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**I. Very Short Answer Type Questions****Ques 1: State the Euclidâ€™s division algorithm.****Sol: **Given two positive integers â€˜aâ€™ and â€˜bâ€™, there exists a unique pair of integers such that

a = (b Ã— q) + r, 0 â‰¤ r < b

**Fig: Euclidâ€™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 â‰ 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} Â· 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} Â· 5^{m}, 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 6 ^{n} is a number such that n is a natural number. Check whether there is any value of n âˆˆ N for which 6^{n} is divisible by 7.**

âˆ´ 6

i.e., the prime factorisation of 6

âˆ´ 150 Ã— 100 = LCM Ã— HCF

â‡’ LCM Ã— 50 = 150 Ã— 100uy

â‡’

**Ques 9: Given that LCM (26, 91) = 182. Find their HCF.****Sol:** âˆµ HCF Ã— LCM = Product of the two numbers

âˆ´ HCF Ã— 182 = 26 Ã— 91

â‡’ **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 â€˜xâ€™.

âˆµ LCM Ã— HCF = Product of the two numbers

âˆ´ 60 Ã— x = 240 Ã— 12

â‡’ **Ques 11. Write 98 as product of its prime factors.****Sol: âˆµ **

âˆ´ 98 = 2 Ã— 7 Ã— 7 â‡’ 98 = 2 Ã— 7^{2}**Ques 12: Write the missing numbers in the following factorisation:**

** ****Sol: **âˆµ 3 Ã— 7 = 21, 21 Ã— 2 = 42

and 42 Ã— 2 = 84

âˆ´ We have

**Ques 13: Without actually performing the long division, state whether will have a terminating or non-terminating repeating decimal expansion. ****Sol: **Let =

âˆµ Prime factors of q are not of the for 2^{n} Â· 5^{m}.

âˆ´ 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:**

âˆ´ Prime factorisation of q is of the form 2^{n} Â· 5^{m}

âˆ´ 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:** âˆµ The denominator of 17/3125 is given by

3125 = 5 Ã— 5 Ã— 5 Ã— 5 Ã— 5

= 1 Ã— 55

= 2^{0 }Ã— 5^{5} |âˆµ 2^{0} = 1

âˆ´

i.e., 17/3125 is a terminating decimal.**Ques 16: Express 156 as a product of its prime factors.****Sol:** âˆµ 156 = 2 Ã— 78

= 2 Ã— 2 Ã— 39

= 2 Ã— 2 Ã— 3 Ã— 13

âˆ´ 156 = 2^{2} Ã— 3 Ã— 13

**Ques 17: If product of two numbers is 20736 and their LCM is 384, find their HCF.****Sol:** âˆµ LCM Ã— HCF = Product of two numbers

âˆ´ 384 Ã— HCF = 20736

â‡’ HCF = 20736 /384 = 54.

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

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

= 105 Ã— 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 Ã— 2 Ã— 5 Ã— 5 Ã— 5 =

2^{2} Ã— 5^{3}

which are in the form 2^{n} Â· 5^{m}

âˆ´ 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: **âˆµ Product of two numbers

= LCM Ã— HCF

âˆ´ 45 Ã— other number

= 9 Ã— 360

â‡’ other number **Ques 23: If p/q is a rational number (q â‰ 0), what is condition of â€˜qâ€™ so that the decimal representation of p/q is terminating? ****Sol: **The required condition is q = 2^{m} Ã— 5^{n}, where m and n are whole numbers.**Ques 24: Write a rational number between âˆš2 and âˆš3 .****Sol:** âˆµ **âˆš2 ** = 1.41 ..... and**âˆš3 ** = 1.73 .....

âˆ´ 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: **âˆµ 3 Ã— 7 = 21

and 21 Ã— 2 = 42

âˆ´ We have:**Ques 26: Write the HCF of the smallest composite number and the smallest prime number.Sol: **âˆµ The smallest composite number = 4

= 2 Ã— 2 = 2

The smallest prime number = 2

= 2

âˆ´ HCF = 2

**Ques 27: Find the HCF of 960 and 432.****Sol:** Applying Euclidâ€™s Algorithm of division for 960 and 432, we have:

960 = 432 Ã— 2 + 96 |96 â‰ 0

432 = 96 Ã— 4 + 48 |48 â‰ 0

96 = 48 Ã— 2 + 0

Since, the remainder = 0

âˆ´ HCF of 960 and 432 is 48**Ques 28: Find the HCF of 72 and 120.Sol:**Using Euclidâ€™s algorithm of division for 72 and 120 we have,

120 = 72 Ã— 1 + 48

72 = 48 Ã— 1 + 24

48 = 24 Ã— 2 + 0

âˆµ The remainder = 0

âˆ´HCF (72, 120) is 24.

130 = 52 Ã— 2 + 26

52 = 26 Ã— 2 + 0

âˆµ The remainder = 0

âˆ´ 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 Ã— 2 Ã— 2 Ã— 3 Ã— 5 = 2^{3 }Ã— 3 Ã— 5

144 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 = 2^{4} Ã— 3^{2}

âˆ´ LCM = 2^{4} Ã— 3^{2} Ã— 5 = 720

HCF = 2^{3} Ã— 3 = 24

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