Real Numbers| Worksheet Solutions| Class 10

Worksheet solutions for chapter: Real Numbers for class 10 are provided below.

 

(1) Use Euclid’s division algorithm to find the HCF of:

(i) 4256 and 56           (ii) 8555 and 145

Ans-

(i) 4256 and 56

4256 > 56 [4256 is dividend and 56 is divisor]

Therefore,

[Divide 4256 by 56, get 76 as a quotient and 0 as remainder]

4256 = 56 × 76 + 0

HCF(56, 4256) = 56

(ii) 8555 and 145

8555 > 145 [8555 is dividend and 145 is divisor]

Therefore,

[Divide 8555 by 145, get 59 as a quotient and 0 as remainder]

8555 = 145 × 59 + 0

HCF(145, 8555) =145

 

(2) Show that any positive odd integer is of the form 8q + 1, or 8q + 3 or 8q + 5, where q is some integer.

Ans-

Let a is positive odd integer

Euclid’s division algorithm

a = bq + r, 0 ≤ r < b.

Here, q and r are quotient and remainder.

a = dividend and

b = divisor

Apply Euclid’s division algorithm with a and b = 8

Since 0 ≤ r < 8, the possible remainders are 0, 1, 2, 3, 4, 5, 6 and 7.

Therefore, possible form of Euclid’s division lemma with remainder 0 and b= 8 is

a = 8q + 0

Here, q is quotient and 8q is divisible by 2.

Hence, a is even positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 1 and b = 8 is

a = 8q + 1

Here, q is quotient and 8q + 1 is not divisible by 2.

Hence, a is odd positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 2 and b =8 is

a = 8q + 2

Here, q is quotient and 8q + 2 is divisible by 2.

Hence, a is even positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 3 and b = 8  is

a = 8q + 3

Here, q is quotient and 8q + 3 is not divisible by 2.

Hence, a is odd positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 4 and b = 8  is

a = 8q + 4

Here, q is quotient and 8q + 4 is divisible by 2.

Hence, a is even positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 5 and b = 8  is

a = 8q + 5

Here, q is quotient and 8q + 5 is not divisible by 2.

Hence, a is odd positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 6 and b = 8  is

a = 8q + 6

Here, q is quotient and 8q + 6 is divisible by 2.

Hence, a is even positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 7 and b = 8  is

a = 8q + 7

Here, q is quotient and 8q + 7 is not divisible by 2.

Hence, a is odd positive integer.

But,a can not be even positive integer. (Given condition)

Therefore, a can not be 8q  and 8q + 2 and  (even integers)

Therefore, any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5

 

(3) Find the LCM and HCF of 210 and 33.

Ans-

Prime factorisation of 210 = 2 × 3 × 5 × 7

Prime fcatorisation of 33 = 3 × 11

HCF (33, 210) = 3

LCM(33, 210) = 2 × 3 × 5 × 7 × 11

= 2310

 

(4) Given that HCF (130, 510) = 10, find LCM(130, 510).

Ans-

(5) Explain why 6 × 5 × 12 + 12 and 6 × 5 × 4 × 3 + 5 are composite numbers.

Ans-

6 × 5 × 12 + 12

= 12(30 + 1)

= 12 × 31

6 × 5 × 12 + 12 is a composite number because it can be written as product of 12 × 31. Composite numbers are those numbers which has two or more than two factors other than 1.

6 × 5 × 4 × 3 + 5

= 5(6 × 4 × 3 + 1)

= 5(73)

6 ×5 × 4×3 + 5 is a composite number because it can be written as product of 5 × 73. Composite numbers are those numbers which has two or more than two factors other than 1.

 

(6) Prove that the following are irrationals:

(i) 3 + √2        (ii) 5√2

Ans-

(i) 3 + √2

Let us assume that 3 + √2 is rational.

So, we can find two coprime integers a and b, where b  0,

Therefore, √2 is rational.

But, this contradicts the fact that √2 is irrational.

Therefore, our assumption that 3 + √2 is rational is incorrect.

So, we conclude that 3 + √2 is irrational.

a = 50c…(2)

Put value of a in equation 1

50 b2 = (50c)2

50 b2 = (50)2c2

b2 = 50 c2

Therefore, 50 divides b2.

So, 50 divides c. (If p divides a2, then p divides a)

Therefore, a and b have at least 50 as a common factor.

But, a and b have no common factor other than 1 because both are coprime numbers. Therefore, it contradicts.

Hence, 5√2 is irrational.

 

(7) Decide whether following decimal expansions are rational or not. If they are rational, and of the form p/q, what can you say about the prime factors of q?

(i) 1.231646              (ii) 2.14135672…

Ans-

(i) 1.231646

It is a terminating decimal expansion. Therefore, it is a rational number.

Hence, prime factor of q will either 2 or 5 or both.

(ii) 2.14135672…

It is non- terminating non- recurring decimal expansion. Therefore, it is not rational number ,ie, irrational number.

 

(8) Without long division, state whether the following rational numbers will have a terminating decimal expansion or non – terminating repeating decimal expansion:

(9) A teacher has to manage 414 students to perform in a function in a group of 42 students. What is the maximum number of group in which they can perform?

Ans-

Use Euclid’s algorithm to find HCF of 414 and 42

414 = 42 × 8 + 36

42 = 36 × 1 + 6

36 = 6 × 6 + 0

Therefore, HCF (414, 42) = 666

Maximum number of group in which students can perform = HCF (414, 42)

=6

Hence, the maximum number of group in which students can perform is 8.

 

(10) Check whether 8n can end with the digit 0 for any natural number n.

Ans

If a number will end up with digit 0,

Therefore, that digit will be divisible by 10 or its prime factors, ie, 5 and 2.

But,

Prime factors of 8n = (2)n

Therefore, 8n can be divisible by only 2.

But, according to Uniqueness of the Fundamental Theorem of Arithmetic, there could not be any other prime factor in the prime factorisation of 8n .

Hence, there is no natural number n for which 8n ends with digit 0.

 

Helping Topics

Real Numbers

NCERT Solutions Class 10