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Two friends, Gabrielle and Hugo, practiced running laps around an oval track. They started together at the starting line. Gabrielle completes each lap in 40 seconds, while Hugo completes each lap in 54 seconds. What is the number of laps Gabrielle has completed the next time they both arrive together at the starting line?

Explanation	Calculations	Help
$We first factor each runner's lap time into prime numbers.$	$40 = 2^{3} \times 5 54 = 2 \times 3^{3}$
$We now find the least common multiple of the two lap times by taking each prime with its highest needed exponent.$	$LCM = 2^{3} \times 3^{3} \times 5 = 1080$	Theory & Tactics Method Card DIV6-A Click to view full details
$Gabrielle's number of laps equals the total time until they meet divided by her lap time.$	$\frac{1080}{40} = \frac{2 ^{3} \times 3 ^{3} \times 5}{2 ^{3} \times 5}$	Theory & Tactics Method Card FDPR2-A Click to view full details
$We cancel out the common factors to obtain Gabrielle's lap count.$	$\frac{2 ^{3} \times 3 ^{3} \times 5}{2 ^{3} \times 5} = \frac{2 ^{3} \times 3 ^{3} \times 5}{2 ^{3} \times 5} = 3^{3} = 27$	Theory & Tactics Method Card FDPR2-A Click to view full details

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C

Two friends, Gabrielle and Hugo, practiced running laps around an oval track. They started together at the starting line. Gabrielle completes each lap in 40 seconds, while Hugo completes each lap in 54 seconds. What is the number of laps Gabrielle has completed the next time they both arrive together at the starting line?

Explanation	Calculations	Help
$We first factor each runner's lap time into prime numbers.$	$40 = 2^{3} \times 5 54 = 2 \times 3^{3}$
$We now find the least common multiple of the two lap times by taking each prime with its highest needed exponent.$	$LCM = 2^{3} \times 3^{3} \times 5 = 1080$	Theory & Tactics Method Card DIV6-A Click to view full details
$Gabrielle's number of laps equals the total time until they meet divided by her lap time.$	$\frac{1080}{40} = \frac{2 ^{3} \times 3 ^{3} \times 5}{2 ^{3} \times 5}$	Theory & Tactics Method Card FDPR2-A Click to view full details
$We cancel out the common factors to obtain Gabrielle's lap count.$	$\frac{2 ^{3} \times 3 ^{3} \times 5}{2 ^{3} \times 5} = \frac{2 ^{3} \times 3 ^{3} \times 5}{2 ^{3} \times 5} = 3^{3} = 27$	Theory & Tactics Method Card FDPR2-A Click to view full details

Scroll horizontally to view all columns

C