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Kindly provide NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 to complete homework.

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Here you will find free NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 PDF to complete your homework on time. Ex 1.1 CBSE Class 10 Maths NCERT Solutions provided here will help you in passing the exam with more marks.

Book NameClass 10 Mathematics NCERT Textbook
ChapterChapter 1 Real Numbers
ExerciseEx 1.1


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

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255.

Solution

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 225 = 135 x 1 + 90

Since remainder 90 = 0, we apply the division lemma to 135 and 90 to obtain 135 = 90 x 1 + 45.

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain 90 = 2 x 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 x 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 x 3 + 102.

Since remainder 102 = 0, we apply the division lemma to 255 and 102 to obtain 255 = 102 x 2 + 51/

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain 102 = 51 x 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.

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

Let a be any positive integer and b = 6.

Then, by Euclid's algorithm, a = 69 + r for some integer q 2 0, and r = 0, 1, 2, 3, 4, 5 because o sr< 6.

Therefore, a = 69 or 69 + 1 or 69 + 2 or 69 + 3 or 69 + 4 or 69 + 5.

Also, 69 + 1 = 2 x 39 + 1 = 2k1 + 1, where ki is a positive integer

69 + 3 = (69 + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

69 + 5 = (69 + 4) + 1 = 2 (39 + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly, 

69 + 1, 69 + 3, 69 + 5 are of the form 2k + 1, where k is an integer. Therefore, 69 + 1, 69 + 3,69 + 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 69 + 1, or 69 + 3, or 69 + 5.

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

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 x 19 + 8 32 = 8 x 4 + 0 The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

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

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q 2 0 And r = 0, 1, 2 because o ≤ r 3.

Therefore, a = 30 or 39 + 1 or 3q + 2 Or,

Where kı, 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.

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

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

a = 31 or 31 + 1 or 31 + 2 Therefore, every number can be represented as these three forms. There are three cases.

Case 1:

When a = 3q

Case 2:

When a = 3q + 1

Case 2:

When a = 3q + 2

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Given CBSE NCERT Solutions for Class 10 Maths Exercise 1.1 can help you in scoring good marks in the class test and board exams.

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