I. Very Short Answer Type Questions
Ques 1: State the Euclid’s division algorithm.
Sol: Given two positive integers ‘a’ and ‘b’, there exists a unique pair of integers such that
a = (b × q) + r, 0 ≤ r < b
Fig: Euclid’s division algorithm
Ques 2: State the fundamental theorem of arithmetic.
Sol: Every composite number can be expressed as product of primes and this decomposition is unique apart from the order in which prime factors occur.
Ques 3: 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.
Ques 4: 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 · 5m, where m and n are non-negative integers.
Ques 5: 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 · 5m, where n, m are non-negative integers.
Ques 6: For any two integers, product of the integers = product of their HCF and LCM. Is this relation true for three or more integers?
Sol: No.
Ques 7: 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.
Ques 8: 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 × 100uy
⇒
Ques 9: Given that LCM (26, 91) = 182. Find their HCF.
Sol: ∵ HCF × LCM = Product of the two numbers
∴ HCF × 182 = 26 × 91
⇒
Ques 10: The LCM and HCF of 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
⇒
Ques 11. Write 98 as product of its prime factors.
Sol: ∵
∴ 98 = 2 × 7 × 7 ⇒ 98 = 2 × 72
Ques 12: Write the missing numbers in the following factorisation:
Sol: ∵ 3 × 7 = 21, 21 × 2 = 42
and 42 × 2 = 84
∴ We have
Ques 13: 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 · 5m.
∴ will have a non-terminating repeating decimal expansion.
Ques 14: Without actually performing the long-division, state will have a terminating or non-terminating repeating decimal expansion.
Sol:
∴ Prime factorisation of q is of the form 2n · 5m
∴ will have a terminating decimal expansion.
Ques 15: 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.
Ques 16: Express 156 as a product of its prime factors.
Sol: ∵ 156 = 2 × 78
= 2 × 2 × 39
= 2 × 2 × 3 × 13
∴ 156 = 22 × 3 × 13
Ques 17: If 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.
Ques 19: Find the HCF × LCM for the numbers 100 and 190.
Sol: HCF × LCM = 1st Number × 2nd Number
= 100 × 190 = 19000.
Ques 20: Find the (HCF × LCM) for the numbers 105 and 120.
Sol: HCF × LCM = 1st number × 2nd number
= 105 × 120 = 12600.
Ques 20: The decimal expansion of the rational number , will terminate after how many places of decimal? (CBSE 2009)
Sol:
Thus, will terminate after 4 places of decimal.
Ques 21: Write whether the rational number will 51/1500 have a terminating decimal expansion or a non-terminating repeating decimal expansion.
Sol: We have, 51/1500 = 17/500
Prime factors of 500
= 2 × 2 × 5 × 5 × 5 =
22 × 53
which are in the form 2n · 5m
∴ It has a terminating decimal expansion.
Ques 22: The HCF and LCM of two numbers are 9 and 360 respectively. If one number is 45, write the other number.
Sol: ∵ Product of two numbers
= LCM × HCF
∴ 45 × other number
= 9 × 360
⇒ other number
Ques 23: If p/q is a rational number (q ≠0), what is condition of ‘q’ so that the decimal representation of p/q is terminating?
Sol: The required condition is q = 2m × 5n, where m and n are whole numbers.
Ques 24: 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 and is 1.5.
Ques 25: Complete the missing entries in the following factor-tree:
Sol: ∵ 3 × 7 = 21
and 21 × 2 = 42
∴ We have:
Ques 26: Write the HCF of the smallest composite number and the smallest prime number.
Sol: ∵ The smallest composite number = 4
= 2 × 2 = 22
The smallest prime number = 2
= 21
∴ HCF = 21 = 2
Ques 27: Find the HCF of 960 and 432.
Sol: Applying Euclid’s Algorithm of division for 960 and 432, we have:
960 = 432 × 2 + 96 |96 ≠0
432 = 96 × 4 + 48 |48 ≠0
96 = 48 × 2 + 0
Since, the remainder = 0
∴ HCF of 960 and 432 is 48
Ques 28: Find the HCF of 72 and 120.
Sol:Using Euclid’s algorithm of division for 72 and 120 we have,
120 = 72 × 1 + 48
72 = 48 × 1 + 24
48 = 24 × 2 + 0
∵ The remainder = 0
∴HCF (72, 120) is 24.
Ques 29: What is the HCF of 52 and 130?
Sol: Using the Euclid’s algorithm of division for 52 and 130, we have:
130 = 52 × 2 + 26
52 = 26 × 2 + 0
∵ The remainder = 0
∴ HCF (130, 52) is 26.
Ques 30: Find the LCM and HCF of 120 and 144 by 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