NCERT Solutions Class 10 Mathematics Chapter 1 Real Numbers

Exercise 1.1

Q.1) Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Sol.1) (i) 135 and 225
We start with the larger number 225
By Euclid’s division algorithm, we have
225 = 1 × 135 + 90
135 = 1 × 90 + 45 = 2 × 45 + 0
HCF(225, 135) = HCF(135, 90) = HCF(90, 45) = 45
Therefore, the HCF of 135 & 225 is 45.

(ii) 196 and 38220
We start with the larger number 38220
By Euclid’s division algorithm, we have
38220 = 196 × 195 + 0
we apply Euclid’s Division Algorithm of divisor 196 & the remainder 0.
Therefore, 196 = 196 × 1 + 0
Therefore, HCF (38220, 196) = 196.

(iii) 867 and 255
Using Euclid’s Division Algorithm
867 = 255 × 3 + 102
Applying Euclid’s Division Algorithm on the divisor 225 & the remainder 102,
We have,
255 = 102 × 2 + 51
Again, applying Euclid’s Division Algorithm on the Divisor 102 & the number 51.
102 = 51 × 2 + 0.
Therefore, HCF(867, 255) = HCF(255,102) = HCF(102,51) = 51.

Q.2) Show that any positive odd integer is of the form 6𝑞 + 1, or 6𝑞 + 3, or 6𝑞 + 5, where 𝑞 is some integer.
Sol.2) Using Euclid’s Division Algorithm, we have
𝑎 = 𝑏𝑞 + 𝑟{𝑟 ≤ 0 < 𝑏} ……….. (1)
Substituting 𝑏 = 6 in equation (1)
𝑎 = 6𝑞 + 𝑟 where 𝑟 ≤ 0 < 6 ⇒ 𝑟 = 0, 1, 2, 3, 4, 5
So total possible forms will 6𝑞 + 0 , 6𝑞 + 1 , 6𝑞 + 2,6𝑞 + 3, 6𝑞 + 4, 6𝑞 + 5
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 a odd number
6𝑞 + 2
6 is divisible by 2 and 2 is also divisible by 2 so it is a even number
6𝑞 + 3
6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number
6𝑞 + 4
6 is divisible by 2 and 4 is also divisible by 2 it is a even number
6𝑞 + 5
6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number
So odd numbers will in form of 6𝑞 + 1, 𝑜𝑟 6𝑞 + 3, 𝑜𝑟 6𝑞 + 5.

Q.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?
Sol.3) For the above problem, the maximum no. of columns would be the HCF of 616 & 32
We can find the HCF of 616 & 32 by using Euclid’s Division Algorithm.
Therefore,
616 = 19 × 32 + 8
32 = 4 × 8 + 0
8 = 8 × 1 + 0
Therefore, HCF(616,32) = HCF of (32, 8) = 8
Therefore, the maximum number of columns in which they can march is 8.

Q.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 the form 3𝑚 or 3𝑚 + 1. ]
Sol.4) 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)
= (3𝑞)² 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, and 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

Q.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.
Sol.5) 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(3𝑞³) = 9𝑚
Where m is an integer such that 𝑚 = 3𝑞³

Case 2: When 𝑎 = 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.

Exercise 1.2

Q.1) Express each number as product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Sol.1) (i) 140
140 = 2 × 70
= 2 × 2 × 35
= 2 × 2 × 2 × 5 × 7
= 2 × 2 × 5 × 7 × 1

(ii) 156
156 = 2 × 78
= 2 × 2 × 39
= 2 × 2 × 13 × 3
= 2 × 2 × 13 × 3 × 1

(iii) 3825
3825 = 3 × 1275
= 3 × 3 × 425
= 3 × 3 × 5 × 85
= 3 × 3 × 5 × 5 × 17
= 3 × 3 × 5 × 5 × 17 × 1

(iv) 5005
5005 = 5 × 1001
= 5 × 7 × 143
= 5 × 7 × 11 × 13
= 5 × 7 × 11 × 13 × 1

(v) 7429
7429 = 17 × 437
= 17 × 19 × 23
= 17 × 19 × 23 × 1

Q.2) 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.2) (i) 26 = 2 × 13
91 = 7 × 13
HCF = 13 LCM =2 × 7 × 13 =182
Product of two numbers 26 × 91 = 2366
Product of HCF and LCM 13 × 182 = 2366
Hence, product of two numbers = product of HCF × LCM

(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.

Q.3) 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.3) (i) 12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3
LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 1 × 17 × 19 × 23 = 11339

(iii) 8 = 1 × 2 × 2 × 2
9 = 1 × 3 × 3
25 = 1 × 5 × 5
HCF =1
LCM = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

Q.4) Given that HCF (306, 657) = 9, find LCM (306, 657).
Sol.4) We have the formula that
Product of LCM and HCF = product of number
𝐿𝐶𝑀 × 9 = 306 × 657
Divide both side by 9 we get LCM = (306 × 657)/9 = 22338

Q.5) Check whether 6𝑛 can end with the digit 0 for any natural number 𝑛.
Sol.5) If any digit has last digit 10 that means
it is divisible by 10 and the factors of 10 = 2 × 5.
So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5
So it can not end with 0

Q.6) Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Sol.6) 7 × 11 × 13 + 13
Taking 13 common, we get
13 (7 x 11 +1 )
13(77 + 1 ) 13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 + 1)
5(1008 + 1) 5(1009)
It is product of two numbers and both numbers are more than 1 so it is a composite number

Q.7) There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Sol.7) They will be meet again after LCM of both values at the starting point.
18 = 2 × 3 × 3 12 = 2 × 2 × 3
𝐿𝐶𝑀 = 2 × 2 × 3 × 3 = 36
Therefore, they will meet together at the starting point after 36 minutes.

Exercise 1.3

Q.1) Prove that √5 is irrational.
Sol.1) Let √5 is a rational number.
Therefore, we can find two integers 𝑎, 𝑏 (𝑏 ≠ 0) such that √5 = 𝑎/𝑏
Let 𝑎 and 𝑏 have a common factor other than 1. Then we can divide them by the common factor, and assume that 𝑎 and 𝑏 are co-prime.
𝑎 = √5𝑏
⇒ 𝑎² = 5𝑏²
Therefore, 𝑎²
is divisible by 5 and it can be said that 𝑎 is divisible by 5.
Let a = 5k, where k is an integer
(5𝑘)² = 5𝑏²
⇒ 5𝑘² = 𝑏²
This means that 𝑏² is divisible by 5 and hence, b is divisible by 5.
This implies that 𝑎 and b have 5 as a common factor.
And this is a contradiction to the fact that 𝑎 and 𝑏 are co-prime.
Hence, √5 cannot be expressed as 𝑝/𝑞 or it can be said that √5 is irrational.

Q.2) Prove that 3 + 2√5 is irrational.
Sol.2) 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.

Q.3) Prove that the following are irrationals:
(i) 1/√2 (ii) 7√5 (iii) 6 + √2
Sol.3) (i) Let take that 1/√2
is a rational number.
So we can write this number as 1/√2 = 𝑎/𝑏
Here 𝑎 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.

Exercise 1.4

Q.1) . Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) 13/3125
(ii) 17/8
(iii) 64/455
(iv) 15/1600
(v) 29/343
(vi) 23/2³5²
(vii) 129/2²5⁷7⁵
(viii) 6/15
(ix) 35/50
(x) 77/210
Sol.1) (i) 13/3125
Factorize the denominator we get
3125 = 5 × 5 × 5 × 5 × 5 = 55

(ii) 17/8
Factorize the denominator we get 8 = 2 × 2 × 2 = 23
So denominator is in form of 2𝑚 so it is terminating.

(iii) 64/455
Factorize the denominator we get 455 = 5 × 7 × 13
There are 7 and 13 also in denominator so denominator is not in form of 2^𝑚 × 5𝑛 so it is not terminating.

(iv) 15/1600
Factorize the denominator we get
1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 26 × 52
so denominator is in form of 2^𝑚 × 5^𝑛
Hence it is terminating

(v) 29/343
Factorize the denominator we get
343 = 7 × 7 × 7 = 73
There are 7 also in denominator
so denominator is not in form of 2^𝑚 × 5^𝑛
Hence it is non-terminating.

(vi) 23/2³ × 5²
Denominator is in form of 2^𝑚 × 5^𝑛
Hence it is terminating.

(vii) 129/2² × 5⁷× 7⁵
Denominator has 7 in denominator
so denominator is not in form of 2^𝑚 × 5^𝑛
Hence it is none terminating.

(viii) 6/15
divide nominator and denominator both by 3 we get 2/5
Denominator is in form of 5m so it is terminating

(ix) 35/50
divide denominator and nominator both by 5 we get 7/10
Factorize the denominator we get 10 = 2 × 5
So denominator is in form of 2^𝑚 × 5^𝑛 so it is terminating.

(x) 77/210
simplify it by dividing nominator and denominator both by 7 we get 11/30
Factorize the denominator we get 30 = 2 × 3 × 5
Denominator has 3 also in denominator
so denominator is not in form of 2^𝑚 × 5^𝑛
Hence it is none terminating

Q.2) Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions

Q.3) The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form 𝑝 , 𝑞 you say about the prime factors of 𝑞?
(i) 43.123456789 (ii) 0.120120012000120000... (iii) 43. ̅1̅2̅̅3̅̅4̅5̅̅6̅̅7̅8̅̅9̅
Sol.3) (i) Since this number has a terminating decimal expansion, it is a rational number of the form 𝑝/𝑞, and 𝑞 is of the form 2^𝑚 × 5^𝑛 .
(ii) The decimal expansion is neither terminating nor recurring.
Therefore, the given number is an irrational number.
(iii) Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form 𝑝/𝑞, and q is not of the form 2^𝑚 × 5^𝑛 .