To solve how long it will take for all three athletes (John, Bheki, and Pete) to have run the same number of laps, we start by determining their lap rates:
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John: He runs 16 laps in 30 minutes, which gives him a lap rate of: \[ \frac{16 \text{ laps}}{30 \text{ minutes}} = \frac{8}{15} \text{ laps per minute}. \]
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Bheki: He runs 12 laps in 30 minutes, giving him a lap rate of: \[ \frac{12 \text{ laps}}{30 \text{ minutes}} = \frac{2}{5} \text{ laps per minute}. \]
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Pete: He runs 8 laps in 30 minutes, giving him a lap rate of: \[ \frac{8 \text{ laps}}{30 \text{ minutes}} = \frac{4}{15} \text{ laps per minute}. \]
Next, we look for the least common multiple (LCM) of the number of laps they run to find when they will all have completed the same number of laps.
The laps each athlete has run after a certain number of minutes \( t \) is given by:
- John's laps: \( \frac{8}{15}t \)
- Bheki's laps: \( \frac{2}{5}t \) can be rewritten as \( \frac{6}{15}t \) (to have a common denominator).
- Pete's laps: \( \frac{4}{15}t \)
To find when they all have run the same number of laps, we set the expressions equal to each other and find a common value.
We are looking for \( t \) such that:
- \( \frac{8}{15}t = \frac{6}{15}t = \frac{4}{15}t = n \text{ (some integer)} \)
To find an integer \( n \) for each function:
- From \( \frac{8}{15}t = n \), we get \( t = \frac{15n}{8} \).
- From \( \frac{6}{15}t = n \), we get \( t = \frac{15n}{6} = \frac{5n}{2} \).
- From \( \frac{4}{15}t = n \), we get \( t = \frac{15n}{4} \).
Now we seek a common \( t \) among all forms.
We express relations in terms of \( n \):
- \( t = \frac{15n}{8} \) implies \( n \) must be a multiple of 8.
- \( t = \frac{5n}{2} \) implies \( n \) must be a multiple of 2.
- \( t = \frac{15n}{4} \) implies \( n \) must be a multiple of 4.
Finding the least common multiple of coefficients on \( n \): To find a common \( n \):
- The LCM of \( 8, 2, \text{ and } 4 \) is \( 8 \).
Thus, substituting \( n = 8 \): \[ t = \frac{15 \times 8}{8} = 15 \text{ minutes}. \] \[ t = \frac{5 \times 8}{2} = 20 \text{ minutes}. \] \[ t = \frac{15 \times 8}{4} = 30 \text{ minutes}. \]
Contrast results, we realize that they can all run the same distance after each completes a full set sequence relatively.
Thus, checking:
- When \( n=8 \), that resolves to their respective boundary limits which all meets when n = 8.
To find the time:
- Each athlete finishes till completion of a least traversal @ 120 Laps existing.
The time duration for all of them to be equal will be: \[ \text{15 minutes will be the interval sequence,} \text{ confirming 30 indeed meets lifers.} \]
In closing: The answer is \( 60 \text{ minutes}\) will mean all three athletes having run an equal amount which verifies \( 30-60\). Each can run collectively to match again following that natural run sequence.
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