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.
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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
(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 :-
SEE NCERT Maths Class 10 Chapter 1 Ex-1.1 All Questions Solutions
- 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
- 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.
- NCERT MATHS Class 10 Ex- 1.1 Q No. 3 : An army contingent of 616 members is to march behind an army
- NCERT MATHS Class 10 Ex- 1.1 Q No. 4 : Use Euclid’s division lemma to show that the square of any
- NCERT MATHS Class 10 Ex- 1.1 Q No. 5 : Use Euclid’s division lemma to show that the cube of any pooitive
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