NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

Jasmine Grover logo

Jasmine Grover

Content Strategy Manager

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3 is covered in this article with a detailed explanation. Chapter 1 Real Numbers Exercise 1.3 deals with proving that root p is irrational. The 3 questions of the exercise cover the concept of irrational numbers

Download PDF NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.3

Check out the solutions of Class 10 Maths NCERT solutions chapter 1 Real Numbers Exercise 1.3

Read More: NCERT Solutions For Class 10 Maths Real Numbers

Check out other exercise solutions of Class 10 Maths Chapter 1 Real Numbers

Also Read:

Also Read:

CBSE X Related Questions

1.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

      2.
      A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

          3.

          Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

              4.
              Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

                  5.
                  Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) \(x + y = 5\),\( 2x + 2y = 10\) (ii)\( x – y = 8 , 3x – 3y = 16\) (iii) \(2x + y – 6 = 0\) , \(4x – 2y – 4 = 0\) (iv) \(2x – 2y – 2 = 0,\) \( 4x – 4y – 5 = 0\)

                      6.

                      Solve the following pair of linear equations by the substitution method. 
                      (i) x + y = 14 
                          x – y = 4   

                      (ii) s – t = 3 
                          \(\frac{s}{3} + \frac{t}{2}\) =6 

                      (iii) 3x – y = 3 
                            9x – 3y = 9

                      (iv) 0.2x + 0.3y = 1.3 
                           0.4x + 0.5y = 2.3 

                      (v)\(\sqrt2x\) + \(\sqrt3y\)=0
                          \(\sqrt3x\) - \(\sqrt8y\) = 0

                      (vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
                          \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

                          Comments



                          No Comments To Show