Class 9 Maths Chapter 1 Question Answers - Number System

Q1.Simplify the following expressions:

(i) (4 + √7) (3 + √2)
(ii) (√5 – √3)²
(iii) (√5 -2)( √3 – √5)

Sol. 

(i) (4 + √7) (3 + √2)

= 12 + 4√2 + 3√7 + √14

(ii)  (√5 – √3)²

= (√5)² + (√3)² – 2(√5)( √3)
= 5 + 3 – 2√15
= 8 – 2√15
(iii) (√5 -2)( √3 – √5)
= √15 – √25 – 2√3 + 2√5
= √15 – 5 – 2√3 + 2√5

Q2. Rationalise the denominator: 2 + √5√3

Sol. Multiply both the numerator and denominator with the same number to rationalise the denominator.

√2 + √5√3 = (√2 + √5) × √3√3 × √3

= √6 + √153

Q3. If 'a' and 'b' are rational numbers and:

3 + √83 − √8 = a + b√8

Find the value of 'a' and 'b':

Solution:
Rationalizing the fraction, we get:

3 + √83 − √8 = (3 + √8) × (3 + √8)(3 − √8) × (3 + √8)

= (3 + √8)²3² − (√8)²

= 9 + 8 + 6√89 − 8 = 17 + 6√81

= 17 + 6√8

Now:

3 + √83 − √8 = a + b√8

Equating a and b both sides
⇒ a + b√8 = 17 +6√8
⇒ a = 17and b = 6

Q4:Find five rational numbers between 3/5 and 4/5.

Sol: We have to find five rational numbers between 3/5 and 4/5.

So, let us write the given numbers by multiplying with 6/6, (here 6 = 5 + 1)

Now,

3/5 = (3/5) × (6/6) = 18/30

4/5 = (4/5) × (6/6) = 24/30

Thus, the required five rational numbers will be: 19/30, 20/30, 21/30, 22/30, 23/30

Q5: $\mathrm{Show}\mathrm{ that }0.3333\dots =0.3¯\mathrm{can}\mathrm{ be}\mathrm{ expressed}\mathrm{ in}\mathrm{ the }\mathrm{form}p/q,\mathrm{where}\mathrm{ p}\mathrm{ and}\mathrm{ q }\mathrm{are}\mathrm{ integers}\mathrm{ and}\mathrm{ q}\ne 0.$

Sol:

Let x = 0.3333…. 

Multiply with 10,

10x = 3.3333…

Now, 3.3333… = 3 + x (as we assumed x = 0.3333…)

Thus, 10x = 3 + x

10x – x = 3

9x = 3

x = 1/3

Therefore, 0.3333… = 1/3. Here, 1/3 is in the form of p/q and q ≠ 0.