If a person has walked 1 and 2/5 miles and they have gone 5/8 of the way around track how many miles is the track? I know the answer and I know how I figured it out, but I cannot explain it to my son. Can anyone help explain this as it might have been explained to a 6th grader?
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Distance / Fraction Problem
- Thread starter walimaas
- Start date
walimaas said:If a person has walked 1 and 2/5 miles and they have gone 5/8 of the way around track how many miles is the track? I know the answer and I know how I figured it out, but I cannot explain it to my son. Can anyone help explain this as it might have been explained to a 6th grader?
5/8 of the way around the track is 1 2/5 or 7/5 miles.
1/5 of 5/8 is 1/8 and 1/5 of 7/5 is 7/25.
1/8 of the way around the track is 7/25 miles.
8 times 1/8 is 1 and 8 times 7/25 is 56/25.
1 (all) of the way around the track is 56/25 miles.
56/25 miles is 2 6/25 miles.
Using algebra...
Let x represent the distance around the track.
5/8 of x is 1 2/5 miles.
5/8 times x = 7/5
\(\displaystyle \frac{5}{8}x=\frac{7}{5}\)
\(\displaystyle \frac{40}{1}\cdot \frac{5}{8}x = \frac{40}{1}\cdot \frac{7}{5}\)
\(\displaystyle 25x = 56\)
\(\displaystyle x=\frac{56}{25}=2\frac{6}{25}\) miles.
If I tried to explain that to a 6th grader, I'd make up a simpler similar problem; like:
"If a person has walked 2 and 1/2 miles and has gone 1/3 of the way around track...";
then draw a circle to illustrate...once the 6th grader understands, then do the tougher problem.
"If a person has walked 2 and 1/2 miles and has gone 1/3 of the way around track...";
then draw a circle to illustrate...once the 6th grader understands, then do the tougher problem.