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**Very Short Answer Type Questions**

**Ques 1: State 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 the 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 2^{n} · 5^{m}, 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 2^{n} · 5^{m}, where n, m are non-negative integers.**Ques 6: For any two integers, the product of the integers = the product of their HCF and LCM. Is this relation true for three or more integers?****Sol: **No.**Ques 7: If 6 ^{n} is a number such that n is a natural number. Check whether there is any value of n ∈ N for which 6^{n} is divisible by 7.**

∴ 6

i.e., the prime factorisation of 6

∴ 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 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

⇒ **Ques 11. Write 98 as the product of its prime factors.****Sol: ∵ **

∴ 98 = 2 × 7 × 7 ⇒ 98 = 2 × 7^{2}**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 2^{n} · 5^{m}.

∴ 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 2^{n} · 5^{m}

∴ 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

= 2^{0 }× 5^{5} |∵ 2^{0} = 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 = 2^{2} × 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 18: Find the LCM and HCF of 120 and 144 by Fundamental Theorem of Arithmetic.****Sol: **We have 120 = 2 × 2 × 2 × 3 × 5 = 2^{3 }× 3 × 5

144 = 2 × 2 × 2 × 2 × 3 × 3 = 2^{4} × 3^{2}

∴ LCM = 2^{4} × 3^{2} × 5 = 720

HCF = 2^{3} × 3 = 24

**Ques 19: Find the HCF × LCM for the numbers 100 and 190.****Sol:** HCF × LCM = 1^{st} Number × 2^{nd }Number

= 100 × 190 = 19000.**Ques 20: Find the (HCF × LCM) for the numbers 105 and 120.****Sol: **HCF × LCM = 1^{st} number × 2^{nd number}

= 105 × 120 = 12600.**Ques 21: 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 22: 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 =

2^{2} × 5^{3}

which are in the form 2^{n} · 5^{m}

∴ It has a terminating decimal expansion.**Ques 23: The HCF and LCM of the two numbers are 9 and 360 respectively. If one number is 45, write the other number. ****Sol: **∵ Product of two numbers

= LCM × HCF

∴ 45 × another number

= 9 × 360

⇒ another number **Ques 24: 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 = 2^{m} × 5^{n}, where m and n are whole numbers.**Ques 25: 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.**Ques 26: Complete the missing entries in the following factor-tree:****Sol: **∵ 3 × 7 = 21

and 21 × 2 = 42

∴ We have:**Ques 27: Write the HCF of the smallest composite number and the smallest prime number.Sol: **∵ The smallest composite number = 4

= 2 × 2 = 2

The smallest prime number = 2

= 2

∴ HCF = 2

**Ques 28: 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 29: 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.

130 = 52 × 2 + 26

52 = 26 × 2 + 0

∵ The remainder = 0

∴ HCF (130, 52) is 26.

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