NCERT Solutions for class 10 Maths chapter 1 Real Numbers

                                                              EXERCISE  1.1



Question 1: Use Euclid’s division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

 Solution 1:

 (i) 135 and 225 Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
 225 = 135 × 1 + 90
Since remainder 90 ≠ 0,
 we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
 We consider the new divisor 90 and new remainder 45,
 and apply the division lemma to obtain
 90 = 2 × 45 + 0
Since the remainder is zero,
 the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45

(ii)196 and 38220
 Since 38220 > 196,
 we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
 Since the divisor at this stage is 196,
 Therefore, HCF of 196 and 38220 is 196.

  (iii)867 and 255 Since 867 > 255,
 we apply the division lemma to 867 and 255 to obtain
 867 = 255 × 3 + 102
 Since remainder 102 ≠ 0,
 we apply the division lemma to 255 and 102 to obtain
 255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51,
 and apply the division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
 Since the divisor at this stage is 51,
 Therefore, HCF of 867 and 255 is 51.


Question 2: Show that any positive odd integer is of the form, 6q + 1 or 6q + 3, or 6q + 5 , where q is some integer.

 Solution 2:
 Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0,
 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r< 6.
Therefore,
if  r = 0
then a = 6q + 0 = 6q = 2 × 3q = 2 × m = 2m { where m = 3q }
if r = 1
then a = 6q + 1 = 2 × 3q + 1 = 2m + 1,{ where 'm' =   3q }
if r = 2
then a = 6q + 2 = 2 ×(3q + 1)  = 2  ×  m = 2m  { where m = 3q + 1 }
if r = 3
then a = 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2m + 1,  { where m = 3q + 1 }
if r = 4
then a =  6q + 4 =   2 (3q + 2)  = 2m   { where m = 3q + 2 }
if r = 5
then a = 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2m + 1, { where m = 3q + 2 }
 Clearly,
 6q + 1, 6q + 3, 6q + 5 are of the form 2m + 1, where m is an integer.
 Therefore,
6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
 Hence, these expressions of numbers are odd numbers.
 And therefore,
any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5



  Question 3: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution 3:
HCF (616, 32) will give the maximum number of columns in which they can march.
 We can use Euclid’s algorithm to find the HCF.
 616 = 32 × 19 + 8
 32 = 8 × 4 + 0
 The HCF (616, 32) is 8. Therefore, they can march in 8 columns each.


 Question 4: Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

[ Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2.
 Now square each of these
 and show that they can be rewritten in the form 3m or 3m + 1.]
Solution 4 :




  Question 5:

 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
 Solution 5:




                                                        EXERCISE 1.2  


Question 1: Express each number as product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution 1:
 (i) 140 = 2 × 2 × 5 × 7
 (ii) 156 = 2 × 2 × 3 × 13
 (iii) 3825 = 3 × 3 × 5 × 5 × 17
 (iv) 5005 = 5 ×7× 11 × 13
 (v) 7429 = 17 × 19 × 23


Question 2: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Solution 2:

 (i) 26 and 91
 26 = 2 × 13
 91 = 7 × 13
 HCF = 13
 LCM = 2 × 7 × 13 = 182
 Product of the two numbers = 26 × 91 = 2366
 HCF × LCM = 13 × 182 = 2366
 Hence,
 product of two numbers = HCF × LCM


 (ii) 510 and 92
 510 = 2 × 3 × 5 × 17
 92 = 2 × 2 × 23
 HCF = 2
 LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of the two numbers = 510 × 92 = 46920
 HCF × LCM = 2 × 23460 = 46920
 Hence,
product of two numbers = HCF × LCM


 (iii) 336 and 54
336 = 2 × 2 × 2 × 2 × 3 × 7
  54 = 2 × 3 × 3 × 3
 HCF = 2 × 3 = 6
 LCM =  2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
Product of the two numbers = 336 × 54 = 18144
 HCF × LCM = 6 × 3024 = 18144
 Hence,
 product of two numbers = HCF × LCM


  Question 3: Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12,15 and 21 (ii) 17,23 and 29 (iii) 8,9 and 25


Solution 3:

 (i) 12,15 and 21
12 = 2× 2× 3
 15 = 3 × 5
 21 = 3 × 7
 HCF = 3
 LCM = 2× 2× 3 × 5 × 7 = 420


 (ii) 17,23 and 29
 17 = 1 × 17
23 = 1 × 23
 29 = 1 × 29
 HCF = 1
 LCM = 17 × 23 × 29 = 11339


 (iii) 8, 9 and 25
 8 = 2 × 2 × 2
 9 = 3 × 3
 25 = 5 × 5
 HCF = 1
 LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800



  Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).

 Solution 4:

 HCF (306, 657) = 9
We know that, LCM × HCF = Product of two numbers
 ∴LCM × HCF = 306 × 657
    LCM  × 9   =  306 × 657
     LCM = 201042/9
      LCM =  22338


Question 5: Check whether 6n can end with the digit 0 for any natural number n.


Solution 5:

 If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5 Prime factorisation of 6n = (2 ×3)n It can be observed that 5 is not in the prime factorisation of 6n. Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.


  Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution 6:
 Numbers are of two types - prime and composite.
A prime number can be divided by 1 and only itself,
 whereas a composite number have factors other than 1 and itself.
It can be observed that
 7 × 11 × 13 + 13
 = 13 × (7 × 11 + 1)
 = 13 × (77 + 1)
= 13 × 78
 = 13 ×13 × 6 The given expression has 6 and 13 as its factors.
 Therefore, it is a composite number.
 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 ×1009 1009 cannot be factorised further.
 Therefore,
the given expression has 5 and 1009 as its factors.
 Hence, it is a composite number.


                            EXERCISE 1.3 

1. Prove that is irrational.

Ans. Let us prove irrational by contradiction.

Let us suppose that is rational. It means that we have co-prime integers a and b (b ≠ 0) such that 

⇒ b=a

Squaring both sides, we get

… (1)

It means that 5 is factor of 

Hence, 5 is also factor of a by Theorem. … (2)

If, 5 is factor of a, it means that we can write a = 5c for some integer c.

Substituting value of a in (1),

It means that 5 is factor of .

Hence, 5 is also factor of b by Theorem. … (3)

From (2) and (3), we can say that 5 is factor of both a and b.

But, a and b are co-prime.

Therefore, our assumption was wrong.  cannot be rational. Hence, it is irrational.

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NCERT Solutions class-10 Maths Exercise 1.3

2. Prove that (3 + 2) is irrational.

Ans. We will prove this by contradiction.

Let us suppose that (3+2) is rational.

It means that we have co-prime integers a and b (b ≠ 0) such that

 ⇒ 

⇒ 

⇒ … (1)

a and b are integers.

It means L.H.S of (1) is rational but we know that  is irrational. It is not possible. Therefore, our supposition is wrong. (3+2) cannot be rational.

Hence, (3+2) is irrational.

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NCERT Solutions class-10 Maths Exercise 1.3

3. Prove that the following are irrationals.

(i) 

(ii) 

(iii) 

Ans. (i) We can prove irrational by contradiction.

Let us suppose that  is rational.

It means we have some co-prime integers a and b (b ≠ 0) such that

=ab

⇒  … (1)

R.H.S of (1) is rational but we know that  is irrational.

It is not possible which means our supposition is wrong.

Therefore,  cannot be rational.

Hence, it is irrational.

(ii) We can prove  irrational by contradiction.

Let us suppose that  is rational.

It means we have some co-prime integers a and b (b ≠ 0) such that

⇒  … (1)

R.H.S of (1) is rational but we know that  is irrational.

It is not possible which means our supposition is wrong.

Therefore,  cannot be rational.

Hence, it is irrational.

(iii) We will prove  irrational by contradiction.

Let us suppose that () is rational.

It means that we have co-prime integers a and b (b ≠ 0) such that

⇒ 

⇒  … (1)

a and b are integers.

It means L.H.S of (1) is rational but we know that is irrational. It is not possible.

Therefore, our supposition is wrong. () cannot be rational.

Hence, () is irrational.

                            EXERCISE 1.4

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating decimal expansion.

(i) 

(ii) 

(iii) 

(iv) 

(v) 

(vi) 

(vii) 

(viii) 

(ix) 

(x) 

Ans. According to Theorem, any given rational number of the form where p and q are co-prime, has a terminating decimal expansion if q is of the form , where m and n are non-negative integers.

(i) 

q = 3125 = 

Here, denominator is of the form , where m = 5 and n = 0.

It means rational number  has a terminating decimal expansion.

(ii) 

q = 8 = 

Here, denominator is of the form , where m = 0 and n = 3.

It means rational number  has a terminating decimal expansion.

(iii) 

q = 455 = 

Here, denominator is not of the, where m and n are non-negative integers.

It means rational number  has a non-terminating repeating decimal expansion.

(iv) 

q = 320 = = 

Here, denominator is of the form , where m = 1 and n = 6.

It means rational number  has a terminating decimal expansion.

(v) 

q = 343 = 

Here, denominator is not of the form , where m and n are non-negative integers.

It means rational number  has non-terminating repeating decimal expansion.

(vi) 

q = 

Here, denominator is of the form , where m = 2 and n = 3 are non-negative integers.

It means rational number  has terminating decimal expansion.

(vii) 

q = 

Here, denominator is not of the form , where m and n are non-negative integers.

It means rational number  has non-terminating repeating decimal expansion.

(viii) 

Here, denominator is of the form , where m = 1 and n = 0.

It means rational number  has terminating decimal expansion.

(ix) 

q = 10 = 

Here, denominator is of the form , where m = 1 and n = 1.

It means rational number  has terminating decimal expansion.

(x) 

Here, denominator is not of the form , where m and n are non-negative integers.

It means rational number  has non-terminating repeating decimal expansion.

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NCERT Solutions for Class 10 Maths Exercise 1.4

2. Write down the decimal expansions of those rational numbers in Question 1 which have terminating decimal expansions.

Ans. (i) 

(ii) 

(iv) 

(vi) 

(viii) 

(ix) 

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NCERT Solutions for Class 10 Maths Exercise 1.4

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If, they are rational, and of the form , what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.1201120012000120000…

(iii) 

Ans. (i) 43.123456789

It is rational because decimal expansion is terminating. Therefore, it can be expressed in  form where factors of q are of the form where n and m are non-negative integers

(ii) 0.1201120012000120000…

It is irrational because decimal expansion is neither terminating nor non-terminating repeating.

(iii) 

It is rational because decimal expansion is non-terminating repeating. Therefore, it can be expressed in  form where factors of q are not of the form where n and m are non-negative integers.