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Class 10 Maths – Solved Questions

Class 10 Maths — Solved Set

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Chapter 1

Real Numbers

Question 1

Find the LCM of the smallest prime number and the smallest odd composite number.

The smallest prime number is 2.
The smallest odd composite number is 9.
LCM(2, 9) = 18.

Answer: 18

Question 15

Find the LCM of the three numbers 28, 44, and 132.

28 = 2² × 7
44 = 2² × 11
132 = 2² × 3 × 11
LCM = 2² × 3 × 7 × 11 = 924.

Answer: 924

Question 18

Find the greatest number which divides 281 and 1249, leaving remainders 5 and 7 respectively.

Compute HCF(281−5, 1249−7) = HCF(276, 1242).
1242 = 276×4 + 138; 276 = 138×2 + 0 → HCF = 138.

Answer: 138

Question 28(a)

Prove that √3 is an irrational number.

Classic contradiction: assume √3 = a/b in lowest terms → 3b² = a² ⇒ 3 | a ⇒ let a = 3c ⇒ b² = 3c² ⇒ 3 | b, contradicting coprime. Hence irrational.

Question 28(b)

Prove that ( √2 + √3 )² is irrational, given √6 is irrational.

( √2 + √3 )² = 5 + 2√6; if this were rational then √6 would be rational — contradiction. Therefore irrational.

Chapter 2

Polynomials

Question 13

If α and β are the zeroes of p(x)=kx² − 30x + 45k and α+β = αβ, find k.

α+β = 30/k, αβ = 45. So 30/k = 45 ⇒ k = 2/3.

Answer: 2/3

Question 19

Assertion–Reason on degree of zero/non-zero constant polynomial.

A: True (degree not defined for zero polynomial). R: True (degree of non‑zero constant is 0). R does not explain A. Option (B).

Question 24

If α,β are zeroes of 5x²−6x+1, find α+β+αβ.

α+β = 6/5, αβ = 1/5 ⇒ sum = 7/5.

Chapter 3

Pair of Linear Equations in Two Variables

Question 17

Type of lines for 3x+4y=5 and 6x+8y=7?

a₁/a₂=b₁/b₂=1/2 but c₁/c₂=5/7 ⇒ no solution ⇒ Parallel.

Question 38 (Case Study)

Ticket charges ₹150 (child) & ₹250 (adult); 300 tickets; ₹55,000 collected.

x + y = 300; 3x + 5y = 1100

Solve → y=100, x=200. If 250 children & 100 adults visit → Amount = ₹62,500.

Chapter 4

Quadratic Equations

Question 8

Find the ratio (Sum of roots) : (Product of roots) of 5x²−6x+21=0.

Sum = 6/5, Product = 21/5 → Ratio = (6/5)/(21/5) = 2:7.

Question 33(a)

(k+1)x² − 6(k+1)x + 3(k+9)=0 has real and equal roots. Find k ≠ −1.

D=0 ⇒ 36(k+1)² − 12(k+1)(k+9)=0 ⇒ k=3.

Question 33(b)

Age word problem (man: 2×square of son's age).

Let son = x → 2x² + 8 = 3(x+8)+4 ⇒ 2x²−3x−20=0 ⇒ x=4. Son: 4 yrs; Man: 32 yrs.

Chapter 5

Arithmetic Progressions (A.P.)

Question 4

Find the next (4th) term of the A.P. √18, √50, √98, …

Simplify: 3√2, 5√2, 7√2 → d=2√2 → 9√2 = √162. Option (C).

Question 9

Find the 14th term from the end of A.P. −11, −8, −5, …, 49.

Reverse: 49, …, −11 with d=−3. a₁₄=49+13(−3)=10. Option (B).

Question 31(a)

S₁₄=1050, a=10. Find 20th term and general term.

From S_n=n/2[2a+(n−1)d] → d=10. a₂₀=200; a_n=10n.

Question 31(b)

Given a=5, last term l=45, S_n=400. Find n and d.

400 = n/2 (5+45) ⇒ n=16. Then 45=5+(16−1)d ⇒ d=8/3.

Chapter 6

Triangles

Question 5 (options mismatch noted)

In ΔABC, DE ∥ BC. If AD=2.4 cm, DB=4 cm and AE=2 cm, find AC.

Using BPT: AD/DB = AE/EC ⇒ 2.4/4 = 2/EC ⇒ EC = 10/3 cm. Hence AC = AE+EC = 2 + 10/3 = 16/3 cm. (Original options appeared inconsistent; value of EC is 10/3 cm.)

Question 20

Assertion–Reason on trapezium proportionality.

Both A and R true; R correctly explains A. Option (A).

Question 32(a)

Given ΔFEC ≅ ΔGDB and ∠1 = ∠2. Prove ΔADE ∼ ΔABC.

From congruence, DB=EC and by converse of BPT, DE ∥ BC ⇒ corresponding angles equal ⇒ ΔADE ∼ ΔABC (AAA).

Question 32(b)

If AB:AC:AD and PQ:PR:PM are proportional (with medians), show ΔABC ∼ ΔPQR.

Construct parallelograms via median extension; prove ΔABE ∼ ΔPQN and ΔACE ∼ ΔPRN by SSS; deduce ∠A=∠P and use SAS to get ΔABC ∼ ΔPQR.

Chapter 7

Coordinate Geometry

Question 23(a)

Find the ratio in which P(−4,6) divides AB where A(−6,10), B(3,−8).

Section formula on x‑coord: −4 = (3k − 6)/(k+1) ⇒ k = 2/7 ⇒ ratio 2:7 (internally).

Question 23(b)

Show that (3,0), (6,4), (−1,3) form an isosceles triangle.

AB = AC = 5; BC = 5√2 ⇒ two equal sides ⇒ isosceles.

Case Study 36

Square garden with ΔPQR, A as origin.

(i) P(4,6), Q(3,2), R(6,5).
(ii)(a) PQ = √17, QR = 3√2. (ii)(b) Point dividing PR in 2:1 internally: (16/3, 16/3).
(iii) PR = √5; all sides unequal ⇒ not isosceles.

Chapter 8

Introduction to Trigonometry

Question 3 (options typo noted)

If 5 tan θ − 12 = 0, find sin θ.

tan θ = 12/5 ⇒ Right triangle (12,5,13) ⇒ sin θ = 12/13. (Given options mismatch.)

Question 12

If x/3 = 2 sin A and y/3 = 2 cos A, find x²+y².

x=6 sin A, y=6 cos A ⇒ x²+y²=36.

Question 25

Evaluate [2 tan²30° · sec²60° · tan45°] / [1 − sin²60°].

= (2·(1/3)·4·1) / (1−3/4) = (8/3)/(1/4) = 32/3.

Question 26

Prove (tanθ/(1−cotθ)) + (cotθ/(1−tanθ)) = 1 + secθ·cosecθ.

Convert to sin/cos and simplify via a³−b³ factorization to obtain RHS. (Paper’s printed identity had a typo.)

Chapter 9

Applications of Trigonometry

Question 10

Shadow length is √3 times the tower height. Find angle of elevation of Sun.

tan θ = 1/√3 ⇒ θ = 30°.

Question 34

From a 15 m high window: elevation to top = 30°, depression to foot = 45°. Find opposite house height.

Street width x = 15 m. Extra height h = 15/√3 = 5√3 ≈ 8.66 m. Total = 23.66 m.

Chapter 10

Circles

Question 6 (assumption noted)

Two tangents RJ and RL. If ∠JRL = 42°, find ∠JOL.

Right angles at tangency points ⇒ quadrilateral sum gives ∠JOL = 138°.

Question 16

A chord of radius 10 cm subtends 90° at centre. Find chord length.

Right isosceles triangle: AB = √(10²+10²) = 10√2 cm.

Question 22(a)

Two tangents from P include 60°; radius 3 cm. Find each tangent length.

In ΔOAP, tan30° = 3/PA ⇒ PA = 3√3 cm.

Question 22(b)

Prove tangents at ends of a diameter are parallel.

Radii ⟂ tangents; equal alternate interior angles ⇒ lines are parallel.

Question 30

Prove: A parallelogram circumscribing a circle is a rhombus.

Use equal tangents from same external points to show adjacent sides equal ⇒ rhombus.

Chapter 11

Areas Related to Circles

Question 7

Perimeter of a sector (r=21 cm, θ=60°).

Arc = 22 cm; Perimeter = 22 + 2·21 = 64 cm.

Question 14

Arc length 10π, r=12 cm. Find central angle.

10π = (θ/360)·2π·12 ⇒ θ = 150°.

Question 27

Sector from r=21 cm with θ=150°. Find arc length and area.

Arc L = 55 cm; Area A = 577.5 cm².

Chapter 12

Surface Areas & Volumes

Question 35

Cylindrical glass: inner diameter 5.6 cm, height 10 cm, bottom has hemispherical raised portion. Find apparent & actual capacity.

r = 2.8 cm, h = 10 cm.
Apparent (cylinder): V = πr²h = 246.4 cm³.
Hemi raised: V = (2/3)πr³ ≈ 45.99 cm³.
Actual: 246.4 − 45.99 ≈ 200.41 cm³.

Chapter 13

Statistics

Question 2

If the mean of the first n natural numbers is (5/9)n, find n.

(n+1)/2 = 5n/9 ⇒ n = 9.

Question 37 (Case Study)

Time to run 100 m (grouped data). Find median class, mean, mode, and students with < 60s.

Time (s)	f	cf	x_i (mid)	f·x_i
0–20	8	8	10	80
20–40	10	18	30	300
40–60	13	31	50	650
60–80	6	37	70	420
80–100	3	40	90	270
Total	40			1720

(i) Median class: N/2=20 ⇒ cumulative just > 20 is 31 ⇒ class 40–60.
(ii)(a) Mean: 1720/40 = 43 s.
(ii)(b) Mode: Modal class 40–60. Using formula gives 46 s.
(iii) < 60 s: 8+10+13 = 31 students.

Chapter 14

Probability

Question 11

Probability that a random number from 1…15 is a multiple of 4.

Multiples: 4,8,12 (3 of 15) ⇒ P = 3/15 = 1/5. (Given option (C) 3/15.)

Question 21

From deck after removing K,Q,A of Clubs & Diamonds: find P(clubs) and P(red).

Remaining cards: 46. Clubs left: 10 ⇒ P = 10/46 = 5/23. Red left: 23 ⇒ P = 23/46 = 1/2.

Question 29

Three unbiased coins tossed.

Sample space size 8.
(i) At least one head: 7/8. (ii) Exactly one tail: 3/8. (iii) Two heads & one tail: 3/8.

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