
sin 60° = height/ladder length
Ladder length = 15/sin 60° = 15/(√3/2) = 30/√3 = 10√3 m ≈ 17.32 m

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

Answer :

Distance = height/tan 60° = 15/√3 = 5√3 m ≈ 8.66 m

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

Answer (a) :

The diagram is a right triangle: the building AB is vertical with height 15 m, the top of the building is A, its foot is B; the ladder now reaches from A down to a point D at the base of the opposite wall, with the ladder making an angle of 30° with the ground at D. BD is the width of the road, which is the side adjacent to the 30° angle, with AB (15 m) opposite to it.

Width of road, BD = height/tan 30° = 15/(1/√3) = 15√3 m ≈ 25.98 m

So, the width of the road is 15√3 m.

OR

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

Answer (b) :

Ladder length = height/sin 30° = 15/(1/2) = 30 m

So, the length of the ladder used in this case is 30 m.

Teacher's Note:
Always identify which side (opposite, adjacent or hypotenuse) is known and which is required before choosing sin, cos or tan.
The same building height (15 m) is reused across all parts, only the angle with the ground changes.

Case Study – 2

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 alternately at A and B, starting with centre at A, of radii 50 cm, 100 cm, 150 cm, ....... as shown in the figure. Spiral 1 has 10 flowers, Spiral 2 has 20 flowers, Spiral 3 has 30 flowers and so on.

37. (i) [1 mark] What is the radius of the 13^th spiral ?

Answer :

The radii form an AP: 50, 100, 150, ... with first term a = 50 and common difference d = 50.
Radius of nth spiral = a + (n−1)d
Radius of 13th spiral = 50 + (13−1) × 50 = 50 + 600 = 650 cm

(ii) [1 mark] If the radius of the n^th spiral is 500 cm, find the value of n.

Answer :

50 + (n−1) × 50 = 500
(n−1) × 50 = 450
n − 1 = 9
n = 10

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

Answer (a) :

The number of flowers per spiral also forms an AP: 10, 20, 30, ... with first term a = 10 and common difference d = 10.

Sum of n terms, S_n = (n/2)[2a + (n−1)d]
S₁₁ = (11/2)[2(10) + (11−1)(10)] = (11/2)[20 + 100] = (11/2)(120) = 660

So, the total number of saplings till the 11th spiral is 660.

OR

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

Answer (b) :

S_n = (n/2)[2(10) + (n−1)(10)] = (n/2)(10)(n+1) = 5n(n+1)

Setting S_n = 450:
5n(n+1) = 450
n(n+1) = 90
n² + n − 90 = 0
n = [−1 ± √(1+360)]/2 = [−1 ± 19]/2

n = 9 (rejecting the negative root n = −10)

So, till the 9th spiral there will be a total of 450 saplings.

Teacher's Note:
Both the radii and the number of flowers per spiral form separate arithmetic progressions - identify a and d for each carefully.
For "total till nth term" questions, use the AP sum formula Sₙ = (n/2)[2a+(n−1)d].

Case Study – 3

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. Two fountains are installed at points P and Q on AB such that AP = PQ = QB.

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

Answer :

Since AB is a diameter of the circular park passing through the centre C, C is the mid-point of AB.
C = ((10+50)/2, (20+50)/2) = (60/2, 70/2) = (30, 35)

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

Answer :

AB = √[(50−10)² + (50−20)²] = √[40² + 30²] = √(1600+900) = √2500 = 50

Radius = AB/2 = 50/2 = 25

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

Answer (a) :

Since AP = PQ = QB, point P divides AB in the ratio AP : PB = 1 : 2 (measuring from A).

Using the section formula, P = [(1×50 + 2×10)/(1+2), (1×50 + 2×20)/(1+2)]
= [(50+20)/3, (50+40)/3]
= [70/3, 90/3]
= [70/3, 30]

So, the coordinates of P are (70/3, 30).

OR

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

Answer (b) :

Since AP = PQ = QB, point Q divides AB in the ratio AQ : QB = 2 : 1 (measuring from A).

Using the section formula, Q = [(2×50 + 1×10)/3, (2×50 + 1×20)/3] = [110/3, 120/3] = [110/3, 40]

Distance AQ = √[(110/3 − 10)² + (40 − 20)²] = √[(80/3)² + 20²] = √[6400/9 + 3600/9] = √[10000/9] = 100/3

So, the distance of the fountain at Q from gate A is 100/3 units (≈ 33.33 units). (This also matches directly with AQ = (2/3) × AB = (2/3) × 50 = 100/3.)

Teacher's Note:
When points divide a segment into three equal parts, they correspond to section ratios of 1:2 and 2:1 from the starting point.
Always double-check coordinate answers using the distance formula as a cross-check, as done above.

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