NCERT MATHS Class 10 Ex- 1.1 Q No. 1 : Use Euclid’s division algorithm to find the HCF of 135 and 225 , 196 and 38220,867 and 255

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Hy Friends Welcome On NCERT MATHS SOLUTIONS !! Today we are going to solve the Question : Use Euclid’s division algorithm to find the HCF of :  135 and 225 , 196 and 38220, 867 and 255 of Exercise 1.1 Question No 1 of Class 10th, which will prove to be very helpful for you.

If you are a student of CBSE, today we are going to give you the CBSE / NCERT Chapter : Real Numbers Exercise1.1 Question No-1 Solutions . Hope you like this post about Class 10th Maths Solution.

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Real Numbers CHAPTER : 1

Symbol of real numbers (ℝ) In mathematics, the real number is the value presented to any amount corresponding to the simple line. Actual numbers include all rational numbers such as -5 and fractional numbers such as 4/3 and all irrational numbers such as √2 (1.41421356 …, square root of 2, an unregulated algebraic number). By incorporating the ample numbers in the actual numbers, they can be presented from the eternal points that can be attributed on a line in the form of a real number line.

Class 10 Exercise 1.1 Question No. 1 : Use Euclid’s division algorithm to find the HCF of:

  • 135 and 225

  • 196 and 38220

  • 867 and 255

Class 10 Exercise 1.1 Question No. 1 Solutions

(i) Use Euclid’s division algorithm to find the HCF of : 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Since remainder 90 ≠ 0,

we apply the division lemma to 135 and 90 to obtain

135 = 90 × 1 +45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

90 = 2 × 45 + 0

Since the remainder is Zero, the process stop.

Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii) Use Euclid’s division algorithm to find the HCF of : 196 and 38220

Since 38220 > 196, we apply the divisio lemma to 38220 and 196 to obtain

38220 = 196 × 195 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 196,

Therefor, HCF of 196 and 38220 is 196.

(iii) Use Euclid’s division algorithm to find the HCF of : 867 and 255

Since  867 > 255, we apply the division lemma to 867 and 255 to obtain

867 =255 × 3 +102

Since remainder 102 ≠ 0, we apply the divison lemma to 255 and 102 to obtain

255  = 102 × 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

102 = 51 ×2+ 0

Since the remainder is zero, the process stop.

Since the divisor at this stage is 51, therefore,

HCF of 867 and 255 is 51.

CLICK ON RED BOX FOR CBSE Maths Class 10 Exercise 1 Real Numbers Next Question Answer :-

NCERT MATHS Class 10 Ex- 1.1 Q No. 2 : Show that any positive add integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

SEE NCERT Maths Class 10 Chapter 1 Ex-1.1 All Questions Solutions 

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