CBSE Class 10 Mathematics Basic Maths Objective Questions Assignment Set 18

Answer: (d) 50. HCF × LCM = product of the numbers, so 11 × 253R = 253 × 550 gives R = 50.

Teacher's Note:
1. Use the relation HCF × LCM = product of the two numbers.
2. Cancel the common factor 253 first to keep the arithmetic short.

Answer: (d) √3 + √5. The first three simplify to 7, 7 and 5, which are rational, while √3 + √5 cannot be written as a fraction.

Teacher's Note:
1. Simplify every perfect square root before deciding.
2. A sum of two different non-square surds is irrational.

Answer: (d) a = 1, b = −6. Sum of zeroes = −a = −1 so a = 1, and product = b = −6.

Teacher's Note:
1. Sum of zeroes is −(coefficient of x) and product is the constant term.
2. Check the sign of a carefully.

Answer: (b) −1. α + β = 5 and αβ = 6, so 5 − 6 = −1.

Teacher's Note:
1. Use the sum and product relations directly.
2. There is no need to find the zeroes.

Answer: (d) 18. 2cot²θ + 2 = 2cosec²θ = 2 × 3² = 18.

Teacher's Note:
1. Take 2 common and use 1 + cot²θ = cosec²θ.
2. cosec θ is the reciprocal of sin θ.

Answer: (c) 3 cm. AC = 17 cm, and r = (AB + BC − AC)/2 = (8 + 15 − 17)/2 = 3 cm.

Teacher's Note:
1. Find the hypotenuse using the Pythagoras theorem first.
2. For a right triangle, r = (sum of the legs − hypotenuse)/2.

Answer: (a) 1 : 4. The sides of triangle DEF are half the sides of ABC and the triangles are similar, so the area ratio is (1/2)² = 1 : 4.

Teacher's Note:
1. The mid-point theorem makes each side of DEF half of the parallel side of ABC.
2. Area ratio is the square of the side ratio.

Answer: (c) 20 cm. DE/AB = 2 is the scale factor, so EF = 2 × BC = 20 cm.

Teacher's Note:
1. Corresponding sides of similar triangles are in the same ratio.
2. Match EF with BC.

Answer: (c) −63. The area of the triangle is zero: x(−4 + 5) + (−3)(−5 − 2) + 7(2 + 4) = x + 21 + 42 = 0, so x = −63.

Teacher's Note:
1. Three points are collinear when the area of the triangle formed is zero.
2. Expand the determinant carefully with signs.

Answer: (d) 2 : 3. Let the ratio be k : 1, then (7k + 2)/(k + 1) = 4, which gives k = 2/3, i.e. 2 : 3.

Teacher's Note:
1. Apply the section formula to the x-coordinate alone.
2. Convert k = 2/3 into a ratio of whole numbers.

Answer: 1. A certain event always happens.

Teacher's Note:
1. Probability lies between 0 and 1.
2. A sure event is at the upper end of the scale.

Answer: \( \dfrac{1}{6} \). The only even prime is 2, so there is 1 favourable outcome out of 6.

Teacher's Note:
1. Remember that 2 is the only even prime number.
2. Probability = favourable outcomes / total outcomes.

Answer: 216. Each die has 6 outcomes, so 6 × 6 × 6 = 216.

Teacher's Note:
1. Multiply the outcomes of the independent dice.
2. 6³ = 216.

Answer: −2a. The sum of the roots is −(2a), so a + b = −2a.

Teacher's Note:
1. Sum of roots is −(coefficient of x) divided by the coefficient of x².
2. The sum here is expressed in terms of a.

Answer: 1005. a₂₀₀ = 10 + 199 × 5 = 1005.

Teacher's Note:
1. Use aₙ = a + (n − 1)d.
2. Here a = 10 and d = 5.

Answer: There are 6 red face cards (3 hearts and 3 diamonds), so P = 6/52 = \( \dfrac{3}{26} \).

Teacher's Note:
1. Face cards are the jack, queen and king of each suit.
2. Only hearts and diamonds are red.

Answer: D = b² − 4ac = 64 − 4(3)(−5) = 64 + 60 = 124.

Teacher's Note:
1. Identify a = 3, b = 8, c = −5.
2. Subtracting a negative product adds.

Answer: Put x = 1: 3 + 2k − 2 = 0, so 2k = −1 and k = −1/2.

Teacher's Note:
1. A zero satisfies the polynomial, so substitute it.
2. Solve the resulting linear equation.

Answer: First term a = 6; common difference d = 2 − 6 = −4.

Teacher's Note:
1. Subtract any term from the next term to get d.
2. d is negative when terms decrease.

Answer: 3 + (n − 1)5 = 98 gives n − 1 = 19, so n = 20; 98 is the 20th term.

Teacher's Note:
1. Set aₙ equal to 98 and solve for n.
2. n must be a positive whole number.