# Video: AQA GCSE Mathematics Higher Tier Pack 3 • Paper 3 • Question 1

The variable 𝑝 is inversely proportional to the square of the variable 𝑞. Circle the correct equation given that 𝑘 is a constant. [A] 𝑝 = 𝑘/𝑞 [B] 𝑝 = 𝑘/𝑞² [C] 𝑝 = 𝑘𝑞 [D] 𝑝 = 𝑘𝑞².

02:06

### Video Transcript

The variable 𝑝 is inversely proportional to the square of the variable 𝑞. Circle the correct equation given that 𝑘 is a constant. The options are 𝑝 equals 𝑘 over 𝑞, 𝑝 equals 𝑘 over 𝑞 squared, 𝑝 equals 𝑘𝑞, and 𝑝 equals 𝑘𝑞 squared.

So what I’m gonna do is look at each of our possible answers in turn. The first one is 𝑝 equals 𝑘 over 𝑞. So 𝑝 equals 𝑘 over 𝑞 can also be written as 𝑝 and then the proportionality sign then one over 𝑞. And what this means is that 𝑝 is inversely proportional to 𝑞.

Well, if we look at the question, we want the variable 𝑝 to be inversely proportional to 𝑞. So that part is correct. However, we want it to be inversely proportional to the square of the variable 𝑞 and that’s not the case in this first answer. So therefore, this first answer is not the correct one.

Well, if we take a look at the second answer, we’ve got 𝑝 again is inversely proportional. So that’s the first part correct. And then, it says “to 𝑞 squared” because we have 𝑞 squared as the denominator. Well, this is correct as well because we were looking for the variable 𝑝 to be inversely proportional to the square of the variable 𝑞. So therefore, this looks like it’s gonna be the correct answer. But we’ll double check the last two just to make sure.

Well, we can deal with the last two together. Because if we look at them both, we have 𝑝 equals 𝑘𝑞 and 𝑝 equals 𝑘𝑞 squared. Well, both of these are not one over. So they’re not inversely proportional. They’re both in fact directly proportional.

So as we’re looking for an inversely proportional relationship, we can definitely rule these two out. So therefore, we can say that the second equation is definitely the correct equation to show that the variable 𝑝 is inversely proportional to the square of the variable 𝑞.