Real Number : Test 1 - Class 10 MCQ

We know that LCM = Product of the greatest power of each prime factor, involved in the numbers.
So, LCM(a, b) = p³q²

We know that HCF = Product of the smallest power of each common prime factor in the numbers. So, HCF(a, b) = pq

9 2n – 4 2n is of the form a 2n — b 2n.It is divisible by both a - b and a + b. So, 9 2n – 4 2n is divisible by both 9 - 4 = 5 and 9 + 4 = 13.

For any n ∈ N, e and 5n end with 6 and 5 respectively. Therefore, 6n – 5n always ends with 6 - 5 = 1.

If any number ends with the digit 0, it should be divisible by 10, i.e. it will be divisible by 2 and 5. Prime factorization of n is given as 23 × 34 × 54 × 7.It can be observed that there is (2 × 5) × (2 × 5) × (2 × 5) ⇒ 10 × 10 × 10 = 1000 Thus, there are 3 zeros in n.

We know that HCF = Product of the smallest power of each common prime factor in the numbers.

The prime factorization of

95 = 5 × 19

And 152 = 23 × 19

∴ HCF = 19

Apply

Apply

The prime factors of a + b

= 3 + 7 = 10 = 2 × 5

So the least prime factor is 2.

Since, the given number is non-terminating recurring,It is a rational number.