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**SHORT ANSWER TYPE QUESTIONS**

**Q.1. Find a rational number between √2 and √3. [Delhi 2019]****Ans:** Let p be rational number between √2 and √3

∴ √2 < p < √3

On squaring throughout, we have

2 < p^{2} < 3

The perfect squares which lie between 2 and 3 are 2 < 2.25 < 2.56 < 2.89 < 3.

Taking square root throughout

The rational numbers between √2 and √3 are 1.5, 1.6 1.7 and more.**Q.2. Prove that √2 is an irrational number. [Delhi 2019]****Ans: **Let √2 = a/b where a and b are co prime integer, b ≠ 0.

Squaring both sides, we get

Multiplying with b on both sides, we get

Multiplying with b on both sides, we get

LHS = 2 x b = Integer

∴ LHS ≠ RHS

∴ Our supposition is wrong.

⇒ √2 is irrational.**Q.3. Prove that 2 + 5 √3 is an irrational number, given that √3 is an irrational number. [CBSE, Allahabad 2019]****Ans: **Let 2 + 5 √3 be a rational number. (p and q are co-prime positive integers, q ≠ 0)

∴

⇒

which is a contradiction because √3 is an irrational number and is a rational.**Q.4. The HCF of two numbers a and b is 5 and their LCM is 200. Find the product ab. [CBSE 20 19 (30/5/2)]****Ans: **We know that

Product of two numbers = product of their LCM and HCF.

ab = 200 x 5

= 1,000**Q.5. A number N when divided by 14 gives the remainder 5. What is the remainder when the same number is divided by 7? [CBSE 2019 (30/2/1)]****Ans:** 5, because 14 is multiple of 7.**Q.6. Find after how many places the decimal form o f the numberwill terminate. [CBSE 2019 (30/3/3)****Ans: **

So, the decimal form will end after four decimal places.**Q.7. Express 429 as product of its prime factors. ****[CBSE 2019 (30/3/3)]****Ans:**

∴ 429 = 3 x 13 x 11**Q.8. Find the HCF of 612 and 1314 using prime factorisation. [CBSE 2019 (30/5/3)]****Ans:** 612 = 2 x 2 x 3 x 3 x 17

1314 = 2 x 3 x 3 x 73

HCF = 2 x 3 x 3 = 18**Q.9. Find the HCF o f 1260 and 7344 using Euclid’s algorithm. [NCERT, CBSE 2019 (30/1/1)]****OR****Use Euclid’s division algorithm to find the HCF o f 255 and 867. [CBSE 2019 (30/4/1)]Ans:** Since 7344 > 1260 and from Euclid’s algorithm

a = bq + r

OR

Solution is similar as above.

Only values are changed

Ans:

Ans:

65 = 5 x 13

117 = 13 x 3 x 3

⇒ HCF =13

According to question,

65n - 117 = 13

65n = 130

n = 2

Ans:

306 = 2 x 3

657 = 3

LCM - 2 X 3

i.e., the possible remainders are 0, 1, 2, 3, 4, 5.

Thus, a can be of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is some quotient.

Since a is odd integer, so a cannot be of the form 6q or x 6q + 2 or 6q + 4 (since they are even).

Thus, a is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

Hence, any odd positive integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

Let a be any positive odd integer and b = 6. Then, by Euclid’s algorithm, a = 6m + r, for some integer m > 0 and 0 < r < 6.

i.e., the possible remainders are 0, 1, 2, 3, 4, 5.

Thus, a can be of the form 6m, or 6m + 1 or 6m + 2 or 6m + 3 or 6m + 4 or 6m + 5, where m is some quotient.

Since a is odd integer, so a cannot be of the form 6m or 6m + 2 or 6m + 4 (since they are even). Thus, a is of the form 6m + 1, 6m + 3 or 6m + 5, where m is some integer.

Hence, any odd positive integer is of the form 6m + 1 or 6m + 3 or 6m + 5, where m is some integer.

HCF of 404 and 9

404 = 2 x 2 x 101

96 = 2 x 2 x 2 x 2 x 2 x 3

Common factor = 2 x 2 = 4

∴ HCF = 4

LCM of 404 and 96

= 2 x 2 x 101 x 2x 2x 2 x 3 = 9696

HCF x LCM = 4 x 9696 = 38784

Product o f two numbers = 404 x 96 = 38784

Clearly, HCF x LCM = Product of two numbers

Hence verified

(rational number)

Ans:

The smallest composite number = 4

∴ HCF of 2 and 4 = 2

Ans:

⇒

For any values of p and q(q ≠ 0), RHS = is rational, but LHS =

This contradicts the fact. So, our assumption is wrong.

∴ 5 + 3 √2 is an irrational number.

∴ it can be written in the form a/b, where b ≠ 0 and a and b are coprime.

So

Subtracting 2 from both sides

This contradicts.

LHS is an irrational number because

So, LHS is equal to RHS not possible.

Hence, our assumption is wrong.

∴ (2 + √3 is an irrational number.

Smallest prime number = 2

So, HCF (4, 2) = 2

**LONG ANSWER TYPE QUESTIONS**

**Q.1. Prove that √5 is an irrational number. [CBSE 2019 (30/5/1)]****Ans:** Let us assume, to the contrary, that Vs is a rational number.

Then, there exist co-prime positive integers a and b such that

So, a = √5 b

Squaring both sides, we have

a^{2} = 5b^{2} ......(i)

⇒ 5 divides a^{2 }⇒ 5 divides a

So, we can write

a = 5c (where c is any integer)

Putting the value of a = 5c in (i), we have

25c^{2} = 5b^{2} => 5c^{2} = b^{2 }

It means 5 divides b^{2} and so 5 divides b.

So, 5 is a common factor of both a and b which is a contradiction.

So, our assumption that √5 is a rational number is wrong.

Hence, we conclude that √5 is an irrational number.

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