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

5592 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

(b) In 2013, the price for each unit of electricity was 13.5 cents. Over the next 3 years, this price increased exponentially at a rate of 8% per year. Calculate the price for each unit of electricity after 3 years


Solve x^2+10x-3=0 by completing the square, simplify your answer.


Make "a" the subject of the following equation: 2b = (3a+4)/(c-a)


How do I use Pythagoras' Theorem to calculate the hypotenuse of a triangle?


We're here to help

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