CBSE Class 10 Mathematics Standard Maths Long Answer Questions Assignment Set 19

Answer: Let a be any positive integer. By Euclid's lemma with divisor 3, a = 3q, 3q + 1 or 3q + 2. Case 1: a² = 9q² = 3(3q²) = 3m. Case 2: a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1. Case 3: a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1. Hence every square is of the form 3m or 3m + 1.

Teacher's Note:
1. Write the integer in the three possible forms 3q, 3q + 1 and 3q + 2.
2. Square each form and take 3 common.

Answer: α + β = 7/6 and αβ = 2/6 = 1/3. New sum = 1/α + 1/β = (α + β)/αβ = (7/6) × 3 = 7/2. New product = 1/(αβ) = 3. The polynomial is k[y² − (7/2)y + 3]; taking k = 2 gives 2y² − 7y + 6.

Teacher's Note:
1. Express the new sum and product using the old ones.
2. Multiply by 2 to clear the fraction.

Answer: Since b² − b − 20 = (b − 5)(b + 4) and (b − 5) + (b + 4) = 2b − 1, the equation factorises as (x − (b − 5))(x − (b + 4)) = 0. So x = b − 5 or x = b + 4.

Teacher's Note:
1. Factorise the constant term so that the factors add up to the coefficient of x.
2. The roots are expressions in b.

Answer: Let the width be w m, so length = w + 4. Half perimeter: w + (w + 4) = 36, so w = 16. The garden is 16 m wide and 20 m long.

Teacher's Note:
1. Half the perimeter equals length plus width.
2. Verify that 16 + 20 = 36.

Answer: Let the ratio be k : 1. The y-coordinate on the x-axis is 0, so (7k − 6)/(k + 1) = 0 and k = 6/7. The ratio is 6 : 7. x = (6 × (−1) + 7 × (−4))/13 = −34/13. The point is (−34/13, 0).

Teacher's Note:
1. A point on the x-axis has y = 0.
2. Use the ratio found to calculate the x-coordinate.

Answer: a₁ = S₁ = 8. S₂ = 12 + 10 = 22, so a₂ = 14 and d = 6. The AP is 8, 14, 20, …. a₁₅ = 8 + 14 × 6 = 92.

Teacher's Note:
1. The first term equals S₁ and the second term is S₂ − S₁.
2. Use aₙ = a + (n − 1)d for the 15th term.

Answer: Let the numbers be a − d, a, a + d. The sum 3a = 3 gives a = 1. Then (1 − d)(1 + d) = −35, so 1 − d² = −35 and d = ±6. The numbers are −5, 1, 7 (or 7, 1, −5).

Teacher's Note:
1. Choosing a − d, a, a + d makes the sum simple.
2. Both signs of d give the same set of numbers.

Answer: AP : PB = 3 : 4. By the section formula, x = (3 × 2 + 4 × (−2))/7 = −2/7 and y = (3 × (−4) + 4 × (−2))/7 = −20/7. So P is (−2/7, −20/7).

Teacher's Note:
1. AP = 3/7 of AB means PB = 4/7 of AB.
2. Apply the section formula with m : n = 3 : 4.

Answer: Let the tower height be h and the distance to the foot be d. tan 45° = h/d gives d = h. tan 60° = (h + 7)/d gives √3 h = h + 7, so h = 7/(√3 − 1) = 7(√3 + 1)/2 ≈ 9.56 m.

Teacher's Note:
1. Use two right triangles with the same base.
2. Rationalise the denominator at the end.

Answer: sin 30° = 1/2 = cos 60°. So 40° + x = 60° and x = 20°.

Teacher's Note:
1. Rewrite the right side as a cosine of a known angle.
2. Then equate the angles.

Answer: Divide numerator and denominator by sin A: LHS = (cot A − 1 + cosec A)/(cot A + 1 − cosec A). Using 1 = cosec²A − cot²A in the numerator: cot A + cosec A − (cosec A − cot A)(cosec A + cot A) = (cosec A + cot A)(1 − cosec A + cot A). The factor (1 − cosec A + cot A) cancels with the denominator, so LHS = cosec A + cot A.

Teacher's Note:
1. Divide through by sin A to bring cosec and cot into view.
2. Replace 1 by cosec²A − cot²A in the numerator.

Answer: Squaring: 1 + 2 sin θ cos θ = 2, so sin θ cos θ = 1/2. tan θ + cot θ = (sin²θ + cos²θ)/(sin θ cos θ) = 1/(1/2) = 2.

Teacher's Note:
1. Squaring the given relation brings out sin θ cos θ.
2. Write tan θ + cot θ over a common denominator.

Answer: Equal roots mean D = 0: (2mc)² − 4(1 + m²)(c² − a²) = 0. So 4m²c² − 4(c² − a² + m²c² − m²a²) = 0, which gives a² + m²a² − c² = 0. Hence c² = a²(1 + m²).

Teacher's Note:
1. Set the discriminant to zero and expand carefully.
2. The m²c² terms cancel.

Answer: Let the speed be x km/h. 360/x − 360/(x + 5) = 48/60 = 4/5. So 1800/(x(x + 5)) = 4/5, giving x² + 5x − 2250 = 0, i.e. (x + 50)(x − 45) = 0. Since speed is positive, the original speed is 45 km/h.

Teacher's Note:
1. Convert 48 minutes into hours.
2. Reject the negative root.

Answer: a₃ and a₇ are roots of t² − 6t + 8 = 0, so they are 2 and 4. Case 1: a₃ = 2, a₇ = 4 gives d = 1/2 and a = 1, so S₁₆ = 8[2 + 15/2] = 76. Case 2: a₃ = 4, a₇ = 2 gives d = −1/2 and a = 5, so S₁₆ = 8[10 − 15/2] = 20.

Teacher's Note:
1. Two possibilities arise because the order of the terms is not given.
2. Use Sₙ = (n/2)[2a + (n − 1)d].

Answer: The middle terms are the 18th, 19th and 20th, so 3a₁₉ = 225 and a + 18d = 75. The last three terms are the 35th, 36th and 37th, so 3a₃₆ = 429 and a + 35d = 143. Subtracting, 17d = 68, so d = 4 and a = 3. The AP is 3, 7, 11, 15, …

Teacher's Note:
1. Three consecutive terms have the middle one as their average.
2. Subtract the two equations to eliminate a.

Answer: Let PA and PB be tangents from P to a circle with centre O, touching it at A and B. Join OA, OB and OP. Radius is perpendicular to the tangent, so ∠OAP = ∠OBP = 90°. In triangles OAP and OBP: OA = OB (radii), OP is common, and the right angles are equal. By RHS congruence, the triangles are congruent, hence PA = PB.

Teacher's Note:
1. Join the centre to the points of contact and to the external point.
2. Use the RHS congruence rule.

Answer: Cards left = 52 − 6 = 46. (i) Face cards left are the 6 red ones: 6/46 = 3/23. (ii) Red queens: 2/46 = 1/23. (iii) The 2 black aces are not face cards, so they remain: 2/46 = 1/23. (iv) Clubs left = 13 − 3 = 10: 10/46 = 5/23.

Teacher's Note:
1. Black face cards are the jack, queen and king of spades and clubs.
2. The denominator changes to 46 for every part.