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
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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}
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