Prove that 0.5757... (recurring) = 19/33. Hence, write 0.3575757... (recurring) as a fraction in its lowest terms.

Two parts to the question. Let's focus on part one:Let x = 0.575757... (1)This means that 100x = 57.575757... (2)If you subtract (1) from (2), we get: 99x = 57Divide both sides by 99: x = 57/99Simplify: x = 19/33
In the second part of the question, we see the key word HENCE which means that we most likely need to use our previous result.Observe that 0.3575757... = 0.3 + 0.0575757... = 0.3 + (0.575757...)/10If we substitute our previous result in and change 0.3 to a fraction:= 0.3 + (19/33)/10= 3/10 + (19/33)/10Evaluate by making all fractions have same common denominator:= 3/10 + 19/330= 99/330 + 19/330= 118/330Then express in lowest terms:= 59/165

Answered by Oliver V. Maths tutor

4939 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

In a class there are 57 students. Of these 32 study Spanish, 40 study German and 12 students study neither. How many students study Spanish but not German?


How can I find x and y?


Claire drove from Manchester to London, it took her 4 hours at an average speed of 85 km/h. Matt drove from Manchester to London, it took him 5 hours. Assuming he took the same route as Claire and took no breaks, work out his average speed in km/h.


How do you factorise and know if it is a difference of two squares ?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences