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

Let the number be x 
divident = divisor×quotient + remainder
x = 61×27 + 32
= 1679
so, the number will be 1679

a = pq²

b = p³q

LCM (a,b) = p³q²

Using the result, HCF × LCM = Product of two natural numbers ⇒ the other number =  product of two no. = LCM *HCF
Let unknown no. = X
>> X * 27 = 9*459
X = 9*459/27
X = 153

So X= 

So option B is correct answer. 

HCF of 95 and 152 = 19



 

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.

A number with terminal decimal expansions have the denominator  in the form,
2^m 5ⁿ where m & n ∈ W.
The number 
Which the denominator is in the form,
 with m = 2 , n = 3.
Hence, it has terminal decimal expansion.

Let a be any positive integer and b = 2
Then by applying Euclid’s Division
Lemma, we have, a = 2q + r where 0 ⩽ r < 2 r = 0 or 1
Therefore, a = 2q or 2q+1 
Therefore, it is clear that a = 2q i.e., 
a is an even integer in the form of 2q

• Factors of 867 = 1, 3, 17, 51, 289, 867
• 255 = 1, 3, 5, 15, 17, 51, 85,255.
• Highest common factor = 51
• So, HCF (867,255) = 51

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

Since, HCF of 56 and 72, by Euclid’s divsion lemma,
72 = 56 × 1 + 16 ……….(i)
56 = 16 ×3 + 8 ……….(ii)
16 = 8× 2 + 0 ……….(iii) 
∴ HCF of 56 and 72 is 8. 
∴ 8 = 56 – 16× 3
8 = 56 – (72 – 56 ×1) ×3
[From eq. (i) : 16 = 72 – 56× 1]
8 = 56 – 3 ×72 + 56× 3
8 = 56 × 4 + (–3) × 72 
∴ x = 4,y = −3

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

 Since, 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers 65 = (70-5),
117 = (125 – 8), which is divisible by the required number.
Now, required number = HCF of 65,117                                     [for the largest number]
For this, 117 = 65 × 1 + 52 [∵ dividend = divisior × quotient + remainder]
⇒ 65 = 52 × 1 + 13
⇒ 52 = 13 × 4 + 0
∴ HCF = 13 
Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8.

Let a be any positive integer and b = 2.
Then by applying Euclid’s Division 
Lemma,
we have, a = 2q+r
where 0 ⩽ r < 2 ⇒ r = 0 or 1 ∴ a = 2q or 2q+1
Therefore, it is clear that a = 2q i.e., a is an even integer.
Also 2q and 2q+1 are two consecutive integers, therefore, 2q+1 is an odd integer.

L.C.M. of 2³×3³ and 2²×3² is the product of all prime numbers with greatest power of every given number = 2³×3³

Smallest prime number = 2 and smallest composite number = 4 
∴ HCF (2, 4) = 2

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.

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

Here, in the denominator of the given fraction the highest power of prime factor 5 is 3, therefore, the decimal expansion of the rational number  will terminate after 3 decimal places.

The correct answer is:

2. a = bq + r, 0 ≤ r < b.

Explanation:

This is the division algorithm, a fundamental concept in number theory.

The division algorithm states that for any two positive integers aaa (the dividend) and b (the divisor), there exist unique whole numbers q (the quotient) and r (the remainder) such that:

a=bq+r

Where:

• q is the quotient (how many times b fits into a),
• r is the remainder (what is left over after dividing),
• 0≤r<b, meaning the remainder is always less than the divisor b but greater than or equal to zero.

Example:

Suppose a=23 and b=5. When you divide 23 by 5:

• The quotient q=4 (because 5 fits into 23 four times),
• The remainder r=3 (because 23−(5×4)=3).

So, the equation becomes:

23=5×4+3

Where a=23, b=5, q=4, and r=3, which fits the form of the division algorithm.

Thus, option 2 is the correct answer because it accurately represents the division algorithm.

Least positive integer divisible by 20 and 24 is LCM (20, 24). 20
= 2²×5 24=2³×3
∴ LCM (20, 24) = 2³x 3 x 5 = 120

When 245 and 1029 are divided by the required number then there is left 5 as remainder. It means that 245 - 5 = 240 and 1029 - 5 = 1024 will be completely divisible by the required number.
Now we determine the HCF of 240 and 1024 by Euclid's Algorithm.
1024 = 240 x 4 + 64
240 = 64 x 3 + 48
64 = 48 x 1 + 16
48 = 16 x 3 + 0
Since remainder comes zero with last divisor 16,
required number = 16

LCM x HCF = First number x Second number 
∴ Required number 

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

 
Here, in the denominator of the given fraction the highest power of prime factor 2 is 3, therefore, the decimal expansion of the rational numberwill terminate after 3 decimal places. 

Find the HCF of 65 and 117,

117 = 1×65 + 52

65 = 1× 52 + 13

52 = 4 ×13 + 0

∴ HCF of 65 and 117 is 13.

65m - 117 = 13

65m = 117+13 = 130

∴m =130/65 = 2