RBSE Class 10th Board Mathematics Paper 2024 pdf with Solutions

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RBSE Class 10 Mathematics — Board Exam 2024 Solutions (Part A)

Here are the solved MCQs, Fill in the blanks, and One-liner very short answer questions from the RBSE Class 10 Board Mathematics Paper (2024). Click See Answer to view solutions with explanation.

Q1 — Multiple Choice Questions (i–xv)

Q1 (i)

HCF of two numbers is 27 and LCM is 162. If one number is 54, find the other number.

We know: HCF × LCM = Product of numbers.

27 × 162 = 54 × N

So, N = (27×162)/54 = 81.

Answer: 81

Q1 (ii)

Find the zeroes of polynomial x² − 2x − 8.

Solve x² − 2x − 8 = 0.

Factorise: (x−4)(x+2)=0 ⇒ x=4 or x=−2.

Answer: −2, 4

Q1 (iii)

Translate into an equation: Twice the first number is 5 less than 7 times the second.

Equation: 2x = 7y − 5.

Q1 (iv)

Find the nth term of the arithmetic progression 2, 5, 8, ...

First term a=2, common difference d=3.

Tn = a + (n−1)d = 2 + (n−1)3 = 3n−1

Answer: 3n−1

Q1 (v)

Find the sum of the first five multiples of 3.

Multiples: 3,6,9,12,15. Sum=45.

Answer: 45

Q1 (vi)

In triangle PQR, if PQ² + QR² = PR², then which angle is 90°?

By Pythagoras, ∠Q = 90°.

Q1 (vii)

Find the distance of the point (0,3) from the x-axis.

Distance = |y| = 3.

Answer: 3 units

Q1 (viii)

Find the value of 2 sin45° cos45°.

Formula: 2 sinA cosA = sin2A.

Here: sin90°=1.

Answer: 1

Q1 (ix)

If the shadow of a tree is 1/√3 times its height, find angle of elevation of the sun.

tan θ = h / (h/√3) = √3 ⇒ θ=60°.

Answer: 60°

Q1 (x)

Tangents drawn at the ends of a diameter of a circle are ______.

Answer: Parallel

Q1 (xi)

Circumference of a circle is 176 cm. Find its radius.

2πr=176 ⇒ r=28 cm.

Q1 (xii)

Volume of a sphere is 36π cm³. Find its radius.

(4/3)πr³=36π ⇒ r³=27 ⇒ r=3 cm.

Q1 (xiii)

Find mean of first 11 whole numbers.

Numbers 0–10. Sum=55. Mean=55/11=5.

Q1 (xiv)

Find mode of 3,3,5,3,4,2,8,4,3,4,3.

3 occurs most (5 times). Mode=3.

Q1 (xv)

If P(E)=0.05, then find P(not E).

P(not E)=1−0.05=0.95.

Q2 — Fill in the Blanks (i–vii)

Q2 (i)

Find the 11th term of the AP −3, −1, 1, 3, ...

a=−3, d=2 ⇒ T11=−3+20=17.

Q2 (ii)

cos²60° − sin²60° = ?

1/4−3/4=−1/2.

Q2 (iii)

Area of circle=154 cm². Radius?

r=7 cm.

Q2 (iv)

Find slant height of cone (r=6,h=8).

l=√(36+64)=10 cm.

Q2 (v)

3 Median = 2 Mean + ______.

Mode.

Q2 (vi)

Class mark of 60–75.

(60+75)/2=67.5.

Q2 (vii)

Probability of number >6 on dice roll.

0.

Q3 — One-liner Very Short Answers (i–x)

Q3 (i)

If x=2³·3² and y=2²·3², find LCM(x,y).

LCM=2³·3²=72.

Q3 (ii)

Find the zeroes of x²−3.

Zeroes=±√3.

Q3 (iii)

Solve equations: x+y=14, x−y=4.

x=9,y=5.

Q3 (iv)

— (Question missing in scanned paper)

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Q3 (v)

Find distance between (−a,a) and (−a,−a).

Distance=2a.

Q3 (vi)

If shadow of a vertical pole equals its height, find angle of elevation.

tanθ=1 ⇒ θ=45°.

Q3 (vii)

From a point on ground 100 m away from base of tower, angle of elevation=60°. Find tower height.

Height=100√3 m.

Q3 (viii)

Radius=7 cm, central angle=60°. Find arc length.

Arc=7π/3 cm.

Q3 (ix)

Total surface area of cube=1014 m². Find side length.

Side=13 m.

Q3 (x)

Find surface area of sphere radius 7 cm.

4πr²=616 cm².

Q4. Prove that √5 is irrational.

Show by contradiction that √5 cannot be expressed as p/q where p, q are integers with no common factor.

Proof:

Assume √5 is rational. Then √5 = p/q where p and q are integers with gcd(p,q)=1 and q ≠ 0.

√5 = p/q  ⇒  5 = p²/q²  ⇒  p² = 5q²

So p² is divisible by 5 ⇒ p is divisible by 5. Let p = 5k for some integer k.

p² = (5k)² = 25k² = 5q²  ⇒  q² = 5k²

Thus q² divisible by 5 ⇒ q divisible by 5. So both p and q are divisible by 5, contradicting gcd(p,q)=1.

Therefore √5 is irrational. □

Q5. Find the zeroes of the quadratic polynomial 2x² − x − 6.

Solve 2x² − x − 6 = 0 using quadratic formula:

x = [1 ± √(1 + 48)] / 4 = [1 ± √49] / 4 = [1 ± 7] / 4

So x = (1 + 7)/4 = 2 or x = (1 − 7)/4 = −6/4 = −3/2.

Zeroes: 2 and −3/2.

Q6. Two numbers differ by 26. One number is three times the other. Find the numbers.

Let smaller number = x. Then larger = 3x. Difference: 3x − x = 26 ⇒ 2x = 26 ⇒ x = 13.

Numbers: 13 and 39.

Q7. In triangle ABC, DE ∥ BC. AD = x, DB = x − 2, AE = x + 2, EC = x − 1. Find x.

Because DE ∥ BC, triangles ADE and ABC are similar. So corresponding sides are proportional:

AD/AB = AE/AC

Compute AB = AD + DB = x + (x − 2) = 2x − 2 and AC = AE + EC = (x + 2) + (x − 1) = 2x + 1.

x / (2x − 2) = (x + 2) / (2x + 1)

Cross-multiply:

x(2x + 1) = (x + 2)(2x − 2)

Left: 2x² + x. Right: (x + 2)(2x − 2) = 2x² + 2x − 4.

Equate: 2x² + x = 2x² + 2x − 4 ⇒ x = 4.

x = 4 (check lengths positive).

Q8. If the distance between the points (x, 3) and (5, 7) is 5 units, find x.

Use distance formula:

√[(x − 5)² + (3 − 7)²] = 5

Square both sides: (x − 5)² + (−4)² = 25 ⇒ (x − 5)² + 16 = 25 ⇒ (x − 5)² = 9.

So x − 5 = ±3 ⇒ x = 8 or x = 2.

Q9. If tan A = 1, find the value of 2 sin A cos A.

tan A = 1 ⇒ A = 45° (principal). Then 2 sin A cos A = sin 2A = sin 90° = 1.

Answer: 1.

Q10. Two vertical pillars of heights 20 m and 14 m stand on level ground. The distance between their tops measured along a straight wire that makes 30° with the horizontal is asked. (Find the length of the wire.)

Let horizontal distance between bases = d. Vertical difference = 20 − 14 = 6 m. The wire connecting the tops makes 30° with horizontal:

tan 30° = (vertical difference) / d = 6 / d ⇒ d = 6 / tan30° = 6 / (1/√3) = 6√3

Length of wire L = √(d² + (vertical difference)²) = √((6√3)² + 6²) = √(108 + 36) = √144 = 12 m.

Wire length = 12 m.

Q11. Two concentric circles have radii 5 cm and 3 cm. A chord of the larger circle is tangent to the smaller circle. Find the length of the chord.

Let O be center, distance from O to chord = radius of smaller circle = 3 cm. Let half chord = x. For larger circle radius 5:

5² = 3² + x² ⇒ x² = 25 − 9 = 16 ⇒ x = 4

Chord length = 2x = 8 cm.

Q12. A sector of a circle has radius 21 cm and central angle 60°. Find the area of the sector. (Use π = 22/7 if numeric answer required.)

Area of sector = (θ/360) × πr² = (60/360) × π × 21² = (1/6) × π × 441 = 73.5π cm².

Using π = 22/7 numeric: 73.5 × 22/7 = 231 cm².

Area = 73.5π cm² (exact) = 231 cm² (π ≈ 22/7).

Q13. A right circular cylinder has base area 154 cm² and height 15 cm. Find its curved surface area.

Base area = πr² = 154 ⇒ r² = (154 × 7) / 22 = 49 ⇒ r = 7 cm.

Curved surface area (CSA) = 2πrh = 2π × 7 × 15 = 210π cm².

Numerical (π = 22/7): 210 × 22/7 = 660 cm².

Q14. Find the mean of the following distribution: x = 5,6,7,8,9,10 and f = 8,6,12,7,5,6.

Compute Σf and Σfx:

Σf = 8+6+12+7+5+6 = 44
Σfx = 5×8 + 6×6 + 7×12 + 8×7 + 9×5 + 10×6 = 321

Mean = Σfx / Σf = 321 / 44 ≈ 7.295454... ≈ 7.30.

Q15. A loaded die has faces A, B, C, D, E, A (i.e., A appears on two faces). Find P(A) and P(D).

Total faces = 6. A appears 2 times ⇒ P(A) = 2/6 = 1/3. D appears once ⇒ P(D) = 1/6.

Q16. How many multiples of 4 lie between 10 and 250?

Smallest multiple of 4 greater than 10 is 12. Largest multiple ≤ 250 is 248.

Sequence: 12,16,...,248 with a = 12, d = 4. Let n be number of terms:

12 + (n−1)4 = 248 ⇒ (n−1)4 = 236 ⇒ n−1 = 59 ⇒ n = 60

60 multiples.

Q17. Points A(6,1), B(8,2), C(9,4) and D(p,3) are vertices of a parallelogram ABCD in order. Find p.

Diagonals of a parallelogram bisect each other. Midpoint of AC = midpoint of BD.

Midpoint AC = ((6+9)/2, (1+4)/2) = (15/2, 5/2)
Midpoint BD = ((8+p)/2, (2+3)/2) = ((8+p)/2, 5/2)

Equate x-coordinates: 15/2 = (8+p)/2 ⇒ 15 = 8 + p ⇒ p = 7.

Q18. Prove that a parallelogram which circumscribes a circle is a rhombus.

If a circle is inscribed in a parallelogram, tangents from a vertex are equal. Let tangent lengths around be x,y,z,w in order. Then sides are:

AB = x+y,  BC = y+z,  CD = z+w,  DA = w+x

Parallelogram ⇒ AB = CD and BC = DA ⇒ x+y = z+w and y+z = w+x. Subtract one from the other:

(x+y) − (y+z) = (z+w) − (w+x) ⇒ x − z = z − x ⇒ 2x = 2z ⇒ x = z

Similarly y = w. Thus AB = x+y and BC = y+x ⇒ AB = BC ⇒ all sides equal. Hence parallelogram with equal sides is a rhombus.

Q19. Find the median for the distribution: 0–10:4, 10–20:28, 20–30:42, 30–40:20, 40–50:6.

Total frequency N = 4+28+42+20+6 = 100. N/2 = 50. Cumulative frequencies: 4, 32, 74, ... So median class = 20–30.

For class 20–30: l = 20, h = 10, f = 42, c.f. before = 32.

Median = l + [(N/2 − c.f.before)/f] × h
= 20 + [(50 − 32)/42] × 10 = 20 + (18/42)×10 = 20 + (3/7)×10 = 20 + 30/7 = 20 + 4.2857 = 24.2857

Median ≈ 24.29.

Q20. In a right triangle, the altitude is 7 cm less than the base. The hypotenuse is 13 cm. Find the other two sides.

Let base = b and altitude = b − 7. By Pythagoras:

b² + (b − 7)² = 13² = 169

⇒ b² + b² − 14b + 49 = 169 ⇒ 2b² − 14b − 120 = 0 ⇒ b² − 7b − 60 = 0.

Solve: b = [7 ± √(49 + 240)]/2 = [7 ± √289]/2 = [7 ± 17]/2. Positive root: (7 + 17)/2 = 12.

So base = 12 cm, altitude = 12 − 7 = 5 cm. (Check: 5² + 12² = 25 + 144 = 169 = 13².)

Q21. Prove the identity: (cosec A − sin A)(sec A − cos A)(tan A + cot A) = 1.

Rewrite each term in sin/cos:

cosec A − sin A = (1/sin A) − sin A = (1 − sin²A)/sin A = cos²A / sin A
sec A − cos A = (1/cos A) − cos A = (1 − cos²A)/cos A = sin²A / cos A
tan A + cot A = (sin A / cos A) + (cos A / sin A) = (sin²A + cos²A)/(sin A cos A) = 1/(sin A cos A)

Multiply all three:

(cos²A / sin A) × (sin²A / cos A) × (1/(sin A cos A))
= (cos²A · sin²A) / (sin A · cos A · sin A · cos A) = 1

Hence the identity is proved.

Q22. Using assumed mean 17.5, find the mean of the frequency distribution with class intervals 0–5(3),5–10(7),10–15(15),15–20(24),20–25(16),25–30(5).

Mid-points x: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5. Frequencies f: 3,7,15,24,16,5. Total f = 70. Take A = 17.5.

Compute d = x − A: −15, −10, −5, 0, 5, 10 (note these are in units of 1 but correspond to midpoints shifted by 17.5; alternatively use d in original units: −15, −10, −5, 0, 5, 10).

Σf d = 3(−15) + 7(−10) + 15(−5) + 24(0) + 16(5) + 5(10)
= −45 −70 −75 + 0 + 80 + 50 = −60

Mean = A + (Σf d) / Σf = 17.5 + (−60)/70 = 17.5 − 6/7 ≈ 17.5 − 0.85714 = 16.64286 ≈ 16.64.