Exercise 1.2
Class X
Q.No. 1. Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Solution:
(i)
140
Prime factors of 140 = 2x2x5x7 Ans.
(ii)
156
Prime factors of 156 = 2x2x3x13 Ans.
(iii)
3825
Prime factors of 3825 = 3x3x5x5x17 Ans.
(iv)
5005
Prime factors of 5005 = 5x7x11x13 Ans.
(v)
7429
Prime factors of 7429 = 17x19x23 Ans.
Q.No. 2. Find the LCM and HCF of the following pairs of integers and
verify that LCMxHCF = Product of the two numbers.
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Solution:
(i)
26 and 91
Prime factors of 26 = 2x13
Prime factors of 91 = 13x7
LCM = 2x7x13 = 182
HCF = 13
LCM x HCF = 182x13=2366
Product of number = 26x91 = 2366
LCMxHCF = Product of number is verified.
(ii)
510 and 92
Prime factors of 510 = 2x3x5x17
Prime factors of 92 = 2x2x23
LCM = 2x2x3x5x17x23 = 23460
HCF = 2
LCMxHCF = 23460x2 = 46920
Product of number = 46920
Hence, LCMxHCF = Product of number is verified.
(iii)
336 and 54
Prime factors of 336 = 2x2x2x2x3x7
Prime factors of 54 = 2x3x3x3
LCM = 2x2x2x2x3x3x3x7 = 3024
HCF = 2x3 = 6
LCMxHCF = 18144
Product of number = 336x54 = 18144
Hence, LCMxHCF = Product of number is verified.
Q.No. 3. Find the LCM and HCF of the following integers by applying
the prime factorisation method.
(i) 12, 15and 21 (ii) 17, 23 and 29 (iii) 8,9 and 25
Solution:
(i)
12, 15 and 21
Prime factorisation of 12 = 2x2x3
Prime factorisation of 15 = 3x5
Prime factorisation of 21 = 3x7
LCM = 2x2x3x5x7 = 420
HCF = 3
(ii)
17, 23 and 29
Prime factorisation of 17 = 17
Prime factorisation of 23 = 23
Prime factorisation of 29 = 29
LCM = 17x23x29 = 11339
HCF = 1
(iii)
8, 9 and 25
Prime factorisation of 8 = 2x2x2 = 8
Prime factorisation of 9 = 3x3
Prime factorisation of 25 = 5x5
LCM = 2x2x2x3x3x5x5 = 1800
HCF = 1
Q.No. 4.Given that HCF (306, 657) = 9 , find LCM (306, 657)
Solution:
LCMxHCF = Product of numbers
or, LCM = Product of numbers/HCF
or, LCM = 306x657/9
or, LCM = 22338 Ans.
Q.No. 5. Check whether
can end with the digit 0 for
any natural number n.
Solution:
Prime factorisation of 6 = 2x3
So, the prime factorisation of 6 is not divisible by 5.
Hence,
cannot end with 0 for any natural number n according to Fundamental
Theorem of Arithmetic.
Q.No. 6. Explain why 7x11x13 + 13 and 7x6x5x4x3x2x1 + 5 are
composite numbers.
Solution:
These numbers can be expressed as the form of the factorisation of
prime numbers.
So, these numbers are composite numbers.
Q.No. 7. There is 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 the same direction. After how many minutes will they meet again at the
starting point.
Solution:
Prime factorisation of 18 = 2x3x3
Prime factorisation of 18 = 2x2x3x3
LCM of 12 and 18 = 2x2x3x3 = 36
So, they will meet again after 36 minutes. Ans.
