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Exercise 1.1 - Chapter 1 Real Numbers class 10 ncert solutions Maths - Free PDF Download 2022 - 2023

Question 1:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1784/Chapter%201_html_m28d8c39e.gif

Answer:

(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 https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1791/Chapter%201_html_6a224ceb.gif, or https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1791/Chapter%201_html_m26c9acdc.gif, or https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1791/Chapter%201_html_m7c05f18e.gif, where q is some integer.

Answer:

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k 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?

Answer:

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.]

Answer:

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1802/Chapter%201_html_5b274860.gif

Where k1k2, 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.

 

Question 5:

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

Answer:

Let a be any positive integer and b = 3

a = 3q + r, where ≥ 0 and 0 ≤ r < 3

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1807/Chapter%201_html_m4d1951f9.gif

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1807/Chapter%201_html_4ca7dbe6.gif

Where m is an integer such that m = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/128/1807/Chapter%201_html_3652ee16.gif

Case 2: When a = 3q + 1,

a= (3q +1)3

a3 = 27q27q9q + 1

a3 = 9(3q3q+ q) + 1

a9m + 1

Where m is an integer such that m = (3q3q+ q)

Case 3: When a = 3q + 2,

a= (3q +2)3

a3 = 27q54q36q + 8

a3 = 9(3q6q4q) + 8

a9m + 8

Where m is an integer such that m = (3q6q4q)

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

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