## Xpert Solutions For Class X Math Chapter 1: Real Numbers

Excercise 1.1

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) Check Greater Value 225 > 135 ,Always greater value divide with small number

Divide 225 by 135 get 1 quotient and 90 remainder as
225 = 135× 1 + 90

Divide 135 by 90 get 1 quotient and 45 remainder as
135 = 90×1 + 45

Divide 90 by 45 get 2 quotient and NO remainder as
90 = 45 × 2 + 0

As there is no remainder and divisor is 45,so HCF is 45

(ii) 38220 > 196

Divide  38220 by 196 then we get quotient 195 and no remainder
38220 = 196 × 195 + 0

So no remainder, 196 is divisor therefore HCF is 196

(iii) 867 > 255

Divide 867 by 255 we get quotient 3 and remainder 102 ,so
867 = 255 × 3 + 102

Now, divide 255 by 102,then we get quotient is 2 and remainder 51 as
255 = 102 × 2 + 51

Divide 102 by 51, then we get quotient 2 and no remainder,so

102 = 51 × 2 + 0

As there is no remainder ,so divisor is 51 is HCF

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

Sol: We know Euclid’s Algorithm is
a =  bq + r , Given equation is : 6q + 1,

so b(Divisor) =6, Now we get equation that 6q +r ,r is remainder and q(Quotient) is equal to or more than 0, r = 1,2,3,4,5 because  0 ≤ r ≤ 6

So possible form is

6q + 0,      6 is divisible by 2,so it is a even number

6q+ 1,       6 is divisible by 2 but 1 is not divisible by 2, so it is odd number

Therefore 6q + 5,  6 is divisible by 2 but 5 is not divisible by 2, so it is odd number

3. An army contingent of 616 members is to march behind an army band of 32 members is 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?

Sol: By HCF (616,32) we can find maximum number of column in which they can march. (HCF means Highest Common Fector)

By Euclid’s  Algorithm to find HCF

616 = 32 × 19 + 8
35 = 8 × 4 + 0

The HCF(616,32) is 8.
Therefore, they can march in 8 column each.

4. Use Euclid’s division lemma to show that the square of any 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 the form 3m or 3m + 1.]
Sol:
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a² = (3q)² or (3q + 1)² or (3q + 2)²
a² = 9q² or 9q² + 6q + 1 or 9q² + 12q + 4
= 3 × (3q²) or 3(3q² + 2q) + 1 or 3(3q² + 4q + 1) + 1
= 3k₁ or 3k₂ + 1 or 3k₃ + 1

Where k₁, k₂, k₃ are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1

1. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9+ 1 or 9m+ 8.

Sol:

Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,
Where m is an integer such that m = 3q3

Case 2: When = 3q + 1,
a3 = (3q +1)3
a3= 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)

Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3= 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)

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

Excercise 1.2

1. Express each number as product of its prime factors:

(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Sol:

(i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23

1. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Sol:
(i) 26 = 2 × 13
91 =7 × 13
HCF = 13  (HCF is Higher Common Factor. Choose values those exist both sides)
LCM =2 × 7 × 13 =182    (LCF is Least common Factor)
Product of two numbers 26 × 91 = 2366
Product of HCF and LCM 13 × 182 = 2366
Hence, product of two numbers = product of HCF × LCM

Other way to find LCM is:- LCM :-   13 × 2 × 7 = 182  ( But Our first priority should by divide by 2 then 3 and so on other prime numbers)
(ii) 510 = 2 × 3 × 5 × 17
92 =2 × 2 × 23
HCF = 2
LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of two numbers 510 × 92 = 46920
Product of HCF and LCM 2 × 23460 = 46920
Hence, product of two numbers = product of HCF × LCM

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 2 × 3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024
Product of two numbers 336 × 54 =18144
Product of HCF and LCM 6 × 3024 = 18144
Hence, product of two numbers = product of HCF × LCM.

1. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Sol:
(i) 12 = 2 × 2 × 3
15 =3 × 5
21 =3 × 7
HCF = 3
LCM = 2 × 2 × 3 × 5 × 7 = 420

Excercise 1.3

1. Prove that √5 is irrational.

Sol:

Let take √5 as rational number
If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get
b√5 = a
To remove root, Squaring on both sides, we get
5b2 = a2 …  (i)

Therefore, 5 divides a2 (i.e b₂ = a₂ / 5) and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also.
That means 5 will divide a. So we can write
a = 5c
Putting value of a in equation (i) we get
5b2 = (5c)2
5b2 = 25c2
Divide by 25 we get

b2/5 = c2

Similarly, we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it contradicts.
Hence √5 is not a rational number, it is irrational.

1. Prove that 3 + 2√5 is irrational.

Sol:

Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a-3b)/b
Now divide by 2, we get
√5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.
Hence, 3 + 2√5 is a irrational number.

1. Prove that the following are irrationals:
(i) 1/√2 (ii) 7√5 (iii) 6 + √2

Sol:

(i) Let take that 1/√2 is a rational number.
So we can write this number as
1/√2 = a/b
Here and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get
1 = (a√2)/b
Now multiply by b
b = a√2
divide by a we get
b/a = √2
Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts.
Hence, 1/√2 is a irrational number

(ii) Let take that 7√5 is a rational number.
So we can write this number as
7√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get
√5 = a/(7b)
Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts. Hence, 7√5 is a irrational number.

(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
√2 = a/b – 6
√2 = (a-6b)/b
Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number. But √2 is a irrational number so it contradicts.
Hence, 6 + √2 is a irrational number.