Class 10 Maths Chapter 1 Question Answers - Real Numbers

Q1: Given that HCF (150, 100) = 50. Find LCM (150, 100).
Sol: LCM × HCF  = Product of the two numbers
∴  150 × 100 =  LCM × HCF
⇒ LCM × 50  =  150 × 100
⇒  

Q2: Given that LCM (26, 91) = 182. Find their HCF.
Sol: ∵  HCF × LCM  =  Product of the two numbers
∴   HCF × 182  =  26 × 91
⇒  

Q3: The LCM and HCF of the two numbers are 240 and 12 respectively. If one of the numbers is 60, then find the other number.
Sol: Let the required number be ‘x’.
∵  LCM × HCF   =  Product of the two numbers
∴ 60 × x  = 240 × 12
⇒

Q4: The decimal expansion of the rational number, will terminate after how many places of decimal?  
Sol:
 

 Thus,   will terminate after 4 places of decimal.    

Q5: What is the exponent of 3 in the prime factorisation of 864.
Sol:

Making prime factors of 864. ,⇒ 864 = 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2= 3³ × 2⁵
∴ Exponent of 3 in prime factorisation of 864 = 3.

Q6: State the fundamental theorem of arithmetic.
Sol: Every composite number can be expressed as the product of primes and this decomposition is unique apart from the order in which prime factors occur.

Q7:  Define an irrational number.
Sol: Those numbers which neither terminate in their decimal expansion nor can be expressed as recurring decimals are irrational numbers i.e., the numbers which cannot be expressed as p/q form (q ≠ 0), are called irrational numbers.

Q8: Write the condition for a rational number which can have a terminating decimal expansion.
Sol: A rational number x = p/q can have a terminating decimal expansion if the prime factorisation of q is of the form of 2ⁿ · 5^m, where m and n are non-negative integers.

Q9: Write the condition for a rational number which has a non-terminating repeating decimal expansion.
Sol: A rational number x = p/q can have a non-terminating repeating decimal expansion if the prime factorisation of q is not of the form 2ⁿ · 5^m, where n, m are non-negative integers.

Q10: Can two numbers have 24 as their HCF and 7290 as their LCM? Give reasons.
Sol: No, because HCF always divides LCM but here 24 does not divide 7290.

Q11: If 6ⁿ is a number such that n is a natural number. Check whether there is any value of n ∈ N for which 6ⁿ is divisible by 7.
Sol: ∵ 6 = 2 × 3
∴  6ⁿ =  (2 × 3)ⁿ = 2ⁿ × 3ⁿ
 i.e., the prime factorisation of 6ⁿ does not contain the prime number 7 thus the number 6ⁿ is not divisible by 7.

Q12: Write 98 as the product of its prime factors.
Sol:  ∵ 

 ∴ 98 = 2 × 7 × 7 ⇒  98 = 2 × 7²

Q13: Without actually performing the long division, state whether  will have a terminating or non-terminating repeating decimal expansion. 
Sol: Let  =   
∵ Prime factors of q are not of the for 2ⁿ · 5^m.
∴  will have a non-terminating repeating decimal expansion.

Q14: Without actually performing the long division, state whether 17/3125 will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
Sol: ∵ The denominator of 17/3125 is given by
3125 =    5 × 5 × 5 × 5 × 5
=  1 × 55
= 2⁰ × 5⁵    |∵ 2⁰ = 1
∴  

i.e., 17/3125 is a terminating decimal.

Q15: Express 156 as a product of its prime factors.
Sol:  ∵  156 = 2 × 78
= 2 × 2 × 39
=  2 × 2 × 3 × 13
∴ 156 = 2² × 3 × 13

Q16: If the product of two numbers is 20736 and their LCM is 384, find their HCF.
Sol: ∵  LCM × HCF = Product of two numbers
∴  384 × HCF =  20736
⇒ HCF  = 20736 /384   = 54.

Q17: Find the LCM and HCF of 120 and 144 by the Fundamental Theorem of Arithmetic.
Sol: We have 120 =  2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5
144 =  2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²
∴ LCM  =  2⁴ × 3² × 5 = 720
HCF  =  2³ × 3 = 24

Q18: Find the HCF × LCM for the numbers 100 and 190.
Solution:  HCF × LCM  =  1^st Number × 2^nd Number
=  100 × 190 = 19000.

Q19: Find the (HCF × LCM) for the numbers 105 and 120.
Solution:  HCF × LCM =  1^st number × 2^{nd number}
= 105 × 120 = 12600.

Q20: Write a rational number between √2 and √3.
Sol: ∵ √2  =  1.41 ..... and
√3  =  1.73 .....
∴ one rational number between 1.41 .....and 1.73 ..... is 1.5
i.e., one rational number between √2 and √3 is 1.5.