RS Aggarwal Test: Real Numbers - 1 - Class 10 MCQ

Assume that √7 be a rational number.
i.e. √7 = p/q, where p and q are co prime.
⇒ 7 = p²/q² 
⇒ 7q² = p²    ...(1)
⇒ p² is divisible by 7, i.e. p is divisible by 7
⇒ For any positive integer c, it can be said that p = 7c, p² = 49c²

Equation (1) can be written as: 7q² = 49c²
⇒ q² = 7c²
This gives that q is divisible by 7.

Since p and q have a common factor 7 which is a contradiction to the assumption that they are co-prime.
Therefore, √7 is an irrational number.

Greatest 5-Digit number = 99999

LCM of 8 and 9,

8 = 2 × 2 × 2

9 = 3 × 3

LCM = 2 × 2 × 2 × 3 × 3 = 72

Now, dividing 99999 by 72, we get

Quotient = 1388

Remainder = 63

So, the greatest 5-digit number divisible by 8 and 9 = 99999 - 63 = 99936

Required number = 99936 + 5 = 99941

Here, a = x³y² and b = xy³.
⇒ a = x _* x _* x _* y _* y and b = xy _* y _* y
∴ LCM(a, b) = x _* x _* x _* y _* y _* y = x³ _* y³ = x³y³
LCM = x³y³

As per question, we have,
p = ab² = a × b × b
q = a³b = a × a × a × b
So, their Least Common Multiple (LCM) = a³ × b²

- The product of a rational number and an irrational number is irrational.
- Exception: If the rational number is zero, the product is zero, which is rational.

L.C.M. of 3, 4, 5, 6, 8 = 2 × 2 × 2 × 3 × 5 = 120 
Pair of 2, 3 and 5 is not completed.
To make it a perfect square, the number should be multiplied by 2, 3, 5.

Required number = 120 x 2 x 3 x 5 = 3600.

Factors are following:
5 = 5 x 1
15 = 5 x 3
20 = 2 x 2 x 5

LCM = 5 x 3 x 2 x 2 = 60
HCF = 5
Ratio = LCM/HCF = 60/5 = 12/1 = 12:1

Therefore, the correct answer is C