NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers Ex 1.1

Here we are giving you the solution of NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1 . You can also download Free NCERT Solutions for Class 10 Maths Chapter 1 PDF on our website . NCERT Maths class 10 chapter 1 exercise 1.1 solutions This post has been prepared by the experienced teachers of EducationForIndia.com . Detailed answers of all the questions in Class 10th Maths Chapter 1 provided in NCERT TextBook .

NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers 

EXERCISE 1.1

Exercise 1.1 Class 10 Maths Question 1

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

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Solutions:

(i) 135 and 225

In the above question, 225 is greater than 135. So, according to Euclid’s division algorithm, we will take

225 = 135 x 1 + 90

As the remainder 90 ≠ 0 so we have to use the division lemma again for 90, to get

135 = 90 x 1 + 45

After that, since 45 ≠ 0 so we have to repeat the same process for 45

90 = 45 x 2 + 0

Here we have got the remainder 0 so we will stop the process over here. As our last divisor is 45

Therefore, HCF of (5) = (15, 90) = (90, 45) is 45

Ans. The HCF of 135 and 225 is 45.

(ii) 196 and 38220

According to the above question, 38220 is greater than 196, so after applying Euclid’s division algorithm and taking 38220 as the divisor we will get the equation

38220 = 196 x 195 + 0

As in the very first step we have got the remainder 0, therefore we will stop here

HCF of (38220, 196) = 196

Ans. The HCF of 196 and 38220 is 196.

(iii) 867 and 255

Here the number 867 is greater than 255. After applying Euclid’s division algorithm on 867, we will get,

867 = 255 x 3 + 102

The Remainder 102 ≠ 0, then by taking 255 as the divisor and applying the division lemma method, we will get,

255 = 102 x 2 + 51

Now 51 ≠ 0, so we will consider 102 as the new divisor, after that we have to repeat the same step and we will get,

102 = 51 x 2 + 0

As of now, we have the remainder zero, so the process will stop here. In the very last step, the divisor is 51, meanwhile, HCF of (867,255) = (255, 102) = (102,51) = 51.

Ans. The HCF of 867 and 255 is 51.

Ex 1.1 Class 10 Maths Question 2

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

Solution:

Let’s take a as any positive integer and take b = 6. After that according to Euclid’s algorithm a = 6q + r and for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6

Let’s substitute the value of r we will get,

If r = 0 than a = 6q

As same as for r = 1, 2, 3, 4, and 5, and the value of a is 6q + 1, 6q +2, 6q +3, 6q +4, and 6q +5.

If a is having a = 6q, 6q +2, 6q +4 then we can conclude that a is an even number and is divisible by 2. We know that a positive integer can be even or odd. Hence, any positive integer is of the form of 6q +1, 6q +3, 6q +5 where q is an integer.

Class 10 Maths Chapter 1 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:

Here, the trick into the question is in the point that the army band members and army contingent members have to march into the same particular columns and that the number of columns must be the maximum possible number.

HCF states the highest number that can be divided exactly into each of two or more numbers. In other words, the HCF of two numbers is the highest number that will divide both the numbers.

Thus, we have to find the HCF of the army band members and the army contingent.

HCF of 616 and 32 will give the maximum number of columns where they will march and for this, we will use Euclid’s algorithm to find the H.C.F:

As 616 >32 so we will divide 616 by 32

We have to divide 32 by 8 now because the remainder is not 0 it is 8

Therefore, the possible maximum number of columns is 8 where they can march

Ans. The HCF of 616 and 32 is 8.

Class 10 Maths Ex 1.1 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 form 3m or 3m + 1.]
Solution:

Let us consider ‘a’ as a positive integer, then

We know a=b q + r, 0≤r<b

Now, a=3q+r, 0≤r<3

The possible remainders could be either 0 or 1 or 2

Now let’s check out the following cases,

We know the thing that, a = 3q or 3q + 1 or 3q + 2 

(a)² = (3q)² or (3q + 1)² or (3q + 2)²

a² = 3(3q²) or (9q² + 6q + 1) or (9q² + 12q + 4)

Case I: a² = 3(3q²) where, m = 3q²

Case II: a² = 3(3q² + 2q) + 1 where, m = (3q² + 2q)

Case III: a² = 3(3q² + 4q +1) + 1 where, m = (3q² + 4q +1)

Thus, can conclude and see from the above cases that a² is of the form 3m or 3m + 1 where, m can be any positive integer.

Ans. So at last, we can say that the square of any positive integer is either of form 3m or 3m + 1.

Exercise 1.1 Class 10 Maths NCERT Solutions 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’s consider a positive integer ‘a’, then

According to Euclid’s division lemma if a and b will be two positive integers then

a = b q + r, where 0 ≤ r < b

Let’s take b = 3, then 0 ≤ r < 3, that can be, r = 0 or 1 or 2 but it can’t be 3 because r is smaller than 3.

That is why the possible values for ‘a’ are 3q or 3q + 1 or 3q + 2.

Now we have to find the cube for all the possible values of ‘a’.

where q is any positive integer, then its cube which we will take as ‘m’ will also be a positive integer.

Now see carefully that the cube of all the positive integers is either of form 9m or 9m + 1 or 9m + 1 for some integer m.

Let’s take ‘a’ as any positive integer and q = 3. Then, a = 3q + r for some integer q ≥ 0 and 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Now let’s check out the following cases

Case I: If a = 3q

(a)³ = (3q)³ = 27q³

= 9 (3q³)

= 9m, where m is an integer such that m = 3q³

Case II: If a = 3q + 1

(a)³ = (3q + 1)³

(a)³ = 27q³ + 27q² + 9q + 1

(a)³ = 9(3q³ + 3q² + q) + 1

(a)³ = 9m + 1, where m is an integer such that m = 3q³ + 3q² + q

Case III: If a = 3q + 2

(a)³ = (3q + 2)³

(a)³ = 27q³ + 54q² + 36q + 8

(a)³ = 9(3q³ + 6q² + 4q) + 8

(a)³ = 9m + 8, where m is an integer such that m = 3q³ + 6q² + 4q

Ans. Hence, it can be seen that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.