# NCERT Solutions for Class 10 Math Chapter 1 Real Numbers Exercise 1.1

In this article we have given NCERT Solutions for Class 10 Math Chapter 1 Exercise 1.1 with pdf. These NCERT solutions are solved by subjects experts from latest edition book.

## NCERT Textbook Class 10 Math Exercise 1.1

Question 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) We have 135 and 225
Since, 225>135, applying Euclid’s Division Lemma to 225 and 135,
We get, 225=135X1+90
Since, remainder 90 ≠0, so applying Euclid’s division lemma to 135 and 90,
We get 135=90X1+45
Since, remainder 45≠0, so applying Euclid’s division lemma to 90 and 45 and, we get
90=45X2+0
Since remainder =0
∴HCF (135, 225) is 45.

(ii) We have 196 and 38220
Since 38220> 196, applying Euclid’s division
Lemma to 38220 and 196, we get
38220=196X195+0
Since remainder=0
∴HCF (196, 38220) is 196.

(iii) We have 867 and 255
Since 867>255, applying Euclid’s division lemma to 867 and 102,
We get 867=255X3+102
Since remainder 102=0, applying Euclid’s division lemma of 102 and 51,
we get 102=51X2+0
∴ HCF (867, 255) is 51.

Question 2. Show that any positive odd integer is of the from6q+1, or 6q+3, 6q+5, where q is form integer.
Solutions: let a be any positive odd integer and b=6, then by Euclid’s division lemma
a=6q+r, r=0,1,2,3,4,5
When ,

Values of a in (i), (ii) and (iii) is not possible as a is odd +ve integer.
Therefore, any +ve odd integer is of the form
(6q+1), (6q+3) or (6q+50).

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: Let a = 616 and b = 32
Applying Euclid’s division lemma to 616 and 32, we get
616=32×19+ 8
Now, applying Euclid’s division lemma of 32 and 8, we get
32=8×4+0
Since, remainder=0
HCF (612, 32) =8
Therefore, maximum number of columns=8.

## Class 10 Math Chapter 1 Exercise 1.1 NCERT Solution

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.
[Hint: Let x be any positive integer then it is of the form 3q, 3q+1 or 3g +2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]
Solution: Let a be ay positive integer and b=3, then by Euclid’s division lemma a=3q+r and r=0, 1, 2
When,

Squaring eqn. (i) on both side, we get
a2 =9q2
a2=3X3q2
=3m, where m=3q2

Squaring eqn. (iii) on Both sides, we get
a2 = (3q+1)2
= 9q2 +1+6q
= 3(3q2 +2q)+1
= 3m+1, where m=3q2

Squaring eqn. (iii) on both side
a2 =(3q+2)2
=9q2 +4+12q
=9q2+3+1+12q
=3(3q2+1+4q) +1
=3m+1,
where m=3q2+1+4q.

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: let a be any +ve integer and be=3, then by Euclid’s division Lemma a=3q+r and r=0,1,2
When,

Taking cube on both side in eqn. (i) we get
a3 =(3q)3
=27q3
=9X3q3=9m, where m=3q3.

Taking cube on both sides in eqn. (ii), we get
a3=(3q +1)3
=27q3 = 1+27q2+9q
=9(3q3+3q2 +q) +1=9m+1,
Where m=3q3+3q3+q

Taking cube on both sides in eqn. (iii), we get
a3=(3q+2)3
=27q3+8+54q2+36q
=27q3+54q2 +36q+8
=9(3q3+6q2+4q)+8
=9m+8,
Where m=3q3+6q2+4q.

## NCERT Class 10 Math Exercises 1.1 Pdf Download

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