Real Numbers MCQ Test — Class 10 CBSE

Solution:

Find the factors of each number:

Factors of 12: — 12 = 2 x 2 x 3

Factors of 18: — 18 = 2 x 3 x 3

Common factors: 2 and 3 — HCF = 2 x 3 = 6

Solution:

Prime factorisation:

4 = 2 x 2 and 6 = 2 x 3

Take the highest power of each prime: — LCM = 2^2 x 3 = 12

Solution:

3/7 is rational (ratio of two integers)

0.25 = 1/4 is rational

0.333... = 1/3 is rational (repeating decimal)

sqrt(5) cannot be written as p/q, so it is irrational

Solution:

In Euclid's division algorithm, we divide the larger number by the smaller:

455 / 42 gives quotient 10 and remainder 35 — 455 = 42 x 10 + 35

We then apply the algorithm to 42 and 35

Solution:

Divide step by step: — 56 / 2 = 28

Continue: — 28 / 2 = 14

Continue: — 14 / 2 = 7

7 is prime, so: — 56 = 2^3 x 7

Solution:

When HCF of two numbers is 1, they share no common factor other than 1.

Such numbers are called co-prime (or relatively prime) numbers.

Example: 8 and 15 are co-prime since HCF(8, 15) = 1

Solution:

Perform long division: — 7 / 8 = 0.875

Since 8 = 2^3, the denominator has only 2 as a prime factor.

Hence the decimal terminates.

Solution:

For a terminating decimal, the denominator must be of the form 2^m x 5^n

40 = 2^3 x 5 which is of the required form — 13/40 = 0.325

3 has factor 3, 12 = 2^2 x 3, 6 = 2 x 3 -- all have 3 as factor, so they don't terminate.

Solution:

We know the fundamental relation: — HCF x LCM = Product of two numbers

Substitute: — 6 x LCM = 1080

Solve: — LCM = 1080 / 6 = 180

Solution:

The Fundamental Theorem of Arithmetic states:

Every composite number can be expressed (factorised) as a product of primes.

This factorisation is unique, apart from the order in which the prime factors occur.