CBSE Class 10 Mathematics Real Numbers VBQs Set 05

Question. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Answer:
(i) \( 140 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7 \)
(ii) \( 156 = 2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13 \)
(iii) \( 3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17 \)
(iv) \( 5005 = 5 \times 7 \times 11 \times 13 \)
(v) \( 7429 = 17 \times 19 \times 23 \)

Question. Find the LCM and HCF of the following integers and verify that LCM \( \times \) HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Answer:
(i) Here \( 26 = 2 \times 13 \) and \( 91 = 7 \times 13 \)
\( \therefore \) HCF = 13 and LCM \( = 2 \times 7 \times 13 = 182 \)
Now, \( 26 \times 91 = 2366 \)
and HCF \( \times \) LCM \( = 13 \times 182 = 2366 \)
(ii) \( 510 = 2 \times 3 \times 5 \times 17 \)
\( 92 = 2 \times 2 \times 23 \)
\( \therefore \) HCF = 2 and LCM = 23460
Now \( 510 \times 92 = 46920 \)
Also HCF \( \times \) LCM \( = 2 \times 23460 = 46920 \)
\( \therefore 510 \times 92 = \text{HCF} \times \text{LCM} \)
(iii) \( 336 = 2^4 \times 3 \times 7 \)
\( 54 = 2 \times 3^3 \)
\( \therefore \) HCF \( = 2 \times 3 = 6 \)
LCM = 3024
Now \( 336 \times 54 = 18144 \)
and HCF \( \times \) LCM \( = 6 \times 3024 = 18144 \)

\( \implies 336 \times 54 = \text{HCF} \times \text{LCM} \)

Question. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Answer:
(i) Given numbers are 12, 15 and 21
\( 12 = 2^2 \times 3, 15 = 3 \times 5, 21 = 3 \times 7 \)
\( \therefore \) HCF = 3, LCM \( = 3 \times 2^2 \times 5 \times 7 = 420 \)
(ii) Given numbers are 17, 23 and 29
\( 17 = 17 \times 1, 23 = 23 \times 1, 29 = 29 \times 1 \)
HCF = 1
LCM \( = 17 \times 23 \times 29 = 11339 \)
(iii) Given numbers are 8, 9 and 25
\( 8 = 2^3, 9 = 3^2, 25 = 5^2 \)
\( \therefore \) HCF = 1, LCM \( = 2^3 \times 3^2 \times 5^2 = 1800 \)

Question. Given that HCF (306, 657) = 9, find LCM (306, 657). Or Write the smallest number which is divisible by both 306 and 657.
Answer: HCF (306, 657) = 9
using HCF \( \times \) LCM = Product of two numbers
we get \( 9 \times \text{LCM} = 306 \times 657 \)

\( \implies \text{LCM} = \frac{306 \times 657}{9} = 22338 \)

Question. Check whether \( 6^n \) can end with the digit 0 for any natural number n.
Answer: Here \( 6^n = (2 \times 3)^n = 2^n \times 3^n \)
Prime in the factorisation of \( 6^n \) are 2 and 3. For \( 6^n \) to end in digit zero prime factorisation of \( 6^n \) should contain the prime 5. Therefore, there is no natural number n for which \( 6^n \) ends with digit zero.

Question. Explain why \( 7 \times 11 \times 13 + 13 \) and \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) are composite numbers.
Answer: Here,
(i) \( 7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times 78 \)
\( = 13 \times 13 \times 2 \times 3 = 2 \times 3 \times 13^2 \)
Factorisation of the number contain more than one prime
\( \therefore \) number is composite.
(ii) \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \)
\( = 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) \)
\( = 5(1008 + 1) = 5 \times 1009 \)
Product of two numbers
\( \therefore \) It is not a prime
\( \therefore 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 \) is a composite number.

Question. 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?
Answer: They will meet again at the starting point in the time which will be equal to the LCM of 18 and 12
Now \( 18 = 2 \times 3^2, 12 = 2^2 \times 3 \)
\( \therefore \) LCM \( = 2^2 \times 3^2 = 36 \)
\( \therefore \) They will meet after 36 minutes.

Question. Prove that \( \sqrt{5} \) is an irrational.
Answer: Let \( \sqrt{5} \) is a rational number and \( \sqrt{5} = \frac{a}{b} \), where a and b are coprime and \( b \neq 0 \),
Now, \( (\sqrt{5})^2 = \left(\frac{a}{b}\right)^2 \)

\( \implies 5b^2 = a^2 \) ...(i)

\( \implies 5 \) is a factor of \( a^2 \)
\( \therefore \) a is also divisible by 5.
Let \( a = 5c \), where c is some integer.
Substituting \( a = 5c \) in (i)
we get \( 5b^2 = (5c)^2 \)
\( \implies 5b^2 = 25c^2 \)
\( \implies b^2 = 5c^2 \)

\( \implies 5 \) is a factor of \( b^2 \)
\( \therefore 5 \) is a factor of b.
\( \therefore 5 \) is a common factor of a and b
This contradicts the fact that a and b are coprime so, our assumption is wrong.
Hence, \( \sqrt{5} \) is irrational.

Question. Prove that \( 3 + 2\sqrt{5} \) is irrational.
Answer: As \( \sqrt{5} \) is irrational.
Let us assume that \( 3 + 2\sqrt{5} \) is rational such that
\( 3 + 2\sqrt{5} = \frac{a}{b} \)
where a and b are coprime and \( b \neq 0 \)

\( \implies 2\sqrt{5} = \frac{a}{b} - 3 \)
\( \implies 2\sqrt{5} = \frac{a - 3b}{b} \)
\( \implies \sqrt{5} = \frac{a - 3b}{2b} \)
\( \because a \) and b are integers
\( \therefore \frac{a - 3b}{2b} = \text{Rational} \)
\( \because \frac{a - 3b}{2b} = \sqrt{5} \)
\( \implies \sqrt{5} \) is also rational.
But this contradicts the fact that \( \sqrt{5} \) is irrational.
Hence our assumption is wrong.
\( \therefore 3 + 2\sqrt{5} \) is irrational.

Question. Prove that the following are irrationals. (i) \( \frac{1}{\sqrt{2}} \) (ii) \( 7\sqrt{5} \) (iii) \( 6 + \sqrt{2} \)
Answer:
(i) As \( \sqrt{2} \) is irrational.
Now let us assume that \( \frac{1}{\sqrt{2}} \) is a rational number and \( \frac{1}{\sqrt{2}} = \frac{a}{b} \), where a and b are coprime and \( b \neq 0 \)

\( \implies \sqrt{2} = \frac{b}{a} \), \( \because a, b \) are integers
\( \therefore \frac{b}{a} \) is a rational. So \( \sqrt{2} \) is also rational
This contradicts the fact that \( \sqrt{2} \) is irrational.
\( \therefore \) Our assumption is wrong.
Hence \( \frac{1}{\sqrt{2}} \) is irrational.
(ii) As \( \sqrt{5} \) is irrational.
Now let us assume that \( 7\sqrt{5} \) is rational

\( \implies 7\sqrt{5} = \frac{a}{b} \), where a and b are coprime, \( b \neq 0 \).
Now, \( 7\sqrt{5} = \frac{a}{b} \)
\( \implies \sqrt{5} = \frac{a}{7b} \)
\( \because a \) and b are integers.
\( \therefore \frac{a}{7b} \) is rational \( \implies \sqrt{5} \) is also rational.
This contradicts the fact that \( \sqrt{5} \) is irrational.
\( \therefore \) Our assumption is wrong.
Hence \( 7\sqrt{5} \) is irrational.
(iii) As \( \sqrt{2} \) is irrational.
Now let us assume that \( 6 + \sqrt{2} \) is a rational number and \( 6 + \sqrt{2} = \frac{a}{b} \) where a and b are coprime, \( b \neq 0 \)

\( \implies \sqrt{2} = \frac{a}{b} - 6 \)
\( \implies \sqrt{2} = \frac{a - 6b}{b} \)
Now a and b are integers.

\( \implies \frac{a - 6b}{b} \) is rational. So \( \sqrt{2} \) is also rational.
This contradicts the fact that \( \sqrt{2} \) is irrational.
\( \therefore \) Our assumption is wrong.
Hence, \( 6 + \sqrt{2} \) is irrational.

Question. If two positive integers a and b are written as \( a = x^3y^2 \) and \( b = xy^3 \), where x, y are prime numbers, then HCF(a, b) is
(a) xy
(b) \( xy^2 \)
(c) \( x^3y^3 \)
(d) \( x^2y^2 \)
Also, find LCM of (a, b).
Answer: (b) \( xy^2 \)
Sol. Here, \( a = x^3y^2 \) and \( b = xy^3 \)

\( \implies a = x \times x \times x \times y \times y \)
and \( b = x \times y \times y \times y \)
\( \therefore \) HCF(a, b) \( = x \times y \times y = x \times y^2 = xy^2 \)
LCM \( = x^3y^3 \)

Question. If two positive integers p and q can be expressed as \( p = ab^2 \) and \( q = a^3b \); where a, b being prime numbers, then LCM (p, q) is equal to
(a) ab
(b) \( a^2b^2 \)
(c) \( a^3b^2 \)
(d) \( a^2b^3 \)
Answer: (c) \( a^3b^2 \)
Sol. LCM (p, q) \( = a^3b^2 \)

Question. Find the HCF and LCM of 6, 72 and 120 using fundamental theorem of arithmetic.
Answer: \( 6 = 2 \times 3 \); \( 72 = 2 \times 2 \times 2 \times 3 \times 3 \); \( 120 = 2 \times 2 \times 2 \times 3 \times 5 \)
Common factors of 6, 72 and 120 are 2 and 3.
HCF \( = 2 \times 3 = 6 \)
LCM \( = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \)
\( \therefore \) LCM = 360

Question. Can two numbers have 18 as their HCF and 380 as their LCM? Give reason.
Answer: No, because HCF is always a factor of LCM. But 18 is not a factor of 380.

Question. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(a) 10
(b) 100
(c) 504
(d) 2520
Answer: (d) 2520
Sol. The prime factorisation of numbers from 2 to 10:
2 = 2
3 = 3
4 = 2 \( \times \) 2
5 = 5
6 = 2 \( \times \) 3
7 = 7
8 = 2 \( \times \) 2 \( \times \) 2
9 = 3 \( \times \) 3
10 = 2 \( \times \) 5
\( \therefore \) hence (1, 2, 3, ..., 10) \( = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2520 \)