NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Ans : (i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 x 1 + 90 
Since remainder \( 90 \neq 0 \), we apply the division lemma to 135 and 90 to 
obtain 
135 = 90 x 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the 
division lemma to obtain 
90 = 2 x 45 + 0
Since the remainder is zero, the process stops. 
Since the divisor at this stage is 45, 
Therefore, the HCF of 135 and 225 is 45. 

(ii) 196 and 38220 
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain 
38220 = 196 x 195 + 0
Since the remainder is zero, the process stops. 
Since the divisor at this stage is 196, 
Therefore, HCF of 196 and 38220 is 196. 

(iii) 867 and 255 
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain 
867 = 255 x 3 + 102 
Since remainder \(102 \neq 0\) we apply the division lemma to 2S5 and 102 to obtain 
255 = 102 x 2+ 51 
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain 
102 = 51 x 2+0 
Since the remainder is zero, the process stops. 
Since the divisor at this stage is 51, 
Therefore, HCF of 867 and 255 is 51. 

Ans : Let a be any positive integer and b = 6. Then, by Euclid's algorithm, 
a = 6q + r for some integer q 0, and r = 0, 1, 2, 3, 4, 5 because 0 \(\leq\) r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 x 3q +1 = \(2 k_{1}+1, \text { where } k_{1}\) is a positive integer
\(6 q+3=(6 q+2)+1=2(3 q+1)+1=2 k_{2}+1, \text { where } k_{2}\) is an integer
\(6 q+5=(6 q+4)+1=2(3 q+2)+1=2 k_{3}+1, \text { where } k_{3}\) is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. 
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. 
Hence, these expressions of numbers are odd numbers. 
And therefore, any odd integer can be expressed in the form 6q + 1,
or 6q + 3,
or 6q + 5 

Ans : HCF (616, 32) will give the maximum number of columns in which they can march. 
We can use Euclid's algorithm to find the HCF. 
616 = 32 x 19 + 8 
32 = 8 x 19 +8 
The HCF (616, 32) is 8. 
Therefore, they can march in 8 columns each. 

Ans : Let a be any positive integer and b = 3. 
Then a = 3q + r for some integer \(q \geq 0\)
And r = 0, 1, 2 because \(0 \leq r<3\)
Therefore, a = 3q or 3q + 1 or 3q + 2 
Or,
\(\begin{array}{l}{a^{2}=(3 q)^{2} \text { or }(3 q+1)^{2} \text { or }(3 q+2)^{2}} \\ {a^{2}=\left(9 q^{2}\right) \text { or } 9 q^{2}+6 q+1 \text { or } 9 q^{2}+12 q+4} \\ {=3 \times\left(3 q^{2}\right) \text { or } 3\left(3 q^{2}+2 q\right)+1 \text { or } 3\left(3 q^{2}+4 q+1\right)+1} \\ {=3 k_{1} \text { or } 3 k_{2}+1 \text { or } 3 k_{3}+1}\end{array}\)
Where \(k_{1}, k_{2}, \text{and } k_{3} \) 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 

Ans : Let a be any positive integer and b = 3 
\(\begin{array}{l}{a=3 q+r, \text { where } q \geq 0 \text { and } 0 \leq r<3} \\ {\therefore a=3 q \text { or } 3 q+1 \text { or } 3 q+2}\end{array}\)
Therefore, every number can be represented as these three forms. 
There are three cases. 
Case 1: When a = 3q, 
\(a^{3}=(3 q)^{3}=27 q^{3}=9\left(3 q^{3}\right)=9 m\),
Where m  \( \text { is an integer such that } m=3 q^{3}\)

\(\begin{array}{l}{\text {Case } 2 : \text { when } a=3 q+1} \\ {a^{3}=(3 q+1)^{3}} \\ {a^{3}=27 q^{3}+27 q^{2}+9 q+1} \\ {a^{3}=9\left(3 q^{3}+3 q^{2}+q\right)+1} \\ {a^{3}=9 m+1} \\ {\text { Where } m \text { is an integer such that } m=\left(3 q^{3}+3 q^{2}+q\right)}\end{array}\)
\(\begin{array}{l}{\text {Case } 3 : \text { when } a=3 q+2} \\ {a^{3}=(3 q+2)^{3}} \\ {a^{3}=27 q^{3}+54 q^{2}+36 q+8} \\ {a^{3}=9\left(3 q^{3}+6 q^{2}+4 q\right)+8} \\ {a^{3}=9 m+8} \\ {\text { Where } m \text { is an integer such that } m=\left(3 q^{3}+6
q^{2}+4 q\right)}\end{array}\)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8. 

Ans : \(\begin{array}{l}{\text { (i) } 140=2 \times 2 \times 5 \times 7=2^{2} \times 5 \times 7} \\ {\text { (ii) } 156=2 \times 2 \times 3 \times 13=2^{2} \times 3 \times 13} \\ {\text { (iii) } 3825=3 \times 3 \times 5 \times 5 \times 17=3^{2} \times 5^{2} \times 17} \\ {\text { (iv) } 5005=5 \times 7 \times 11 \times 13} \\ {\text { (v) } 7429=17 \times 19 \times 23}\end{array}\) 

Ans : (i) 26 and 91 
26 = 2 x 13 
91 = 7 x 13
HCF = 13
LCM = 2x7x13 = 182 
Product of the two  numbers 26 x 91 = 2366 
HCF x LCM = 13 x 182 = 2366 
Hence, product of two numbers = HCF x LCM 

(ii) 510 and 92
510 = 2 x 3 x 5 x 17
92 = 2 x 2 x 23
HCF = 2 
LCM = 2 x 2 x 3 x 5 x 17 x 23 = 23460
Product of the two numbers 510 x 92 = 46920 
HCF x LCM = 2 x 23460
= 46920
Hence, product of two numbers = HCF x LCM M

(iii) 336 and 54
336 = 2 x 2 x 2 x 2 x 3 x 7
\(\begin{array}{l}{336=2^{4} \times 3 \times 7} \\ {54=2 \times 3 \times 3 \times 3} \\ {54=2 \times 3^{3}} \\ {\text { HCF }=2 \times 3=6} \\ {\text { LCM }=2^{4} \times 3^{3} \times 7=3024} \\ {\text { Product of the two numbers }=336 \times 54=18144} \\ {\text { HCF } \times L C M=6 \times 3024=18144}\end{array}\)
Hence, product of two numbers  = HCF x LCM 

Ans : (i) 12, 15 and 21
\(12=2^{2} \times 3\)
15 = 3 x 5 
21 = 3 x 7
HCF = 3 
LCM = \(2^{2} \times 3 \times 5 \times 7=420\)

(ii) 17,23 and 29
 17 = 1 x 17
 23 = 1 x 23
 29 = 1 x 29
 HCF = 1
 LCM = 17 x 23 x 29 = 11339

(iii) 8,9 and 25
 8 = 2 x 2 x 2
 9 = 3 x 3
 25 = 5 x 5
HCF = 1
LCM = 2 x 2 x 2 x 3 x 3 x 5 x 5= 1800 

Ans : H.C.F (306, 657) = 9 
We know that, LCM x  HCF = Product of two numbers 
Therefore, LCM x HCF = 306 x 657 
\(\begin{array}{l}{\mathrm{LCM}=\frac{306 \times 657}{\mathrm{HCF}}=\frac{306 \times 657}{9}} \\ {\mathrm{LCM}-22338}\end{array}\) 

Ans : If any number ends with the digit 0, it should be divisible by 10 or in 
other words, it will also be divisible by 2 and 5 as 10 = 2 x 5
Prime factorisation of \(6^{n}=(2 \times 3)^{n}\)
It can be observed that 5 is not in the prime factorisation of  \(6^{n}\)
Hence, for any value of n, \(6^{n}\) will not be divisible by 5. 
Therefore, \(6^{n}\) cannot end with the digit 0 for any natural number n. 

Ans : Numbers are of two types - prime and composite. Prime numbers can 
be divided by 1 and only itself, whereas composite numbers have 
factors other than 1 and itself. 
It can be observed that 
7 x 11 x 13 + 13 = 13 x (7 x 11 +1) =13 x (77 + 1)
= 13 x 78 
= 3 x 13 x 6 
The given expression has 6 and 13 as its factors. Therefore, it is a 
composite number. 
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3  x 2 x 1 + 1)
= 5 x (1008 + 1) 
= 5 x 1009
1009 cannot be factorised further. Therefore, the given expression has 
5 and 1009 as its factors. Hence, it is a composite number. 

Ans : It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular 
path respectively i.e., LCM of 18 minutes and 12 minutes.
18 = 2 x 3 x 3  
And, 12 = 2 x 2 x 3 
LCM of 12 and 18 = 2 x 2 x 3 x 3 = 36 
Therefore, Ravi and Sonia will meet together at the starting point after 
36 minutes. 

Ans : Let \(\sqrt{5}\)  is a rational number. 
Therefore, we can find two integers a, b \((b \neq 0) \text { such that }{\sqrt{5}}=\frac{a}{b}\)
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. 
\(\begin{array}{l}{a=\sqrt{5} b} \\ {a^{2}=5 b^{2}}\end{array}\)
Therefore, \(a^{2}\) is divisible by 5 and it can be said that a is divisible by 5. 
Let a = 5k, where k is an integer.
\((5 k)^{2}=5 b^{2}\)
\(b^{2}=5 k^{2}\) This means that \(b^{2}\) is divisible by 5  and hence, b is divisible by 5.
This implies that a and b have S as a common factor. 
And this is a contradiction to the fact that a and b are co-prime. 
\(\begin{array}{l}{\text { Hence, } \sqrt{5} \text { cannot be expressed as } \frac{p}{q} \text { or it can be said that } \sqrt{5} \text { is }} \\ {\text { irrational. }}\end{array}\) 

Ans : Let \(3+2 \sqrt{5}\) is rational.
\(\begin{array}{l}{\text { Therefore, we can find two integers } a, b(b \neq 0) \text { such that }} \\ {3+2 \sqrt{5}=\frac{a}{b}} \\ {2 \sqrt{5}=\frac{a}{b}-3} \\ {\sqrt{5}=\frac{1}{2}\left(\frac{a}{b}-3\right)}\end{array}\)
Since a and b are integers, \(\frac{1}{2}\left(\frac{a}{b}-3\right)\) will also be rational and therefore ,\(\sqrt{5}\)  is rational.
This contradicts the fact that \(\sqrt{5}\)  is irrational. Hence, our assumption 
that\(3+2 \sqrt{5}\) is rational is false. Therefore, \(3+2 \sqrt{5}\) is irrational. 

Ans : (i) \(\frac{1}{\sqrt{2}}\)
Let \(\frac{1}{\sqrt{2}}\)  is rational. 
Therefore, we can find two integers a, b \((b \neq 0)\) such that 
\(\begin{array}{l}{\frac{1}{\sqrt{2}}=\frac{a}{b}} \\ {\sqrt{2}=\frac{b}{a}} \\ {\frac{b}{a} \text { is rational as a and } b \text { are integers. }}\end{array}\).
Therefore,\(\sqrt{2}\) is rational which contradicts to the fact that \(\sqrt{2}\) irrational. 
Hence, our assumption is false and  \(\frac{1}{\sqrt{2}}\)is irrational. 

(ii) \(7 \sqrt{5}\)
Let  \(7 \sqrt{5}\) is rational.
\(\begin{array}{l}{\text { Therefore, we can find two integers } a, b(b \neq 0) \text { such that }} \\ {7 \sqrt{5}=\frac{a}{b} \text { for some integers a and } b} \\ {\therefore \sqrt{5}=\frac{a}{7 b}}\end{array}\) 
\(\frac{a}{7 b}\) is rational as a and b are integers.
Therefore,  \(\sqrt{5}\) should be rational .
This contradicts  the fact that \(\sqrt{5}\) is irrational. Therefore ,our assumption that \(7 \sqrt{5}\) is irrational .

(iii) \(6+\sqrt{2}\)
Let \(6+\sqrt{2}\) be rational .
Therefore we can find two integers a, b \((b \neq 0)\) such that 
\(\begin{array}{l}{6+\sqrt{2}=\frac{a}{b}} \\ {\sqrt{2}=\frac{a}{b}-6}\end{array}\) 
Since a and b are integers, \(\frac{a}{b}-6\) is also rational and hence,\(\sqrt{2}\) should  
be rational. This contradicts the fact that \(\sqrt{2}\)  is irrational. Therefore, 
our assumption is false and hence, \(6+\sqrt{2}\) is irrational. 

Ans : (i) \( \frac{13}{3125} \)
\( 3125=5^{5} \)
The denominator is of the form \( 5^{m} \)
Hence, the decimal expansion of \( \frac{13}{3125} \) is terminating.

(ii) \( \frac{17}{8} \)
\( 8=2^{3} \)
The denominator is of the form \( 2^{m} \)
Hence, the decimal expansion of \( \frac{17}{8} \) is terminating.

(iii) \(\frac{64}{455}\)
455 = 5 x 7 x 13 
Since the denominator is not in the form \(2^{m} \times 5^{n}\), and it also contains 
7 and 13 as its factors, its decimal expansion will be non-terminating repeating. 

(iv) \(\begin{array}{l}{\frac{15}{1600}} \\ {1600=2^{6} \times 5^{2}}\end{array}\)
The denominator is of the form \(2^{m} \times 5^{n}\)
Hence, the decimal expansion of \(\frac{15}{1600}\) is terminating. 

(v) \(\begin{array}{l}{\frac{29}{343}} \\ {343=7^{3}}\end{array}\)
Since the denominator is not in the form \(2^{m} \times 5^{n}\), and it has 7 as its factor, the decimal expansion of \(\frac{29}{343}\)is non-terminating repeating.  

\((vi)^{\frac{23}{2^{3} \times 5^{2}}}\)
The denominator is of the form \(2^{m} \times 5^{n}\). 

Hence, the decimal expansion of \(\frac{23}{2^{3} \times 5^{2}}\) is terminating. 

(vii) \(\frac{129}{2^{2} \times 5^{7} \times 7^{5}}\)
Since the denominator is not of the form \(2^{m} \times 5^{n}\), and it also has 7 as 
its factor, the decimal expansion of \(\frac{129}{2^{2} \times 5^{7} \times 7^{5}}\) is non-terminating repeating. 

(viii) \(\frac{6}{15}=\frac{2 \times 3}{3 \times 5}=\frac{2}{5}\)
The denominator is of the form \(5^{n}\). 
Hence, the decimal expansion of \(\frac{6}{15}\) is terminating. 

(ix) \(\begin{array}{l}{\frac{35}{50}=\frac{7 \times 5}{10 \times 5}=\frac{7}{10}} \\ {10=2 \times 5}\end{array}\)
The denominator is of the form \(2^{m} \times 5^{n}\). 
Hence, the decimal expansion of \(\frac{35}{50}\) is terminating. 

(x)\(\begin{array}{l}{\frac{77}{210}=\frac{11 \times 7}{30 \times 7}=\frac{11}{30}} \\ {30=2 \times 3 \times 5}\end{array}\)
Since the denominator is not of the form \(2^{m} \times 5^{n}\), and it also has 3 as 
its factors, the decimal expansion of \(\frac{77}{210}\) is non-terminating repeating. 

Ans : (i) \(\quad \frac{13}{3125}=0.00416\)

(ii) \(\quad \frac{17}{8}=2.125\)

(iv)\(\quad \frac{15}{1600}=0.009375\)

(vi) \(\quad \frac{23}{2^{3} \times 5^{2}}=\frac{23}{200}=0.115\)

(viii) \(\frac{6}{15}=\frac{2 \times 3}{3 \times 5}=\frac{2}{5}=0.4\)

(ix) \(\quad \frac{35}{50}=0.7\)
 

Ans : (i) 43.123456789 
Since this number has a terminating decimal expansion, it is a rational number of the form \(\frac{p}{q}\) and q  is of the form \(2^{m} \times 5^{n}\)
i.e., the prime factors of q will be either 2 or 5 or both. 

(ii) 0.120120012000120000 
The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number. 

(iii) \( 43.\overline{ 1 2 3 4 5 6 7 8 9 } \)
Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form and q is not of the form \(\frac{p}{q}\) and q is not of  the form  \(2^{m} \times 5^{n}\) i.e., the prime factors of q will also have a factor other than 2 or 5.