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

OV
Answered by Oliver V. Maths tutor

6701 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Sue has 2 cats. Each cat eats 1 4 of a tin of cat food each day. Sue buys 8 tins of cat food. Has Sue bought enough cat food to feed her 2 cats for 14 days? You must show how you get your answer.


The straight line L1 passes through the points with coordinates (4, 6) and (12, 2) The straight line L2 passes through the origin and has gradient -3. The lines L1 and L2 intersect at point P. Find the coordinates of P.


y = p x q^(x - 1), When x = 1, y = 10, and when x = 6, y = 0.3125. Find the value of 'y' when x = 3


Solve 67x – 5 = 12x + 13


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning