CBSE Class 10 Mathematics Real Numbers MCQs Set 19

Answer: (b) 3 × 11 × 13

Explanation:
1. Divide 429 by 3: 429 = 3 × 143.
2. 143 = 11 × 13, so 429 = 3 × 11 × 13.

Teacher's Note:
1. Test divisibility by 2, 3, 5, 7, 11 in order.
2. The digit sum 4 + 2 + 9 = 15 shows 429 is divisible by 3.

Answer: (d) 91

Explanation:
1. HCF × LCM = product of the two numbers.
2. 13 × 182 = 26 × other number.
3. Other number = 2366 ÷ 26 = 91.

Teacher's Note:
1. Use HCF × LCM = a × b for two numbers only.
2. Divide step by step to avoid big multiplications.

Answer: (b) 4, 9696

Explanation:
1. 404 = 2² × 101 and 96 = 2⁵ × 3.
2. HCF = 2² = 4.
3. LCM = (404 × 96) ÷ 4 = 9696.

Teacher's Note:
1. HCF takes the lowest powers of common primes.
2. Get the LCM from the product formula once the HCF is known.

Answer: (c) 625

Explanation:
1. Subtract the remainders: 1250, 9375 and 15625.
2. 1250 = 2 × 5⁴, 9375 = 3 × 5⁵, 15625 = 5⁶.
3. The required number is their HCF = 5⁴ = 625.

Teacher's Note:
1. Remove the remainders first, then take the HCF.
2. The divisor must be larger than every remainder (625 is).

Answer: (d) 9:30 a.m.

Explanation:
1. Required interval = LCM of 3, 4, 8, 12, 15.
2. 8 = 2³, 12 = 2² × 3, 15 = 3 × 5, so LCM = 2³ × 3 × 5 = 120 minutes = 2 hours.
3. 7:30 a.m. + 2 hours = 9:30 a.m.

Teacher's Note:
1. Events repeating together again means LCM.
2. Convert minutes to hours before adding to the clock time.

Answer: (c) 12

Explanation:
1. The number per room must divide 60, 84 and 108 exactly, and we want the largest such number.
2. 60 = 2² × 3 × 5, 84 = 2² × 3 × 7, 108 = 2² × 3³.
3. HCF = 2² × 3 = 12 (this needs 5 + 7 + 9 = 21 rooms).

Teacher's Note:
1. The word maximum with equal grouping points to HCF.
2. Always read whether the question asks for the number per room or the number of rooms.

Answer: (c) 288

Explanation:
1. Minimum number divisible by both 32 and 36 is their LCM.
2. 32 = 2⁵ and 36 = 2² × 3².
3. LCM = 2⁵ × 3² = 288.

Teacher's Note:
1. Minimum common multiple means LCM.
2. Check: 288 ÷ 32 = 9 and 288 ÷ 36 = 8.

Answer: (b) a²b²

Explanation:
1. p = a × b² and q = a² × b.
2. LCM takes the highest power of each prime: a² and b².
3. LCM = a²b².

Teacher's Note:
1. For LCM, use the highest power of every prime present.
2. Here HCF would be ab.

Answer: (c) 12

Explanation:
1. 48 = 2⁴ × 3, 72 = 2³ × 3², 108 = 2² × 3³.
2. HCF = 2² × 3 = 12.

Teacher's Note:
1. HCF takes the smallest power of each common prime.
2. The HCF cannot exceed the smallest number (48).

Answer: (b) 432

Explanation:
1. 48 = 2⁴ × 3, 72 = 2³ × 3², 108 = 2² × 3³.
2. LCM = 2⁴ × 3³ = 16 × 27 = 432.

Teacher's Note:
1. LCM takes the highest power of each prime.
2. The LCM must be divisible by 108; 432 ÷ 108 = 4.

Answer: (a) 8:27:12 hrs

Explanation:
1. They change together after the LCM of 48, 72, 108 = 432 seconds.
2. 432 seconds = 7 minutes 12 seconds.
3. 8:20:00 + 7 min 12 s = 8:27:12 hrs.

Teacher's Note:
1. Convert the LCM in seconds to minutes and seconds.
2. Add the interval to the starting time.

Answer: (d) 5184

Explanation:
1. HCF = 12 and LCM = 432.
2. HCF × LCM = 12 × 432 = 5184.

Teacher's Note:
1. For three numbers HCF × LCM is not equal to the product of the numbers.
2. Here the product of the numbers is 373248, which is different.

Answer: (b) 2³ × 7 × 11

Explanation:
1. 616 = 2 × 308 = 2² × 154 = 2³ × 77.
2. 77 = 7 × 11, so 616 = 2³ × 7 × 11.

Teacher's Note:
1. Keep dividing by 2 first.
2. In prime factorisation every factor must itself be prime (14 is not).

Answer: (c) 2464

Explanation:
1. 32 = 2⁵ and 616 = 2³ × 7 × 11.
2. LCM = 2⁵ × 7 × 11 = 2464.

Teacher's Note:
1. LCM uses the highest power of 2, which is 2⁵.
2. Check that 2464 ÷ 616 = 4.

Answer: (b) 8

Explanation:
1. The number of columns must divide both 616 and 32.
2. HCF(616, 32) = 2³ = 8.

Teacher's Note:
1. The maximum common number of columns is the HCF.
2. 616 ÷ 8 = 77 and 32 ÷ 8 = 4 per column.

Answer: (d) 8

Explanation:
1. 1000 = 2³ × 5³ and 56 = 2³ × 7.
2. HCF = 2³ = 8.

Teacher's Note:
1. Only the prime 2 is common.
2. Take its smaller power, which is 3.

Answer: (a) 2³ × 5³

Explanation:
1. 1000 = 10³ = (2 × 5)³ = 2³ × 5³.
2. 8 × 125 uses non-prime factors, so the prime factorisation is 2³ × 5³.

Teacher's Note:
1. A prime factorisation uses only primes.
2. Options (b) and (c) are equal to 1000 but are not products of primes.

Answer: (b) 13915

Explanation:
1. 253 = 11 × 23, so z = 23.
2. 2783 = 253 × y gives y = 11.
3. x = 5 × 2783 = 13915.

Teacher's Note:
1. Work from the bottom of the tree upwards.
2. Multiply the two factors at each branch to get the number above.

Answer: (c) 11

Explanation:
1. 253 = 11 × 23, so z = 23.
2. 2783 ÷ 253 = 11, so y = 11.

Teacher's Note:
1. y is the missing factor of 2783.
2. Check: 5 × 11 × 11 × 23 = 13915.

Answer: (d) Four

Explanation:
1. The denominator is 2 × 5⁴, so the decimal terminates.
2. Number of places = max(1, 4) = 4.

Teacher's Note:
1. Take the larger exponent.
2. Check: multiply by 2³ to get a denominator of 10⁴.

Answer: (a) 2⁰ × 5¹, 0.6

Explanation:
1. The denominator is 5 = 2⁰ × 5¹.
2. 3 ÷ 5 = 0.6.

Teacher's Note:
1. Write the missing prime with exponent 0.
2. Check the decimal by direct division.

Answer: (c) 2⁴ × 5¹, 0.1625

Explanation:
1. 80 = 16 × 5 = 2⁴ × 5¹.
2. 13 ÷ 80 = 0.1625.

Teacher's Note:
1. Factorise 80 as 16 × 5.
2. Check: 0.1625 × 80 = 13.

Answer: (d) 2² × 5³, 0.254

Explanation:
1. 500 = 4 × 125 = 2² × 5³.
2. 127 ÷ 500 = 0.254.

Teacher's Note:
1. Factorise the denominator first.
2. Check: 0.254 × 500 = 127.

Answer: (c) \( \frac{635}{1998} \)

Explanation:
1. Let x = 0.3178178... Then 10x = 3.178178... and 10000x = 3178.178178...
2. Subtract: 9990x = 3175.
3. x = 3175 ÷ 9990 = 635/1998.

Teacher's Note:
1. Multiply by powers of 10 so that the repeating parts line up.
2. Always reduce the fraction to lowest terms (divide by 5).

Answer: (d) non-terminating repeating

Explanation:
1. 144 = 2⁴ × 3² has no factor of 7, so 7 stays in the denominator.
2. A denominator with a prime other than 2 and 5 gives a decimal that is non-terminating repeating.

Teacher's Note:
1. Cancel common factors first.
2. A rational number can never be non-terminating non-repeating.

Answer: (d) 2 and 5

Explanation:
1. 327.7081 = 3277081/10⁴.
2. Since 10⁴ = 2⁴ × 5⁴ the denominator q has prime factors 2 and 5.

Teacher's Note:
1. A terminating decimal always has q of the form 2ⁿ5ᵐ.
2. Cancelling can only remove factors, never add new ones.

Answer: (d) All of these

Explanation:
1. 26.0, 8.83 and 3.75 all terminate.
2. Terminating decimals are rational, so the answer is all of these.

Teacher's Note:
1. Every terminating decimal is rational.
2. Rational: 26 = 26/1, 8.83 = 883/100, 3.75 = 15/4.

Answer: (c) \( \frac{15}{4} \)

Explanation:
1. 3.75 = 375/100.
2. Dividing by 25 gives 15/4.

Teacher's Note:
1. Write over a power of 10 and reduce.
2. Check: 15 ÷ 4 = 3.75.

Answer: (d) 11.10203040...

Explanation:
1. Only 11.10203040... lies between 8.83 and 11.2 (the others are below 8.83 or above 11.2).
2. Its digits neither terminate nor repeat, so it is irrational.

Teacher's Note:
1. Check the position of each number first.
2. A non-terminating, non-repeating decimal is irrational.

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

Explanation:
1. A is a standard result and is true.
2. The negative of an irrational number is irrational, so R is false.

Teacher's Note:
1. If r is rational and r + s were rational, then s would be rational.
2. Negative of √2 is −√2, still irrational.