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Marla starts running around a circular track at the same tim
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24 Jun 2020, 09:27
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Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 32 laps around the track per hour and Nick completes 12 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick?
Marla starts running around a circular track at the same tim
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24 Jun 2020, 09:31
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Carcass wrote:
Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 32 laps around the track per hour and Nick completes 12 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick?
(A) 5 (B) 8 (C) 12 (D) 15 (E) 20
Let t = the time (in HOURS) that it takes Marla to complete 4 more laps than Nick. So, after t hours, we can write: (Marla's lap count) = (Nick's lap count) + 4
Now that we have a "word equation" we need only fill in the missing information
Marla completes 32 laps per hour We can think of 1 lap as being a unit of distance. So, 32 laps per hour is Marla's speed.
Distance = (speed)(time) So, after t hours, Marla's lap count = 32t
Nick completes 12 laps around the track per hour So, after t hours, Nick's lap count = 12t
We can now plug the above values into the word equation. We get: 32t = 12t + 4 Subtract 12t from both sides to get: 20t = 4 Solve: t = 4/20 = 1/5 HOURS
Marla starts running around a circular track at the same tim
[#permalink]
24 Jun 2020, 09:34
1
Carcass wrote:
Marla starts running around a circular track at the same time Nick starts walking around the same circular track. Marla completes 32 laps around the track per hour and Nick completes 12 laps around the track per hour. How many minutes after Marla and Nick begin moving will Marla have completed 4 more laps around the track than Nick?
(A) 5 (B) 8 (C) 12 (D) 15 (E) 20
Let t = the time (in HOURS) that it takes Marla to complete 4 more laps than Nick. So, after t hours, we can write: (Marla's lap count) = (Nick's lap count) + 4
Now that we have a "word equation" we need only fill in the missing information
Marla completes 32 laps per hour We can think of 1 lap as being a unit of distance. So, 32 laps per hour is Marla's speed.
Distance = (speed)(time) So, after t hours, Marla's lap count = 32t
Nick completes 12 laps around the track per hour So, after t hours, Nick's lap count = 12t
We can now plug the above values into the word equation. We get: 32t = 12t + 4 Subtract 12t from both sides to get: 20t = 4 Solve: t = 4/20 = 1/5 HOURS