**Notes**
**Ncert Solutions**
**Assignments**
**Videos**
**Revision sheet**
In this page we have *NCERT Solutions for Class 9th Maths: Chapter 1 NumberSystems * for
Exercise 1.5 and 1.6 . Hope you like them and do not forget to like , social share
and comment at the end of the page.Free download also available

**Exercise 1.5**

**Question 1 **
Classify the following numbers as rational or irrational:

(i) 2 - √5

(ii) (3 + √23) - √23

(iii) 2√7/7√7

(iv) 1/√2

(v) 2π

**Solution**
**Question 2 **
Simplify each of the following expressions:

(i) (3 + √3) (2 + √2)

(ii) (3 + √3) (3 - √3)

(iii) (√5 + √2)

^{2}
(iv) (√5 - √2) (√5 + √2)

**Solution**
(i) (3 + √3) (2 + √2)

= 3 × 2 + 2 + √3 + 3√2+ √3 ×√2

= 6 + 2√3 +3√2 + √6

(ii) (3 + √3) (3 - √3)

Now as (

*a* +

*b*) (

*a* -

*b*) =

*a*^{2} -

*b*^{2}
= 3

^{2} - (√3)

^{2}
= 9 - 3= 6

(iii) (√5 + √2)

^{2 }
Now as (

*a* +

*b*)

^{2} =

*a*^{2} +

*b*^{2} + 2

*ab*
= (√5)

^{2} + (√2)

^{2} + 2 ×√5 × √2

= 5 + 2 + 2 × √5× 2

= 7 +2√10

(iv) (√5 - √2) (√5 + √2)

Now as (

*a* +

*b*) (

*a* -

*b*) =

*a*^{2} -

*b*^{2}
= (√5)

^{2} - (√2)

^{2}
= 5 – 2= 3

**Question 3**
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

**Solution**
There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857...

**Question 4. **
Represent √9.3 on the number line.

**Question 5**
Rationalize the denominators of the following:

(i) 1/√7

(ii) 1/ (√7-√6)

(iii) 1/ (√5+√2)

(iv) 1/ (√7-2)

**Solution**
**Exercise 1.6**

**Question 1. **
Find:

(i) 64

^{1/2}
(ii) 32

^{1/5}
(iii) 125

^{1/3}
**Solution**
i) 64

^{1/2} =(2

^{6})

^{1/2 } =2

^{6 X1/2 } = 2

^{3} =8

ii) 32

^{1/5} =(2

^{5})

^{1/5} = 2

iii) 125

^{1/3} = (5

^{3})

^{1/3} =5

**Question 2**
Find:

(i) 9

^{3/2}
(ii) 32

^{2/5}
(iii) 16

^{3/4}
(iv) 125

^{-1/3}
**Solution**
i) 9

^{3/2} = (3

^{2})

^{3/2} = 27

ii) 32

^{2/5} = (2

^{5})

^{2/5} =4

iii) 16

^{3/4} = (2

^{4})

^{3/4} =8

iv) 125

^{-1/3} = 1/125

^{1/3} = 1/(5

^{3})

^{1/3} =1/5

**Question 3. **
Simplify:

(i) 2

^{2/3}.2

^{1/5}
(ii) (1/3

^{3})

^{7}
(iii) 11

^{1/2}**/**11

^{1/4}
(iv) 7

^{1/2}.8

^{1/2}
**Solution**
i) 2

^{2/3}.2

^{1/5} =2

^{ 2/3+ 1/5} = 2

^{10+3/15} =2

^{13/15}
ii) (1/3

^{3})

^{7}
=1/3

^{3X7} = 1/3

^{21} =3

^{ -21}
iii) 11

^{1/2}**/**11

^{1/4}
= 11

^{1/2 -1/4 } = 11

^{1/4}
iv) 7

^{1/2}.8

^{1/2}
=(7X8)

^{1/2} = 56

^{1/2}
Download Number System NCERT Solutions 1.5 and 1.6 as pdf

Class 9 Maths
Class 9 Science