CBSE Class 10 Mathematics Real Numbers Worksheet Set G

Points to Remember

• Euclid’s division lemma : Given positive integers \( a \) and \( b \), there exist unique integers \( q \) and \( r \) satisfying \( a = bq + r \), \( 0 \le r < b \).
• Let \( x = \frac{p}{q} \) be a rational number, such that the prime factorisation of \( q \) is of the form \( 2^m \times 5^n \), where \( m, n \) are non-negative integers. Then \( x \) has a decimal expansion which terminates.
• For any two positive integers \( a \) and \( b \), \( \text{HCF} (a, b) \times \text{LCM} (a, b) = a \times b \)

Multiple Choice Questions

Question. Euclid‘s division Lemma states that for two positive integers \( a \) and \( b \), there exists unique integer \( q \) and \( r \) satisfying \( a = bq + r \), and 
(a) \( 0 < r < b \)
(b) \( 0 < r \le b \)
(c) \( 0 \le r < b \)
(d) \( 0 \le r \le b \)
Sol. Euclid‘s division Lemma states that for two positive integers \( a \) and \( b \), there exist unique integers \( q \) and \( r \) satisfying \( a = bq + r \), and \( 0 \le r < b \).
\ Option (c) is correct.
Answer: (c)

Question. The sum of exponents of prime factors in the prime factorisation of 196 is 
(a) 3
(b) 4
(c) 5
(d) 2
Sol. Prime factors of \( 196 = 2^2 \times 7^2 \)
So, sum of exponents \( = 2 + 2 = 4 \)
\ Option (b) is correct.
Answer: (b)

Question. Has the rational number \( \frac{567}{2^2 5^7 7^2} \) a terminating or non-terminating decimal representation ?
(a) Terminating
(b) Non-terminating
(c) Neither
(d) Both
Sol. The rational number \( \frac{567}{2^2 5^7 7^2} \) has a non-terminating decimal representation because \( (2^2 5^7 7^2) \neq (2^m \times 5^n) \)
So, the correct option is (b).
Answer: (b)

Question. The decimal expansion of the rational number \( \frac{43}{2^4 5^3} \) will terminate after how many places of decimal ?
(a) 1
(b) 2
(c) It is interminable
(d) 4
Sol. \( \frac{43}{2^4 5^3} = \frac{43 \times 5}{2^4 \times 5^4} = \frac{215}{(2 \times 5)^4} = \frac{215}{10^4} = 0.0215 \)
Thus, the rational number terminates after 4 decimal places.
So, the correct option is (d).
Answer: (d)

Question. Complete the missing entries in the following factor tree :
(a) 42 and 21
(b) 24 and 12
(c) 7 and 3
(d) 84 and 42
Sol. Since, \( 3 \times 7 = 21 \)
and \( 2 \times 21 = 42 \)
So, the correct option is (a).
Answer: (a)

Question. Which one of the following is a non-terminating repeating decimal ?
(a) \( \frac{35}{14} \)
(b) \( \frac{14}{35} \)
(c) \( \frac{1}{7} \)
(d) \( \frac{7}{8} \)
Sol. \( \frac{35}{14} = \frac{5 \times 7}{2 \times 7} = \frac{5}{2} = \frac{5}{2^1 \times 5^0} \)
\( \frac{14}{35} = \frac{2 \times 7}{5 \times 7} = \frac{2}{5} = \frac{2}{2^0 \times 5^1} \)
\( \frac{7}{8} = \frac{7}{2^3 \times 5^0} \)
But in \( \frac{1}{7} \), the denominator is not of the form \( 2^m \times 5^n \).
Hence, \( \frac{1}{7} \) is a non-terminating repeating decimal.
So, the correct option is (c).
Answer: (c)

Fill in the blanks

Question. The decimal expansion of \( \frac{1268}{2^3 \times 5} \) will terminate after ............. decimal places.
Answer: 3

Question. If \( p = 2^2 \times 3^3 \times 5 \) and \( q = 2^3 \times 3^2 \times 7 \), then HCF \( (p, q) = \) ............. .
Answer: \( 2^2 \times 3^2 = 36 \)

Question. If \( \text{HCF} (a, 9) = 6 \) and \( \text{LCM} (a, 9) = 9 \), then the value of \( a \) is ..............
(a) 3
(b) 36
(c) 6
Answer: 36

True / False

Question. If \( p \) is a prime and \( p \) divides \( a^2 \), then \( p \) also divides \( a \), where \( a \) is a positive integer.
Answer: True

Question. Let \( x = \frac{p}{q} \) be a rational number, and prime factorisation of \( q \) is of the form \( 2^m 7^n \), then \( x \) has a decimal expansion which terminates.
Answer: False (Prime factorisation of \( q \) should be of the form \( 2^m 5^n \).)

Very Short Answer Type Questions

Question. Write whether \( \frac{2\sqrt{45} + 3\sqrt{20}}{2\sqrt{5}} \) on simplification gives a rational or an irrational number.
Sol. We have, \( \frac{2\sqrt{45} + 3\sqrt{20}}{2\sqrt{5}} = \frac{2(3\sqrt{5}) + 3(2\sqrt{5})}{2\sqrt{5}} = \frac{6\sqrt{5} + 6\sqrt{5}}{2\sqrt{5}} = \frac{12\sqrt{5}}{2\sqrt{5}} = 6 \)
(a rational number)
Thus, on simplification, \( \frac{2\sqrt{45} + 3\sqrt{20}}{2\sqrt{5}} \) gives a rational number.
Answer: Rational number

Question. Show that \( 3\sqrt{7} \) is an irrational number.
Sol. Let us assume, to the contrary, that \( 3\sqrt{7} \) is rational.
That is, we can find co-primes \( a \) and \( b \) (\( b \neq 0 \)) such that \( 3\sqrt{7} = \frac{a}{b} \)
On rearranging, we get \( \sqrt{7} = \frac{a}{3b} \)
Since 3, \( a \) and \( b \) are integers, therefore, \( \frac{a}{3b} \) can be written in the form of \( \frac{p}{q} \), so \( \frac{a}{3b} \) is rational, and so \( \sqrt{7} \) is rational.
But this contradicts the fact that \( \sqrt{7} \) is irrational.
So, we conclude that \( 3\sqrt{7} \) is irrational.
Hence Proved.

Question. Express the number \( 0.31\overline{78} \) in the form of rational number \( \frac{a}{b} \).
Sol. Let \( x = 0.31783178.... \) ...(i)
Multiplying by 10000, we get
\( 10000x = 3178.31783178..... \) ...(ii)
Subtracting (i) from (ii), we get
\( 9999x = 3178 \)
\( \Rightarrow x = \frac{3178}{9999} \)
Answer: \( \frac{3178}{9999} \)

Question. What is the HCF of smallest prime number and the smallest composite number ?
Sol. Smallest prime \( = 2 \)
Smallest composite \( = 4 \)
\( \text{HCF}(2, 4) = 2 \)
The HCF of the smallest prime and smallest composite is 2.
Answer: 2

Question. If \( \text{HCF} (336, 54) = 6 \), find \( \text{LCM} (336, 54) \).
Sol. Given, \( \text{HCF} (336, 54) = 6 \)
We know, \( \text{HCF} \times \text{LCM} = \text{one number} \times \text{other number} \)
\( \Rightarrow 6 \times \text{LCM} = 336 \times 54 \)
\( \Rightarrow \text{LCM} = \frac{336 \times 54}{6} \)
\( = 336 \times 9 = 3024 \)
Answer: 3024

Question. Find a rational number between \( \sqrt{2} \) and \( \sqrt{3} \). 
Sol. As \( \sqrt{2} = 1.414.... \) and \( \sqrt{3} = 1.732 ..... \)
So, a rational number between \( \sqrt{2} \) and \( \sqrt{3} \) is 1.5 or we can take any number between 1.414 and 1.732.
Answer: 1.5

Question. Write whether the rational number \( \frac{441}{2^2 \times 5^7 \times 7^2} \) a terminating or a non-terminating decimal representation.
Sol. We have, \( \frac{441}{2^2 \times 5^7 \times 7^2} = \frac{9 \times 7^2}{2^2 \times 5^7 \times 7^2} = \frac{9}{2^2 \times 5^7} \)
i.e., denominator is of the form \( 2^m \times 5^n \).
Hence, the given rational number has terminating decimal expansion.
Answer: Terminating decimal expansion

Question. The decimal expansion of the rational number \( \frac{89}{2^3 \times 5^2} \) will terminate after how many decimal places ?
Sol. We have, \( \frac{89}{2^3 \times 5^2} = \frac{89 \times 5}{2^3 \times 5^2 \times 5} = \frac{445}{(2 \times 5)^3} = \frac{445}{10^3} = 0.445 \)
Hence, the given rational number will terminate after three decimal places.
Answer: 3 decimal places

Question. Find the least number of square tiles required to pave the ceiling of a room 15 m 17 cm long and 9 m 2 cm broad.
Sol. We have, Length \( = 15\text{m } 17\text{cm} = 1517\text{ cm} \)
Breadth \( = 9\text{m } 2\text{cm} = 902\text{ cm} \)
Side of each square tile \( = \text{HCF} (1517, 902) = 41\text{ cm} \)
\( \therefore \text{Required number of tiles} = \frac{1517 \times 902}{41 \times 41} = 814 \)
Answer: 814

Question. If \( a \) and \( b \) are two positive integers such that \( a = 14b \), then find the HCF of \( a \) and \( b \).
Sol. We have, \( a = 14b \)
Now, \( a = 1 \times a = 1 \times 14 \times b \)
and \( b = 1 \times b \)
So, \( \text{HCF} (a, b) = 1 \times b = b \)
Answer: \( b \)

Short Answer Type Questions-I

Question. Atul, Ravi and Tarun go for a morning walk. They step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ?
Sol. We have,
\( 40 = 2^3 \times 5 \)
\( 42 = 2 \times 3 \times 7 \)
\( 45 = 3^2 \times 5 \)
\( \therefore \text{LCM of 40, 42 and 45} = 2^3 \times 3^2 \times 5 \times 7 = 2520 \)
So, after the walk of 2520 cm each can cover the same distance in complete steps.
Answer: 2520 cm

Question. The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 750. If one number is 250, then find the other number.
Sol. Let HCF be ‘H’.
Then LCM = 14 H.
Sum of LCM and HCF is 750
\( \therefore 14\text{H} + \text{H} = 750 \)
\( \Rightarrow 15\text{H} = 750 \)
\( \Rightarrow \text{H} = \frac{750}{15} = 50 \)
\( \therefore \text{LCM} = 14\text{H} = 14 \times 50 = 700 \)
We know,
Product of LCM and HCF = Product of two numbers.
Let the other number be \( y \).
\( 700 \times 50 = 250 \times y \)
\( \Rightarrow y = \frac{700 \times 50}{250} = 140 \)
Hence, the other number is 140.
Answer: 140

Question. Three alarm clocks ring at intervals of 4, 12 and 20 minutes respectively. If they start ringing together, after how much time will they next ring together ?
Sol. To find the time when the clocks will next ring together, we have to find the LCM of 4, 12 and 20 minutes.
\( 4 = 2^2 \)
\( 12 = 2^2 \times 3 \)
\( 20 = 2^2 \times 5 \)
\( \therefore \text{LCM of 4, 12 and 20} = 2^2 \times 3 \times 5 = 60 \text{ minutes.} \)
So, the clocks will ring together again after 60 minutes or one hour.
Answer: 60 minutes (or 1 hour)

Question. Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.
Sol. Let the required number be \( x \).
Now, required number, \( x = \text{HCF of } (2011 – 9) \text{ and } (2623 – 5) \)
\( = \text{HCF of 2002 and 2618} \)
Now, \( 2002 = 2 \times 7 \times 11 \times 13 \)
and \( 2618 = 2 \times 7 \times 11 \times 17 \)
\( \therefore \text{Required number, } x = 2 \times 7 \times 11 = 154 \)
Answer: 154

Question. Show that every positive even integer is of the form \( 2q \), and that every positive odd integer is of the form \( 2q + 1 \), where \( q \) is some integer. 
Sol. Let \( a \) be any positive integer and \( b = 2 \).
Then, by Euclid’s algorithm, \( a = 2q + r \), for some integer \( q \ge 0 \), and \( r = 0 \) or \( r = 1 \), because \( 0 \le r < 2 \).
So, \( a = 2q \) or \( a = 2q + 1 \)
Now, if \( a \) is of the form \( 2q \), then \( a \) is an even integer.
Therefore, any positive odd integer is of the form \( 2q + 1 \).
Hence Proved.

Question. Given that \( \sqrt{3} \) is an irrational number, prove that \( (2 + \sqrt{3}) \) is an irrational number.
Sol. Let us assume that \( 2 + \sqrt{3} \) is a rational number.
Then, \( 2 + \sqrt{3} \) can be written in the form, \( 2 + \sqrt{3} = \frac{p}{q} \), where \( p \) and \( q \) are co-primes.
or \( \sqrt{3} = \frac{p}{q} - 2 = \frac{p - 2q}{q} \)
\( \frac{p - 2q}{q} \) is in rational form of \( \frac{p}{q} \), so \( \sqrt{3} \) is a rational number. But we are already given that \( \sqrt{3} \) is an irrational number. So, our assumption is wrong.
Hence, \( 2 + \sqrt{3} \) is an irrational number.
Hence Proved.

Question. Show that \( 5 - 2\sqrt{3} \) is an irrational number.
Sol. Let \( 5 - 2\sqrt{3} = \frac{p}{q} \) be a rational number.
\( \therefore 5 - \frac{p}{q} = 2\sqrt{3} \) or \( \frac{5q - p}{2q} = \sqrt{3} \), which is contradiction because \( \sqrt{3} \) is an irrational number and \( \frac{5q - p}{2q} \) is a rational.
\( \therefore \) Our supposition is wrong and hence, \( 5 - 2\sqrt{3} \) is irrational.
Hence Proved.

Question. Check whether \( 4^n \) can end with the digit 0 for any natural number \( n \). 
Sol. If the number \( 4^n \) were to end with the digit zero, then it would be divisible by 2 and 5 both as, \( 10 = 2 \times 5 \).
Now, \( 4^n = (2 \times 2)^n \)
i.e., there is no factor of 5.
Hence, there is no natural number \( n \) for which \( 4^n \) ends with the digit zero.
Answer: No natural number \( n \) exists.

Short Answer Type Questions-II

Question. Find the largest positive integer that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively.
Sol. It is given that on dividing 398 by the required number, there is a remainder of 7. This means that \( 398 – 7 = 391 \) is exactly divisible by the required number. In other words, required number is a factor of 391.
Similarly, required positive integer is a factor of \( 436 – 11 = 425 \) and \( 542 – 15 = 527 \).
Clearly, required number is the \( \text{HCF of 391, 425 and 527} \).
Using the factor tree, we get the prime factorisation of 391, 425 and 527 as follows :
\( 391 = 17 \times 23 \), \( 425 = 5^2 \times 17 \) and \( 527 = 17 \times 31 \)
\( \therefore \text{HCF of 391, 425 and 527 is 17} \).
Hence, required number \( = 17 \).
Answer: 17

Question. Three sets of English, Hindi and Sociology books dealing with cleanliness have to stacked in such a way that all the books are stored topic-wise and height of each stack is the same. The number of English books is 96, number of Hindi books is 240 and the number of Sociology books is 336. Assuming that the books are of same thickness, determine the number of stacks of English, Hindi and Sociology books.
Sol. In order to arrange the books as required, we have to find the largest number that divides 96, 240 and 336 exactly.
Now, largest number \( = \text{HCF of 96, 240 and 336} \)
\( 96 = 2^5 \times 3 \)
\( 240 = 2^4 \times 3 \times 5 \)
and \( 336 = 2^4 \times 3 \times 7 \)
\( \therefore \text{HCF} = 2^4 \times 3 = 16 \times 3 = 48 \)
So, there must be 48 books in each stack.
\( \therefore \text{No. of stacks of English books} = \frac{96}{48} = 2 \)
\( \text{No. of stacks of Hindi books} = \frac{240}{48} = 5 \)
\( \text{No. of stacks of Sociology books} = \frac{336}{48} = 7 \)
Answer: English: 2, Hindi: 5, Sociology: 7

Question. Using Euclid’s division algorithm, find the HCF of the numbers 867 and 255. 
Sol. We have, \( 867 > 255 \)
So, by Euclid’s division lemma
\( 867 = 255 \times 3 + 102 \)
\( 255 = 102 \times 2 + 51 \)
\( 102 = 51 \times 2 + 0 \)
So, the HCF of two numbers 867 and 255 is 51.
Answer: 51

Question. Using Euclid‘s Algorithm, find the HCF of 2048 and 960.
Sol. Step I : Here \( 2048 > 960 \) so, on applying Euclid‘s algorithm, we get
\( 2048 = 960 \times 2 + 128 \)
Step II : Because remainder \( 128 \neq 0 \), so, on applying Euclid‘s algorithm between 960 and 128, we get
\( 960 = 128 \times 7 + 64 \)
Step III : Again remainder \( 64 \neq 0 \), so
\( 128 = 64 \times 2 + 0 \)
Here remainder is 0. So, process ends here.
And dividend is 64 so, required HCF is 64.
Answer: 64

Question. 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 \( m \). 
Sol. Let \( a \) be any positive integer.
Then by the principle of Euclid’s division algorithm corresponding to the positive integers \( a \) and 3, there exist non-negative integers such that \( a = 3q + r \) where \( 0 \le r < 3 \)
Case 1 : If \( r = 0 \), then
\( a = 3q \)
or \( a^2 = 9q^2 = 3(3q^2) = 3m \)
where \( m = 3q^2 \) and is an integer.
Case 2 : If \( r = 1 \), then
\( a = 3q + 1 \)
or \( a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1 \)
where \( m = 3q^2 + 2q \) and is also an integer.
Case 3 : If \( r = 2 \), then
\( a = 3q + 2 \)
or \( a^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 = 3m + 1 \)
where \( m = 3q^2 + 4q + 1 \) and is also an integer
Hence, the square of any positive integer is either of the form \( 3m \) or \( (3m + 1) \) for some integer \( m \).
Hence Proved.

Question. Show that any positive odd integer is of the form \( 4q + 1 \) or \( 4q + 3 \) where \( q \) is some integer.  
Sol. Let \( a \) be any positive odd integer and \( b = 4 \).
\( \therefore \) By applying Euclid’s division lemma, we have
\( a = bq + r \) where \( 0 \le r < b \)
or \( a = 4q + r \) where \( r = 0, 1, 2 \) and 3
Hence, \( a = 4q \) when \( r = 0 \) and is a positive even integer.
\( \Rightarrow a = 4q + 1 \) when \( r = 1 \) and is a positive odd integer.
\( \Rightarrow a = 4q + 2 \) when \( r = 2 \) and is a positive even integer.
\( \Rightarrow a = 4q + 3 \) when \( r = 3 \) and is a positive odd integer.
Hence, any positive odd integer is of the form \( 4q + 1 \) or \( 4q + 3 \) where \( q \) is some integer.
Hence Proved.

Question. Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
Sol. Required number \( = \text{HCF of } (445 – 4), (572 – 5) \text{ and } (699 – 6) \)
\( = \text{HCF of 441, 567 and 693} \)
Now, \( 693 > 567 \), so by Euclid’s division lemma
\( 693 = 567 \times 1 + 126 \)
\( 567 = 126 \times 4 + 63 \)
\( 126 = 63 \times 2 + 0 \)
So, \( \text{HCF of 693 and 567 is 63} \).
Now, \( 441 > 63 \), so by Euclid’s division lemma,
\( 441 = 63 \times 7 + 0 \)
So, \( \text{HCF of 441, 567 and 693} = 63 \)
Hence, the required greatest number is 63.
Answer: 63

Question. Prove that \( \frac{2 + \sqrt{3}}{5} \) is an irrational number, given that \( \sqrt{3} \) is an irrational number. 
Sol. Let \( \frac{2 + \sqrt{3}}{5} \) is a rational number
\( \therefore \frac{2 + \sqrt{3}}{5} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime numbers.
or \( 2 + \sqrt{3} = \frac{5a}{b} \)
\( \sqrt{3} = \frac{5a}{b} - 2 \)
\( \sqrt{3} = \frac{5a - 2b}{b} \)
In R.H.S., \( a, b, 2 \) and 5 are integers.
\( \therefore \) R.H.S. is a rational number but L.H.S. \( = \sqrt{3} \), which is given that \( \sqrt{3} \) is an irrational.
So, it is a contradiction.
Hence, \( \frac{2 + \sqrt{3}}{5} \) is an irrational number.
Answer: Irrational number

Question. Find HCF and LCM of 404 and 96 and verify that \( \text{HCF} \times \text{LCM} = \text{Product of the two given numbers.} \)
Sol. Numbers: 404, 96. To find: HCF and LCM.
\( 404 = 2^2 \times 101 \)
\( 96 = 2^5 \times 3 \)
\( \text{HCF = greatest common factor} = 2^2 = 4 \).
\( \text{LCM = all factors (least power)} = 2^5 \times 3 \times 101 = 9696 \).
Product of two numbers \( = 404 \times 96 = 38784 \).
Product of \( \text{HCF} + \text{LCM} = 9696 \times 4 = 38784 \).
Hence, \( \text{HCF} \times \text{LCM} = \text{product of two numbers.} \)
Answer: HCF=4, LCM=9696

Passage Based Questions

A teacher told 8 students to write a real number on their notebook. Students wrote:
1. 2.14114111411114 .....
2. \( 4.\overline{37} \)
3. \( \frac{2}{7} \)
4. 6.25
5. \( \frac{25}{10} \)
6. \( \frac{37}{15} \)
7. \( \frac{343}{35} \)
8. \( \frac{368}{19} \)

Question. Out of the following, how many numbers are terminating decimal rational numbers?
Answer: 3

Question. Write a non-terminating non-recurring decimal number.
Answer: 2.14114111411114 ......

Long Answer Type Questions

Question. Show that one and only one out of \( n, n + 2 \) and \( n + 4 \) is divisible by 3, where \( n \) is any positive integer.
Sol. We know that any positive integer is of the form \( 3q \) or \( 3q + 1 \) or \( 3q + 2 \) for some integer \( q \) and one and only one of these possibilities can occur.
So, we have following cases :
Case I : When \( n = 3q \)
In this case, we have \( n = 3q \), which is divisible by 3
Now, \( n = 3q \Rightarrow n + 2 = 3q + 2 \), leaves remainder 2 when divided by 3 \( \Rightarrow n + 2 \) is not divisible by 3
Again, \( n = 3q \Rightarrow n + 4 = 3q + 4 = 3(q + 1) + 1 \), leaves remainder 1 when divided by 3 \( \Rightarrow n + 4 \) is not divisible by 3.
Thus, \( n \) is divisible by 3 but \( n + 2 \) and \( n + 4 \) are not divisible by 3.
Case II : When \( n = 3q + 1 \)
In this case, we have \( n = 3q + 1 \), leaves remainder 1 when divided by 3 \( \Rightarrow n \) is not divisible by 3.
Now, \( n = 3q + 1 \Rightarrow n + 2 = (3q + 1) + 2 = 3(q + 1) \), \( \Rightarrow n + 2 \) is divisible by 3
Again, \( n = 3q + 1 \Rightarrow n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 \), leaves remainder 2 when divided by 3 \( \Rightarrow n + 4 \) is not divisible by 3.
Thus, \( n + 2 \) is divisible by 3 but \( n \) and \( n + 4 \) are not divisible by 3.
Case III : When \( n = 3q + 2 \)
In this case, we have \( n = 3q + 2 \), leaves remainder 2 when divided by 3 \( \Rightarrow n \) is not divisible by 3
Now, \( n = 3q + 2 \Rightarrow n + 2 = 3q + 2 + 2 = 3(q + 1) + 1 \), leaves remainder 1 when divided by 3 \( \Rightarrow n + 2 \) is not divisible by 3
Again, \( n = 3q + 2 \Rightarrow n + 4 = 3q + 2 + 4 = 3(q + 2) \), \( \Rightarrow n + 4 \) is divisible by 3
Thus, \( n + 4 \) is divisible by 3 but \( n \) and \( n + 2 \) are not divisible by 3.
Hence Proved.

Question. Show that one and only one out of \( n, (n + 1) \) and \( (n + 2) \) is divisible by 3, where \( n \) is any positive integer.
Sol. Let \( n, n + 1, n + 2 \) be three consecutive positive integers. We know that \( n \) is of the form \( 3q, 3q + 1 \) or \( 3q + 2 \).
Case I : When \( n = 3q \)
In this case, \( n \) is divisible by 3, but \( n + 1 \) and \( n + 2 \) are not divisible by 3.
Case II : When \( n = 3q + 1 \)
In this case \( n + 2 = (3q + 1) + 2 = 3q + 3 = 3(q + 1) \)
\( \Rightarrow (n + 2) \) is divisible by 3
But \( n \) and \( n + 1 \) are not divisible by 3.
Case III : When \( n = 3q + 2 \)
In the case, \( n + 1 = (3q + 2) + 1 = 3q + 3 = 3(q + 1) \)
\( \Rightarrow (n + 1) \) is divisible by 3
But \( n \) and \( n + 2 \) are not divisible by 3.
Hence, only one out of \( n, n + 1 \) and \( n + 2 \) is divisible by 3.
Hence Proved.

Question. Prove that \( \sqrt{2} \) is an irrational number. 
Sol. To prove that \( \sqrt{2} \) is an irrational number, let us assume its opposite i.e., \( \sqrt{2} \) is a rational number.
Also let it be \( \frac{a}{b} \) in its simplest form where \( a \) and \( b \) are co-prime numbers having HCF of 1. Also let \( b \neq 0 \).
\( \therefore \sqrt{2} = \frac{a}{b} \)
\( \Rightarrow (\sqrt{2})^2 = (\frac{a}{b})^2 \)
\( \Rightarrow 2 = \frac{a^2}{b^2} \)
\( \Rightarrow 2b^2 = a^2 \)
Thus \( 2 \mid a^2 \) [\(\because 2 \mid 2b^2 \) and \( 2b^2 = a^2\)]
\( \Rightarrow 2 \mid a \) …(i)
[From theorem, if \( p \) is a prime member and ‘\( a \)’ is a positive integer then, if \( p \) divides \( a^2 \) then \( p \) divides \( a \)]
Now let \( a = 2c \) for some integer \( c \).
Replacing the value of \( a \) in the equation \( 2b^2 = a^2 \), we have
\( 2b^2 = (2c)^2 \)
\( \Rightarrow 2b^2 = 4c^2 \)
\( \Rightarrow b^2 = 2c^2 \)
\( \Rightarrow 2 \mid b^2 \) [\(\because 2 \mid 2c^2 \) and \( 2c^2 = b^2\)]
\( \Rightarrow 2 \mid b \) ...(ii)
Hence, from (i) and (ii), we find that 2 is a common factor of both \( a \) and \( b \). However, this contradicts the fact that \( a \) and \( b \) have only 1 as their common factor. Such a contradiction arises by considering \( \sqrt{2} \) as a rational number. Hence, it is proved that \( \sqrt{2} \) is an irrational number.
Hence Proved.

Question. Use Euclid’s division lemma to show that the cube of any positive integer is either of the form \( 9m, 9m + 1 \) or \( 9m + 8 \) for some integer \( m \). 
Sol. Let \( a \) any positive number and \( b = 3 \) such that \( a = 3q + r \) where \( r = 0, 1, 2 \)
Case I : When \( r = 0 \)
\( \therefore a = 3q \)
\( a^3 = 27q^3 = 9(3q^3) = 9m \), where \( m = 3q^3 \)
Case II : When \( r = 1 \)
\( \therefore a = 3q + 1 \)
\( \Rightarrow a^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9 (3q^3 + 3q^2 + q) + 1 \)
\( \Rightarrow a^3 = 9m + 1 \), where \( m = 3q^3 + 3q^2 + q \)
Case III : When \( r = 2 \)
\( \therefore a = 3q + 2 \)
\( \Rightarrow a^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 \)
\( \Rightarrow a^3 = 9m + 8 \), where \( m = 3q^3 + 6q^2 + 4q \)
Hence, cube of any integer is either of the form \( 9m, 9m + 1 \) or \( 9m + 8 \) where \( m \) is some integer.
Hence Proved.

Question. Prove that \( \sqrt{5} \) is an irrational number. Hence show that \( 3 + 2\sqrt{5} \) is also an irrational number.
Sol. Let \( \sqrt{5} \) be a rational number.
So, \( \sqrt{5} = \frac{p}{q} \)
Squaring both sides, \( 5 = \frac{p^2}{q^2} \Rightarrow q^2 = \frac{p^2}{5} \)
\( \Rightarrow 5 \) is a factor of \( p^2 \Rightarrow 5 \) is a factor of \( p \).
Now, again let \( p = 5c \).
So, \( \sqrt{5} = \frac{5c}{q} \)
On squaring both sides, \( 5 = \frac{25c^2}{q^2} \Rightarrow q^2 = 5c^2 \Rightarrow c^2 = \frac{q^2}{5} \)
\( \Rightarrow 5 \) is factor of \( q^2 \Rightarrow 5 \) is a factor of \( q \).
Here 5 is a common factor of \( p, q \) which contradicts the fact that \( p, q \) are co-prime.
Hence our assumption is wrong, \( \sqrt{5} \) is an irrational number.
Now, we have to show that \( 3 + 2\sqrt{5} \) is an irrational number. So let us assume that \( 3 + 2\sqrt{5} \) is a rational number.
\( \Rightarrow 3 + 2\sqrt{5} = \frac{p}{q} \Rightarrow 2\sqrt{5} = \frac{p}{q} - 3 \Rightarrow 2\sqrt{5} = \frac{p - 3q}{q} \Rightarrow \sqrt{5} = \frac{p - 3q}{2q} \)
\( \frac{p - 3q}{2q} \) is in the rational form of \( \frac{p}{q} \), so \( \sqrt{5} \) is a rational number but we have already proved that \( \sqrt{5} \) is an irrational number so contradiction arises because we supposed wrong that \( 3 + 2\sqrt{5} \) is a rational number. So, we can say that \( 3 + 2\sqrt{5} \) is an Irrational number.
Hence Proved.

Question. Find the HCF of 105 and 120 using Euclid’ sdivision algorithm. Also, find their LCM andverify that \( \text{HCF} \times \text{LCM} = \text{Product of the two numbers.} \)
Sol. Given numbers are 105 and 120.
Since, \( 120 > 105 \), so by Euclid’s division lemma
\( 120 = 105 \times 1 + 15 \)
\( 105 = 15 \times 7 + 0 \)
So, HCF of 120 and 105 \( = 15 \)
Now, \( 120 = 2^3 \times 3 \times 5 \) and \( 105 = 3 \times 5 \times 7 \)
So, LCM of 120 and 105 \( = 2^3 \times 3 \times 5 \times 7 = 840 \)
Verification : \( \text{HCF} \times \text{LCM} = 15 \times 840 = 12600 \)
Product of the numbers \( = 105 \times 120 = 12600 \)
From (i) and (ii), \( \text{HCF} \times \text{LCM} = \text{Product of the numbers} \)
Hence Proved.

Question. Prove that \( \frac{1}{3 + \sqrt{11}} \) is an irrational number.
Sol. Let us assume that \( \frac{1}{3 + \sqrt{11}} \) is a rational number.
Then, \( \frac{1}{3 + \sqrt{11}} = \frac{p}{q} \), where \( p, q \) are co-prime integers.
\( \Rightarrow q = 3p + p\sqrt{11} \) or \( \sqrt{11} = \frac{q - 3p}{p} \)
Since, \( p \) and \( q \) are co-prime integers, so, \( \frac{q - 3p}{p} \) is a rational number.
\( \Rightarrow \sqrt{11} \) is a rational number.
But we already know that \( \sqrt{11} \) is an irrational number.
It means our assumption is wrong.
Hence, \( \frac{1}{3 + \sqrt{11}} \) is an irrational number.
Hence Proved.

Case Study

A teacher wants to conduct an activity in the class. For this activity the rectangular desk of the classroom are to be arranged such that they form a large square. The dimension of a rectangular desk is 18 cm and 24 cm.

Question. Find the least length of the side of the square ?
(a) 80 cm
(b) 85 cm
(c) 72 cm
(d) 75 cm
Sol. To make a square, \( 18 \times 24 \) desks is arranged in a manner that both dimensions are equal. Least length of side of square \( \text{ABCD} = \text{LCM}(18, 24) = 72\text{ cm} \).
Answer: (c)

Question. Calculate the number of rows of desks used to form a square shape.
(a) 5
(b) 4
(c) 6
(d) 10
Sol. Number of rows \( = \frac{\text{Side of square}}{\text{Length of rectangle}} = \frac{72}{18} = 4 \).
Answer: (b)

Question. Calculate the number of columns used to form a square shape.
(a) 3
(b) 4
(c) 7
(d) 2
Sol. Number of columns \( = \frac{\text{Side of square}}{\text{Breadth of rectangle}} = \frac{72}{24} = 3 \).
Answer: (a)

Question. Complete the formula : \( \text{HCF}(a, b) \times ........ = a \times b \)
(a) \( \text{LCM} (c, d) \)
(b) \( a^2 \)
(c) \( b^2 \)
(d) \( \text{LCM} (a, b) \)
Answer: (d)

Question. If 3 and 8 are two numbers. And their HCF is 1, find their LCM.
(a) 24
(b) 8
(c) 3
(d) 22
Sol. \( \text{LCM} (a, b) \times \text{HCF} (a, b) = a \times b \Rightarrow \text{LCM} \times 1 = 3 \times 8 \Rightarrow \text{LCM} = 24 \).
Answer: (a)

Assertion and Reasoning Based Questions

DIRECTIONS : In the following questions, a statement 1 is followed by statement 2. Mark the correct choice as :
(A) If both statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1.
(B) If both statement 1 and statement 2 are true but statement 2 is not the correct expalanation of statement 1.
(C) If statement 1 is true, but statement 2 is false.
(D) If statement 1 is false and statement 2 is true.

Question. Statement 1 : The expression \( 0.\overline{2356} = 0.2356235623562356 .... \)
Statement 2 : When a number is repeated infinite times, then we need to add the bar for the repeated number.
Sol. In decimal expression, when two numbers are divided, sometimes, we may get infinite numbers as the quotient. In that case, we are unable to write the numbers infinite time. So, we can use bar for the number which are repeated infinite times.
Answer: (A)

Question. Statement 1 : The prime factorisation of a natural number is unique, except for the order of its factors.
Statement 2 : The fundamental theorem of Arithmetic states that all composite numbers cannot be expressed as a product of prime factors.
Sol. By theorem, the prime factorisation of a natural number is unique, except for the orders of its factors. In Fundamental Theorem of Arithmetic, all composite numbers cannot be expressed as a product of prime factors. Hence, statement 1 is true but statement 2 is false.
Answer: (C)

Question. Statement 1 : Any positive integer is of the form \( 3q \) or \( 3q + 1 \) or \( 3q + 2 \) for some interger \( q \).
Statement 2 : Any positive even integer is of the form \( 6q + 1 \) or \( 6q + 3 \) or \( 6q + 5 \) for some integer \( q \).
Sol. Let \( a \) be positive integer and \( b = 3 \). Applying division lemma, we get \( a = 3q + r \). \( a = 3q \) or \( a = 3q + 1 \) or \( a = 3q + 2 \). Hence statement 1 is true. \( 6q + 1, 6q + 3, 6q + 5 \) are positive odd integer for some integer \( q \). Hence, the statement 2 is false.
Answer: (C)

Question. Statement 1 : The full form of HCF is highest common factor and it is the product of the smallest power of each common prime factor in the numbers.
Statement 2 : The full form of LCM is lowest common multiple and it is the product of the lowest power of each prime factor involved in the numbers.
Sol. In fundamental theorem, LCM is the product of highest power of each prime factor involved in the numbers. So, the given statement 2 is false.
Answer: (C)

Archives

Question. Two positive integers \( a \) and \( b \) can be written as \( a = x^3 y^2 \) and \( b = xy^3 \); \( x, y \) are prime numbers. Find LCM \( (a, b) \).
Answer: \( x^3 y^3 \)

Question. Express 23150 as product of its prime factors. Is it unique?
Answer: \( 2 \times 5 \times 5 \times 463 \)

Question. Explain why \( (17 \times 5 \times 11 \times 3 \times 2 + 2 \times 11) \) is a composite number?
Answer: It has factors other than 1 and itself.

Question. State whether the real number 52.0521 is rational or not. If it is rational express it in the form \( \frac{p}{q} \), where \( p, q \) are co-prime integers and \( q \neq 0 \). What can you say about prime factorisation of \( q \)?
Answer: Yes, terminating. Prime factorisation of \( q \) is of form \( 2^m 5^n \).

Question. The length, breadth and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly.
Answer: 25 cm

Question. Find LCM and HCF of 3930 and 1800 by prime factorisation method.
Answer: LCM = 235800, HCF = 30.