CBSE Class 9 Mathematics Question Paper Set 41 with Solutions

Answer: (c) Assertion (A) is true but Reason (R) is false.
A quadrilateral with all four sides equal is, by definition, a rhombus, so A is true. One pair of parallel opposite sides only gives a trapezium (for example a rectangle that is not a square), so R is false.

Teacher's Note:
1. A rhombus needs all four sides equal; parallel sides alone are never enough.
2. Always test a Reason with a quick counter-example such as a rectangle or a trapezium.

Answer: (c) 5√14
√10 × √35 = √350 = √(25 × 14) = 5√14.

Teacher's Note:
1. Multiply the numbers under the roots first, then pull out perfect squares.
2. Do not forget to check each option by squaring: (5√14)² = 350.

Answer: (c) \( \dfrac{11}{30} \)
Let x = 0.3666… Then 10x = 3.666… and 100x = 36.666…
Subtracting, 90x = 33, so x = 33/90 = \( \dfrac{11}{30} \).

Teacher's Note:
1. Multiply by powers of 10 so that the repeating parts line up and cancel on subtraction.
2. Only the digit 6 repeats here, so use 10x and 100x, not 100x and 1000x.

Answer: (c) 10
p(−x) = 2(−x) + 5 = −2x + 5.
p(x) + p(−x) = (2x + 5) + (−2x + 5) = 10.

Teacher's Note:
1. Replace x by −x everywhere, including inside the bracket, before adding.
2. The x terms cancel, which is the quick check that the answer is a constant.

Answer: (b) remains the same
Multiplying or dividing both sides by the same non-zero number gives an equivalent equation, so the solutions do not change.

Teacher's Note:
1. Equivalent equations have exactly the same solutions.
2. The number must be non-zero; multiplying by 0 would destroy the equation.

Answer: (b) (1 − x)(1 + x + x²)
Using a³ − b³ = (a − b)(a² + ab + b²) with a = 1, b = x: 1 − x³ = (1 − x)(1 + x + x²).

Teacher's Note:
1. Learn the signs: a³ − b³ has a minus in the first bracket and plus signs in the second.
2. Multiply the factors back to verify; the x² and x terms should cancel.

Answer: (b) 2x + y = 1
Substituting x = 2, y = −3: 2(2) + (−3) = 4 − 3 = 1, which matches. The others give x + y = −1, x + 2y = −4, x − y = 5, none of which match.

Teacher's Note:
1. A point is a solution only if it satisfies the equation when substituted.
2. Be careful with the sign of y while substituting a negative value.

Answer: (c) on the y-axis
Every point of the form (0, y) is at zero distance from the y-axis, so it lies on the y-axis.

Teacher's Note:
1. Points on the y-axis have x = 0; points on the x-axis have y = 0.
2. Axes do not belong to any quadrant.

Answer: (b) 4x − 3 = x + 1
4x − 3 = x + 1 gives 3x = 4, so x = 4/3, which is of the form x = constant, a vertical line parallel to the y-axis. The other equations all contain both x and y.

Teacher's Note:
1. A line x = k is vertical, hence parallel to the y-axis.
2. Simplify the equation first; it may hide a single-variable equation.

Answer: (b) 50°
1. The diagonals of a rhombus bisect its angles, so ∠ACB = ∠ACD = 40° and ∠BCD = 80°.
2. Opposite angles of a rhombus are equal, so ∠BAD = 80°.
3. In triangle ABD, AB = AD, so ∠ADB = (180° − 80°)/2 = 50°.

Teacher's Note:
1. Use the facts that diagonals bisect the angles and that adjacent sides are equal.
2. Alternatively, the diagonals meet at 90°, so ∠ADB = 90° − 40° = 50° as well.

Answer: (d) 0°
Opposite angles of a parallelogram are equal, so ∠Q = ∠S and ∠Q − ∠S = 0°.

Teacher's Note:
1. Q and S are opposite vertices, not consecutive ones.
2. Consecutive angles add up to 180°, but opposite angles are equal.

Answer: (d) 26
Range = largest value − smallest value = 32 − 6 = 26.

Teacher's Note:
1. Range only needs the maximum and the minimum, so scan the data once for each.
2. Do not subtract the first and last numbers in the list; the data is not sorted.

Answer: (d) 24 cm
l² = r² + h², so h² = 26² − 10² = 676 − 100 = 576 and h = 24 cm.

Teacher's Note:
1. Radius, height and slant height form a right-angled triangle with the slant height as hypotenuse.
2. Recognise the triple 10, 24, 26 (a multiple of 5, 12, 13) to save time.

Answer: 1. Write 101 = 100 + 1 and use (a + b)³ = a³ + 3a²b + 3ab² + b³.
2. (101)³ = 100³ + 3(100)²(1) + 3(100)(1)² + 1³ = 1000000 + 30000 + 300 + 1.
3. (101)³ = 1030301
OR
1. Here a = 28, b = −15, c = −13 and a + b + c = 28 − 15 − 13 = 0.
2. If a + b + c = 0 then a³ + b³ + c³ = 3abc.
3. So the value = 3 × 28 × (−15) × (−13) = 3 × 5460 = 16380.

Teacher's Note:
1. Split the number into a round number plus a small number so that every term is easy to compute.
2. In the OR part, first check that the three numbers add up to zero; that is the whole trick.

Answer: (i) Abscissa is the x-coordinate and ordinate the y-coordinate, so the point is (−5, 4).
(ii) On the y-axis x = 0, so the point is (0, −7).
(iii) On the x-axis y = 0, so the point is (1, 0).
(iv) Distance 3 units from the y-axis on the positive side gives x = 3; abscissa = ordinate gives y = 3, so the point is (3, 3).
OR
1. (−3, a) lies on y = 4x, so a = 4 × (−3) = −12.
2. (b, 4) lies on y = 4x, so 4 = 4b and b = 1.

Teacher's Note:
1. Remember the order (abscissa, ordinate) = (x, y); a point on an axis has the other coordinate zero.
2. For the OR part, a point lies on a line exactly when its coordinates satisfy the equation.

Answer: 1. 3√45 = 3 × 3√5 = 9√5 and √125 = 5√5.
2. √200 = 10√2 and √50 = 5√2.
3. Expression = 9√5 − 5√5 + 10√2 − 5√2 = 4√5 + 5√2.
OR
1. Write the fractions with a common denominator 24: 1/3 = 8/24 and 1/2 = 12/24.
2. Rational numbers between 8/24 and 12/24 are 9/24, 10/24 and 11/24.
3. That is 3/8, 5/12 and 11/24 (any three valid numbers are accepted).

Teacher's Note:
1. Write each surd as (perfect square) × (prime part) and then combine only like surds.
2. For the OR part, make the denominators large enough so that three whole numbers fit between the numerators.

Answer: Given: ABCD is a parallelogram. The bisectors of ∠A and ∠B meet at P, those of ∠B and ∠C at Q, those of ∠C and ∠D at R, and those of ∠D and ∠A at S.
To prove: PQRS is a rectangle.
Proof:
1. AD ∥ BC, so ∠A + ∠B = 180° (consecutive interior angles).
2. In triangle APB, ∠PAB + ∠PBA = ½∠A + ½∠B = ½(180°) = 90°.
3. So ∠APB = 180° − 90° = 90°. Since ∠SPQ is vertically opposite to ∠APB, ∠SPQ = 90°.
4. In the same way, using ∠B + ∠C = 180°, ∠C + ∠D = 180° and ∠D + ∠A = 180°, we get ∠Q = ∠R = ∠S = 90°.
5. PQRS is a quadrilateral with all four angles equal to 90°.
Hence PQRS is a rectangle.

Teacher's Note:
1. The key fact is that consecutive angles of a parallelogram are supplementary, so half of each add to 90°.
2. Show all four angles are 90°, not just one, because one right angle alone does not make a rectangle.

Answer: Classes are 0−5, 5−10, … where the upper limit is excluded (a value 5 goes into 5−10).

Teacher's Note:
1. Tally each value once and then check that the frequencies add up to 40.
2. Follow the stated rule for borders: 5 belongs to 5 − 10, not to 0 − 5.

Answer: 1. Volume is conserved: 27 × \( \dfrac{4}{3}\pi r^3 \) = \( \dfrac{4}{3}\pi r'^3 \), so r′³ = 27r³.
2. (i) r′ = 3r.
3. S = 4πr² and S′ = 4π(3r)² = 36πr², so S : S′ = 4πr² : 36πr².
4. (ii) S : S′ = 1 : 9.
OR
1. Revolving about the side 12 cm gives a cone with height h = 12 cm and base radius r = 5 cm (the other perpendicular side).
2. V = \( \dfrac{1}{3}\pi r^2 h \) = \( \dfrac{1}{3}\pi \times 25 \times 12 \) = 100π cm³.
3. With π = 22/7, V ≈ 314.29 cm³ (that is, 100π cm³).

Teacher's Note:
1. Melting changes the shape but not the volume, so equate total volumes first.
2. For the OR part, the side you revolve about becomes the height and the other leg becomes the radius.

Answer: Radius r = 28/2 = 14 cm.
1. (i) V = \( \dfrac{1}{3}\pi r^2 h \), so 9856 = \( \dfrac{1}{3} \times \dfrac{22}{7} \times 14 \times 14 \times h \) = \( \dfrac{616}{3} h \).
Hence h = 9856 × 3 / 616 = 48 cm.
2. (ii) l = √(r² + h²) = √(196 + 2304) = √2500 = 50 cm.
3. (iii) Curved surface area = πrl = \( \dfrac{22}{7} \times 14 \times 50 \) = 2200 cm².
OR
1. Inner radius r = 10.5/2 = 5.25 cm.
2. Inner curved surface area = 2πr² = 2 × \( \dfrac{22}{7} \) × 5.25 × 5.25 = 173.25 cm².
3. Cost = 173.25 × \( \dfrac{16}{100} \) = ₹27.72.
4. Cost of tin-plating = ₹27.72.

Teacher's Note:
1. Convert the diameter to the radius before using any formula, and find the height before the slant height.
2. A hemispherical bowl has curved surface 2πr² only; do not add the circular rim, because plating is only on the inside curved part.

Answer: Given: ABCD is a quadrilateral whose diagonals AC and BD are equal and bisect each other at O, with AC ⊥ BD.
To prove: ABCD is a square.
Proof:
1. The diagonals bisect each other, so ABCD is a parallelogram.
2. In triangles AOB and AOD: OB = OD (O is the midpoint of BD), ∠AOB = ∠AOD = 90° and AO is common. So △AOB ≅ △AOD (SAS) and AB = AD.
3. In a parallelogram opposite sides are equal, so AB = CD and AD = BC. With AB = AD, all four sides are equal, so ABCD is a rhombus.
4. In triangles ABD and BAC: AB is common, AD = BC and BD = AC (given). So △ABD ≅ △BAC (SSS) and ∠DAB = ∠CBA.
5. In a parallelogram consecutive angles are supplementary: ∠DAB + ∠CBA = 180°. With ∠DAB = ∠CBA, each equals 90°.
6. A parallelogram with one right angle is a rectangle, and all its angles are 90°.
7. A figure with four equal sides and four right angles is a square.
Hence ABCD is a square.

Teacher's Note:
1. Build the proof in stages: parallelogram, then rhombus (perpendicular diagonals), then right angle (equal diagonals).
2. Give a reason in brackets for every congruence, such as SAS or SSS, to earn full marks.

Answer: 1. Write y = \( \dfrac{6 - 3x}{4} \) and make a table of values:

Teacher's Note:
1. Choose x values that give simple y values, and plot at least three points so that you can catch an arithmetic slip if they do not line up.
2. The x-intercept comes from putting y = 0 and the y-intercept from putting x = 0; write both as ordered pairs.