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QUESTION: 1

Pairs of natural numbers whose least common multiple is 78 and the greatest common divisor is 13 are:

Solution:

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 ]

QUESTION: 2

1/√3 is -

Solution:

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.

QUESTION: 3

7√3 is -

Solution:

√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.

QUESTION: 4

5√3 is -

Solution:

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.

QUESTION: 5

Solution:

QUESTION: 6

HCF (p,q) × LCM (p,q) =

Solution:

P=a²b³

q=a³b

HCF (p,q) =a²b

LCM (p.q)=a³b³

As we know that HCF(p,q)xLCM(p,q)=a²bxa³b³

=a⁵b⁴ ---------- equation 1

pq=a²b³xa³b

=a⁵b⁴-----------equation 2

from 1 and 2

we can say that HCF(p,q)xLCM(p,q)=pq

QUESTION: 7

HCF (p,q,r) · LCM (p,q,r) =

Solution:

Since HCF(p,q,r)*LCM(p,q,r) is not equal to pq/r, neither it is equal toqr/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.

QUESTION: 8

Two natural numbers whose sum is 85 and the least common multiple is 102 are:

Solution:

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

QUESTION: 9

0.737373...=

Solution:

⇒ a = 0.737373...

⇒ 100a = 73.737373

= 73 + a

⇒ a = 73/99 = p/q

⇒ p=73 and q = 99 are co-prime.

Here, q=3^{2 }X 11.

QUESTION: 10

If p is a positive prime integer, then √p is –

Solution:

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

⇒ pb^{2} = a^{2} ….(i) [Squaring both the sides]

⇒ a^{2} is divisible by p

⇒ a is divisible by p

So, we can write a = pc for some integer c.

Therefore, a^{2} = p^{2}c^{2} ….[Squaring both the sides]

⇒ pb^{2} = p^{2}c^{2} ….[From (i)]

⇒ b^{2} = pc^{2}

⇒ b^{2} 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.

QUESTION: 11

LCM of three numbers 28, 44, 132 is –

Solution:

**LCM** = 2^{2}. 7^{1}. 11^{1}. 3^{1} = 924

QUESTION: 12

If a is a positive integer and p be a prime number and p divides a^{2}, then

Solution:

If a is a positive integer and p be a prime number and p divides a^{2}

^{So it implies that p divides a . }

QUESTION: 13

If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is

Solution:

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

QUESTION: 14

If *x* = and *y* = , then the value of (*x*^{2} + *y*^{2}) is

Solution:

QUESTION: 15

If x = , then 3x is –

Solution:

QUESTION: 16

Find the value of x then

Solution:

(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

QUESTION: 17

is equal to –

Solution:

QUESTION: 18

The product of 4√6 and is -

Solution:

QUESTION: 19

If two positive integers a and b are written as a = x^{3}y^{2} and b = xy^{3}; x, y are prime numbers, then HCF (a, b) is

Solution:

Given that, a =x^{3}y^{2} = x × x × x × y × y

and b = xy^{3} = x × y × y × y

∴ HCF of a and b = HCF (x^{3}y^{2},xy^{3}) = x × y × y = xy^{2 }

[Since, HCF is the product of the smallest power of each common prime factor involved in the numbers]

QUESTION: 20

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

Solution:

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.

QUESTION: 21

What will be the least possible number of the planks, if three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length?

Solution:

The lengths of three pieces of timber are 42m, 49m and 63m respectively.

We have to divide the timber into equal length of planks.

∴ Greatest possible length of each plank = HCF (42, 49, 63)

Prime factorization:

42 = 2 × 3 × 7

49 = 7 × 7

63 = 3 × 3 × 7

∴HCF = Product of the smallest power of each common prime factor involved in the numbers = 7

Hence, the greatest possible length of each plank is 7m.

QUESTION: 22

The value of i^{37}+^{ }1/i^{67} is –

Solution:

i^{37}+1/i^{6}^{7}

=(i^{2})^{18.}i + 1/(i^{2})^{33}.i

=1.i + 1/-1.i (Since i^{2}=-1)

=i -1/i

=(i^{2 }- 1)/i

=(-1-1)/i

=-2/i

Now, multiplying by i/i,

-2/i . i\i

=-2i/i^{2}

=-2i/-1

=2i

QUESTION: 23

The standard form of (1 + i) (1 + 2i) is –

Solution:

( 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.

QUESTION: 24

The conjugate of 4 – 5i is –

Solution:

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

QUESTION: 25

The multiplicative inverse of 1 – i is –

Solution:

1/1+i

= 1/1+i * 1-i/1-i

= 1-i/1+1

= 1-i/2

= 1/2 - 1/2i

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