Question

Four plumbers estimated the length of the radius of a cylindrical pipe. The estimates made by the plumbers are listed. Which list shows these lengths in order from greatest to least?

Answer 1

Answer:

$\frac{\sqrt{3}}{11}>\frac{\pi }{24}>\frac{3}{25}>\frac{9}{100}$

Step-by-step explanation:

To get the order of the lengths in the order from greatest to least we will make the denominator of the given fractions common and then compare the numerators.

$\frac{3}{25}, \frac{\sqrt{3}}{11}, \frac{9}{100},\frac{\pi }{24}$

LCM of the denominators,

2 |  25   11    100   24

2 |  25    11     50    12

5 |  25    11     25     6  

5 |    5     11     5      6  

      1      11      1       6

LCM = 2×2×5×5×6×11 = 6600

$\frac{3}{25}=\frac{3}{25}\times \frac{264}{264}$

   $=\frac{792}{6600}$

$\frac{\sqrt{3}}{11}=\frac{\sqrt{3} }{11}\times \frac{600}{600}$

    $=\frac{600\sqrt{3}}{6600}=\frac{1039}{6600}$

$\frac{9}{100}=\frac{9}{100}\times \frac{66}{66}$

     $=\frac{594}{6600}$

$\frac{\pi }{24}=\frac{\pi }{24}\times \frac{275}{275}$

    $=\frac{275\pi }{6600}$

    $=\frac{864}{6600}$

Now we can arrange these fractions in the order of greatest to least.

$\frac{1039}{6600}>\frac{864}{6600}>\frac{792}{6600}>\frac{594}{6600}$

$\frac{\sqrt{3}}{11}>\frac{\pi }{24}>\frac{3}{25}>\frac{9}{100}$