Question

2. EXP 0.6+0.7+0.47in the from pq where p and q Integers and q#0
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Answer 1

Answer:

Step-by-step explanation:

Let x = 0.7Bar

⇒ x = 0.777..........  .........(1)

Multiplying (1) by 10 

⇒ 10x = 7.7...... = 7.777 ...........(2)

Subtracting (1) from (2)

⇒ 10x - x = 9x

 

⇒ 7.777 - 0.777 = 7

    

⇒ 9x = 7 

⇒ x = 7/9

Let y = 0.47Bar

Multiplying (1) by 100 

⇒ 100y = 47.4747  ............(2)

Subtracting (1) from (2)

⇒ 100y - y = 99y

⇒ 47.4747 - 0.4747 = 47

⇒ 99y = 47

⇒ y = 47/99

Now,

0.7Bar = 7/9

and 0.47Bar = 47/99

0.6 = 6/10

6/10 + 7/9 + 47/99

Taking L.C.M. of the denominator and then solving it, we get.

⇒ (594 + 770 + 470)/990

⇒ 1834/990

⇒ 917/495

Hence, 0.6 + 0.7Bar + 0.47Bar = 917/495

⇒ y = 0.4747 ..............  ........(0)