NCERT Solutions for class 10 maths chapter 1

Real Numbers

NCERT Solutions for class 10 maths chapter 1 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 225

Answer 1

i. 135 and 225
225 is greater than 135. Therefore, by Euclid’s division algorithm
225 = 135 × 1 + 90
remainder 90 ≠ 0, Again
135 = 90 × 1 + 45
Again, 45 ≠ 0,
90 = 45 × 2 + 0
The remainder is now zero in the last step, the divisor is 45.
Hence, the HCF of 225 and 135 is 45.

ii. 196 and 38220
38220>196,
therefore the by applying Euclid’s division algorithm and taking 38220 as divisor,
38220 = 196 × 195 + 0
remainder as 0 here So
Hence, the HCF of 196 and 38220 is 196.

iii. 867 and 225
867 is greater than 225. Now apply now Euclid’s division algorithm on 867,
867 = 225 × 3 + 102
Again Remainder 102 ≠ 0,
225 = 102 × 2 + 51
Again, 51 ≠ 0.
102 = 51 × 2 + 0
The remainder is now zero in the last step, the divisor is 51,
Hence, the HCF of 867 and 225 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.

Answer 2

Let a be any positive integer and b = 6.
by Euclid’s algorithm,
a = 6q + r, where q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.
If r = 0, then a = 6q
Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively. here a = 6q, 6q+2, 6q+4,
a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 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?

Answer 3

Number of army contingent members=616
Number of army band members = 32
The maximum number of columns in which they can march=HCF(616, 32).
By Using Euclid’s algorithm to find their HCF, we get,
Since, 616>32,
616 = 32 × 19 + 8
Since, 8 ≠ 0, therefore, taking 32 as new divisor
32 = 8 × 4 + 0
Now we have got remainder as 0 So,HCF (616, 32) = 8.
Hence, 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.

Answer 4

Let 'a' be any positive integer and b = 3.
So, a= 3q + r for some integer q≥0 and
r = 0, 1, 2, as r ≥ 0 and r < 3.
Or a = 3q, 3q+1 and 3q+2
by squaring both the sides, we get,
a² = (3q)² or (3q +1)²
=(9q²) or 9q² + 6q +1
Let 3q² = k
Therefore, a²= 3k .........(i)or
a² = (3q + 1)² = (3q)²+1²+2×3q×1 = 9q² + 1 +6q = 3(3q²+2q) +1 3q²+2q =k
a²= 3k + 1 .........(ii)
a²= (3q + 2)² = (3q)²+2²+2×3q×2 = 9q² + 4 + 12q = 3 (3q² + 4q + 1)+1
Again,3q² + 4q + 1 = k,
a²= 3k + 1.....................(iii)
from equation i,ii and iii,
we can say that, 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.

Answer 5

Let a be any positive integer and b= 3. By Euclid’s division algorithm, then, a = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3. Therefore, putting the value of r,
q = 3q
or
3q + 1
or
3q + 2
Now, by taking the cube of all
Case (i): When r = 0, then,
a³= (3q)³ = 27q³= 9(3q³)= 9m; where m = 3q³
Case (ii): When r = 1, then,
a³ = (3q+1)³ = (3q)³ +1³+3×3q×1(3q+1) = 27q³+1+27q²+9q
x³ = 9(3q³+3q²+q)+1
Putting = m, we get,
Putting (3q³+3q³+q) = m, we get ,
a³ = 9m+1
Case (iii): When r = 2, then,
a³ = (3q+2)³= (3q)³+2³+3×3q×2(3q+2) = 27q³+54q²+36q+8
a³=9(3q³+6q²+4q)+8
Putting (3q³+6q²+4q) = m
a³= 9m+8
it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.