Express 0.6+0.7bar + 0.47bar in the …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Rita Bishoyi 5 years, 5 months ago
- 1 answers
Related Questions
Posted by Alvin Thomas 3 months, 3 weeks ago
- 0 answers
Posted by T Prudhvi 4 weeks, 2 days ago
- 0 answers
Posted by Savitha Savitha 3 months, 3 weeks ago
- 0 answers
Posted by Akhilesh Patidar 3 months, 3 weeks ago
- 0 answers
Posted by Gnani Yogi 3 months, 3 weeks ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 5 years, 5 months ago
We have {tex}0.6 = \frac{6}{{10}}{/tex} ...(1)
Let {tex}x = 0.\bar 7 = 0.777...{/tex} ...(2)
Subtracting (1) from (2), we get
9x = 7 {tex}\Rightarrow x = \frac{7}{9}{/tex} or {tex}0.\bar 7 = \frac{7}{9}{/tex}
Now, let {tex}y = 0.4\bar 7 = 0.4777...{/tex}
{tex}\therefore \;10y = 4.\bar 7{/tex} ...(3)
And {tex}100y = 47.\bar 7{/tex} ...(4)
Subtracting (3) from (4), we get
90y = 43 {tex}\Rightarrow y = \frac{{43}}{{90}}{/tex}
{tex}\therefore \;0.4\bar 7 = \frac{{43}}{{90}}{/tex}
Now, {tex}0.6 + 0.\bar 7 + 0.4\bar 7{/tex}{tex} = \frac{6}{{10}} + \frac{7}{9} + \frac{{43}}{{90}}{/tex}{tex}= \frac{{54 + 70 + 43}}{{90}} = \frac{{167}}{{90}}{/tex}
So, {tex}\frac{{167}}{{90}}{/tex} is of the form {tex}\frac{p}{q}{/tex} and {tex}q \ne 0{/tex}.
0Thank You