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Video: Direct and Inverse Proportion Word Problems

Tim Burnham

Learn to apply your knowledge of direct and inverse proportion to questions presented in a wordy format, for which you have to extract relevant information, define suitable variables, then construct and solve your own proportionality equations.

13:28

Video Transcript

Before watching this video, you should already know how to tackle direct an inverse proportion problems. But we’re gonna look at wordy problems in this video. So for example here, a car’s fuel consumption varies inversely with its speed when travelling at forty miles per hour. It’s fuel consumption is sixty miles per gallon. How much fuel would be saved if it travelled a hundred miles at a constant speed of thirty miles per hour compared to seventy miles per hour? So you have to define some variables and then construct an equation of proportionality between those variables and then use that equation to solve the problem.

Let’s have a go at tackling this problem then. So we’ve got the car’s fuel consumption and it’s varying inversely with its speed. So I’m gonna define a variable 𝑐 which represents the fuel consumption in miles per gallon of the car. And I’m gonna let 𝑠 represent the speed in miles per hour of the car. And we’re told that 𝑐 varies inversely with 𝑠. So 𝑐 is directly proportional to one over 𝑠. And when we say that something is directly proportional to something else, we mean that it’s some constant, some simple number, times that something else. So this means that we can come up with the equation 𝑐 equals a constant of proportionality, just a number, let’s call it 𝑘 times one over 𝑠. And we might write that as just 𝑘 over 𝑠. So 𝑐 is equal to 𝑘 over 𝑠.

Now we’re also told in the question that when it’s travelling at forty miles per hour, so when 𝑠 is forty then the fuel consumption is sixty miles per gallon. So 𝑐 is equal to sixty. So putting those two values in for 𝑐 and 𝑠, we’ve got sixty is equal to 𝑘 over forty. Now if I multiply both sides of my equation by forty, I can then eliminate forty from the right-hand side, just leaving me with 𝑘. So the forties cancel and sixty times forty is two thousand four hundred. So this means that 𝑘 is two thousand four hundred, so we can put that 𝑘 back into our original equation. So 𝑐 is equal to two thousand four hundred times one over 𝑠 or 𝑐 is equal to two thousand and four hundred over 𝑠. So that’s the equation which defines the relationship between the fuel consumption and the speed of the car.

And now we’ve got to work out the fuel consumption travelling at a speed of thirty miles an hour and seventy miles an hour and then apply that fuel consumption to travelling a hundred miles at that speed. At thirty miles an hour then, 𝑐 is equal to two thousand four hundred over thirty. Well I can divide the top and the bottom by ten, so we’ve got two thousand and forty divided by three. Well that’s eighty. And at seventy miles an hour, 𝑐 is equal to two thousand four hundred divided by seventy. So again I can divide top and bottom by ten, and I’ve got two hundred and forty over seven. Well it doesn’t divide down nicely, so I’m just gonna leave it as two hundred and forty over seven miles per gallon.

Now with the speed of thirty miles an hour, I can do eighty miles on one gallon of fuel. So if I divide that by eighty then an eightieth of eighty miles is one mile and an eightieth of one gallon is one over eighty, so one mile takes one eightieth of a gallon. But of course I want to go a hundred miles, so I need to multiply that amount by a hundred. So if I had to travel a hundred times as far, I’m gonna use a hundred times as much fuel. So at thirty miles an hour, I use a hundred over eighty gallons of fuel. Now that cancels down to five over four. So let’s make a note of that. A hundred miles at thirty miles per hour uses five over four gallons of fuel, one and a quarter gallons of fuel. So now let’s do the same calculation again but this time at seventy miles an hour.

At this speed, I can do two hundred and forty divided by seven miles with one gallon of fuel. That’s what two hundred and forty over seven miles per gallons means. So if I divide that by two hundred and forty over seven, I’m gonna get the number of gallons that I use for one mile. And of course one divided by two hundred and forty over seven is the same as one over one divided by two hundred and forty over seven. And I can convert that into a multiplication calculation by flipping the second fraction, so one over one divided by two hundred and forty over seven is the same as one over one times seven over two hundred and forty, which is just seven over two hundred and forty.

So one mile takes seven two hundred and fortieths of a gallon. But of course I don’t wanna travel one mile, I want to travel a hundred miles. So if I multiply those by a hundred, one times a hundred is a hundred miles and seven over two hundred and forty times a hundred is seven hundred over two hundred and forty. Now I can divide the numerator and the denominator both by ten to give me seventy over twenty-four. And then I can divide them by two to leave me with thirty-five over twelve gallons. So to travel a hundred miles at seventy miles an hour uses thirty-five over twelve gallons of fuel. So what I need to do is to take one of those away from the other to see how much extra fuel I’m using travelling at seventy miles per hour.

So the calculation I’ve got to do to find out the difference in the fuel that I’ve used is thirty-five over twelve, the amount of gallons that I use at seventy miles an hour minus five over four, which is the number of gallons I use travelling at thirty miles an hour. Now what I need is a common denominator. So if I multiply the second term there by three over three, of course three over three is just one, so I’m multiplying a number by one. So I’m not changing that number, I’m just using a slightly different version of it. And in this case, it’s gonna be an equivalent fraction which has a denominator of twelve. So that calculation becomes thirty-five over twelve minus fifteen over twelve which is twenty over twelve. And dividing the top by four, I get five. Dividing the bottom by four, I get three. So that’s five over three or one and two-thirds gallons difference.

So let’s make our answer nice and clear. We would save one and two-thirds gallons of fuel. So overall, word problems are a little bit more tricky than the straightforward standard problems that you might encounter. First of all, you’ve got to make your way through a paragraph of a text, then you’ve got to define your own variables. Then you’ve got to come up with your relationship between those two variables in the form of a proportional relationship and then an equation. And then you might have some, you know, relatively complicated calculations to do to actually answer the question. It’s not just a case of coming up with an equation and then plugging a value into that equation. You may have to do more things with those numbers.

Right, let’s move on to another question then. A sand timer has twenty cubic inches of sand which falls at a steady rate of one cubic inch per minute. If all the sand is in the bottom bulb of the timer and then I turn the timer over, is the amount of sand remaining in the top half inversely proportional to the amount of time since turning it? So we started off with the all the sand down the bottom. Then we flipped it over, so the bottom half becomes the top half and then all the sand slowly pours down. So the question is: “Is the amount of sand up here inversely proportional to the amount of time since we turned it over?” Well let’s define some variables then. Let 𝑥 be the amount of time since we turned it over in minutes. And let 𝑦 be the volume of sand left up the top half in cubic inches.

Well I can quickly put together a small table of values. So after zero minutes, all twenty cubic inches of sand are still at the top half. After ten minutes at a rate of one cubic inch per minute falling down, we would only have ten cubic inches left. And after twenty minutes, all twenty cubic inches of sand would’ve drained away. So there’d be nothing left up top. So it looks like there is a linear relationship between 𝑥 and 𝑦 and it passes through zero, twenty. When 𝑥 is zero, 𝑦 is twenty, so it doesn’t pass through the origin. So that means it’s not a directly proportional relationship. But could it be an inversely proportional relationship which is what the question is actually asking. Well if that was the case, 𝑦 would be equal to some constant divided by 𝑥. And when 𝑥 is equal to zero, you’d have this constant divided by zero, which would be infinity. But in fact 𝑦 isn’t infinity. It’s twenty, so it looks like it’s not an inversely proportional relationship either.

One last question then. The intensity of the light from a torch varies inversely with the square of the distance from the bulb. The light has an intensity of ten lumens at a distance of seven feet. At what distance would the intensity be fifteen lumens? Give your answer correct to one decimal place. So this is a relationship in which the light from the torch is varying inversely with the square of the distance from the bulb. So let’s define a couple of variables. First of all, 𝑖 is the intensity of the light and the units there are lumens. And 𝑑 is the distance from the bulb and the units there are feet.

Now the question says the intensity varies inversely with the square of the distance from the bulb. That means that 𝑖 is directly proportional to one over the distance squared and we can write that as an equation. 𝑖 is equal to 𝑘 times one over 𝑑 squared or simply 𝑘 over 𝑑 squared. Now the question also tells us that the light has an intensity of ten lumens at a distance of seven feet. So 𝑖 equals ten when 𝑑 is equal to seven. So if we put those values into our calculation, we could work out the value of 𝑘, our constant of proportionality. So that means that ten is equal to 𝑘 over seven squared or seven squared is forty-nine. So in this case ten is equal to 𝑘 over forty-nine. Well if I multiply both sides by forty-nine, I can work out what 𝑘 is.

Well forty-nine times ten is four hundred and ninety and 𝑘 over forty-nine times forty-nine is just 𝑘. So 𝑘 is four hundred and ninety. I can now complete my equation. So for this particular torch following those particular rules, 𝑖, the intensity of the light in lumens, is equal to four hundred and ninety divided by the distance in feet squared. So let’s go on and answer the rest of the question then. At what distance would the intensity be fifteen lumens? Well when 𝑖 is fifteen, that means that fifteen is equal to four hundred and ninety divided by 𝑑 squared. Well let’s multiply both sides by 𝑑 squared, which means that fifteen 𝑑 squared is equal to four hundred and ninety. And now let’s divide both sides by fifteen. Well I can divide both of those by five, and four hundred and ninety divided by five is ninety-eight and fifteen divided by five is three. So 𝑑 squared is equal to ninety-eight over three. So now I could take the square root of both sides and I get 𝑑 is equal to seven root six over three. Now technically there’s also a positive and a negative version of that because negative seven root six over three times negative seven root six over three would also give us ninety-eight over three. And to one decimal place, that’s five point seven feet, so positive or negative five point seven feet to one decimal place.

Now in practical terms in this particular problem, the answer negative five point seven feet doesn’t make a lot of sense. If we’ve got a torch and it’s shining in that direction, yes five point seven feet in that direction, the intensity here will be fifteen lumens. But if we go five point seven feet in the other direction here, well the torch would’ve have been in the way. So I don’t think there’ll be any light there from the bulb at all. So our answer is just the positive five point seven feet to one decimal place. So with this question, although we did reject the negative answer, we showed the possibility of a negative answer in our working out, and we showed the logic for rejecting that by drawing this diagram on the page and then just giving our answer five point seven feet. So that kind of level of detail in your working out is very important.

Now we could’ve made a different assumption about the torch. Maybe it was one of those candle torches, and in that case there would be this kind of part of a sphere with a radius of five point seven to one decimal place feet all around the bulb. It’s quite difficult to draw actually. But imagine this kind of sphere sitting around the bulb with a radius of five point seven feet at that point on the sphere, all points on the surface of that sphere, the intensity of the light would be fifteen lumens.

Anyway, hopefully going through those few questions has just given you the confidence now to use your knowledge of direct an inverse variation or at direct an inverse proportion and to be able to answer wordy problems.