Use Euclid’s division algorithm to | Class 10 Mathematics Chapter Real Numbers, Real Numbers NCERT Solutions

Current NCERT chapter

Academic session 2026-27, Chapter 1 in the current curriculum.

Welcome to the NCERT Solutions for Class 10 Mathematics - Chapter Real Numbers. This page offers a step-by-step solution to the specific question from Exercise 1, Question 1:

Use Euclid’s division algorithm to find the HCF of :
   (i) 135 and 225      (ii) 196 and 38220      (iii) 867 and 255

. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 1:

Use Euclid’s division algorithm to find the HCF of :
   (i) 135 and 225      (ii) 196 and 38220      (iii) 867 and 255

Answer:

(i) Here, we have to find H.C.F of 135 and 225

First divide divide the larger integer smaller integer

Since, 225 > 135

Therefore, by Euclid’s Division algorithm

225 = 135 × 1 + 90                                       (i)

Here 90 ≠ 0, so proceed the same procedure further

Again by E.D.L,              (E.D.L = Euclid’s division algorithm)

135 = 90 × 1 + 45                                         (ii)

As we know, 45 ≠ 0 therefore, again by E.D.L

90 = 45 × 2 + 0                                             (iii)

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 45.

From (i), (ii) and (iii)

H.C.F (225, 135) = H.C.F (135, 90) = H.C.F (90, 45) = 45.

 

(ii) Here, we have to find H.C.F of 38220 and 196

First divide the larger integer smaller integer

Since, 3822 > 196

Therefore by Euclid’s Division Algorithm

38220 = 196 × 195 + 0

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 196.

Hence, H.C.F (38220, 196) = 196.

 

(iii) Here, we have to find H.C.F of 867 and 255

First divide the larger integer smaller integer

Since, 867 > 255

Therefore, by Euclid’s Division algorithm

867 = 255 × 3 + 102                           (i)

Remainder 102 ≠ 0, so proceed the same procedure further using E.D.L

255 = 102 × 2 + 51                            (ii)

Here, 51 ≠ 0 again using E.D.L = 51 × 2

102 = 51 × 2 + 0                                (iii)

Here, r = 0 so we cannot proceed further. The divisor at this Stage is 51.

From (i), (ii) and (iii)

H.C.F (867, 255) = H.C.F (255, 102) = H.C.F (102, 51) = 51.


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Real Numbers.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Welcome to the NCERT Solutions for Class 10 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 1: Use Euclid’s division algorithm to find the HCF of :    (i) 135 and 225   ....

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