CBSE Class 10 Mathematics Real Numbers VBQs Set C

SHORT ANSWER Type Questions

Question. Write the smallest number which is divisible by both 306 and 657.
Answer: Given numbers are 306 and 657. The smallest number divisible by 306 and 657 = LCM(306, 657). Prime factors of \( 306 = 2 \times 3 \times 3 \times 17 \). Prime factors of \( 657 = 3 \times 3 \times 73 \). The LCM of (306, 657) = \( 2 \times 3 \times 3 \times 17 \times 73 = 22338 \). Hence, the smallest number divisible by 306 and 657 is 22,338.

Question. A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form \( \frac{p}{q} \)? Give reasons. 
Answer: As 327.7081 is a terminating decimal number, the denominator of the rational number must be of the form \( 2^m \times 5^n \). Thus, \( 327.7081 = \frac{3277081}{10000} = \frac{3277081}{10^4} = \frac{3277081}{2^4 \times 5^4} = \frac{p}{q} \). So, the prime factors of \( q \) are 2 and 5. Here, \( q \) is of the form \( 2^m \times 5^n \), where \( m \) and \( n \) are natural numbers. The prime factors of \( p \) and \( q \) will be either 2 or 5 or both.

Question. Without actually performing the long division, write the decimal expansion of \( \frac{11725}{2^3 \times 5^4} \).
Answer: \( \frac{11725}{2^3 \times 5^4} = \frac{11725 \div 5}{2^3 \times 5^3} = \frac{2345}{(10)^3} = 2.345 \)

Question. Write any two irrational numbers whose product is a rational number.
Answer: Consider two irrationals as, \( 5 - 2\sqrt{2} \) and \( 5 + 2\sqrt{2} \). Here, \( (5 - 2\sqrt{2})(5 + 2\sqrt{2}) = 5^2 - (2\sqrt{2})^2 = 25 - 8 = 17 \) (a rational number).

Question. Using prime factorisation method, find the HCF and LCM of 210 and 175. 
Answer: The prime factorisations of 210 and 175 are: \( 210 = 2 \times 3 \times 5 \times 7 \) and \( 175 = 5 \times 5 \times 7 \). So, HCF (210, 175) = \( 5 \times 7 = 35 \); and LCM (210, 175) = \( 2 \times 3 \times 5 \times 7 \times 5 = 1050 \).

Question. Prove that the number \( 4^n \), n being a natural number, can never end with the digit 0.
Answer: If \( 4^n \) ends with 0, then it must have 5 as a factor. But, \( (4)^n = (2^2)^n = 2^{2n} \), i.e., the only prime factor of \( 4^n \) is 2. Also, we know from the Fundamental Theorem of Arithmetic that the prime factorisation of each number is unique. \( \therefore 4^n \) can never end with 0.

Question. Find the two numbers which on multiplication with \( \sqrt{360} \) gives a rational number. Are these numbers rational or irrational? 
Answer: \( \sqrt{360} = \sqrt{2 \times 2 \times 2 \times 3 \times 3 \times 5} = 6\sqrt{10} \). If we multiply \( 6\sqrt{10} \) with \( \sqrt{10} \) and 1. We get, \( 6\sqrt{10} \times \sqrt{10} \times 1 = 60 \). Hence, numbers are \( \sqrt{10} \) and 1. Where, 1 is a rational number and \( \sqrt{10} \) is an irrational number.

Question. Prove that \( \sqrt{5} \) is an irrational number. 
Answer: Let us assume, to the contrary, that \( \sqrt{5} \) is a rational number and its simplest form is \( \frac{a}{b} \), where \( a \) and \( b \) are integers having no common factor other than 1 and \( b \neq 0 \).
Now, \( \sqrt{5} = \frac{a}{b} \)
\( \Rightarrow 5 = \frac{a^2}{b^2} \)
\( \Rightarrow 5b^2 = a^2 \) ...(i)
\( \Rightarrow a^2 \) is divisible by 5 [\( \because 5b^2 \) is divisible by 5]
\( \Rightarrow a \) is divisible by 5 [\( \because \) 5 is a prime number and divides \( a^2 \Rightarrow 5 \) divides \( a \)]
Let \( a = 5c \), for some integer ‘c’
On substituting \( a = 5c \) in (i), we get
\( 5b^2 = (5c)^2 \)
\( \Rightarrow 5b^2 = 25c^2 \)
\( \Rightarrow b^2 = 5c^2 \)
\( \Rightarrow b^2 \) is divisible by 5 [\( \because 5c^2 \) is divisible by 5]
\( \Rightarrow b \) is divisible by 5
Since \( a \) and \( b \) are both divisible by 5, 5 is a common factor of \( a \) and \( b \). But this contradicts the fact that \( a \) and \( b \) have no common factor other than 1. This contradiction has arisen because of our incorrect assumption that \( \sqrt{5} \) is a rational number. Hence, \( \sqrt{5} \) is irrational.

Question. Prove that \( 2 + 5\sqrt{3} \) is an irrational number, given that \( \sqrt{3} \) is an irrational number. 
Answer: Let us assume that \( 2 + 5\sqrt{3} \) is a rational number. Therefore: \( 2 + 5\sqrt{3} = \frac{a}{b} \) (where, ‘a’ and ‘b’ are co-primes).
\( \Rightarrow 5\sqrt{3} = \frac{a}{b} - 2 \)
\( \Rightarrow 5\sqrt{3} = \frac{a - 2b}{b} \)
\( \Rightarrow \sqrt{3} = \frac{a - 2b}{5b} \)
Therefore, \( \frac{a - 2b}{5b} \) is in the form of \( \frac{a}{b} \) which is a rational number. But, this contradicts the fact that \( \sqrt{3} \) is an irrational number. Therefore, our assumption is wrong and \( 2 + 5\sqrt{3} \) is an irrational number.

Question. Prove that \( \sqrt{2} \) is an irrational number. 
Answer: Let us assume \( \sqrt{2} \) be a rational number and its simplest form be \( \frac{a}{b} \), \( a \) and \( b \) as coprimes. So, \( \sqrt{2} = \frac{a}{b} \Rightarrow a^2 = 2b^2 \). Thus, \( a^2 \) is a multiple of 2. \( \Rightarrow a \) is a multiple of 2. ...(i)
Let \( a = 2m \) for some integer \( m \). Then, \( (2m)^2 = 2b^2 \Rightarrow 4m^2 = 2b^2 \Rightarrow b^2 = 2m^2 \). Thus, \( b^2 \) is a multiple of 2. \( \Rightarrow b \) is a multiple of 2. ...(ii)
From (i) and (ii), 2 is a common factor of \( a \) and \( b \). This contradicts the fact that \( a \) and \( b \) are co-primes. Hence, \( \sqrt{2} \) is an irrational number.

Question. Find HCF and LCM of 404 and 96 and verify that HCF \( \times \) LCM = Product of the two given numbers. 
Answer: Given, numbers are 404 and 96. Prime factorisation of both the number: \( 404 = 2 \times 2 \times 101 = 2^2 \times 101 \). \( 96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3 \). HCF of 404 and 96 = \( 2^2 = 4 \). LCM of 404 and 96 = \( 2^5 \times 3 \times 101 = 9696 \).
Now, HCF \( \times \) LCM = \( 4 \times 9696 = 38784 \) ...(i)
Product of two numbers = \( 404 \times 96 = 38784 \) ...(ii)
From (i) and (ii) we get HCF \( \times \) LCM = Product of two numbers. Hence, verified.

Question. Write the denominator of rational number \( \frac{257}{5000} \) in the form \( 2^m \times 5^n \), where m, n are non-negative integers. Then, write its decimal expansion without actual division. 
Answer: Denominator of the rational number \( \frac{257}{5000} \) is 5000. Now, \( 5000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = (2)^3 \times (5)^4 \), which is of the type \( 2^m \times 5^n \), where \( m = 3 \) and \( n = 4 \) are non–negative integers. Simplifying the given fraction: \( \frac{257}{5000} = \frac{257}{5^4 \times 2^3} = \frac{257 \times 2}{5^4 \times 2^3 \times 2} = \frac{514}{5^4 \times 2^4} = \frac{514}{10^4} = 0.0514 \). So, 0.0514 is the required decimal expansion of the rational number \( \frac{257}{5000} \).

Question. Three bells toll at intervals of 12 minutes, 15 minutes and 18 minutes respectively, if they start tolling together, after what time will they next toll together?
Answer: The required time is the LCM of 12, 15 and 18. \( 12 = 2 \times 2 \times 3 \), \( 15 = 3 \times 5 \), \( 18 = 2 \times 3 \times 3 \). LCM = \( 2^2 \times 3^2 \times 5 = 180 \). So, next time the bells will ring together after 180 minutes or 3 hours.

Question. Find if \( \frac{987}{10500} \) will have terminating or non–terminating (repeating) decimal expansion. Give reasons for your answer. 
Answer: Yes, it will have a terminating decimal expansion. Simplified denominator has factor in the form of \( 2^m \times 5^n \). \( \frac{987}{10500} = \frac{3 \times 7 \times 47}{2 \times 2 \times 3 \times 5 \times 5 \times 5 \times 7} = \frac{47}{2^2 \times 5^3} = \frac{47 \times 2}{2^2 \times 2 \times 5^3} = \frac{94}{1000} = 0.094 \). And we know, if \( \frac{p}{q} \) is a rational number, such that the prime factorization of \( q \) is of the form \( 2^m \times 5^n \) where \( n \) and \( m \) are non–negative integers, then \( x \) has a decimal expansion which terminates. Hence, it terminates.

Question. On a morning walk, three people 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 covers the same distance in complete steps? 
Answer: We know that the LCM is the product of the greatest power of each prime factor of the numbers. We have to find the LCM of 40, 42 and 45 to get the required minimum distance. For this, we find prime factorisation: \( 40 = 2 \times 2 \times 2 \times 5 \), \( 42 = 2 \times 3 \times 7 \), \( 45 = 3 \times 3 \times 5 \). LCM (40, 42, 45) = \( 2^3 \times 3^2 \times 5 \times 7 = 2520 \). Hence, each person should walk a minimum distance of 2520 cm, so that each of them can cover the same distance in complete steps.

Question. A merchant has 120 litres and 180 litres of two kinds of oil. He wants to sell oil by filling the two kinds of oil in tins of equal volumes. What is the greatest volume of such a tin?
Answer: In order to find volume of such a tin, we need to find the largest number which exactly divides 120 and 180 which is nothing but the HCF (120, 180). \( 120 = 2 \times 2 \times 2 \times 3 \times 5 \), \( 180 = 2 \times 2 \times 3 \times 3 \times 5 \). H.C.F. (120, 180) = \( 2 \times 2 \times 3 \times 5 = 60 \). Hence, the greatest volume of each tin is 60 litres.

Question. Using prime factorisation, find HCF and LCM of 18, 45 and 60. Check if HCF \( \times \) LCM = product of the numbers.
Answer: Here, \( 18 = 2 \times 3^2 \), \( 45 = 3^2 \times 5 \), and \( 60 = 2^2 \times 3 \times 5 \). So, HCF (18, 45, 60) = 3; and LCM (18, 45, 60) = \( 2^2 \times 3^2 \times 5 = 180 \). Clearly, HCF \( \times \) LCM = \( 3 \times 180 = 540 \) whereas, product of numbers = \( 18 \times 45 \times 60 = 48600 \). Hence, HCF \( \times \) LCM \( \neq \) Product of numbers.

Question. Show that the square of any positive odd integer is of the form 8m + 1, for some integer m.
Answer: Let a be any positive integer. So, it is of the form \( 2q + 1 \), for some integer \( q \). i.e. \( a = 2q + 1 \Rightarrow a^2 = (2q + 1)^2 = 4q^2 + 4q + 1 = 4q(q + 1) + 1 \). Now, \( q(q + 1) \) is either 0 or even. So, it is \( 2m \), where \( m \) is a whole number. \( \therefore a^2 = 4 \times 2m + 1 \) i.e. \( 8m + 1 \).

Question. Prove that \( \sqrt{p} + \sqrt{q} \) is irrational, where p and q are primes. 
Answer: Let us suppose that \( \sqrt{p} + \sqrt{q} = a \), where a is a rational number, \( \Rightarrow \sqrt{p} = a - \sqrt{q} \). On squaring both sides, we get \( p = a^2 + q - 2a\sqrt{q} \). \( \Rightarrow \sqrt{q} = \frac{a^2 + q - p}{2a} \). Therefore, the above statement is a contradiction as the right hand side is a rational number, while the left hand side \( \sqrt{q} \) is irrational, since \( q \) is a prime number. So, our assumption is wrong. Hence, \( \sqrt{p} + \sqrt{q} \) is irrational.

Question. Prove that \( 3 + 2\sqrt{5} \) is irrational number.
Answer: Let \( 3 + 2\sqrt{5} = \frac{a}{b} \). Here \( a \) and \( b \) are two co-prime integers and \( b \neq 0 \). Subtracting 3 from both sides, we get \( 2\sqrt{5} = \frac{a}{b} - 3 \Rightarrow 2\sqrt{5} = \frac{a - 3b}{b} \). On dividing both sides by 2, we get \( \sqrt{5} = \frac{a - 3b}{2b} \). Here \( a \) and \( b \) are integers, so \( \frac{a - 3b}{2b} \) is a rational number which implies \( \sqrt{5} \) should be a rational number, but \( \sqrt{5} \) is an irrational number so it is a contradiction. Hence, \( 3 + 2\sqrt{5} \) is an irrational number.

Question. Show that \( 5 + 2\sqrt{7} \) is an irrational number, where \( \sqrt{7} \) is given to be an irrational number. 
Answer: Let us assume, on the contrary, that \( 5 + 2\sqrt{7} \) is a rational number. i.e., \( 5 + 2\sqrt{7} = \frac{a}{b} \), where ‘a’ and ‘b’ are co-prime numbers. \( \Rightarrow 2\sqrt{7} = \frac{a}{b} - 5 \Rightarrow \sqrt{7} = \frac{a - 5b}{2b} \). Since \( \frac{a - 5b}{2b} \) is a rational number, \( \sqrt{7} \) is also a rational number, which is contradiction to the given results. Hence, \( 5 + 2\sqrt{7} \) is irrational.

LONG ANSWER Type Questions 

Question. Show that the square of any positive integer cannot be of the form (5q + 2) or (5q + 3) for any integer q. 
Answer: Let ‘a’ be any positive integer. Then, it is of the form 5p, or 5p + 1 or 5p + 2 or 5p + 3 or 5p + 4.
Case 1 When \( a = 5p \Rightarrow a^2 = 25p^2 = 5(5p^2) = 5q \), where \( q = 5p^2 \).
Case 2 When \( a = 5p + 1 \Rightarrow a^2 = 25p^2 + 10p + 1 = 5(5p^2 + 2p) + 1 = 5q + 1 \), where \( q = 5p^2 + 2p \).
Case 3 When \( a = 5p + 2 \Rightarrow a^2 = 25p^2 + 20p + 4 = 5(5p^2 + 4p) + 4 = 5q + 4 \), where \( q = 5p^2 + 4p \).
Case 4 When \( a = 5p + 3 \Rightarrow a^2 = 25p^2 + 30p + 9 = 5(5p^2 + 6p + 1) + 4 = 5q + 4 \), where \( q = 5p^2 + 6p + 1 \).
Case 5 When \( a = 5p + 4 \Rightarrow a^2 = 25p^2 + 40p + 16 = 5(5p^2 + 8p + 3) + 1 = 5q + 1 \).
Thus, square of any positive integer cannot be of the form 5q + 2 or 5q + 3, for any integer n.

Question. Prove that one of every three consecutive positive integers is divisible by 3. 
Answer: Let n, n + 1 and n + 2 be three consecutive positive integers. Also, we know that a positive integer n is of the form 3q, 3q + 1 or 3q + 2.
Case I: When n = 3q. Here n is clearly divisible by 3. But ( n + 1) and (n + 2) are not divisible by 3.
Case II: When n = 3q + 1. Here \( n + 2 = 3q + 3 = 3(q + 1) \). Clearly, it is divisible by 3. But n and (n + 1) are not divisible by 3.
Case III: when n = 3q + 2. Here, \( n + 1 = 3q + 3 = 3(q + 1) \). clearly, (n + 1) is divisible by 3. But n and (n + 2) are not divisible by 3. Hence, one of every three consecutive positive integers is divisible by 3.

Question. Prove that \( \sqrt{n} \) is not a rational number, if n is not perfect square.
Answer: Let \( \sqrt{n} \) be a rational number. \( \therefore \sqrt{n} = \frac{p}{q} \), where p and q are co-prime integers, \( q \neq 0 \). On squaring both sides, we get \( n = \frac{p^2}{q^2} \Rightarrow p^2 = nq^2 \) ...(i). \( \Rightarrow n \) divides \( p^2 \). [Let p be a prime number. If p divided \( a^2 \) then p divides a, where a is a positive integer]. \( \Rightarrow n \) divides p ...(ii). Let \( p = nm \), where m is any integer. \( \Rightarrow p^2 = n^2 m^2 \). (i) \( \Rightarrow n^2 m^2 = nq^2 \Rightarrow q^2 = nm^2 \Rightarrow n \) divides \( q^2 \Rightarrow n \) divides q ...(iii). From (ii) and (iii), n is a common factor of both p and q which contradicts the assumption that p and q are co-prime integer. So, our supposition is wrong, \( \sqrt{n} \) is an irrational number.

Question. The decimal expansions of some real numbers are given below. In each case, decide whether they are rational or not. If they are rational, write it in the form \( \frac{p}{q} \). What can you say about the prime factors of q?
(A) 0.140140014000140000 ...
(B) \( 0.\bar{16} \)
Answer: (A) We have, 0.140140014000140000... a non-terminating and non-repeating decimal expansion. So it is irrational. It cannot be written in the form of \( \frac{p}{q} \).
(B) We have, \( 0.\bar{16} \) a non-terminating but repeating decimal expansion. So it is rational. Let \( x = 0.\bar{16} \). Then, \( x = 0.1616... \) ...(i), \( 100x = 16.1616... \) ...(ii). On subtracting (i) from (ii), we get \( 100x - x = 16.1616 - 0.1616 \Rightarrow 99x = 16 \Rightarrow x = \frac{16}{99} \). The denominator (q) has factors other than 2 or 5.