πŸ“ Mathematics Β· MCQ Practice

Class 10 Mathematics
MCQ Practice β€” 20 Questions

Test your Maths skills with 20 interactive MCQs covering all major topics. Click any option to instantly check your answer and see the explanation.

πŸ“ 20 Questions πŸ“ All Topics 🎯 CBSE Class 10 βœ… Instant Answers
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Chapter 1–2 Real Numbers & Polynomials
Q1
HCF (36mΒ², 18m) = 18m, where m is a prime number. Which statement about this is correct?
βœ… Explanation
36mΒ² = 2 Γ— 18 Γ— m Γ— m and 18m = 18 Γ— m. The common factors are 18 and m. So HCF = 18m. Note: HCF is always ≀ the smaller number (18m here).
Q2
If the zeroes of a polynomial p(x) are βˆ’3 and 8, then the product of zeroes is:
βœ… Explanation
Product of zeroes = Ξ± Γ— Ξ² = (βˆ’3) Γ— (8) = βˆ’24. Sum of zeroes = βˆ’3 + 8 = 5. For a polynomial axΒ² + bx + c, product of zeroes = c/a and sum = βˆ’b/a.
Q3
The value of m for which 14 βˆ’ 2√3 is proved irrational (given √3 is irrational) uses which method?
βœ… Explanation
We assume 14 βˆ’ 2√3 is rational (= a/b). This leads to √3 = (14bβˆ’a)/2b, which is rational β€” contradicting the given fact that √3 is irrational. This is proof by contradiction.
Chapter 3–4 Linear & Quadratic Equations
Q4
The system 3x βˆ’ 5y + 7 = 0 and βˆ’6x + 10y + 14 = 0 is:
βœ… Explanation
a₁/aβ‚‚ = 3/βˆ’6 = βˆ’1/2, b₁/bβ‚‚ = βˆ’5/10 = βˆ’1/2, c₁/cβ‚‚ = 7/14 = 1/2. Since a₁/aβ‚‚ = b₁/bβ‚‚ β‰  c₁/cβ‚‚, the lines are parallel β†’ no solution β†’ inconsistent.
Q5
For the quadratic equation 3xΒ² βˆ’ 7x + m = 0 to have real and equal roots, the value of m is:
βœ… Explanation
For equal roots, discriminant D = 0. D = bΒ² βˆ’ 4ac = (βˆ’7)Β² βˆ’ 4(3)(m) = 49 βˆ’ 12m = 0. So 12m = 49, giving m = 49/12.
Q6
Two water taps together fill a tank in 8⁸⁄₉ hours. The larger tap takes 4 hours less than the smaller. The time taken by the smaller tap alone is:
βœ… Explanation
Let smaller tap = x hrs, larger = (xβˆ’4) hrs. Together: 80/9 Γ— [1/x + 1/(xβˆ’4)] = 1. This gives 9xΒ² βˆ’ 196x + 320 = 0. Solving: x = 20 (x = 16/9 rejected). Smaller tap = 20 hrs.
Q7
For the A.P. βˆ’15/4, βˆ’10/4, βˆ’5/4, … the value of a₁₆ βˆ’ a₁₂ is:
βœ… Explanation
Common difference d = βˆ’10/4 βˆ’ (βˆ’15/4) = 5/4. a₁₆ βˆ’ a₁₂ = (a + 15d) βˆ’ (a + 11d) = 4d = 4 Γ— 5/4 = 5.
Q8
The sum of first n terms of an AP is Sβ‚™ = 2nΒ² + 13n. The 10th term of this AP is:
βœ… Explanation
aβ‚™ = Sβ‚™ βˆ’ Sₙ₋₁ = (2nΒ² + 13n) βˆ’ [2(nβˆ’1)Β² + 13(nβˆ’1)] = 4n + 11. So a₁₀ = 4(10) + 11 = 51... wait: 40 + 11 = 51? Let me recheck: aβ‚™ = 2nΒ² + 13n βˆ’ 2nΒ² + 4n βˆ’ 2 βˆ’ 13n + 13 = 4n + 11. a₁₀ = 40 + 11 = 51. Hmm but the answer in the board paper was 31. Let me recheck: Sβ‚™ = 2nΒ² + 13n. aβ‚™ = Sβ‚™ βˆ’ Sₙ₋₁ = 2nΒ²+13n βˆ’ [2(n-1)Β²+13(n-1)] = 2nΒ²+13n βˆ’ [2nΒ²βˆ’4n+2+13nβˆ’13] = 2nΒ²+13nβˆ’2nΒ²βˆ’9n+11 = 4n+11. So a₁₀ = 51. The correct answer is 51.
Chapter 6–7 Triangles & Coordinate Geometry
Q9
In β–³ABC, D divides BC in ratio 1:2. With A(1,5), B(βˆ’2,1), C(4,2), the coordinates of D are:
βœ… Explanation
Section formula with m₁:mβ‚‚ = 1:2: x = (1Γ—4 + 2Γ—(βˆ’2))/(1+2) = (4βˆ’4)/3 = 0. y = (1Γ—2 + 2Γ—1)/(1+2) = 4/3. So D = (0, 4/3).
Q10
The line joining P(βˆ’4, βˆ’2) and Q(10, 4) is divided by the y-axis in the ratio:
βœ… Explanation
On y-axis, x = 0. Using section formula: (kΓ—10 + 1Γ—(βˆ’4))/(k+1) = 0 β†’ 10k βˆ’ 4 = 0 β†’ k = 2/5. So ratio = 2:5.
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Chapter 9–12 Circles, Areas & Surface Areas
Q11
PQ and PR are tangents from P to a circle (radius 5 cm, centre O). If OP = 13 cm, then PQ =
βœ… Explanation
Tangent βŠ₯ radius at point of contact. So OQ βŠ₯ PQ. By Pythagoras: PQΒ² = OPΒ² βˆ’ OQΒ² = 13Β² βˆ’ 5Β² = 169 βˆ’ 25 = 144. PQ = 12 cm.
Q12
An arc of length 2.2 cm subtends an angle ΞΈ at the centre of a circle with radius 2.8 cm. The value of ΞΈ is:
βœ… Explanation
Arc length = (ΞΈ/360Β°) Γ— 2Ο€r. So 2.2 = (ΞΈ/360Β°) Γ— 2 Γ— (22/7) Γ— 2.8 = (ΞΈ/360Β°) Γ— 17.6. ΞΈ = (2.2 Γ— 360Β°)/17.6 = 45Β°.
Q13
A conical cavity of maximum volume is carved from a solid hemisphere of radius 10 cm. Curved surface area of the cone is (Ο€ = 3.14):
βœ… Explanation
For max volume cone in hemisphere: height = radius = 10 cm. Slant height β„“ = √(rΒ²+hΒ²) = √(100+100) = 10√2 cm. CSA = Ο€rβ„“ = 3.14 Γ— 10 Γ— 10√2 = 314√2 cmΒ².
Q14
A camping tent is hemispherical with radius 1.4 m and has a door of area 0.50 mΒ². The outer surface area of the tent is:
βœ… Explanation
Outer surface = CSA of hemisphere βˆ’ area of door = 2Ο€rΒ² βˆ’ 0.50 = 2 Γ— (22/7) Γ— 1.4Β² βˆ’ 0.5 = 12.32 βˆ’ 0.5 = 11.82 mΒ².
Chapter 8 & 9 Trigonometry & Heights/Distances
Q15
Simplest form of sec A / √(secΒ²A βˆ’ 1) is:
βœ… Explanation
secΒ²A βˆ’ 1 = tanΒ²A (identity). So sec A / √tanΒ²A = sec A / tan A = (1/cos A) / (sin A/cos A) = 1/sin A = cosec A.
Q16
A wire from point A on ground reaches the top of pole BC at 60° elevation. If AB = 5√3 m, the length of the wire AC is:
βœ… Explanation
cos 60Β° = AB/AC β†’ 1/2 = 5√3/AC β†’ AC = 10√3 m. Alternatively, tan 60Β° = BC/AB β†’ BC = 5√3 Γ— √3 = 15 m. Then AC = √(ABΒ² + BCΒ²) = √(75+225) = √300 = 10√3 m. βœ“
Chapter 13–14 Statistics & Probability
Q17
In step deviation method, xΜ„ = 64, h = 5, a = 62.5. The value of Ε« is:
βœ… Explanation
xΜ„ = a + Ε« Γ— h β†’ 64 = 62.5 + Ε« Γ— 5 β†’ 1.5 = 5Ε« β†’ Ε« = 0.3.
Q18
Two dice are rolled together. The probability of getting outcome (x, y) where x > y is:
βœ… Explanation
Favourable outcomes (x > y): (2,1),(3,1),(3,2),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5) = 15 outcomes. P = 15/36 = 5/12.
Q19
Meena's probability of winning a lottery is 0.08. If 800 tickets were sold, how many did she buy?
βœ… Explanation
P(winning) = tickets bought / total tickets β†’ 0.08 = n/800 β†’ n = 0.08 Γ— 800 = 64 tickets.
Q20
Which of the following cannot be the probability of an event?
βœ… Explanation
Probability must satisfy 0 ≀ P(E) ≀ 1. Option C: 10/0.2 = 50. Since 50 > 1, it cannot be a probability. All other options are between 0 and 1.
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