Question

Arrange the following in ascending order 17 upon 18, 43 upon 45,59upon 60,31 and 36

Answer 1

Given:

$\frac{17}{18},\frac{43}{45},\frac{59}{60}\;and\;\frac{31}{36}$

To arrange :

In ascending order

Solution:

$18=3\times 6$

$36=3\times 12$

$45=3\times 15$

$60=3\times 20$

LCM(18,36,45,60)=$3\times 6\times 12\times 15\times 20=64800$

$\frac{17\times 3600}{64800}=\frac{61200}{64800}$

$\frac{43\times 1440}{45\times 1440}=\frac{61920}{64800}$

$\frac{59\times 1080}{60\times 1080}=\frac{63720}{64800}$

$\frac{31\times 1800}{36\times 1800}=\frac{55800}{64800}$

Therefore,

$\frac{55800}{64800}<\frac{61200}{64800}<\frac{61920}{64800}<\frac{63720}{64800}$

Ascending order is given by

$\frac{31}{36}<\frac{17}{18}<\frac{43}{45}<\frac{59}{60}$