CBSE Class 10 Mathematics MCQs Set 03

Answer: (a) 19.625 m²

Explanation:
1. The horse can graze a quarter circle of radius 5 m.
2. Area = \( \frac{1}{4} \times 3.14 \times 5^2 \) = 19.625 m².

Teacher's Note:
1. At a corner of a square the grazing region is a quarter circle.
2. The rope is shorter than the side, so it stays inside the field.

Answer: (b) 78.5 m²

Explanation:
1. Quarter circle of radius 10 m: \( \frac{1}{4} \times 3.14 \times 100 \) = 78.5 m².

Teacher's Note:
1. 10 m is still less than 15 m, so no part is cut off.
2. Use the quarter-circle area.

Answer: (a) 58.875 m²

Explanation:
1. Area with 10 m rope = 78.5 m² and with 5 m rope = 19.625 m².
2. Increase = 78.5 − 19.625 = 58.875 m².

Teacher's Note:
1. Subtract the two quarter-circle areas.
2. Keep all decimal places.

Answer: (c) 205.37 m²

Explanation:
1. Field = 15 × 15 = 225 m².
2. Grazed = 19.625 m², so not grazed = 225 − 19.625 = 205.375 m² ≈ 205.37 m².

Teacher's Note:
1. Subtract the grazed area from the whole field.
2. 205.375 rounds to 205.37 as given in the option.

Answer: (a) 146.5 m²

Explanation:
1. Not grazed = 225 − 78.5 = 146.5 m².

Teacher's Note:
1. Whole field minus the quarter circle.
2. Check that the quarter circle fits: radius 10 < 15.

Answer: (d) 13/125 is a rational number with a non-terminating repeating decimal

Explanation:
1. 125 = 5³ has only the prime 5, so 13/125 = 0.104 terminates: statement (d) is wrong.
2. 30 = 2 × 3 × 5, 7 and 91 = 7 × 13 contain other primes, so the first three decimals do not terminate.

Teacher's Note:
1. Check the prime factors of the denominator.
2. Only 2 and 5 allow a terminating decimal.

Answer: (a) The area of a circle of radius 6 cm with a sector angle of 60° is \( \frac{132}{14} \) cm² (use π = 22/7)

Explanation:
1. Sector area = \( \frac{60}{360} \times \frac{22}{7} \times 36 \) = 132/7 cm², not 132/14, so (a) is incorrect.
2. (b) \( \frac{300}{360} \times 3.14 \times 196 \) = 512.87 ✓; (c) 10π : 25π = 2 : 5 ✓; (d) ½ × 22 × 6 = 66 ✓.

Teacher's Note:
1. Find the true value before comparing with the statement.
2. Area of sector = ½ × arc length × radius.

Answer: (b) \( \frac{49\pi}{12} \) cm²

Explanation:
1. Radius = 7 cm.
2. Area = \( \frac{30}{360} \times \pi \times 49 \) = 49π/12 cm².

Teacher's Note:
1. Convert the diameter to radius.
2. 30° is one twelfth of the full circle.

Answer: (d) 3600

Explanation:
1. 16 = 2⁴, 20 = 2² × 5, 24 = 2³ × 3, so LCM = 2⁴ × 3 × 5 = 240.
2. To make it a perfect square multiply by 3 × 5: 2⁴ × 3² × 5² = 3600.

Teacher's Note:
1. Every prime in a perfect square has an even power.
2. Check: 3600 = 60².

Answer: (b) 30.81 cm²

Explanation:
1. Sector = \( \frac{45}{360} \times \frac{22}{7} \times 28^2 \) = 308 cm².
2. Triangle = ½ r² sin 45° = ½ × 784 × 0.7071 = 277.19 cm².
3. Segment = 308 − 277.19 = 30.81 cm².

Teacher's Note:
1. Segment area = sector area − triangle area.
2. The triangle area uses ½ r² sin θ.

Answer: (a) 980

Explanation:
1. Take the smallest power of each common prime: 2², 5¹, 7² (3 is missing in the third number).
2. HCF = 4 × 5 × 49 = 980.

Teacher's Note:
1. A prime missing from any number is not in the HCF.
2. Use the lowest exponent.

Answer: (a) 14 cm²

Explanation:
1. Perimeter = 2r + πr/2 = 25, so r(2 + 11/7) = 25 gives r = 7 cm.
2. Minor segment = quarter circle − triangle = \( \frac{1}{4} \times \frac{22}{7} \times 49 - \frac{1}{2} \times 49 \) = 38.5 − 24.5 = 14 cm².

Teacher's Note:
1. First find r from the perimeter.
2. For 90° the triangle has legs equal to r.

Answer: (b) AB² = AD² − BC · DE + ¼BC²

Explanation:
1. In right triangle AEB: AB² = AE² + BE², and in right triangle AED: AD² = AE² + DE².
2. Since AC > AB, E lies on DC's side and BE = BD − DE = ½BC − DE.
3. AB² = AD² − DE² + (½BC − DE)² = AD² − BC · DE + ¼BC².

Teacher's Note:
1. Use Pythagoras theorem in both right triangles with the same height AE.
2. Write BE in terms of BD and DE.

Answer: (b) 1.6 m

Explanation:
1. Distance walked in 4 s = 1.2 × 4 = 4.8 m.
2. Let the shadow be x: by similar triangles \( \frac{0.9}{3.6} = \frac{x}{x + 4.8} \), so 3x = 4.8 and x = 1.6 m.

Teacher's Note:
1. The lamp-post and the girl form similar right triangles.
2. The light ray reaches the ground at the tip of the shadow.

Answer: (b) 346.5 cm²

Explanation:
1. (√3/4)a² = 121√3 gives a = 22 cm, so the wire length is 66 cm.
2. 2πr = 66 gives r = 10.5 cm.
3. Area = \( \frac{22}{7} \times 10.5^2 \) = 346.5 cm².

Teacher's Note:
1. The wire length stays constant.
2. Equilateral area = (√3/4)a².

Answer: (c) 5 second

Explanation:
1. Time per revolution: 1 s, 5/3 s and 5/2 s.
2. LCM of 6/6, 10/6, 15/6 seconds = LCM(6, 10, 15)/6 = 30/6 = 5 s.

Teacher's Note:
1. Convert revolutions per minute to seconds per revolution.
2. LCM of fractions = LCM of numerators over HCF of denominators.

Answer: (d) \( \frac{a+b}{a-b} \)

Explanation:
1. cosec θ = (a² + b²)/(a² − b²) and cos θ = 2ab/(a² + b²), so cot θ = 2ab/(a² − b²).
2. Sum = (a + b)²/((a − b)(a + b)) = (a + b)/(a − b).

Teacher's Note:
1. Find cos θ from a right triangle with sides a² − b² and a² + b².
2. a² + b² + 2ab = (a + b)².

Answer: (a) 14 cm²

Explanation:
1. Ratio of areas = (AB/DE)² = 4.
2. ar(DEF) = 56 ÷ 4 = 14 cm².

Teacher's Note:
1. Square the side ratio.
2. The smaller triangle has the smaller area.

Answer: (d) \( \frac{\sin\theta}{\sin\phi} \)

Explanation:
1. From the first: tan θ − a tan θ cos φ = a sin φ, so a = tan θ/(sin φ + tan θ cos φ) = sin θ/(sin φ cos θ + sin θ cos φ) = sin θ/sin(θ + φ).
2. Similarly b = sin φ/sin(θ + φ).
3. Hence a/b = sin θ/sin φ.

Teacher's Note:
1. Solve each relation for a and b.
2. The common factor sin(θ + φ) cancels.

Answer: (b) 32, 30

Explanation:
1. Let the parts be x and 62 − x: \( \frac{x/4}{2(62 - x)/5} = \frac{2}{3} \).
2. 15x = 16(62 − x), so 31x = 992 and x = 32; the parts are 32 and 30.

Teacher's Note:
1. Cross-multiply the ratio.
2. Check: 8 : 12 = 2 : 3.

Answer: (b) \( \frac{5}{6} \)

Explanation:
1. Same numbers: 6 outcomes out of 36.
2. P(different) = 1 − 1/6 = 5/6.

Teacher's Note:
1. Use the complement of a doublet.
2. Different means not a doublet.

Answer: (b) 1

Explanation:
1. Product of zeroes = 161/23 = 7.
2. 2 × 7 = 14p, so p = 1.

Teacher's Note:
1. Use product = c/a.
2. Divide 14 by 14.

Answer: (c) 67

Explanation:
1. The sum of three odd primes would be odd, so one prime is 2.
2. The other two add up to 98 and differ by 36: 31 and 67.
3. The largest is 67.

Teacher's Note:
1. 100 is even, so 2 must be one of the primes.
2. Check: 2 + 31 + 67 = 100.

Answer: (a) 9

Explanation:
1. Ratio of areas = (2.1/2.8)² = (3/4)² = 9/16.
2. ar(ABC) = 16 × 9/16 = 9 cm².

Teacher's Note:
1. Square the ratio of corresponding sides.
2. 2.1 : 2.8 = 3 : 4.

Answer: (c) 192

Explanation:
1. LCM = product ÷ HCF = 3072 ÷ 16 = 192.

Teacher's Note:
1. HCF × LCM = product of the two numbers.
2. Check: 16 × 192 = 3072.

Answer: (c) 81, 54

Explanation:
1. The numbers are 27a and 27b with a + b = 5 and 27ab = 162, so ab = 6.
2. a, b = 2, 3 gives the numbers 54 and 81: 81, 54.

Teacher's Note:
1. Write the numbers as multiples of the HCF.
2. Check: 81 + 54 = 135.

Answer: (c) 1

Explanation:
1. Co-prime numbers have no common factor except 1.
2. HCF = 1.

Teacher's Note:
1. Co-prime is defined by HCF = 1.
2. Their LCM would be pq.

Answer: (d) equal

Explanation:
1. HCF divides both numbers and divides LCM; if HCF = LCM, each number divides... a = b is forced.
2. The numbers must be equal.

Teacher's Note:
1. HCF ≤ each number ≤ LCM.
2. Equality of both forces the two numbers to be the same.

Answer: (c) p³q²

Explanation:
1. Highest powers: p³ and q².
2. LCM = p³q².

Teacher's Note:
1. LCM takes the highest power of each prime.
2. HCF would be pq.

Answer: (c) 78.5 cm²

Explanation:
1. Minor sector = \( \frac{90}{360} \times 3.14 \times 100 \) = 78.5 cm².

Teacher's Note:
1. 90° is a quarter of the circle.
2. Use π = 3.14.

Answer: (a) 28.5 cm²

Explanation:
1. Minor segment = sector − triangle = 78.5 − ½ × 10 × 10.
2. = 78.5 − 50 = 28.5 cm².

Teacher's Note:
1. The triangle OAB is right-angled with legs 10 cm.
2. Subtract the triangle from the sector.

Answer: (d) 235.5 cm²

Explanation:
1. Whole circle = 3.14 × 100 = 314 cm².
2. Major sector = 314 − 78.5 = 235.5 cm².

Teacher's Note:
1. Major sector = circle − minor sector.
2. It subtends 270° at the centre.

Answer: (a) 285.5 cm²

Explanation:
1. Major segment = circle − minor segment = 314 − 28.5 = 285.5 cm².

Teacher's Note:
1. Use the minor segment area 28.5 cm².
2. Major segment = major sector + triangle = 235.5 + 50.

Answer: (b) 15.71 cm

Explanation:
1. Arc = \( \frac{90}{360} \times 2 \times 3.14 \times 10 \) = 15.7 cm (15.71 with more digits of π).

Teacher's Note:
1. The arc is a quarter of the circumference.
2. Use the formula θ/360 × 2πr.

Answer: (c) x² − 12x + 27

Explanation:
1. α + β = 4 and αβ = 3.
2. New sum = 3(α + β) = 12 and new product = 9αβ = 27.
3. Polynomial: x² − 12x + 27.

Teacher's Note:
1. Form x² − (sum)x + product.
2. Scaling both zeroes by 3 multiplies the product by 9.

Answer: (b) 29

Explanation:
1. −1 = (3x + 4 × 2) ÷ 7 gives x = −5.
2. 2 = (3y + 4 × 5) ÷ 7 gives y = −2.
3. x² + y² = 25 + 4 = 29.

Teacher's Note:
1. Use the section formula with m : n = 3 : 4.
2. Solve for x and y one at a time.

Answer: (c) \( (15/2, 33/2) \)

Explanation:
1. Mid-point = ((6 + 9)/2, (17 + 16)/2).
2. = (15/2, 33/2).

Teacher's Note:
1. Add the coordinates and halve.
2. x of J and I is 6 and 9.

Answer: (b) x − 13 = 0

Explanation:
1. (x − 9)² + (y − 8)² = (x − 17)² + (y − 8)².
2. −18x + 81 = −34x + 289, so 16x = 208 and x = 13.
3. The locus is x − 13 = 0.

Teacher's Note:
1. Equal distances mean the point lies on the perpendicular bisector.
2. QS is horizontal, so the bisector is the vertical line x = 13.

Answer: (a) \( \frac{31}{25} \)

Explanation:
1. (AB + 1)² = AB² + 49 gives AB = 24 and AC = 25.
2. cos A = 24/25 and sin A = 7/25.
3. cos A + sin A = 31/25.

Teacher's Note:
1. Use Pythagoras theorem with AC = AB + 1.
2. 7, 24, 25 is a triplet.