Class 10 Maths Chapter 1 Practice Question Answers - Real Numbers

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(i) 140
Using the division of a number by prime numbers method, we can get the product of prime factors of 140.
Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2² × 5 × 7

(ii) 156
Using the division of a number by prime numbers method, we can get the product of prime factors of 156.
Hence, 156 = 2 × 2 × 13 × 3 = 2² × 13 × 3

(iii) 3825
Using the division of a number by prime numbers method, we can get the product of prime factors of 3825.
Hence, 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17

(iv) 5005
Using the division of a number by prime numbers method, we can get the product of prime factors of 5005.
Hence, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13

(v) 7429
Using the division of a number by prime numbers method, we can get the product of prime factors of 7429.
Hence, 7429 = 17 × 19 × 23 = 17 × 19 × 23

Ans: As we know that,
HCF × LCM = Product of the two given numbers
So,
9 × LCM = 306 × 657
LCM = (306 × 657)/9 = 22338
Therefore, LCM(306,657) = 22338

Ans: Let 3 + 2√5 be a rational number.
Then the co-primes x and y of the given rational number where (y ≠ 0) is such that:
3 + 2√5 = x/y
Rearranging, we get,
2√5 = (x/y) – 3
√5 = 1/2[(x/y) – 3]
Since x and y are integers, thus, 1/2[(x/y) – 3] is a rational number.
Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational.
Thus, our assumption that 3 + 2√5 is a rational number is wrong.
Hence, 3 + 2√5 is irrational.

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Ans: If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5.
Prime factorization of 6ⁿ = (2 × 3)ⁿ
Therefore, the prime factorization of 6n doesn’t contain the prime number 5.
Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.

Ans:The smallest prime number = 2
The smallest composite number = 4
Prime factorisation of 2 = 2
Prime factorisation of 4 = 2 × 2
HCF(2, 4) = 2
Therefore, the HCF of the smallest prime number and the smallest composite number is 2.

Ans: To find the maximum possible length of each smaller piece, we need to calculate the HCF of $72$72, $96$96, and 120.

Ans: Prime factorisation of 404 = 2 × 2 × 101
Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3
HCF = 2 × 2 = 4
LCM = 2⁵ × 3 × 101 = 9696
HCF × LCM = 4 × 9696 = 38784
Product of the given two numbers = 404 × 96 = 38784
Hence, verified that LCM × HCF = Product of the given two numbers.