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

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.