Video Transcript
In an inline speed skating competition, Ethan completes each lap in 20 seconds. Benjamin completes each lap in one and one-third minutes. And Daniel completes each lap in 30 seconds. Given that they all start skating simultaneously, determine how many minutes will pass before they all cross the start line together again.
So let’s have a look at this question and see if we can work out what we’re being asked to do. Here, we have a skating ring with a start line. We’re told that Ethan completes a lap in 20 seconds. That means he begins at the start line, goes around, and is back at the start line by 20 seconds. If we compare the lap time to Daniel’s lap time since he takes 30 seconds. That means that when Ethan is at the start line, Daniel won’t have made it back to the start line for a complete lap. And Benjamin who completes a lap in one and a third minutes will be even further behind at the 20 seconds when Ethan has completed a lap. So we need to calculate how long after the start of the race it will be until Benjamin, Ethan, and Daniel are all crossing over the start line again.
So let’s note down our skaters and their lap times. Starting with Ethan, who does one lap in 20 seconds. This means that he would complete the laps in 20 seconds, another one at 40 seconds, another lap at 60 seconds, another lap at 80 seconds, and so on. Benjamin does one lap in one and a third minutes. But let’s see if we can change one and a third minutes into a value in seconds. Well, in one minute, there are 60 seconds. And in a third of a minute, that’s the same as a third of 60 seconds. That’s the same as 60 divided by three. And that will give us 20 seconds. So in one and a third minutes, there’s 60 plus 20, which is 80 seconds. So for Benjamin, who completes his lap in one and a third minutes, we can also say that that’s a lap in 80 seconds. So he’ll complete a lap in 80 seconds, two laps in 160 seconds, three laps in 240 seconds, and so on. Daniel completes a lap every 30 seconds. So that’s one lap in 30 seconds, two laps in 60 seconds, three laps in 90 seconds, and four laps in 120 seconds.
So to find the time that it will take before our three skaters cross the start line again together, we’re looking for the smallest value that appears in all three lists. This is also called finding the lowest or least common multiple. We have the word multiple since, for example, for Ethan, he does a lap in 20 seconds. Recap the values 20, 40, 60, 80, which are multiples of 20. We’re looking for common multiples. That is, we’re looking for values that appear across the lists. For example, 80 appears in Ethan’s list, multiples of 20. And it also appears in Benjamin’s list, which are multiples of 80. And we use the word lowest or least because we’re looking for the first common multiple. That’s the smallest one.
So to find the lowest common multiple of 20, 30, and 80, we could do this in two ways. The first way would simply be to continue our lists of multiples and then identify the smallest common multiple. However, depending on our values, this can often take quite a long time. So let’s look at a different method. In this case, we’re going to use prime factors. That means we’re going to split 20, 30, and 80 into its prime factors.
So starting with 20, we’re going to think of a pair of factors that multiply to give 20. This could be five and four. As we write our factors, we think about if their prime numbers are not. So starting with five, we know that five is a prime number. So we can circle it. And the branch of our factor tree can start there. We know that four isn’t a prime number, since it can be written as the pair of factors two and two. So we add those to our factor tree. And since two and two are both prime numbers, then we circle them. And then we stop there. So this means that 20 can be split into the prime factors five, two, and two, which we can write as five times two squared since two squared is the same as two times two.
Moving on to 30 then, we’ll split 30 into a pair of factors. We could choose the factors three and 10 and then identify if either of those are prime numbers. Since three is a prime number, we can circle it. And we check 10. The 10 can be split into the factors two and five. And then we check if two and five are prime. And since both two and five are prime, we can circle them and then stop there. And now we can write that 30 is equal to two times three times five.
Moving on then to 80, we can write 80 as the pair of factors four and 20. And then, starting with our left branch before, we can check if it’s prime. And since we know that four is equal to two times two, we can add these to our branch. And since two is prime, then we can circle both of these branches. Then, we can write our value 20 into the factors two and 10. And we know that two again is prime. So we can circle this too. And 10 can be split into the factors of two and five, which are both prime. So we circle those and then stop. And since we have four twos and a five, then this means we can write 80 as two to the fourth power times five. And now, we have split our values 20, 30, and 80 into their prime factors. Let’s see how this will help answer the original question.
Well, to find the lowest common multiple of 20, 30, and 80, we need to multiply the factors that these have in common. Looking at our prime factors, we can see that we have twos in common. We also have a three. And we also have fives. And we need to make sure that we take the highest exponent of each number. So starting with twos, we can see that we have two squared and two and a two to the fourth power. And our highest one there would be two to the fourth power.
Next, checking our threes, well we only have one three. And it’s three to the power of one. So that stays as it is. And checking our fives, each number of 20, 30, and 80 does have a five as a factor. But it’s only the highest power. There’s no five squared, for instance. So the five stays as it is. So this means we can find the lowest common multiple by calculating two to the fourth power times three times five. And since two to the fourth power is the same as two times two times two times two, that’s the same as 16. So we can evaluate this as 16 times three times five. And so our answer is 240 seconds. But let’s see if we can change this into minutes. To change 240 into minutes then, we divide by 60, seeing as there’re 60 seconds in one minute, which gives us the answer four minutes.
So our final answer, then, is that Benjamin, Ethan, and Daniel will all pass the start line together again after four minutes.
