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

√7 is

Solution:

Assume that √7 be a rational number. So, √7 = p/q where p and q are co prime.

7 = p^{2}/q^{2}

7q^{2 }= p^{2} ...(1)

This shows that p^{2} is divisible by 7, then p is divisible by 7, then for any positive integer c, it can be said that p = 7c, p^{2} = 49c^{2}

Then equation (1) can be written as,

7q^{2} = 49c^{2}

q^{2} = 7c^{2}

This gives that q is divisible by 7.

So, p and q has a common factor 7 which is a contradiction to the assumption that they are co prime.

Hence, √7 is an irrational number.

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 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

Ratio = LCM/HCF = 60/5 = 12/1 = 12:1

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