CBSE Class 10 Mathematics Standard Maths Short Answer Questions Assignment Set 20

Answer: HCF = product of the lowest powers of the common primes = 3² × 5 = 45.

Teacher's Note:
1. Take the smaller power of each common prime factor.
2. 3² × 5 = 9 × 5 = 45.

Answer: d = √[(−a − a)² + (−b − b)²] = √(4a² + 4b²) = 2√(a² + b²).

Teacher's Note:
1. Apply the distance formula directly.
2. Take 4 common under the root.

Answer: Equal roots need D = 0: (k√3)² − 4(3)(4) = 0, so 3k² = 48 and k² = 16. Hence k = ±4.

Teacher's Note:
1. Equal roots mean the discriminant is zero.
2. Do not forget the negative value of k.

Answer: By the basic proportionality theorem, CD/DA = CE/EB. So (x + 3)/(2x) = x/(2x − 1), giving (x + 3)(2x − 1) = 2x². This simplifies to 5x − 3 = 0, so x = 3/5.

Teacher's Note:
1. A line parallel to one side divides the other two sides in the same ratio.
2. The x² terms cancel, leaving a linear equation.

Answer: Reverse the AP: the last term −33 becomes the first term and d = 4. The 11th term from the end = −33 + 10 × 4 = 7. (Check: 27 − 5 × 4 = 7, the 6th term from the start of 16 terms.)

Teacher's Note:
1. Counting from the end is the same as an AP with the last term as the first term.
2. Change the sign of the common difference.

Answer: cot A = 8/15, so take base = 8 and perpendicular = 15. Hypotenuse = √(64 + 225) = 17. Hence sin A = 15/17 and sec A = 17/8.

Teacher's Note:
1. Draw a right triangle using cot A = base/perpendicular.
2. Use the Pythagoras theorem for the hypotenuse.

Answer: The radii are 1 cm and 2 cm. Area of the ring = π(2² − 1²) = 3π = 9.43 cm² (approximately, using π = 3.14).

Teacher's Note:
1. The ring area is the difference of the two circle areas.
2. Convert diameters into radii first.

Answer: Non-defective bulbs = 188. P = 188/200 = 47/50.

Teacher's Note:
1. First find the number of good bulbs.
2. You may also use 1 − 12/200.

Answer: 90 = 2 × 3² × 5 and 144 = 2⁴ × 3². HCF = 2 × 3² = 18. LCM = 2⁴ × 3² × 5 = 720.

Teacher's Note:
1. HCF uses the smallest powers of common primes and LCM uses the greatest powers of all primes.
2. Check: 18 × 720 = 90 × 144.

Answer: Since the triangle is right angled at C, AB² = AC² + BC² (Pythagoras theorem). Being isosceles with the right angle at C, AC = BC. So AB² = AC² + AC² = 2AC².

Teacher's Note:
1. The equal sides of a right isosceles triangle are the two legs.
2. Replace BC by AC in the Pythagoras relation.

Answer: a₁₆ = S₁₆ − S₁₅ = (2 × 256 + 48) − (2 × 225 + 45) = 560 − 495 = 65.

Teacher's Note:
1. The nth term is Sₙ − Sₙ₋₁.
2. Calculate both sums separately to avoid errors.

Answer: Let A(3, 0), B(6, 4), C(−1, 3). AB² = 9 + 16 = 25, AC² = 16 + 9 = 25, BC² = 49 + 1 = 50. Since AB² + AC² = BC², the triangle is right angled at A (and also isosceles).

Teacher's Note:
1. Find the squares of all three sides.
2. Check whether the two smaller squares add up to the largest.

Answer: sin(A + B) = 1 gives A + B = 90°; sin(A − B) = 1/2 gives A − B = 30°. Adding, 2A = 120°, so A = 60° and B = 30°.

Teacher's Note:
1. Convert each trigonometric value to an angle.
2. Solve the two linear equations in A and B.

Answer: Arc length l = (θ/360°) × 2πr, so θ = (12 × 360)/(2 × (22/7) × 6) ≈ 114.5°. Area of sector = ½ × l × r = ½ × 12 × 6 = 36 cm².

Teacher's Note:
1. Use the arc length formula to find the angle.
2. Area of a sector can also be found as ½ × arc × radius.

Answer: Area of sector = (120/360) × (22/7) × 196 = 205.33 cm². Area of triangle = ½ r² sin 120° = ½ × 196 × (√3/2) = 49√3 = 84.87 cm². Minor segment = 205.33 − 84.87 = 120.46 cm². Circle = 616 cm², so major segment = 616 − 120.46 = 495.54 cm² (approximately).

Teacher's Note:
1. Minor segment equals sector minus triangle.
2. Major segment equals the whole circle minus the minor segment.

Answer: The modal class is 30–35 (highest frequency 25). l = 30, f₁ = 25, f₀ = 10, f₂ = 7, h = 5. Mode = 30 + (25 − 10)/(2 × 25 − 10 − 7) × 5 = 30 + 75/33 = 32.27 (approximately).

Teacher's Note:
1. The modal class has the highest frequency.
2. Use Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h.