CBSE Previous Year Questions: Real Numbers |

**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 = 1.41

√3 = 1.73

On squaring throughout, we have

2 < p2 < 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 us assume that √2 is a rational number.

So it can be expressed in the form p/q where p, q are co-prime integers and q ≠ 0

√2 = p/q

Here p and q are coprime numbers and q ≠ 0

√2 = p/q

On squaring both the sides we get,

=> 2 = (p/q)2

=> 2q2 = p2……………………………..(1)

p2/2 = q2

So 2 divides p and p is a multiple of 2.

⇒ p = 2m

⇒ p2 = 4m2 ………………………………..(2)

From equations (1) and (2), we get,

2q2 = 4m2

⇒ q2 = 2m2

⇒ q2 is a multiple of 2

⇒ q is a multiple of 2

Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√2 is an irrational number.

**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:**The formula that shows the relation between LCM and HCF of a and b is: LCM (a, b) = (a × b) / HCF (a, b)

HCF (a, b) = 5, and LCM (a, b) = 200

LCM (a, b) = (a × b) / HCF (a, b)

200 = (a × b) / 5

a × b = 200 × 5

a × b = 1000

Therefore, the product of 'a' and 'b' is 1000.**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:** Let the number be X

Since dividing the number gives remainder as 5, X can be written as 14 * Y + 5

where Y will be the quotient on dividing X by 14.

X = 14 * Y + 5

X = 2 * 7 * Y + 5

Let's divide both sides by 7 (since we want to know the remainder when X is divided by 7)

X/7 = (2 * 7 * Y)/7 + 5/7

X/7 = 2 * Y + 5/7

Since Y is a whole number (quotient), thus 2 * Y will also be a whole number and what remains is 5/7.

Thus, the remainder is 5 when X is divided by 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 of 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

**Q.10. Write a rational number between √2 and √3. [CBSE 2019 (30/1/2)]**

Ans:A rational number between √2 and √3 is √2.89 = 1.7 = 17/10.

Ans:

**Q.11. If HCF of 65 and 117 is expressible in the form 65n - 117, find the value of n. [CBSE 2019 (30/3/3)]**

Ans:HCF (65, 117) is given by

Ans:

65 = 5 x 13

117 = 13 x 3 x 3

⇒ HCF =13

According to question,

65n - 117 = 13

65n = 130

n = 2

**Q.12. Write the smallest number which is divisible by both 306 and 657.**

**[CBSE 2019 (30/2/1)]Ans: **Here, to find the required smallest number we will find LCM of 306 and 657.

306 = 2 x 32 x 17

657 = 32 x 73

LCM - 2 X 32 x 17 x 73 = 22338

**Q.13. Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is some integer.**

**[NCERT]**

**OR**

**Show that any positive odd integer is of the form 6 m + 1 or 6 m + 3 or 6 m + 5, where m is some integer. [**

**CBSE 2019(30/5/2)]**

**Ans:**Let a be any positive odd integer and b = 6. Then, by Euclid's algorithm, a = 6q + r, for some integer q

__>__0 and 0

__<__r

__<__6.

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.

**OR**

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.

**Q.14. Find HCF and LCM of 404 and 96 and verify that HCF x LCM = Product of the two given numbers. [CBSE 2018]**

**Ans:**

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

**Verification:**

HCF x LCM = 4 x 9696 = 38784

Product of two numbers = 404 x 96 = 38784

Clearly, HCF x LCM = Product of two numbers

Hence verified

**Q.15. Write whether**

**on simplification gives an irrational or a rational number. [CBSE 2018 (C)]**

**Ans:**

(rational number)

**Q.16. What is the HCF of smallest prime number and the smallest composite number? [CBSE 2018]**

Ans:

Ans:

**Prime number:**A prime number is a number that has exactly two factors i.e. it can be divided by only the number ‘1’ and itself.

The smallest prime number is = 2

**Composite number:**A composite number has more than two factors, which means apart from getting divided by number 1 and itself, it can also be divided by at least one integer or number. We don’t consider the number ‘1’ as a composite number.

The smallest composite number is = 4

2 is the factor of 4

∴ H.C.F of the smallest prime number and the smallest composite number is 2.

**Q.17. Given that √2 is irrational, prove that (5 + 3 √2 ) is an irrational number [CBSE 2018]**

Ans:Let 5 + 3 √2 = p/q be a rational number, where p and q, have no common factor other than 1 and q ≠ 0.

Ans:

⇒

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

**√2 is s irrational.**

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

∴ 5 + 3 √2 is an irrational number.

**Q.18. Given that √3 is an irrational number, prove that (2 + √3 ) is an irrational number. [CBSE 2018 (C)]**

**Ans:**Let (2 + √3 is not an irrational.

∴ 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

**√**3 is an irrational number (given) and RHS is an rational number.

So, LHS is equal to RHS not possible.

Hence, our assumption is wrong.

∴ (2 + √3 is an irrational number.

**Q.19. What is the HCF of the smallest composite number and the smallest prime number? [CBSE 2018]**

**Ans:**The smallest prime number is 2.

The smallest composite number is 4.

To find the HCF of 2 and 4 let us write their prime factorization.

2 = 2

4 = 2 × 2

HCF (2, 4) = 2

Thus, the HCF of the smallest composite number and the smallest prime number is 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

a2 = 5b2 ......(i)

⇒ 5 divides a2 ⇒ 5 divides a

So, we can write

a = 5c (where c is any integer)

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

25c2 = 5b2 => 5c2 = b2

It means 5 divides b2 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.