Real Numbers (Chapter 1, Exercise 1.1) – Class 10 Maths Solutions NCERT

The Chapter 1 of Class 10 Mathematics based on NCERT is about Real Numbers. This page contains the Solution of Exercise 1.1, Chapter 1 of Class 10 Mathematics. The Exercise 1.1 of Class 10 Mathematics consists of 5 questions which are based on Euclid’s Division Algorithm or Lemma.

DescriptionNCERT Maths Solutions
SubjectMathematics
Class10 (X)
Chapter1
Exercise1.1
TopicEuclid’s Division Lemma or Algorithm
SyllabusNCERT

The solution of all the 5 questions of Exercise 1.1 of Class 10 is given below.

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

i. 135 and 225

Solution: Here,

225 > 135

\therefore c=225 and d=135

Using Euclid’s Division Lemma,

c = dq +r

\Rightarrow 225=135\times 1+90

Since the remainder is not equal to zero (0), using Euclid’s Division Lemma,

\Rightarrow 135=90\times 1 +45

Again, the remainder is not equal to zero (0). Using Euclid’s Division Lemma,

\Rightarrow 90=45\times 2+0

The remainder becomes zero (0).

Therefore, the HCF of 135 and 225 is 45.

ii. 196 and 38220

Solution: Here,

38220 > 196

\therefore c=38220 and d=196

Using Euclid’s Division Lemma,

c=dq+r

\Rightarrow 38220=196\times 195+0

The remainder becomes zero (0).

Therefore, the HCF of 196 and 38220 is 196.

iii. 867 and 255

Solution: Here,

867 > 255

\therefore c=867 and d=255

Using Euclid’s Division Lemma,

c = dq +r

\Rightarrow 867=255\times 3+102

Since the remainder is not equal to zero (0), using Euclid’s Division Lemma,

\Rightarrow 255=102\times 2 +51

Again, the remainder is not equal to zero (0). Using Euclid’s Division Lemma,

\Rightarrow 102=51\times 2+0

The remainder becomes zero (0).

Therefore, the HCF of 867 and 255 is 51.

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:

Let a be any integer and b=6

From Euclid’s Division Algorithm,

a=bq+r

\Rightarrow a=6q+r

Case I: When r=1, a=6q+1

If we put, q=1, 2, 3, 4, 5, 6,…

a=7, 13, 19, 25, 31,…

Case II: When r=3, a=6q+3

If we put, q=1, 2, 3, 4, 5, 6,…

a=9, 15, 21, 27, 33,…

Case III: Similarly, When r=5, a=6q+5

If we put, q=1, 2, 3, 4, 5, 6,…

a=11, 17, 23, 29, 35,…

In all the above three cases, the value of a is a positive odd integer.

Therefore, any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

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:

Number of members in the army contingent =616

Number of members in the army band =32

The maximum number of columns in which they can march is the HCF of 616 and 32.

Here, 616 > 32

Using Euclid’s Division Algorithm,

c=dq+r

\therefore 616=32\times 19+8

Since the remainder is not equal to zero (0), using Euclid’s Division Algorithm,

\Rightarrow 32=8\times 4+0

The remainder becomes zero (0).

Therefore, the HCF of 616 and 32 is 8.

The maximum number of columns in which they can march is 8.

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:

Let’s consider a positive integer a.

From Euclid’s Division Lemma, for any two positive integers a and b, there exist unique integers q and r, such that a = bq + r, 0 ≤ r < b

Here, b = 3 and r = 0, 1, 2

\therefore\, a=3q\,or\,3q + 1\,or\,3q + 2

\Rightarrow (a)^2 = (3q)^2 \,or\,(3q + 1)^2\,or\,(3q + 2)^2

\Rightarrow a^2 = 3(3q^2)\,or\,(9q^2 + 6q + 1)\,or\,(9q2 + 12q + 4)

Now,

Case I: a^2 = 3(3q^2)  here, m = 3q^2

Case II: a^2 = 3(3q^2 + 2q) + 1 here, m = (3q^2 + 2q)

Case III: a^2 = 3(3q^2 + 4q +1) + 1 here, m = (3q^2 + 4q +1)

From the above three cases, a^2 is of the form 3m or 3m + 1 where, m is any positive integer.

Therefore, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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 positive integer and b=3.

From Euclid’s Division Lemma we know that

a = bq + r, 0 ≤ r < b

\therefore a = 3q + r and r=0,\,1,\,2

Case I: When r = 0,

a = 3q

\Rightarrow a^3 = (3q)^3

\Rightarrow a^3 = 27q^3

\Rightarrow a^3 = 9(3q^3)

\Rightarrow a^3 = 9m, where m = 3q^3

Case II: When r = 1,

a = 3q + 1

\Rightarrow a^3 = (3q + 1)^3

\Rightarrow a^3 = (3q)^3 + 13 + 3 × 3q × 1(3q + 1)

\Rightarrow a^3 = 27q^3 + 1 + 9q × (3q + 1)

\Rightarrow a^3 = 27q^3 + 1 + 27q^2 + 9q

\Rightarrow a^3 = 27q^3 + 27q^2 + 9q + 1

\Rightarrow a^3 = 9 ( 3q^3 + 3q^2 + q) + 1

\Rightarrow a^3 = 9m + 1, where m = ( 3q^3 + 3q^2 + q)

Case III: When r = 2,

a = 3q + 2

a^3 = (3q + 2)^3

a^3 = (3q)^3 + 23 + 3 × 3q × 2 (3q + 1)

a^3 = 27q^3 + 8 + 54q^2 + 36q

a^3 = 27q^3 + 54q^2 + 36q + 8

a^3 = 9 (3q^3 + 6q^2 + 4q) + 8

a^3 = 9m + 8, where m = (3q^3 + 6q^2 + 4q)

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

I hope the solutions of Class 10, Exercise 1.1, Chapter 1 of Mathematics have helped you. If you have any issues/queries related to the NCERT Solutions of Class 10 Mathematics, Exercise 1.1, feel free to contact me at [email protected] or [email protected] or fill the form here.

Related Posts

4.8/5 - (34 votes)
Love this Post? Share with Friends

Leave a Comment