Question

"Question 3 Express the following in the form p/q, where p and q are integers and q ≠ 0.
(i) 0.6
(ii) 0.47
(iii) 0.001

Class 9 - Math - Number Systems Page 14"

Answer 1

Hi Friend,

Here is your answer,

1. 0.6 

Let us assume that x=0.6= 0.6666............------> (1)

Now, we need to Multiply equation (1) with 10

10x = 10×( 0.666...)

10x=6.666...



Subtracting equation(2) from equation(1)
10x-x=6.6666.... - 0.6666...
9x = 6 

x= 9/6 = 2/3 

2. 0.47 = 0.4777777----> (1) 

Now we need to multiply with equation (1) with 10

10x = 10 × (0.47777)...

10x= 4.7777....

Now we need to subtract equation (2) from equation (1)

10x-x = 4.7777...- 0.47777.....

9x =  4.3000 

x= 4.3 /9 

Therefore x = 43/90 

3) 0.001

Let us assume that x= 0.001 = 0.001001......

Now let us multiply equation with 1000

1000 x = 1000 × ( 0.001001...) 


Now we need to subtract (2) equation (1) equation 

1000x - x  = 1.001001...- 0.001001...

999x = 1

x = 1/999


Hope it helps you!

Answer 2

Non terminating repeating decimal has two types:
Pure Recurring decimals and mixed recurring decimals.

Pure recurring decimals:
Ko
A decimal number in which all the digits after decimal point are repeated. E.g 0.675, 0.45

Mixed recurring decimals:

A decimal number in which at least 1 digits after the decimal point is not repeated and others are repeated. E.g , 0.72, 0. 645 e.t.c

Conversion of non terminating repeating decimal number:

i) put the given decimal number is equal to X.

ii) remove the bar if any and write a repeating digits at least twice..

iii) if the repeating decimal has one place repetition multiply by 10, if there is 2 place repetition multiply by 100 & so on.

iv) subtract the number in step ii from the number obtained step iii.

v) divide both sides of the equation by the coefficient of x.

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Solution:

(i) 0.6 = 0.666…
Let x = 0.666……. (1)
Here only one digit is repeating so multiply by 10 on both sides

10 × x = 10× 0.666….
10x = 6.666…. (2)
On subtracting equation 1 from equation 2

10x- x = 6.666…. - 0.666….

9x = 6

x = 9/ 6 = ⅔

Here , 0.6 = ⅔



(ii)
Let x= 0.47 = 0.4777…. (1)

Here, 1 digit is not repeating so multiply eq. 1 by 10
10 × x = 10 × 0.47777

10 x = 4.7777…… (2)

Now only 1 digit is repeating so multiply eq 2 by 10 we get
10 × 10x = 10 × 4.777….

100 x = 47.777….. (3)

On subtracting equation 2 from equation 3 we get

100 x-10 x= 47.777…. - 4.777….

90 x=43

X=43/90

Here, 0.47 = 43/90



(iii) 0.001 = 0.001001001…

Let x = 0.001001001…. (1)

Here, 3 digit is repeating so multiply by 1000

1000 × x = 1000 × . 001001001….

1000x = 1.001001001…. (2)

On subtracting equation 1 from equation 2

1000x -x = 1.001001001…. - 0.001001001...
999x = 1

x = 1/999

Hence ,0.001 = 1/999

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Hope this will help you...