kuttingchai
A 100-meter sprinting track is marked off in sixths and in eighths. What is the shortest approximate distance, in meters, between any two of the marks that do not overlap?
A 3.87
B 4.16
C 6.25
D 8.06
E 12.50
Convert the fraction to those which have a Common demoninator: \(1/6\) , \(1/8\) => \(1/24\)
List marks in terms of 24ths:
\(\frac{1}{6}\)ths:
\(\frac{4}{24} , \frac{8}{24} , \frac{12}{24} , \frac{16}{24} , \frac{20}{24} , \frac{24}{24}\)
\(\frac{1}{8}\)ths:
\(\frac{3}{24} , \frac{6}{24} , \frac{9}{24} , \frac{12}{24} , \frac{15}{24} , \frac{18}{24} , \frac{21}{24} , \frac{24}{24}\)
Without overlapping, the closest possible would be \(\frac{1}{24}\), we see this occur four times when comparing the two groups: 4-3, 8-9, 16-15, 20-21
The track is 100 meters, so multiply \(\frac{1}{24}\) by 100 --> \(\frac{1}{24}*100=\frac{100}{24}=4.166666\)