RS Aggarwal Test: Real Numbers - 3 - Class 10 MCQ

Let the no.s be a & b,

GCD [ a , b ] = 13;
=> Let a = 13m and b = 13n

Now, LCM [ a , b ] = 78
=> 13m | 78
=> m | 6 and similarly, n | 6; --> Also, m does not divide n

.'. Possible values of (m,n) are :-> 
 [ 1 , 6 ] , [ 2 , 3 ]
.'. Possible values for ( a , b ) are : [ 13 , 78 ] , [ 26 , 39 ]

Let as assume to the contrary that 1/✓3 is rational number. Then, we can find coprime integers a and b (≠ 0) such that :
1/✓3= P/Q { where P and Q are co-prime and Q not equal to 0}
✓3 P =Q .1 
✓3 = Q/P 
 Since Q/P is a Rational number

Since L.H.S = R.H.S

So ✓3 is also rational number.

But this contradicts the fact that ✓3 is an irrational number. This contradiction has arisen because of our incorrect assumption that 1/√3 is a rational number.

So this proves that 1/√3 is an irrational number. 

√3 cannot be expressed in the form p/q where q ≠ 0 which is the form of a rational number. So any number multiplied by it will give an irrational number. So 7√3 is irrational.

To prove : 5-√3 is irrational

Proof: 
Let us assume that 5 -√3 is rational.

Let ,
 5 - √3 = r , where "r" is rational

5 - r = √3
Here,
LHS is purely rational.But,on the other hand ,RHS is irrational.
This leads to a contradiction.
Hence,5-√3 is irrational.

In the 1760s, Johann Heinrich Lambert proved that the number π (pi) is irrational: that is, it cannot be expressed as a fraction a/b, where a is an integer and b is a non-zero integer.

Since HCF(p,q,r)*LCM(p,q,r) is not equal to pq/r, neither it is equal to qr/p and neither is p,q,r. So the correct answer is D . Also, HCF(p,q,r)*LCM(p,q,r) is not equal to p*q*r . This condition only holds for two numbers.

Prime factorisation of 102 = 2 x 3 x 17.
Prime factorisation of 85 = 5 x 17 = (2+3) x 17.
The two numbers are:
1. 2 x 17 = 34.
2. 3 x 17 =51

⇒ a = 0.737373...
⇒  100a = 73.737373
= 73 + a
⇒ a = 73/99 = p/q
⇒  p=73 and q = 99 are co-prime.
Here, q=3² X 11.

Let us assume, to the contrary, that √p is
rational.
So, we can find coprime integers a and b(b ≠ 0)
such that √p = a/b
⇒ √p b = a
⇒ pb² = a² ….(i) [Squaring both the sides]
⇒ a² is divisible by p
⇒  a is divisible by p
So, we can write a = pc for some integer c.
Therefore, a² = p²c² ….[Squaring both the sides]
⇒ pb² = p²c² ….[From (i)]
⇒ b² = pc²
⇒ b² is divisible by p
⇒ b is divisible by p
⇒ p divides both a and b.
⇒ a and b have at least p as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction arises because we have
assumed that √p is rational.
Therefore, √p is irrational.

LCM = 2². 7¹. 11¹. 3¹ = 924

If a is a positive integer and p be a prime number and p divides a²
So it implies that p divides a . 

The correct option is Option A.

x = √3 + 1 / √3 - 1

y = √3 - 1 / √3 + 1

Now, 

  x² + y²

= (√3 + 1 / √3 - 1)² + (√3 - 1 / √3 + 1)²

= (3 + 2√3 + 1 / 3 - 2√3 + 1) + (3  - 2√3 + 1 / 3 + 2√3 + 1)

= (4 + 2√3 / 4 - 2√3) + (4 - 2√3 / 4 + 2√3)

= (4 + 2√3)² + (4 - 2√3)² / (4 - 2√3) × (4 + 2√3)

By solving this, 

= 28 + 16√3 + 28 - 16√3 / 16 - 12

= 56 / 4

= 14

(3/5)^2x-3= (5/3)^x-3
(3/5)^2x-3=(3/5)^3-x
2x-3=3-x
2x+x=6
3x=6
x=2

Given that, a =x³y² = x × x × x × y × y
and b = xy³ = x × y × y × y
∴ HCF of a and b = HCF (x³y²,xy³) = x × y × y = xy² 
[Since, HCF is the product of the smallest power of each common prime factor involved in the numbers]

The least number divisible by all the numbers from 1 to 10 will be the LCM of these numbers.

We have,

1 = 1

2 = 2 × 1

3 = 3 × 1

4 = 2 × 2

5 = 5 × 1

6 = 2 × 3

7 = 7 × 1

8 = 2 × 2 × 2

9 = 3 × 3

10 = 2 × 5

So, LCM of these numbers = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520

Hence, least number divisible by all the numbers from 1 to 10 is 2520.

i³⁷+1/i⁶⁷
=(i²)^18.i + 1/(i²)³³.i
=1.i + 1/-1.i         (Since i²=-1)
=i -1/i
=(i² - 1)/i
=(-1-1)/i
=-2/i
Now, multiplying by i/i,
-2/i . i\i 
=-2i/i²
=-2i/-1
=2i

( 1 + i ) (1 + 2i )

=1 + 2i + i + 2i^2

=1 + 3i -2  ( Because i^2 = -1)

= -1 + 3i 

= a + ib = -1 + 3i

Here a = -1 and b = 3 

So the standard form of  (1 + i) (1 + 2i) is : 

-1 + 3i

So option D is correct answer. 

The conjugate of the complex number a+bi is a−bi.
In order to find the complex conjugate, you just reverse the sign of the imaginary part of the number ; so it is 4+5i