Video Transcript
In this video, we’re gonna look at a puzzle based on a visit to your granny’s house.
One day you decide to go and visit granny. There’s a straight flat path from your home front door to granny’s front door. And you walk to a house without stopping along the way. After you arrive, you have a lovely time. You have a great chat. She gives you some of her tasty chocolate cake. You help her to edit some photos on her computer. And then all too soon, it’s time to go home. As you’re setting off home, she says, “How many times faster than your average speed on the way here will your average speed on the way home need to be in order for your overall average traveling speed to be twice your average speed on the way here?”
So, let’s say we walk to granny’s house really slowly. If you run home really fast, then your average speed for the whole journey there and back will be greater than just the average speed from home to granny’s. It took less time than it would have done if you just walked there and back. So, your average speed for both journeys together is faster than the average speed for just the first journey to granny’s when you walked really slowly.
But exactly, how many times faster do you have to travel on the way home in order to make the overall average twice as fast as the average speed on the way there? Why not pause the video now and have a think about this problem and try to come up with an answer yourself? Then, we can go through some thoughts and ideas about how to approach it together. Okay, then, let’s think about this.
First of all, I should say that this is a math puzzle. So, we do need to recognize that we’re in math world and granny’s do say that kind of thing, and it doesn’t seem at all weird. Next, it’s probably worth adding a few clarifying details about the problem. The path between your house and granny’s is straight, horizontal, remains at a fixed length throughout the problem, and has a finite length greater than zero. The length isn’t zero. Granny doesn’t live at your house. And the path isn’t infinite in length; it’s not a trick question.
Now, it’s often easier to start by thinking about a specific case with actual numbers to help you collect your thoughts. In fact, it’s great to start with the easiest numbers you can to help you work things out. If we walk to granny’s at an average speed of one kilometer an hour, then we want the overall average speed there and back to be two kilometers an hour. It’s tempting to think that because we’re traveling the same distance along the same path each way, then we’d need to travel at three kilometers an hour on the way back. The average of one and three is one plus three over two, which is four over two, which is two.
But wait! The average speed is the total distance traveled divided by the total time taken. Let’s say that the houses are six kilometers apart. That means that the total distance there and back is two times six. That’s 12 kilometers. The journey there at one kilometer an hour takes six hours. Every one hour, you travel one kilometer, so it takes six hours to travel six kilometers. The journey back at three kilometers an hour takes two hours. Every one hour, you travel three kilometers. So, it takes two lots of one hour to travel two lots of three kilometers. That’s six kilometers.
In total, you’ve traveled 12 kilometers in six hours plus two hours. That’s eight hours. This means your average speed there and back is 12 divided by eight. That’s one and a half kilometers an hour. One and a half kilometers an hour isn’t twice as fast as one kilometer an hour; it’s only one and a half times as fast. We’re gonna have to look at this another way. Now, we thought about it using actual numbers. Let’s bring in some algebra as well and consider a general solution to the problem.
Let’s say that the houses are at distance of 𝑑 kilometers apart and you travel at an average speed of 𝑎 kilometers an hour when walking to granny’s. And let’s also say that your average speed on the way home is 𝑏 kilometers an hour. Right, we know that speed is the distance traveled divided by the time taken. We can rearrange this equation to get an expression for time. Time equals distance over speed. We can use this formula to work out the time taken to get to granny’s and the time taken to get back home. Then, we can add those times together to find the total time taken there and back.
Going to granny’s, we traveled 𝑑 kilometers at a speed of 𝑎 kilometers an hour. So, it takes 𝑑 over 𝑎 hours. Coming home, we again traveled 𝑑 kilometers, but this time at a speed of 𝑏 kilometers an hour. So that takes 𝑑 over 𝑏 hours. The total time taken then is 𝑑 over 𝑎 plus 𝑑 over 𝑏 hours. By multiplying each term in that expression by one, we’ve still got the same total time. Multiplying something by one doesn’t change its magnitude. The version of one that we’re gonna multiply the first term by is 𝑏 over 𝑏 and the version of one that we’re going to multiply the second term by is 𝑎 over 𝑎.
Now, we’ve got 𝑑 times 𝑏 over 𝑎 times 𝑏 plus 𝑑 times 𝑎 over 𝑏 times 𝑎. But 𝑎 times 𝑏 is the same as 𝑏 times 𝑎. So, we’ve got a common denominator. So, we can write that expression as a single term 𝑑𝑏 plus 𝑑𝑎 over 𝑎𝑏. Then, we spot we’ve got a common factor on the numerator. So, factoring out the 𝑑 gives us 𝑑 times 𝑏 plus 𝑎 over 𝑎𝑏. But remember, 𝑏 plus 𝑎 is the same as 𝑎 plus 𝑏. So, let’s write this as 𝑑 times 𝑎 plus 𝑏 over 𝑎𝑏. So, the time taken to get to granny’s and back is 𝑑 times 𝑎 plus 𝑏 all over 𝑎𝑏 hours, and the distance there and back was two lots of 𝑑, two 𝑑 kilometers.
So, we can write down an expression for the overall average speed. It’s the total distance traveled two 𝑑 kilometers divided by the total time taken 𝑑 times 𝑎 plus 𝑏 over 𝑎𝑏 hours. Remember, this is total distance divided by total time. So, we can write it as total distance divided by total time. We write two 𝑑 as two 𝑑 over one, just to make the next bit of the calculation a little bit easier. We’ve turned it into a fraction division. And dividing fractions is easy as pie. Flip the second and multiply, which means we can write it like this. Then, we spot that the 𝑑s cancel from the numerator and the denominator. And since one times 𝑎 plus 𝑏 is just 𝑎 plus 𝑏, we’ve now got two 𝑎𝑏 over 𝑎 plus 𝑏.
Remember, granny said that the overall average speed needed to be twice the average speed of the journey to her house, which was 𝑎 kilometers an hour. So, the overall average speed will need to be two 𝑎 kilometers an hour. Now, we can equate our two expressions for the overall average speed. And we know that two 𝑎 is equal to two 𝑎𝑏 over 𝑎 plus 𝑏. All we need to do now is rearrange this equation to find 𝑏 in terms of 𝑎. In other words, arrange it so that 𝑏 is on its own on one side and all the other terms involving 𝑎 are on the other side. But first, let’s just clear a little bit of space so we can show our calculations.
First, we’ll need to multiply both sides by 𝑎 plus 𝑏. So, we can cancel the 𝑎 plus 𝑏 in the denominator on the right-hand side. So, that’s two 𝑎 times 𝑎 plus 𝑏 is equal to two 𝑎𝑏. Let’s distribute the two 𝑎 on the left-hand side. Two 𝑎 times 𝑎 is two 𝑎 squared and two 𝑎 times 𝑏 is two 𝑎𝑏. So, we’ve got two 𝑎 squared plus two 𝑎𝑏 is equal to two 𝑎𝑏. Now, let’s try to eliminate 𝑏 from the left-hand side by subtracting two 𝑎𝑏 from both sides.
On the left, two 𝑎𝑏 minus two 𝑎𝑏 is nothing. But unfortunately, on the right, we’ve also got two 𝑎𝑏 minus two 𝑎𝑏. We haven’t got an equation with 𝑏 as the subject. We’ve just eliminated it altogether. Well, still, this says that two 𝑎 squared is zero. And this means that 𝑎 must be zero. If we traveled at zero kilometers an hour to granny’s, we wouldn’t have gotten there at all. Okay, let’s take a step back.
What if we’ve factored out the 𝑏 to get 𝑏 times two 𝑎 minus two 𝑎 at this point? Then, we could divide both sides by two 𝑎 minus two 𝑎 and then cancel it from the right-hand side to leave us with two 𝑎 squared over two 𝑎 minus two 𝑎 equals 𝑏. For now, let’s put aside the issue that we’ve just divided something by zero, which is a real no-no in math. This puzzle isn’t a divide-by-zero problem, but I just wanted to take a look at something.
Now, we’ve got an expression for 𝑏 in terms of 𝑎. But if you look at it, two 𝑎 minus two 𝑎 is zero. So, it means that 𝑏 is equal to something divided by zero. 𝑎 is a positive nonzero speed. So, if we square it, then double the answer, we certainly have a positive number. Now, we’ve got a positive number divided by zero, which is undefined. Oh dear! If instead of being zero in the denominator though, we thought about smaller and smaller positive values. We can see that this expression for 𝑏 is going to give us larger and larger values.
For example, 16 divided by eight is two. But as we make the denominator smaller, closer to zero, the answer gets bigger and bigger and bigger. And similarly, as two 𝑎 minus two 𝑎 gets closer to zero, then the value of 𝑏 gets larger and larger. Taking this to the extreme, we can say that the limit of 𝑏 as two 𝑎 minus two 𝑎 gets closer and closer to zero from the positive direction is positive ∞. To make the overall average speed twice as fast as the speed on the way to granny’s, we’ll need to travel infinitely fast on the way home.
So, what’s going wrong? Well, let’s take another look at the puzzle. We travel from home to granny’s at 𝑎 kilometers an hour. Then, we say we want the average speed for the whole journey, which is exactly twice as far, to be twice that of the first part. Now, traveling twice the speed means you get twice as far in the same time. Granny was tricking you. However fast or slow you traveled to granny’s, if the average speed of going there and back is twice the speed of going there, then you’d be traveling twice as far in the same time. That means it would take exactly the same amount of time to travel to granny’s and back as it did to travel just there, which means you’d have to do the return part of the journey in zero time or infinite speed.
Granny’s puzzle really sounded like it was something we were gonna be able to answer. But after a bit of working and a bit of thinking, we can see that it isn’t possible to run home fast enough to make the overall average speed there and back twice as fast as the average speed on the way there.
