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  • Express 0.6 + 0.bar 7 +0.bar47 in the form p by q where p and q are integers and q not equal to 0.

Express 0.6 + 0\cdot \bar{7}+0\cdot \overline{47} in the form \frac{p}{q} where p and q are integers and q\neq 0.

Answers (1)

Answer. \frac{167}{90}
Solution. Let x = 0.6
Multiply by 10 on LHS and RHS
10x = 6
x= \frac{6}{10}

x= \frac{3}{5}
So, the \frac{p}{q} from of 0.6 = \frac{3}{5}
Let y = 0\cdot \bar{7}
Multiply by 10 on LHS and RHS

10y = 7.7777 …….
10y – y = 7\cdot \bar{7}-0\cdot \bar{7}

          = 7.77777 ….. – 0.77777 ……
9y = 7

y= \frac{7}{9}
So the \frac{p}{q} from of 0.7777 = \frac{7}{9}
Let z = 0.47777…
Multiply by 10 on both side
10z = 4.7777 ….
10z – z = 4\cdot \bar{7}-0\cdot4\bar{7}

9z = 4.3

z\approx \frac{4\cdot 3}{9}
z= \frac{43}{90}
So the \frac{p}{q}from of 0.4777 …. = \frac{43}{90}
Therefore, \frac{p}{q} form of 0.6 + 0\cdot \bar{7}+0\cdot4\bar{7} is,

x+y+z= \frac{3}{5}+\frac{7}{9}+\frac{43}{90}
= \frac{\left ( 54+70+43 \right )}{90}
= \frac{167}{90}
Hence the answer is \frac{167}{90}

 

Posted by

infoexpert27

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