CBSE Class 10 Mathematics Real Numbers Important Questions Set 03

Answer: HCF × LCM = product of the two numbers = 100 × 190 = 19000.

Teacher's Note:
1. For two positive integers, HCF × LCM always equals their product.
2. So no factorisation is needed here.

Answer: HCF = (26 × 169) ÷ 338 = 4394 ÷ 338 = 13.

Teacher's Note:
1. Use HCF × LCM = product of the numbers.
2. Here 26 = 2 × 13 and 169 = 13², which also gives HCF 13.

Answer: 107 = 4 × 26 + 3, so q = 26.

Teacher's Note:
1. Divide 107 by 4 to get quotient 26 and remainder 3.
2. This is Euclid's division lemma a = bq + r with 0 ≤ r < b.

Answer: Using (a + b)(a − b) = a² − b², the product is 2 − 3 = −1, which is rational.

Teacher's Note:
1. Two irrational factors can still have a rational product.
2. Always simplify before deciding.

Answer: All three are primes, so HCF = 1 and LCM = 17 × 23 × 29 = 11339.

Teacher's Note:
1. Distinct primes have no common factor other than 1.
2. Their LCM is simply the product.

Answer: 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ has no factor 5. A number ending in 0 must be divisible by both 2 and 5, so no.

Teacher's Note:
1. Ending in zero needs the prime factors 2 and 5 together.
2. Check the prime factorisation for 5.

Answer: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

Teacher's Note:
1. This is the statement the board expects word for word.
2. Uniqueness is what lets us find HCF and LCM by prime factorisation.

Answer: 2658 = 2 × 1329 = 2 × 3 × 443. Since 443 is not divisible by any prime up to 19, it is prime. So 2658 = 2 × 3 × 443.

Teacher's Note:
1. Divide by the smallest primes in turn.
2. Test a prime only up to its square root, here about 21.

Answer: 5 × 3 × 11 + 11 = 11(15 + 1) = 11 × 16 = 176, and 5 × 7 + 7 × 3 = 7(5 + 3) = 7 × 8 = 56. Each has more than two factors, so both are composite.

Teacher's Note:
1. Take the common factor out and see that the other factor is greater than 1.
2. A number with a factor other than 1 and itself is composite.

Answer: (b) 150. LCM = 1800 ÷ 12 = 150.

Teacher's Note:
1. Use HCF × LCM = a × b.
2. Divide the product by the HCF.

Answer: (c) 2√3. A non-zero rational times an irrational is irrational, while √25 = 5, 3/7 and 0.25 are rational.

Teacher's Note:
1. Simplify each option first.
2. Perfect-square roots and fractions are rational.

Answer: (c) 2520. LCM = 2³ × 3² × 5 × 7 = 2520.

Teacher's Note:
1. Take the highest power of each prime up to 10.
2. Multiply 8, 9, 5 and 7.

Answer: 1288 = 575 × 2 + 138
575 = 138 × 4 + 23
138 = 23 × 6 + 0
The last non-zero divisor is 23, so HCF = 23.

Teacher's Note:
1. Keep dividing the previous divisor by the previous remainder.
2. Stop when the remainder is zero.

Answer: 6 = 2 × 3; 72 = 2³ × 3²; 120 = 2³ × 3 × 5.
HCF = 2 × 3 = 6; LCM = 2³ × 3² × 5 = 360.

Teacher's Note:
1. HCF takes the smallest power of each common prime.
2. LCM takes the greatest power of every prime that appears.

Answer: 26 = 2 × 13 and 91 = 7 × 13, so HCF = 13 and LCM = 2 × 7 × 13 = 182.
HCF × LCM = 13 × 182 = 2366 = 26 × 91. Verified.

Teacher's Note:
1. Factorise both numbers into primes first.
2. The verification works for any two positive integers.

Answer: 117 = 65 × 1 + 52; 65 = 52 × 1 + 13; 52 = 13 × 4 + 0, so HCF = 13.
13 = 65 − 52 = 65 − (117 − 65) = 2 × 65 − 1 × 117.
So m = 2, n = −1.

Teacher's Note:
1. Work backwards through the steps of the algorithm.
2. The values of m and n are not unique.

Answer: The required number divides 122 − 5 = 117, 150 − 7 = 143 and 115 − 11 = 104 exactly.
117 = 9 × 13, 143 = 11 × 13, 104 = 8 × 13, so the HCF is 13.

Teacher's Note:
1. Subtract each remainder from its number first.
2. The answer is the HCF of the results.

Answer: Maximum books per stack = HCF(336, 240, 96) = 48.
Stacks = 336 ÷ 48 + 240 ÷ 48 + 96 ÷ 48 = 7 + 5 + 2 = 14.

Teacher's Note:
1. Equal heights with the fewest stacks need the HCF.
2. Add the stacks needed for each subject.

Answer: HCF(96, 112) = 16 fruits per box.
Boxes = 96 ÷ 16 + 112 ÷ 16 = 6 + 7 = 13.

Teacher's Note:
1. The fewest boxes means the most fruit per box, which is the HCF.
2. The concept used is HCF.

Answer: z = 2 × 17 = 34; y = 2 × 34 = 68; x = 2 × 68 = 136.

Teacher's Note:
1. Work upwards from the bottom of the tree.
2. Each number is the product of the two below it.

Answer: 72 = 2³ × 3², 126 = 2 × 3² × 7, 168 = 2³ × 3 × 7.
HCF = 2 × 3 = 6; LCM = 2³ × 3² × 7 = 504.
HCF × LCM = 3024, but 72 × 126 × 168 = 1524096, so they are not equal.

Teacher's Note:
1. The product rule holds only for two numbers.
2. With three numbers the two sides differ.

Answer: Let the integer be a = 3q + r with r = 0, 1 or 2.
r = 0: a² = 9q² = 3(3q²). r = 1: a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1. r = 2: a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1.
Hence a² is 3m or 3m + 1.

Teacher's Note:
1. Divide the integer by 3 and consider each possible remainder.
2. Square it and group the multiples of 3.

Answer: Let a = 6q + r with r = 0, 1, 2, 3, 4, 5.
The values 6q, 6q + 2 and 6q + 4 are even, being divisible by 2. So an odd integer must be 6q + 1, 6q + 3 or 6q + 5.

Teacher's Note:
1. Write all six remainders when dividing by 6.
2. Remove the ones that give even numbers.

Answer: Write n = 3q + r, r = 0, 1, 2.
r = 0: n is divisible by 3; n + 2 = 3q + 2 and n + 4 = 3q + 4 are not.
r = 1: n + 2 = 3q + 3 is divisible; n = 3q + 1 and n + 4 = 3q + 5 are not.
r = 2: n + 4 = 3q + 6 is divisible; n = 3q + 2 and n + 2 = 3q + 4 are not.
So exactly one of n, n + 2, n + 4 is divisible by 3.

Teacher's Note:
1. Check each of the three possible remainders on division by 3.
2. In every case exactly one number is a multiple of 3.

Answer: Suppose √3 is rational, so √3 = p/q with p, q co-prime integers and q ≠ 0.
Then 3q² = p², so 3 divides p² and hence 3 divides p. Write p = 3m.
Then 3q² = 9m², so q² = 3m², and 3 divides q. Now 3 is a common factor of p and q, which contradicts that they are co-prime.
Hence √3 is irrational.

Teacher's Note:
1. The proof is by contradiction on the assumption that p and q are co-prime.
2. It uses the fact that if a prime divides p² then it divides p.

Answer: Suppose 5 + 3√2 = a/b is rational with b ≠ 0. Then √2 = (a − 5b)/(3b), which is rational.
But √2 is irrational, a contradiction.
Hence 5 + 3√2 is irrational.

Teacher's Note:
1. Isolate the known irrational number on one side.
2. Rational numbers are closed under addition, subtraction and division by a non-zero number.

Answer: Assume √5 = p/q with p, q co-prime and q ≠ 0. Then p² = 5q², so 5 divides p² and therefore 5 divides p. Let p = 5m.
Then 25m² = 5q², so q² = 5m² and 5 divides q. So p and q have the common factor 5, a contradiction.
Hence √5 is irrational.

Teacher's Note:
1. The steps are the same as for √2 and √3 with 5 in place of the prime.
2. Always state the contradiction clearly in the last line.

Answer: Any odd positive integer is 4q + 1 or 4q + 3 (Euclid's lemma with b = 4).
(4q + 1)² = 16q² + 8q + 1 = 8(2q² + q) + 1.
(4q + 3)² = 16q² + 24q + 9 = 8(2q² + 3q + 1) + 1.
So the square is of the form 8m + 1.

Teacher's Note:
1. Use the odd forms 4q + 1 and 4q + 3.
2. Take out 8 as a common factor from all terms but the last.

Answer: Let a = 3q + r, r = 0, 1, 2.
r = 0: a³ = 27q³ = 9(3q³).
r = 1: a³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1.
r = 2: a³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8.
Hence the cube is 9m, 9m + 1 or 9m + 8.

Teacher's Note:
1. Expand the cube of 3q + r for each remainder.
2. Every term except the constant is a multiple of 9.