Question

If $0.6+0.\overline{7}+0.4\overline{7}$ is equal to $\frac{p}{90}$, then find the value of $p$.

Answer 1

$0.6+0.\overline{7}+0.4\overline{7}$
Let $x=0.6,y=0.\overline{7}$ and $z=0.4\overline{7}$
Therefore, $x=\frac{6}{10}$
Now,
$y=0.\overline{7}$.....(1)
$10y=10\times 0.\overline{7}$....(2)
Subtracting (1) from (2)
$10y-y=10\times 0.\overline{7}-0.\overline{7}$
$9y=7$
$y=\frac{7}{9}$
Now,
$z=0.4\overline{7}$.....(1)
$10z=10\times 0.4\overline{7}$....(2)
$100y=100\times 0.4\overline{7}$....(3)
Subtracting (2) from (3)
$100z-10z=100\times 0.4\overline{7}-10\times 0.4\overline{7}$
$90z=47.\overline{7}-4.\overline{7}$
$z=\frac{43}{90}$
Therefore, $x+y+z=\frac{6}{10}+\frac{7}{9}+\frac{43}{90}$
$=$ $\frac{54+70+43}{90}$
$=$ $\frac{167}{90}$

$\therefore$ $P=167$