एनसीईआरटी कक्षा 10वी के गणित अध्याय 1 वास्तविक संख्याएं प्रश्नवली 1.1 का समाधान इंग्लिश और हिंदी में | NCERT Solutions Class 10th Math Chapter 1 Real Numbers Exercise 1.1 In English With Videos and pdf | Exercise 1.1 Class 10 Maths NCERT Solutions in English | Class 10 math exercise 1.1 solution in English and Hindi | Class 10 Maths Chapter 1 Solutions in English |

In this blog :- NCERT Solutions Class 10th Math Chapter 1 Real Numbers Exercise 1.1 In English With Videos and pdf | Class 10th Math Chapter One exercise 1.1 solution in English with Video and pdf | Class 10 math exercise 1.1 solution in English and Hindi | Real Numbers Class 10 PDF Solutions | Class 10 Maths Chapter 1 Solutions in Hindi | Class 10 Maths Chapter 1 PDF | Class 10 Maths Chapter 1 Solutions PDF download | Class 10 Maths Chapter 1 Solutions | Exercise 1.1 Class 10 Maths NCERT Solutions in Hindi English | एनसीईआरटी कक्षा 10वी के गणित प्रश्नवली 1.1 का समाधान इंग्लिश और हिंदी में | Exercise 1.1 Class 10 Maths NCERT Solutions in English | प्रश्नावली 1.1 कक्षा 10 गणित | Class 10 Maths Chapter 1 Exercise 1.1 Solutions | कक्षा 10 गणित अध्याय 1 | प्रश्नावली 1.1 का पहला सवाल | प्रश्नावली 1 कक्षा 10 गणित | 10 वीं कक्षा गणित अध्याय 1 हल हिंदी व अंग्रेजी में | Exercise 1.1 Class 10 Maths NCERT Solutions in english |

Class 10th Math Chapter One exercise 1.1 solution in English

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NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 Question 1 in Hindi and English with videos and pdf download | 

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Real Numbers Exercise 1.1

Ques 1:-  Use Euclid's algorithm to find the HCF of the following numbers.

i) 135 and 225

To obtain the HCF of two positive integers, say c and d, with c > d,

we follow the steps below:

Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, 

q and r such that c = dq + r, 0 ≤ r < d.

Step 2: If r = 0, d is the HCF of c and d. 

If r ≠ 0, apply the division lemma to d and r.

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Solution :-  (i) We know that, = 225>135

Applying Euclid’s division algorithm:

(Dividend = Divisor × Quotient + Remainder)

225 = 135 ×1+90

Here remainder = 90,

So, Again Applying Euclid’s division algorithm

135 = 90×1+45

Here remainder = 45, 

So, Again Applying Euclid’s division algorithm

90 = 45×2+0

Remainder = 0,

Hence, HCF of (135, 225) = 45

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ii) 196 and 38220

Solution:- (ii) We know that, 38220>196

So, Applying Euclid’s division algorithm

38220 = 196×195+0 (Dividend = Divisor × Quotient + Remainder)

Remainder = 0

Hence, HCF of (196, 38220) = 196

iii) 867 and 255

(iii) We know that, 867>255

So, Applying Euclid’s division algorithm

867 = 255×3+102 (Dividend = Divisor × Quotient + Remainder)

Remainder = 102

So, Again Applying Euclid’s division algorithm

255 = 102×2+51

Remainder = 51

So, Again Applying Euclid’s division algorithm

102 = 51×2+0

Remainder = 0

Hence,

(HCF 0f 867 and 255) = 51

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NCERT Solutions Class 10 Maths Chapter 1 Exercise 1.1 Question 2 in Hindi and English with videos and pdf download |

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

To Prove: Any Positive odd integer is of the form 6q + 1, 6q + 3, 6q + 5

Proof: To prove the statement by Euclid's lemma we have to consider.....

divisor as 6 and then find out the possible remainders when divided by 6 By taking, " a " as any positive integer and b = 6.

Applying Euclid’s algorithm a = 6 q + r

As divisor is 6 the remainder can take only 

6 values from 0 to 5

Here, r = remainder = 0, 1, 2, 3, 4, 5 and 

q ≥ 0 

So, total possible forms are

6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5, 

6q + 0, (6 is divisible by 2, its an even number)

6q + 1, (6 is divisible by 2 but 1 is not divisible by 2, its an odd number)

6q + 2, (6 and 2 both are divisible by 2, its an even number)

6q + 3, (6 is divisible by 2 but 3 is not divisible by 2, its an odd number)

6q + 4, (6 and 4 both are divisible by 2, its an even number)

6q + 5, (6 is divisible by 2 but 5 is not divisible by 2, its an odd number)

Therefore, odd numbers will be in the form 

6q + 1, or 6q + 3, or 6q+5

Hence, Proved.

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NCERT Solutions Class 10th Math Chapter 1 Real Numbers Exercise 1.1 ques no 3 In English and Hindi With Videos and pdf |

Ques 3:- In a parade, an army contingent of 616 members is to march behind an army band of 32 members. Both the groups are to march in the same number of columns what is the maximum number of columns in which they can march.

Solution:  Suppose, both groups are arranged in 'n' columns, for completely filling each column,

The maximum no of columns in which they can march is the highest common factor of their number of members. 

i.e. n = HCF(616, 32)

By using, Euclid’s division algorithm

616 = 32×19+8

Remainder ≠ 0

So, again Applying Euclid’s division algorithm.

32 = 8×4+0

HCF of (616, 32) is 8.

So, They can march in 8 columns each.

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NCERT Solutions Class 10th Math Chapter 1 Real Numbers Exercise 1.1 ques no 4 In English and Hindi With Videos and pdf |

Ques 4:- Using Euclid's division lemma, show that the square of a positive integer is of the form 3m or 3m+1 for some integer m.
[ Hint:- Let x be any positive integer. Then it can be written as 3q, 3q+1 or 3q+2. Square each of these and show that these squares can be written in the form 3m or 3m+1. ]

Solution: To Prove: Square of any number is of the form 3 m or 3 m +1

Proof: To prove this statement from Euclid's division lemma, take any number as a divisor, in question we have 3m and 3m + 1 as the form.

So, By taking, ‘a’ as any positive integer and b = 3.

Applying Euclid’s algorithm a = bq + r.

a = 3q + r

Here, r = remainder = 0, 1, 2 and q ≥ 0 as the divisor is 3 there can be only 3 remainders, 0, 1 and 2.

So, putting all the possible values of the remainder in, 

a = 3q + r

a = 3q + 0 or 3q+1 or 3q+2

And now squaring all the values,

When a= 3q Squaring both sides we get,

a² = (3q)²

a² = 9q²

a² =3 (3q²)

a² = 3 k1

Where k1 = 3q²

When a = 3q+1, Squaring both sides we get,

a² = (3q + 1)²

a² = 9q² + 6q + 1

a² =3( 3q² + 2q )+ 1

a² = 3k² + 1

Where k2 = 3q² + 2q

When a = 3q+2, Squaring both sides we get, 

a² = (3q + 2)²

a² = 9q2 + 12q + 4

a² = 9q2 + 12q + 3+1

a² = 3(3q2 + 4q + 1) +1

a² = 3k3 + 1

Where k3 = 3q² + 4q + 1

Where k1, k2 and k3 are some positive integers.

Hence, it can be said that the square of any positive integer is either of

the form 3m or 3m+1.

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NCERT Solutions Class 10th Math Chapter 1 Real Numbers Exercise 1.1 ques no 5 In English and Hindi With Videos and pdf |

Ques 5:- Using Euclid's division lemma, show that the cube of a positive integer is of the form 9m, 9m+1 or 9m+8.

Solution: Let a be any positive integer. Then, it is of the form.....

3q + 0 or, 3q + 1 or, 3q + 2.

We know that according to Euclid's division lemma: a = bq + r So, we have the following cases:-

Case I .....When a = 3q

In this case, we have

a³ = (3q)³ = 27q³ = 9(3q)³ = 9m, 

where m = 3q³

Case II..... When a = 3q + 1

In this case, we have

a³ = (3q + 1)³

⇒27q³ + 27q² + 9q + 1

⇒9q(3q² + 3q + 1) + 1

⇒a³ = 9m + 1, 

where m = q(3q² + 3q + 1)

Case III .....When a = 3q + 2

In this case, we have

a³ = (3q + 1)³

⇒27q³ + 54q² + 36q + 8

⇒9q(3q² + 6q + 4) + 8

⇒a³ = 9m + 8, 

where m = q(3q² + 6q + 4)

Hence, a³ is the form of 9m or, 9m + 1 or, 9m + 8

Chapter 1 (Real Numbers) Exercise 1.2 Solution

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