NCERT Class 10th Math chapter 1 Exercise 1.1 Question Answer. NCERT Solution for Class 10 Maths Chapter 1 Exercise 1.1 of real numbers NCERT Solution for CBSE, HBSE, RBSE and Up Board and all other chapters of class 10 math.
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NCERT Class 10 Math chapter 1 Real Numbers Exercise 1.1 Question Answer Solution.
Class 10 Math Chapter 1 Ex. 1.1 Question Answer
1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
Ans. Since 225 > 135, we apply the division lemma to 225 and 135, to get
225 = 135 × 1 + 90
Since the remainder 90 ≠ 0, we apply the division lemma to 135 and 90, to get
135 = 90 × 1 + 45
We consider the new dividend 90 and new divisor 45 to get
90 = 45 × 2 + 0
As we get remainder as zero, In this stage our divisor is 45.
Therefore, the HCF of 225 and 135 is 45.
(ii) 196 and 38220
Ans. Since 38220 > 196, we apply the division lemma to 38220 and 196, to get
38220 = 196 × 195 + 0
As we get remainder as zero, In this stage our divisor is 196
Therefore, the HCF of 38220 and 196 is
(iii) 867 and 255
Ans. Since 867 > 255, we apply the division lemma to 867 and 255, to get
867 = 255 × 3 + 102
Since the remainder 102≠ 0, we apply the division lemma to 255 and 102, to get
255 = 102 × 2 + 51
We consider the new dividend 102 and new divisor 51 to get
102 = 51 × 2 + 0
As we get remainder as zero, In this stage our divisor is 51.
Therefore, the HCF of 867 and 255 is 51.
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Ans. Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b= 6.
By applying Euclid’s algorithm a = 6q + r
Since 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.
That is, a can be 6q, or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is the quotient .
However, a is odd, a cannot be 6q or 6q + 2 or 6q + 4.
Therefore, any odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5.
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?
Ans. Maximum number of columns = HCF (616, 32)
Now, let us use Euclid’s algorithm to find their HCF. We have
616 = 32 × 19 + 8
32 = 8 × 4 + 0
So, the HCF of 616 and 32 is 8.
Therefore, the maximum number of columns in which they can march will be 8.
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 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1 .
Ans. Let a be any positive integer and b = 3.
By Euclid’s algorithm a = 3q + r, for some integer q≥0 and 0 ≤ r < 3.
Since 0 ≤ r < 3, the possible remainders are 0, 1 and 2.
So a = 3q, or 3q + 1, or 3q + 2.
Now taking the squares of all these cases.
Case (i) When r = 0
a = 3q ( squaring both sides )
a2 = (3q)2 = 9q2
taking 3 as a common factor
a2 = 3(3q2)
Putting m = 3q2
we get a2 = 3m
Case (ii) When r = 1
a = 3q + 1 (squaring both sides)
a2 = (3q + 1)2 = (3q)2 + 12 + 2(3q)(1) = 9q2 + 1 + 6q
Taking 3 as a common factor, we get
a2 = 3(3q2 + 2q) + 1
putting m = 3q2 + 2q
We get a2 = 3m + 1
Case (iii) When r = 2
a = 3q + 2 ( squaring both sides )
a2 = (3q + 2)2 = (3q)2 + 22 + 2(3q)(2) = 9q2 + 4 + 12q = 9q2 + 12q + 3 + 1
Taking 3 as a common factor, we get
a2 = 3(3q2 + 4q + 1) + 1
putting m = 3q2 + 4q + 1
We get a2 = 3m + 1
From all the cases we get that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
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.
Ans. Let a be any positive integer and b = 3.
By Euclid’s algorithm a = 3q + r
Since 0 ≤ r < 3, the possible remainders are 0, 1 and 2.
That is, a can be 3q, or 3q + 1, or 3q + 2, where q is the quotient.
Now taking the cubes of all these expressions.
Case (i) when r = 0
a = 3q (cubing both side)
a3 = (3q)3 = 27q3 = 9( 3q3 ) = 9m, where 3q3 = m
Case (ii) when r = 1
a = 3q + 1 ( cubing both side)
Identity used ( x + y )3 = x3 + y3 + 3xy( x + y )
a3 = (3q + 1)3 = (3q)3 + 13 + 3(3q)(1)(3q + 1) = 27q3 + 1 + 27q2 + 9q
Taking 9 as a common factor, we get
a3 = 9(3q3 + 3q2 + q) + 1
Therefore, a3 = 9m + 1, where ,m = 3q3 + 3q2 + q
Case (iii) When r = 2
a = 3q + 2 (cubing both sides)
a3 = (3q + 2)3 = (3q)3 + 23 + 3(3q)(2)(3q + 2) = 27q3 + 8 + 54q2 + 36q
Taking 9 as a common factor, we get
a3 = 9(3q3 + 6q2 + 4q) + 8
Therefore, a3 = 9m + 8, where m = 3q3 + 6q2 + 4q.
From all the cases above it is showed that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.