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 ,

r=0,a=6q ….(i)
r=1,a=6q+1
r=2,a=6q+2 ….(ii)
r=3,a=6q+3
r=4,a=6q+4 …..(iii)
r=5,a=6q+5

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,

r=0,a=3q ….(i)
r=1,a=3q+1 ….(ii)
r=2,a=3q+2 ….(iii)

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,

r=0,a=3q ….(i)
r=1,a=3q+1 ….(ii)
r=2,a=3q+2 ….(iii)

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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