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This mock test of MCQ : Real Numbers - 1 for Class 10 helps you for every Class 10 entrance exam.
This contains 10 Multiple Choice Questions for Class 10 MCQ : Real Numbers - 1 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

√7 is

Solution:

Lets assume that √7 is rational number. ie √7=p/q.

suppose p/q have common factor then

we divide by the common factor to get √7 = a/b were a and b are co-prime number.

that is a and b have no common factor.

√7 =a/b co- prime number

√7= a/b

a=√7b

squaring

a²=7b² .......1

a² is divisible by 7

a=7c

substituting values in 1

(7c)²=7b²

49c²=7b²

7c²=b²

b²=7c²

b² is divisible by 7

that is a and b have atleast one common factor 7. This is contradicting to the fact that a and b have no common factor.This happend because of our wrong assumption.

hence,√7 is irrational

QUESTION: 2

For positive integers a and 3, there exist unique integers q and r such that a = 3q + r, where r must satisfy:

Solution:

Euclid’s division Lemma:

It tells us about the divisibility of integers. It states that any positive integer ‘a’ can be divided by any other positive integer ‘ b’ in such a way that it leaves a remainder ‘r’.

Euclid's division Lemma states that for any two positive integers ‘a’ and ‘b’ there exist two unique whole numbers ‘q’ and ‘r’ such that , a = bq + r, where 0≤ r < b.

Here, a= Dividend, b= Divisor, q= quotient and r = Remainder.

Given : a = 3q+r

In this question ,

b = 3

The values 'r’ can take 0 ≤ r < 3.

QUESTION: 3

For some integer q, every odd integer is of the form

Solution:

We know that, odd integers are 1, 3, 5, ...

So, it can be written in the form of 2q + 1,

where, q = integer = Z

or q = ..., -1, 0,1,2,3, ...

∴ 2q + 1 = ..., -3, -1, 1, 3,5, ...

**Alternate Method**

Let 'a' be given positive integer. On dividing 'a' by 2, let q be the quotient and r be the remainder. Then, by Euclid's division algorithm, we have

a = 2q + r, where 0 ≤ r < 2

⇒ a = 2q + r, where r = 0 or r = 1

⇒ a = 2q or 2q + 1

when a = 2q + 1 for some integer q, then clearly a is odd.

QUESTION: 4

Find the greatest number of 5 digits, that will give us remainder of 5, when divided by 8 and 9 respectively.

Solution:

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

QUESTION: 5

The product of two consecutive integers is divisible by

Solution:

As the case is of 2 consecutive integers, one will be odd and the other one even, so their product will always be even which eventually will be divisible by an even no., i.e. 2, here.

QUESTION: 6

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

Solution:

Here, a = x^{3}y^{2} and b = xy^{3}.

⇒ a = x _{*} x _{*} x _{*} y _{*} y and b = xy _{*} y _{*} y

∴ LCM(a, b) = x _{* }x _{* }x _{* }y _{*} y* *_{*} y = x^{3} _{*} y^{3} = x^{3}y^{3}

LCM = x^{3}y^{3}

QUESTION: 7

n^{2} - 1 is divisible by 8 if n is

Solution:

If n is even then it is not possible for n=2.

If n is natural number then it is not possible for n=1,2.

If n is integer similarly n²- 1 is not divisible by 8.

If n is odd the possible for n is 3,5,7. Thus the n is odd number.

QUESTION: 8

If two positive integers p and q can be expressed as p = ab^{2} and q = a^{3}b; where a, b being prime numbers, then LCM (p, q) is equal to

Solution:

As per question, we have,

p = ab^{2} = a × b × b

q = a^{3}b = a × a × a × b

So, their Least Common Multiple (LCM) = a^{3} × b^{2}

QUESTION: 9

The least perfect square number which is divisible by 3, 4, 5, 6 and 8 is

Solution:

L.C.M. of 3, 4, 5, 6, 8 = 120 ;

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

QUESTION: 10

The ratio between the LCM and HCF of 5,15, 20 is:

Solution:

5, 15 = 5 x 3, 20 = 2 x 2 x 5

LCM(5, 15, 20) = 5 x 3 x 2 x 2 = 60

HCF(5, 15, 20) = 5

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