RD Sharma Test: Real Numbers - 1 - Class 10 MCQ

If possible let a√b be rational. Then a√b = p/q, 
where p and q are non-zero integers, having no common factor other than 1.
Now, a√b = p/q ⇒ √b = p/aq……….(i)
But, p and aq are both rational and aq ≠ 0. 
∵ p/aq is rational.
Therefore, from eq. (i), it follows that √b is rational.
The contradiction arises by assuming that a√b is rational.
Hence, a√b is irrational.

Using the result, HCF × LCM = Product of two natural numbers
 ⇒ LCM (a, b) = 1800/12 = 150

The quotient of a rational number and a nonzero irrational number is irrational.

Here’s why this statement holds true:

• A rational number can be expressed as a fraction, like a/b, where a and b are integers and b is not zero.
• An irrational number cannot be written as a simple fraction. Examples include √2 and π.
• Dividing a rational number by a nonzero irrational number results in a value that cannot be simplified to a fraction of integers.
• This is due to the irrational number adding a non-repeating, non-terminating decimal part, making the result irrational.

For instance, dividing a rational number like 1 by an irrational number like √2 gives 1/√2, which is irrational.

Therefore, the statement is confirmed: the quotient of a rational number and a nonzero irrational number is indeed irrational.

• The decimal expansion of a fraction is terminating if the denominator is only powers of 2 and 5.
• 2²×5³=500, which consists of only 2’s and 5’s.
• Hence, 315/500=0.63 is a terminating decimal.

Since √3 and √5 both are irrational number therefore (√3+√5)² is an irrational number.

• Formula: HCF×LCM=Product of the numbers
• Given, HCF = 18, LCM = 540.
• Product = 18×540=9720

• A number ends in 0 if it has at least one factor of 5.
• 6ⁿ=(2×3)ⁿ only contains 2 and 3, never 5.
• Hence, 6ⁿ can never end with 0.

• Prime factorization: 864 = 2⁵ × 3³.
• For a number to be a perfect square, the exponent of every prime factor must be even.
• In 864, both the exponent 5 for 2 and the exponent 3 for 3 are odd.
• To make these exponents even, you need to remove one factor of 2 and one factor of 3.
• Multiplying these factors, 2 × 3, gives 6, which is the smallest number by which 864 should be divided to yield a perfect square.
• On dividing 864 by 6, we get 144, which is a perfect square

The LCM of 2³×3² and 2²×3³ is calculated as follows:

The LCM is 2³×3³.

• Identify the prime factors of each number:
  • First number: 2³×3²
  • Second number: 2²×3³
• For the LCM, take the highest power of each prime factor:
  • For 2: the highest power is 2³
  • For 3: the highest power is 3³
• Combine these to find the LCM:

• The HCF is obtained by taking minimum powers of common factors.
• Common factor: 2² 

• If r is rational and i is irrational, then r+i is always irrational.
• Example: 2 + √3 is irrational.

H.C.F.(p,q,r)× L.C.M.(p,q,r) ≠ p×q×r. This condition is only applied on HCF and LCM of two numbers.

• A number has a rational square root only if it is a perfect square.
• 225 = 15², so √225 ​=15 is rational.

Since √2 and √5 are both irrational numbers, therefore  is an irrational number.

• Prime factorization: 392=2³×7²
• For a perfect cube, all exponents must be multiples of 3.
• Multiply by 7 to make exponent of 7 = 3.

• Formula: HCF × LCM
• HCF × 70 = 14 × 35
• HCF = (14 × 35)/70 = 7

• A rational number has a terminating decimal if its denominator has only 2 or 5 as prime factors.
• 125/500 simplifies to 1/4, which has denominator = 2²

The least positive integer divisible by 20 and 24 is LCM (20, 24). 

20 = 2² × 5 
24 = 2³ × 3

∴ LCM (20, 24) = 2³ x 3 x 5 = 120

• Any two consecutive odd numbers are coprime.
• HCF of coprime numbers is always 1.

• Between any two real numbers, there are infinitely many irrational numbers.
• Example: 3.142857...,3.141592...,3.73205...3.142857..., 3.141592..., 3.73205...3.142857...,3.141592...,3.73205... (approximations of irrationals).

 = (a−b) 
Since a and b both are positive rational numbers, therefore the difference is also rational.

• Formula: HCF × LCM = a × b
• 4 × 120 = a × b ► a × b = 480
• Among given options, 12 × 40 = 480 satisfies this condition
• Although (16, 30) gives a product of 480, its HCF is 2, not 4, so it does not satisfy the given conditions.

To find the remainder when 5¹⁰⁰ is divided by 4, follow these steps:

• Calculate 5¹⁰⁰ mod 4.
• Since 5 is congruent to 1 modulo 4 (because 5 - 4 = 1), we can simplify:
• 5¹⁰⁰ mod 4 = (1¹⁰⁰) mod 4.
• This results in 1 since any number raised to a power remains the same when taken modulo 4.

Thus, the remainder when 5¹⁰⁰ is divided by 4 is 1.

• Step 1: Analyze each condition
  x ≡ 1 (mod 2) means when x is divided by 2, remainder is 1 → x = 2k + 1

  x ≡ 2 (mod 3) means x = 3m + 2

  x ≡ 3 (mod 4) means x = 4n + 3

  Step 2: Check candidates for x from the largest modulus condition
  The numbers that satisfy x ≡ 3 (mod 4) are:

  3, 7, 11, 15, 19, 23, 27, 31, 35, …

  Check these numbers one by one against the other two conditions:

  x = 3:
  3 mod 2 = 1 
  3 mod 3 = 0 ✘ (needs remainder 2)

  x = 7:
  7 mod 2 = 1 
  7 mod 3 = 1 ✘ (needs remainder 2)

  x = 11:
  11 mod 2 = 1 
  11 mod 3 = 2 

  So, x = 11 satisfies all three conditions.

  Answer:
  The smallest positive integer is 11