At Saralstudy, we are providing you with the solution of Class 10 Mathematics Real Numbers according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 10 Mathematics Real Numbers so that you can prepare for the exam according to your own pace and your speed.
(i) Here, we have to find H.C.F of 135 and 225
First divide divide the larger integer smaller integer
Since, 225 > 135
Therefore, by Euclid’s Division algorithm
225 = 135 × 1 + 90 (i)
Here 90 ≠ 0, so proceed the same procedure further
Again by E.D.L, (E.D.L = Euclid’s division algorithm)
135 = 90 × 1 + 45 (ii)
As we know, 45 ≠ 0 therefore, again by E.D.L
90 = 45 × 2 + 0 (iii)
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 45.
From (i), (ii) and (iii)
H.C.F (225, 135) = H.C.F (135, 90) = H.C.F (90, 45) = 45.
(ii) Here, we have to find H.C.F of 38220 and 196
First divide the larger integer smaller integer
Since, 3822 > 196
Therefore by Euclid’s Division Algorithm
38220 = 196 × 195 + 0
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 196.
Hence, H.C.F (38220, 196) = 196.
(iii) Here, we have to find H.C.F of 867 and 255
First divide the larger integer smaller integer
Since, 867 > 255
Therefore, by Euclid’s Division algorithm
867 = 255 × 3 + 102 (i)
Remainder 102 ≠ 0, so proceed the same procedure further using E.D.L
255 = 102 × 2 + 51 (ii)
Here, 51 ≠ 0 again using E.D.L = 51 × 2
102 = 51 × 2 + 0 (iii)
Here, r = 0 so we cannot proceed further. The divisor at this Stage is 51.
From (i), (ii) and (iii)
H.C.F (867, 255) = H.C.F (255, 102) = H.C.F (102, 51) = 51.
Let a be any positive integer b = 6. Then by Euclid’s algorithm,
a = 6q + r for some integer q ≥ 0
So, values of r we get, r = 0, 1, 2, 3, 4, 5
When, r = 0, then, a = 6q + 0,
similarly for r = 1, 2, 3, 4, 5 the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5 respectively.
If a = 6q, 6q+2, 6q+4, then a is even and divisible by 2.
Therefore, any positive odd integer is in the form of 6q+1, 6q+3, 6q+5
Where q is some integer.
Total no. of army contingent members = 616
No. of army band members = 32
To find max. numbers of the same columns in which the both groups march. We have to find it.
Their highest common factor. since, 616 > 32 then, by Euclid’s algorithm
616 = 32 × 19 + 8 (i)
Here 8 ≠ 0 then by again using Euclid algorithm
32 = 8 × 4 + 0 (ii)
Here, r =0 so we cannot proceed further. The divisor at this Stage is 8.
So the no. of columns is 8.
Let x be any positive integer and b = 3.
Then, by euclid’s algorithm
x = 3q + r, where q ≥ 0 and r = 0, 1, 2 [0 ≤ r ≤ b]
Case (i) : For r = 0, x = 3q, = x2 = 9q2, taking 3 as common,
x2 = 9q2 = 3 (3q2), which is of the form 3m, where m = 3q2.
Case (ii) : For r = 1, x = 3q + 1
x2 = 9q2 + 1 + 6q, taking 3 as common,
= 3 (3q2 + 2q) + 1, which is of the form 3m + 1, where m = 3q2 + 2q
Case (iii) : For r = 2, 3q + 2
x2 = 9q2 + 4 + 12q = (9q2 + 12q + 3) + 1, taking 3 as common,
= 3 (3q2 + 4q + 1) + 1, which is of the form 3m +1, where m = 3q2 + 4q + 1
Hence, x2 is either of the form 3m, 3m + 1 for some integer m.
Let a be any positive integer and b = 3. Then, by euclid’s algorithm
a = 9q + r, where q ≥ 0 and r = 0, 1, 2 [ 0 ≤ r ≤ b ]
For r = 0, x = 3q, or
For r = 1, x = 3q +1
For r = 2, x = 3q + 2
Now by taking the cube of all the three above terms, we get,
Case (i) : when r = 0, then,
x3 = (3q)3 = 27q3 = 9 (3q3) = 9m; where m = 3q
Case (ii) : when r = 1, then,
x3 = (3q +1)3 = (3q)3 + 13 +3 × 3q × 1 (3q + 2 ) = 27q3 + 1 +27q2 + 9q
Taking 9 as common factor, we get,
x3 = 9 (3q3 +3q2 + q) +1
Putting (3q3 + 3q2 + q) = m, we get,
x3 = 9m + 1
Case (iii) : when r = 2, then,
x3 = (3q + 2)3 = (3q)3 + 23 + 3 × 3q × 2 (3q + 2) = 27q3 + 54q2 + 36q + 8
Taking 9 as common factor , we get
x = 9 (3q + 6q + 4q) + 8
Putting (3q + 6q + 4q) = m, we get,
x = 9m + 8,
Therefore, it is proved that the cube of any positive integer is of the form 9m, 9m + 1, 9m + 8.
(i) 140
By taking the L.C.M of 140, we will get the product of its prime factors.
Therefore, 140 = 2 × 2 × 5 × 7 ×1 = 22 × 5 × 7
(ii) 156
By taking the L.C.M of 156, we will get the product of its prime factors.
Therefore, 156 = 2 × 2 × 13 × 3 × 1 = 22 × 13 × 3
(iii) 3825
By taking the L.C.M of 3825, we will get the product of its prime factors.
Therefore, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 32 × 52 × 17 × 1
(iv) 5005
By taking the L.C.M of 5005, we will get the product of its prime factors.
5005 = 5 × 5 × 11 × 13 × 1 = 5 × 7 × 11 × 1
(v) 7429
By taking the L.C.M of 7429, we will get the product of its prime factors.
Therefore, 7429 = 17 × 19 × 13 × 1
(i) 26 and 91
First we have to find the L.C.M and H.C.F of 26, 91 using fundamental theorem of arithmetic,
26 = 13 × 2
91 = 13 × 7
For L.C.M - list all the prime factors (only once) of 26, 91 with their greatest power.
L.C.M (26, 91) = 13 × 2 × 7 = 182
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (26, 91) = 13
Verification : L.C.M (26, 91) × H.C.F (26, 91) = 26 × 91
182 × 13 = 2366, => 2366 = 2366
Hence, proved.
(ii) 510 and 92
First we have to find the L.C.M and H.C.F of 510, 92 using fundamental theorem of arithmetic,
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
For L.C.M - list all the prime factors (only once) of 510, 92 with their greatest power.
L.C.M (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (510, 92) = 2
Verification : L.C.M (510, 92) × H.C.F (510, 92) = 510 × 92
23460 × 2 = 46920, => 46920 = 46920
Hence, proved.
(iii) 336 and 54
First we have to find the L.C.M and H.C.F of 336, 54 using fundamental theorem of arithmetic,
336 = 2 × 2 × 2 × 2 × 3 × 7 = 24 × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 33
For L.C.M - list all the prime factors (only once) of 336, 54 with their greatest power.
L.C.M (336, 54) = 3024
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (336, 54) = 2 × 3 = 6
Verification : L.C.M (336, 54) × H.C.F (336, 54) = 336 × 54
3024 × 6 = 18144 => 18144 = 18144
Hence, proved.
(i) 12, 15 and 21
12 = 2 × 3 × 2
15 = 3 × 5
21 = 3 × 7
L.C.M (12, 15, 21) = 2 × 3 × 2 × 5 × 7 = 420
H.C.F (12, 15, 21) = 3
(ii) 17, 23 and 29
17 = 17 × 1
23 = 23 × 1
29 = 29 ×1
L.C.M (17, 23, 29) = 11339
H.C.F (17, 23, 29) = 1
(iii) 8, 9 and 25
8 = 2 × 2 × 2 × 1
9 = 3 × 3 × 1
25 = 5 × 5 × 1
L.C.M (8, 9, 25) = 1800
H.C.F (8, 9, 25) = 1
Given,
H.C.F (306, 657) = 9
First number (a) = 306
Second number (b) = 657
As we know, H.C.F × L.C.M = a × b
9 × L.C.M = 306 × 657
L.C.M = (306 × 657) / 9 = 22338
Hence, L.C.M (306, 657) = 22338
Suppose the number 6n for any n € N ends with zero. In this case 6n is divisible by 5.
Prime factorization of 6n = (2 × 3)n
Therefore, the prime factorization of 6n doesn’t contain prime number 5.
Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.
Composite numbers are factors other than 1 and itself.
From here we can observe that :
(a) 7 × 11 × 13 + 13 = 13 (7 × 11 × 1 + 1) = 13 (77 + 1)
= 13 × 78
= 13 × 13 × 6
Therefore, this expression has 6 and 13 as its factors. Hence, it is a composite number.
(b) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 (7 × 6 × 1 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
1009 can not be factored further. Therefore, this expression has 5 and 1009 as its factors.
Hence, it is a composite number.
Time taken by Sonia to drive one round = 18 min
Time taken by Ravi to drive one round = 15 min
Let they both meet after time ‘t’. Here, ‘t’ is L.C.M of 18, 12
Now, 18 = 2 × 3 × 3
12 = 2 × 3 × 2
L.C.M (18, 12) = 22 × 32 = 36
Thus, they both meet after 36 minutes at the starting point.
Let us consider √5 is a rational number.
Therefore, √5 = pq, where p and q are integers and q ≠ 0
If pa and q have any common factor ,then dividing by that common factor,
We have, √5 = ab, where a and b are co primes.
a = √5b
On squaring, a² = 5b² ........... (1)
Due to the presence of 5 on RHS, we say that 5 is a factor of a².
5 divides a²
Since, a is prime , therefore 5 divides a [ theorem ]
Therefore , a = 5k , where k is an integer.
Putting a = 5k in (1) , we have
25k² = 5b²
b² = 5k²
This shows that 5divides b². But b is a prime no and so 5 divides b also.
Thus 5 is a common factor of a and b . This is contradiction to the fact that
a and b have no common factor other than one.
Thus our consideration is wrong and so √ 5 is not a rational number.
Hence, proved.
Let us assume 3 + 2√5 is a rational number.
Therefore, 3 + 2√5 = p/q where p and q are co primes and q ≠ 0.
3 + 2√5 = ab
On solving, 2√5 =(a/b) - 3
√5 =1/2 (a/b - 3)
Since a, b are integers and 1/2 (a/b-3 ) is also a rational number.
But we know √5 is an irrational number.
Thus our assumption is wrong. 3 + 2√5 is not a rational number.
Hence proved.
(i) Let us assume 1/√2 is a rational number.
1/√2 = p/q , where q ≠ 0 and p and q are co primes.
On reciprocal,
√2 = qp ................(1)
Since, q and p are integers and q/p is also a rational number
As we know √2 is an irrational number.
From (1)
√2 ≠ q/p
Thus our assumption is wrong 1/√2 is not a rational number.
Hence, proved
(ii) Let us suppose 7√5 is a rational number.
7√5 = p/q, where p and q are co primes and q ≠ 0
On solving , √5 = (p/q)7 .....................(1)
Since p, q and 7 integers and (p/q)7 is also a rational number.
And we know √5 is an irrational number.
From (1)
√5 ≠ (p/q) / 7
So our supposition Is wrong 7√5 is not a rational number.
Hence, proved.
(iii) Let us suppose 6 + √2 is a rational number.
6 + √2 = a/b, where a, b are co primes and b ≠ 0.
On solving,
√2 = a/b - 6 .....................(1)
Since a, b and 6 are integers and a/b - 6 is also a rational number.
And we know that √2 is an irrational number.
From (1)
√2 ≠ a/b - 6
Thus our Superposition is wrong 6√2 is not a rational number.
Hence, proved.
Note: - If x be a rational number, then x can be expressed in the form p/q where p and q are Co- primes. Then, if the prime factorization of denominator (q) is in the form of 2m × 5n where n, m are non- negative integers, then x has a decimal expansion which terminates. If the prime factorization is not in this form of 2m × 5n then x has a decimal expansion which is non terminating.
(i) 13 / 3125
= 5 × 5 × 5 × 5 × 5 = 5⁵ (on factorization of q)
Since, its factorization contains only power of 5.
Therefore, it has a terminating decimal expansion.
(ii) 17 / 5
= 2 × 2 × 2 = 2³ (on factorization of q)
Since, its factorization contains only power of 2.
Therefore, it has a terminating decimal expansion.
(iii) 64 / 455
= 5 × 7 × 13 = 5¹ × 7 × 13 (on factorization of q)
Since the factorization of not in the 2m ×5n
Therefore, it has a non -terminating decimal expansion.
(iv) 15 / 1600
= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2⁶ × 5² (on factorization of q)
Since the factorization of q is in the form 2m ×5n
Therefore, it has a terminating decimal expansion
(V) 29 / 343
= 7 × 7 × 7 = 7³ (on factorization of q)
Since the factorization of denominator is not in the form 2m × 5n
Therefore it has a non -terminating repeating decimal
(vi) 23 / 23 52
Since factorization of q already given and it is in the form 2m × 5n
Therefore it has a terminating decimal expansion.
(vii) 129 / 22 57 75
Since factorization of q already given but it is not in the form 2m × 5n
Therefore it has a terminating decimal expansion.
(viii) 6 / 15 = 2/5
5 = 5 × 1 (on factorization of q)
Since 5 is the only factor in denominator.
Therefore it has a non -terminating repeating decimal expansion.
(ix). 35 / 50 = 7 / 10
10 = 2 × 5 = 2¹ × 5 (on factorization of q)
Since the factorization of denominator is in the form 2m × 5n
Therefore it has a terminating decimal expansion.
(x) 77 / 210 = 11/30
30 = 2 × 3 × 5 × 7 = 2¹ × 3 × 5 × 7. (on factorization of q)
Since the factorization of denominator is not in the form 2m × 5n
Therefore it has a non -terminating repeating decimal expansion.
(i) 13 / 3125
(ii) 17 / 8
(iii) 64/455 is non terminating repeating decimal expansion.
(iv) 15 / 1600
(v) 29/343 is non terminating repeating decimal expansion.
(vi) 23 / 23 52
(vii) 120 / 23 57 75 is non terminating repeating decimal expansion.
(viii) 6/15 = 2/5
Exercise 4 ( Page No. : 18 )
Note: A rational number has a decimal expansion which is either terminating or non -Terminating repeating otherwise the given number is irrational.
Since, the given decimal expansion is terminating. Therefore it is a rational number and we know that the prime factorization of denominator of rational 2 and 5 or both.
Since, the given decimal expansion is non- terminating non- repeating. Therefore it is irrational number.
Since, the given decimal expansion is non- terminating repeating. Therefore it is a rational number and we know that the prime factorization of denominator of rational 2 and 5 or both.
