CBSE Class 10 Mathematics MCQs Set 02

Answer: (d) \( \pm 4 \)

Explanation:
1. 1. Sum of zeros = \( -\dfrac{b}{a} = \dfrac{2}{k^2-14} \).
2. 2. Given sum = 1, so \( k^2 - 14 = 2 \), that is \( k^2 = 16 \).
3. 3. Hence \( k = \pm 4 \).

Teacher's Note:
1. Use sum of zeros = -b/a with the correct signs.
2. Remember that k^2 = 16 gives two values of k, positive and negative.

Answer: (d) \( \dfrac{5\pi}{7} \)

Explanation:
1. 1. Let \( \angle A = x \). Since AQ = PQ, \( \angle QPA = x \), so the exterior angle \( \angle PQB = 2x \).
2. 2. Since PQ = BP, \( \angle PBQ = 2x \). Then the exterior angle \( \angle BPC = x + 2x = 3x \).
3. 3. Since BC = BP, \( \angle C = \angle BPC = 3x \), and AB = AC gives \( \angle B = 3x \).
4. 4. Angle sum: \( x + 3x + 3x = \pi \), so \( x = \dfrac{\pi}{7} \).
5. 5. In \( \triangle AQP \), \( \angle AQP = \pi - 2x = \dfrac{5\pi}{7} \).

Teacher's Note:
1. Mark each equal pair of sides and write the base angles one by one.
2. Exterior angle of a triangle equals the sum of the two opposite interior angles.

Answer: (d) becomes four times

Explanation:
1. 1. Circumference \( 2\pi r \): \( 2\pi r = 4\pi \) gives \( r = 2 \); \( 2\pi r = 8\pi \) gives \( r = 4 \).
2. 2. Area changes from \( \pi(2)^2 = 4\pi \) to \( \pi(4)^2 = 16\pi \).
3. 3. The ratio is 16 : 4, so the area becomes four times.

Teacher's Note:
1. Area depends on the square of the radius.
2. When the radius doubles, the area becomes four times.

Answer: (c) 25 cm

Explanation:
1. 1. Perimeter = OA + OB + arc AB.
2. 2. Arc AB = \( \dfrac{1}{4} \times 2\pi r = \dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7 = 11 \) cm.
3. 3. Perimeter = 7 + 7 + 11 = 25 cm.

Teacher's Note:
1. Do not forget the two straight radii in the perimeter of a sector.
2. The arc of a quadrant is one-fourth of the circumference.

Answer: (a) 5

Explanation:
1. 1. Let \( AM/AC = t \). BNMP is a parallelogram.
2. 2. \( \triangle ANM \sim \triangle ABC \) with area ratio \( t^2 \), and \( \triangle MPC \sim \triangle ABC \) with area ratio \( (1-t)^2 \).
3. 3. Area of BNMP = \( 1 - t^2 - (1-t)^2 = 2t(1-t) = \dfrac{5}{18} \), so \( t^2 - t + \dfrac{5}{36} = 0 \) and \( t = \dfrac{5}{6} \) or \( \dfrac{1}{6} \).
4. 4. M is closer to C, so \( t = \dfrac{5}{6} \) and \( \dfrac{AM}{MC} = \dfrac{5/6}{1/6} = 5 \).

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides.
2. Subtract the two corner triangles from the whole triangle to get the parallelogram.

Answer: (c) 90°

Explanation:
1. 1. Reflection in AB gives \( \angle QAB = \angle PAB \); reflection in AC gives \( \angle RAC = \angle PAC \).
2. 2. So \( \angle QAR = \angle QAB + \angle PAB + \angle PAC + \angle RAC = 2(\angle PAB + \angle PAC) = 2\angle A \).
3. 3. Q, A, R are collinear, so \( \angle QAR = 180^\circ \) and \( \angle A = 90^\circ \).

Teacher's Note:
1. A reflection keeps the angle between the mirror and the line unchanged.
2. A straight angle is 180 degrees.

Answer: (b) \( x^2 - \dfrac{a}{b}x - \dfrac{1}{b} \)

Explanation:
1. 1. \( \alpha + \beta = -a \) and \( \alpha\beta = -b \).
2. 2. Sum of new zeros = \( \dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{-a}{-b} = \dfrac{a}{b} \).
3. 3. Product of new zeros = \( \dfrac{1}{\alpha\beta} = -\dfrac{1}{b} \).
4. 4. Polynomial: \( x^2 - \dfrac{a}{b}x - \dfrac{1}{b} \).

Teacher's Note:
1. Use sum and product of zeros instead of finding the zeros.
2. Write the polynomial as x^2 - (sum)x + (product).

Answer: (a) \( \text{Area}(\triangle ABD) = \tfrac{1}{2}\text{Area}(\triangle CAE) \)

Explanation:
1. 1. Let AB = BC = a. Then \( AC = a\sqrt{2} \).
2. 2. Equilateral triangles are similar, so the ratio of their areas is the square of the ratio of sides: \( \dfrac{\text{Area}(ABD)}{\text{Area}(CAE)} = \dfrac{a^2}{(a\sqrt{2})^2} = \dfrac{1}{2} \).
3. 3. So \( \text{Area}(\triangle ABD) = \tfrac{1}{2}\text{Area}(\triangle CAE) \).

Teacher's Note:
1. Any two equilateral triangles are similar.
2. Find AC from the Pythagoras theorem first.

Answer: (a) 4

Explanation:
1. 1. Let the numbers be 33a and 33b with a and b co-prime.
2. 2. \( 33(a+b) = 528 \), so \( a + b = 16 \).
3. 3. Co-prime pairs with sum 16: (1, 15), (3, 13), (5, 11), (7, 9). (2, 14), (4, 12), (6, 10), (8, 8) are not co-prime.
4. 4. So there are 4 pairs of numbers.

Teacher's Note:
1. Write the numbers as HCF times co-prime factors.
2. Count each unordered pair once.

Answer: (a) 3

Explanation:
1. 1. Sum of zeros = \( -\dfrac{b}{a} = \dfrac{12}{4} = 3 \).

Teacher's Note:
1. Compare with ax^2 + bx + c.
2. Sum of zeros is -b/a.

Answer: (d) \( \dfrac{c}{a} \)

Explanation:
1. 1. Product of zeros = \( \dfrac{\text{constant term}}{\text{coefficient of } x^2} = \dfrac{c}{a} \).

Teacher's Note:
1. Memorise: sum = -b/a, product = c/a.

Answer: (c) \( \beta = 4\alpha \)

Explanation:
1. 1. One zero is four times the other, so taking \( \alpha \) as the smaller zero, \( \beta = 4\alpha \).

Teacher's Note:
1. Translate the sentence into an equation.

Answer: (d) \( \dfrac{3}{5} \)

Explanation:
1. 1. \( \alpha + \beta = 3 \) and \( \beta = 4\alpha \), so \( 5\alpha = 3 \).
2. 2. \( \alpha = \dfrac{3}{5} \).

Teacher's Note:
1. Substitute the relation into the sum of zeros.

Answer: (b) \( -\dfrac{56}{75} \)

Explanation:
1. 1. \( \alpha = \dfrac{3}{5} \), \( \beta = \dfrac{12}{5} \).
2. 2. Product of zeros = \( \alpha\beta = \dfrac{36}{25} = \dfrac{3k+8}{4} \).
3. 3. \( 3k + 8 = \dfrac{144}{25} \), so \( 3k = -\dfrac{56}{25} \) and \( k = -\dfrac{56}{75} \).

Teacher's Note:
1. Use the product of zeros after finding both zeros.

Answer: (b) (3, 4)

Explanation:
1. 1. P is in column 3 and row 4, so P = (3, 4).

Teacher's Note:
1. Read the x-coordinate (column) first, then the y-coordinate (row).

Answer: (a) \( 3\sqrt{2} \) units

Explanation:
1. 1. \( PQ = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{9+9} = 3\sqrt{2} \).

Teacher's Note:
1. Use the distance formula.

Answer: (c) 6 units

Explanation:
1. 1. P and R have the same y-coordinate, so \( PR = 9 - 3 = 6 \) units.

Teacher's Note:
1. For points on a horizontal line, subtract the x-coordinates.

Answer: (a) square

Explanation:
1. 1. \( PQ = QR = RS = SP = 3\sqrt{2} \) (each uses a difference of 3 in x and in y).
2. 2. Diagonals \( PR = 6 \) and \( QS = 7 - 1 = 6 \) are equal.
3. 3. All sides equal and diagonals equal, so PQRS is a square.

Teacher's Note:
1. A quadrilateral with equal sides and equal diagonals is a square.
2. Compare the diagonals to separate a square from a rhombus.

Answer: (d) (6, 4)

Explanation:
1. 1. Mid-point = \( \left(\dfrac{6+6}{2}, \dfrac{7+1}{2}\right) = (6, 4) \).

Teacher's Note:
1. Use the mid-point formula.

Answer: (b) (0, 1)

Explanation:
1. 1. AB is vertical and BC is horizontal, so \( \angle B = 90^\circ \).
2. 2. The hypotenuse AC is a diameter of the circle, so the centre is the mid-point of AC.
3. 3. Centre = \( \left(\dfrac{-2+2}{2}, \dfrac{5-3}{2}\right) = (0, 1) \).

Teacher's Note:
1. Angle in a semicircle is a right angle, so the hypotenuse of a right triangle is a diameter.
2. Check the centre is equidistant from all three vertices.

Answer: (d) \( \tan^2 A \)

Explanation:
1. 1. \( 1 + \tan^2 A = \sec^2 A \) and \( 1 + \cot^2 A = \operatorname{cosec}^2 A \).
2. 2. \( \dfrac{\sec^2 A}{\operatorname{cosec}^2 A} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A \).

Teacher's Note:
1. Convert numerator and denominator with the two identities.
2. Simplify sin^2 over cos^2 as tan^2.

Answer: (a) \( \dfrac{9}{25} \)

Explanation:
1. 1. Total outcomes = 25.
2. 2. Primes from 1 to 25: 2, 3, 5, 7, 11, 13, 17, 19, 23, that is 9.
3. 3. Probability = \( \dfrac{9}{25} \).

Teacher's Note:
1. 1 is neither prime nor composite.
2. List the primes in order so none is missed.

Answer: (b) \( \dfrac{8}{25} \)

Explanation:
1. 1. Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, that is 8.
2. 2. Probability = \( \dfrac{8}{25} \).

Teacher's Note:
1. Count with 25 divided by 3, ignoring the remainder.

Answer: (c) \( \dfrac{12}{25} \)

Explanation:
1. 1. Multiples of 2: 2, 4, ..., 24, that is 12.
2. 2. Probability = \( \dfrac{12}{25} \).

Teacher's Note:
1. The number of multiples of 2 up to 25 is the integer part of 25/2.

Answer: (a) \( \dfrac{16}{25} \)

Explanation:
1. 1. Multiples of 2: 12; multiples of 3: 8; multiples of both (that is, of 6): 6, 12, 18, 24, so 4.
2. 2. Multiples of 2 or 3 = 12 + 8 - 4 = 16.
3. 3. Probability = \( \dfrac{16}{25} \).

Teacher's Note:
1. Numbers divisible by both are counted twice, so subtract them once.

Answer: (d) \( \dfrac{312}{25} \)

Explanation:
1. 1. Divide numerator and denominator by \( \cos^2\theta \): the expression becomes \( \dfrac{2\tan\theta}{1 - \tan^2\theta} \).
2. 2. With \( \tan\theta = \dfrac{12}{13} \): \( \dfrac{24/13}{1 - 144/169} = \dfrac{24/13}{25/169} = \dfrac{24 \times 13}{25} = \dfrac{312}{25} \).

Teacher's Note:
1. Dividing by cos^2 turns every term into tan.
2. Keep the fractions together to avoid arithmetic slips.

Answer: (a) \( \sqrt{2q+1} \) cm

Explanation:
1. 1. Third side \( = \sqrt{p^2 - q^2} = \sqrt{(p-q)(p+q)} \).
2. 2. \( p - q = 1 \) and \( p + q = 2q + 1 \), so third side \( = \sqrt{2q+1} \) cm.

Teacher's Note:
1. Factorise the difference of squares.
2. Use p = q + 1 to write p + q in terms of q.

Answer: (a) \( 8PT^2 = 3PR^2 + 5PS^2 \)

Explanation:
1. 1. Let \( PQ = y \) and \( QS = ST = TR = x \).
2. 2. By Pythagoras: \( PS^2 = y^2 + x^2 \), \( PT^2 = y^2 + 4x^2 \), \( PR^2 = y^2 + 9x^2 \).
3. 3. \( 3PR^2 + 5PS^2 = 3y^2 + 27x^2 + 5y^2 + 5x^2 = 8y^2 + 32x^2 = 8PT^2 \).

Teacher's Note:
1. Apply the Pythagoras theorem in each right triangle with Q as the right angle.
2. Match the coefficients of y^2 and x^2.

Answer: (a) 8

Explanation:
1. 1. The product is \( (7^{95})^2 - (3^{58})^2 = 7^{190} - 3^{116} \).
2. 2. Unit digits of powers of 7 repeat as 7, 9, 3, 1; \( 190 = 4 \times 47 + 2 \), so \( 7^{190} \) ends in 9.
3. 3. Unit digits of powers of 3 repeat as 3, 9, 7, 1; \( 116 = 4 \times 29 \), so \( 3^{116} \) ends in 1.
4. 4. The difference ends in \( 9 - 1 = 8 \).

Teacher's Note:
1. Use the identity a^2 - b^2 to reduce the work.
2. Powers of 3 and 7 have unit digits that repeat in a cycle of 4.

Answer: (b) 20 minutes

Explanation:
1. 1. \( \dfrac{2\pi}{3} \) radians \( = \dfrac{2 \times 180^\circ}{3} = 120^\circ \).
2. 2. The minute hand turns \( 360^\circ \) in 60 minutes, that is \( 6^\circ \) per minute.
3. 3. Time = \( \dfrac{120}{6} = 20 \) minutes.

Teacher's Note:
1. Convert radians to degrees with pi = 180 degrees.
2. The minute hand moves 6 degrees every minute.

Answer: (c) 3 or -9

Explanation:
1. 1. Let the other end be \( (10, y) \). Then \( (10-2)^2 + (y+3)^2 = 10^2 \).
2. 2. \( (y+3)^2 = 36 \), so \( y + 3 = \pm 6 \).
3. 3. \( y = 3 \) or \( y = -9 \).

Teacher's Note:
1. Both the positive and negative square roots give valid points.

Answer: (b) (7, -14)

Explanation:
1. 1. Slope of the line through the two points = \( \dfrac{-5-4}{2+3} = -\dfrac{9}{5} \).
2. 2. For (7, -14): slope from (2, -5) is \( \dfrac{-14+5}{7-2} = -\dfrac{9}{5} \), the same, so it is collinear.
3. 3. For (0, 0): slope from (2, -5) is \( \dfrac{5}{-2} \); for (0, -1): \( \dfrac{4}{-2} \); for (3, 1): \( \dfrac{6}{1} \). None equals \( -\dfrac{9}{5} \).

Teacher's Note:
1. Points are collinear when all slopes between them are equal.
2. Test each option against the slope of the given line.

Answer: (b) 48 km/h

Explanation:
1. 1. Let the speed of the super fast train be \( x \) km/h. The slower train runs at \( (x-16) \) km/h.
2. 2. \( \dfrac{192}{x-16} - \dfrac{192}{x} = 2 \), so \( 192 \times 16 = 2x(x-16) \).
3. 3. \( x^2 - 16x - 1536 = 0 \), that is \( (x - 48)(x + 32) = 0 \).
4. 4. Speed cannot be negative, so \( x = 48 \) km/h.

Teacher's Note:
1. Time taken = distance divided by speed.
2. Reject the negative root of the quadratic equation.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. 1. The two tangents from an external point are equal, and the two triangles formed with the centre are congruent (SSS), so they subtend equal angles at the centre. A is true.
2. 2. A parallelogram circumscribing a circle has AB + CD = BC + DA, which gives AB = BC, so it is a rhombus. R is true.
3. 3. R does not explain A, so the answer is (b).

Teacher's Note:
1. Tangent segments from an external point are equal.
2. Check whether the reason actually uses the idea in the assertion.

Answer: (c) A is true, but R is false.

Explanation:
1. 1. Opposite angles of a cyclic quadrilateral add up to 180°, so the opposite angle is 180° - 40° = 140°. A is true.
2. 2. R says the sum is 360°, which is false (it is 180°).
3. 3. So A is true but R is false.

Teacher's Note:
1. Opposite angles of a cyclic quadrilateral are supplementary.
2. 360 degrees is the sum of all four angles, not of a pair of opposite angles.

Answer: (d) A is false, but R is true.

Explanation:
1. 1. OA and OB are perpendicular to the tangents, so \( \angle APB + \angle AOB = 180^\circ \).
2. 2. \( \angle APB = 180^\circ - 110^\circ = 70^\circ \), not 90°. A is false.
3. 3. R is a standard true statement.

Teacher's Note:
1. In quadrilateral OAPB the angles at A and B are right angles.
2. The angle between two tangents is supplementary to the angle at the centre.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 1. Shadow length \( = \dfrac{h}{\tan\theta} \), so as \( \theta \) decreases, \( \tan\theta \) (the slope of the sun's rays) decreases and the shadow increases. A is true.
2. 2. R states exactly this cause, so R is true and explains A.

Teacher's Note:
1. Shadow length = height / tan(angle of elevation).
2. A smaller angle of elevation means a longer shadow.

Answer: (d) A is false, but R is true.

Explanation:
1. 1. Moving closer to the object makes the line of sight steeper, so the angle of elevation increases. A is false.
2. 2. The object subtends a larger angle at the eye when we are closer. R is true.

Teacher's Note:
1. Draw two positions of the observer to see the change.
2. A closer observer looks up more steeply.

Answer: (a) a sphere and a cylinder

Explanation:
1. 1. A surahi has a round (spherical) body with a long narrow neck, which is a cylinder.
2. 2. So it is the combination of a sphere and a cylinder.

Teacher's Note:
1. Picture the shape of the object and split it into standard solids.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. 1. Volume of the cuboid = \( 10 \times 5.5 \times 3.5 = 192.5 \) cm³.
2. 2. Volume of one coin = \( \dfrac{22}{7} \times (0.875)^2 \times 0.2 = 0.48125 \) cm³ (r = 0.875 cm, h = 0.2 cm).
3. 3. Number of coins = \( \dfrac{192.5}{0.48125} = 400 \). A is true.
4. 4. R gives the two volume formulas used above, so it explains A.

Teacher's Note:
1. Convert thickness 2 mm to 0.2 cm before using the formula.
2. Number of coins = volume of cuboid divided by volume of one coin.