Exercise 1.1 Class 10 Maths – Real Numbers Complete Solutions

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exercise 1.1 class 10 mahs

Exercise 1.1 Class 10 Maths

Class 10 Maths Chapter 1 Exercise 1.1

Class 10 Maths Chapter 1 Exercise 1.1 : We are providing Class 10 Maths solutions  for 1st chapter “Real Numbers” textbook provided by NCERT for SEBA, CBSE etc complete exercise solutions with Additional Most Important Question in our website – online, absolutely for FREE for everyone. No need to do any sign in process, just browse and enjoy all of the Class 10 Maths Chapter 1 exercise 1.1 solutions from us. From the First exercise “Real Numbers”, there are total 4 exercise given in bellow table. Click on the exercise number to browse the complete solutions from ex 1.1 class 10 ..

You are at Real Numbers Exercise 1.1

Go to Real Numbers Exercise 1.2

Go to Real Numbers Exercise 1.3

Go to Real Numbers Exercise 1.4

Sub-Contents of “Chapter 1 Real Numbers” : There are  total 6 sub contents exists in Class 10 maths chapter 1 Real Numbers. We need to study all of the sub-contents first. The Sub Contents are-

  1. Introduction
  2. Euclid’s Division algorithm
  3. The Fundamental Theorem of Arithmetic
  4. Revisiting Irrational Numbers
  5. Revisiting Rational Numbers & Their Decimal Expansion
  6. Summary
FROM LATEST UPDATED NCERT SYLLABUS

 


Exercise 1.1 Class 10 Maths

 Real Numbers

Class 10 Maths Ex 1.1 Question 1 

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

  1. 135 and 225
  2. 196 and 38220
  3. 867 and 225
  4. 272 and 1032
  5. 405 and 2520
  6. 155 and 1385
  7. 384 and 1296
  8. 1848 and 4058

Solutions: (i) Here, 225 > 135

So, 225 = 135 × 1 + 90

⟹ 135 = 90 × 1 + 45

⟹ 90 = 45 × 2 + 0
 

Therefore, HCF( 135, 225 ) = 45

(ii) Here, 38220 > 196

So, 38220 = 196 × 195 + 0

Therefore, 196 is the greatest common divisor

So, HCF( 196, 38220 )  =  196

(iii) Here, 867 > 225

Now,   867 = 225 × 3 + 192

⇒225 = 192 × 1 + 33

⇒ 192 = 33 × 5 + 27

⇒ 33 = 27 × 1 + 6

⇒ 27 = 6 × 4 + 3

⇒ 6 =  3 × 2 + 0

So, HCF( 867, 225 ) = 3

(iv) Here, 1032 > 272

So, 1032 = 272 × 3 + 216

⇒272 = 216  ×  1 + 56

⇒216 = 56  ×  3 + 48

⇒56 = 48 ×  1 + 8

⇒48 = 8  ×  6 + 0

So, HCF(272, 1032) = 8

(v) Here, 2520 > 405

So, 2520 = 405 × 6 + 90

⇒ 405 = 90 × 4 + 45

⇒ 90 = 45 × 2 + 0

So, HCF(2520, 405) = 45

(vi) Here, 1385 = 155 × 8 + 145

⇒ 155 = 145 × 1 + 10

⇒ 145 = 10 × 14 + 5

⇒ 10 = 5 × 2 + 0

So, HCF(155, 1385) = 5

(vii) Here, 1296 > 384

So, 1296 = 384 × 3 + 144 

⇒ 384 = 144 × 2 + 96

⇒ 144 = 96 × 1 + 48

⇒ 96 = 48 × 2 +   0

So, HCF(384, 1296) = 48

(viii) Here, 4058 > 1848

So, 4058 = 1848 ×  2 + 362 

⇒ 1848 = 362 × 5 + 38

⇒ 362 = 38 × 9 + 20

⇒ 38 = 20 × 1 + 18

⇒ 20 = 18 × 1 + 2

⇒ 18 = 2 × 9 + 0

So, HCF(4058, 1848) = 2

[ Step by Step Hints/Rules are given in bellow portion. Please check if not understand ]


Class 10 Maths Chapter 1 Exercise 1.1 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: For any two integers a & b, from Euclid’s Lemma, we can get another two integers q & r, so that,

a = bq + r  where 0 ≤ r < b

Suppose, b = 6, then r = 0, 1, 2, 3, 4, 5

For r = 0,  a = 6q, which is even

For r = 1, a = 6q + 1, which is odd

For r = 2, a = 6q + 2, which is even

For r = 3, a = 6q + 3, which is odd

For r = 4, a = 6q + 4, which is even

For r = 5, a = 6q + 5, which is odd

Finally, we see that, at r = 1, 3, 5 , the values of “a” are odd.

So, any positive odd integers is of the form 6q + 1, 6q + 2 or 6q + 3

Class 10 Math Ex 1.1 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: Here, The highest common factor of 616 & 32 will be the maximum number of columns in which they can march. So, we need to find the HCF( 616, 32 )

So, 616 = 32 × 19 + 8

⇒ 32 = 8 × 4 + 0

So, HCF( 616, 32) = 8

So, maximum no of columns in which they can march is 8

Class 10 Math Ex 1.1 Question 4

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

Solution: For any two integers a & b, from Euclid’s Lemma, we can get another two integers q & r, so that,

a = bq + r  where 0 ≤ r < b

Let, b = 3, then r = 0, 1, 2

For r = 0, a = 3q

⇒a² = (3q)² = 3.3q² = 3m

Where m = 3q²

For r = 1, a = 3q + 1

⇒a² = (3q + 1)² = 9q² + 6q + 1

⇒a² = 3(3q² + 2q) + 1

⇒a² = 3m + 1, where m = 3m + 1

For, r = 2 , a = 3q + 2

⇒ a² = (3q + 2)² = 9q² + 12q + 4

⇒a² = 9q² + 12q + 3 + 1

⇒a² = 3(3q² + 4q + 1) + 1

⇒a² = 3m + 1, where m = 3q² + 4q + 1

We see that, for r = 0, a² = 3m & for r = 1 & 2, a² = 3m + 1

That means, the square of any positive integer is of the form 3m or 3m + 1.

[ All the formulae used here are given in bellow portion, Please check ]

Class 10 Math Ex 1.1 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: For any two integers a & b, from Euclid’s Lemma, we can get another two integers q & r, so that,

a = bq + r  where 0 ≤ r < b

Let, b = 3, then r = 0, 1, 2

For r = 0, a = 3q

⇒a³ = (3q)³ = 9.3q³  = 9m

Where m = 3q³

For r = 1, a = 3q + 1

⇒a³ = (3q + 1)³

⇒a³ = (3q)³ + 3.(3q)².1 + 3.3q.1 + 1³

⇒a³ = 27q³ + 27q² + 9q + 1

⇒a³ = 9(3q³ + 3q² + q) + 1

⇒a³ = 9m + 1, where m = 3q³ + 3q² + 1

For r = 3, a = 3q + 2

⇒a³ = (3q + 2)³

⇒a³ = (3q)³ + 3.(3q)².2 + 3.3q.2² + 2³

⇒a³ = 27q³ + 54q² + 36q + 8

⇒a³ = 9(3q³ + 6q + 4q) + 8

⇒a³ = 9m + 8, where m = 3q³ + 6q + 4q

We see that, at r = 0, a³ = 9m; at r = 1, a³ = 9m + 1 and at r = 2, a³ = 9m + 8

That means, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8

[ All the formulae used here are given in bellow portion, Please check ]

Chapter 10 Maths Ex 1.1 Question 6

Himadri has a collection of 625 indian postal stumps and 325 international postal stumps. She wants to display them in identical groups of indian and international stumps with non stumps left out. What is the greatest number of groups Himadri can display the stumps?

Solution: Clearly, The highest common factor of 625 and 325 will be the greatest number of groups that Himadri can display the stumps.

Here, 625 > 325

so, 625 = 325 ×  1 + 300

⇒ 325 = 300 × 1 + 25

⇒ 300 = 25 × 12 + 0

so. HCF( 625, 325) = 25

So, the greatest number of groups that Himadri can display the stumps is 25.

Chapter 10 Maths Ex 1.1 Question 7

Two ropes are of length 64 cm and 80 cm. Both are to be cut into pieces of
equal length. What should be the maximum length of the pieces?

Solution: The HCF of 64 and 80 will be the maximum lenghts of the pieces.

Here 80 > 64

So, 80 = 64 × 1 + 16

⇒ 64 = 16 × 4 + 0

So, HCF(80, 60) = 16

Therefore, Maximum lenghts of the pieces is 20cm.


Formulae Used in Exercise 1.1

Some other algebraic formulae that are used in this ex 1.1 maths to solve or prove the problems are given bellow:

(i) Maths Ex 1.1 Question 4

Formula Used:

(a + b)² = a² + 2ab + b²

(ii) Maths Ex 1.1 Question 5

Formula Used:

(a + b)³ = a³ + 3a²b + 3ab² + b³

Steps To Find HCF for Ex 1.1 Question 1

To solve the problems given in Class 10 maths exercise 1.1 Question 1 from Class 10 Math Textbook, we need to follow some steps, which are given bellow:

  1. Firstly, Compare the numbers and take the biggest number.
  2. Secondly, Write the biggest number in LHS and now divide this by the small number
  3. Write whole this system using the Euclid’s Lemma Method which is:
Dividend = Divisor × Quotient + Remainder

 

4. Repeat the step until the remainder becomes “0”

5. The divisor of the step, when the remainder become “0”, is the final solution, means the HCF

NCERT SOLUTIONS FOR CLASS 10 MATHS :

We are providing complete Class 10 Maths Solution . You can get complete and updated solution from our website. Please click on your prefered chapter bellow from NCERT CLASS 10 MATHS and see the solutions. You can get complete ncert solutions for class 10 from here. 

1 Chapter 1  Real Numbers
2 Chapter 2  Polynomials
3 Chapter 3  Pair of Linear Equations in Two Variables
4 Chapter 4  Quadratic Equations
5 Chapter 5  Arithmetic Progressions
6 Chapter 6  Triangles
7 Chapter 7  Co-ordinate Geometry
8 Chapter 8  Introduction to trigonometry
9 Chapter 9  Application of Trigonometry
10 Chapter 10  Circle
11 Chapter 11  Constructions
12 Chapter 12  Area Related to Circles
13 Chapter 13  Surface area and volumes
14 Chapter 14 Statistics
15 Chapter 15  Probability

 

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