Video: AQA GCSE Mathematics Higher Tier Pack 4 β€’ Paper 2 β€’ Question 19

𝑧 is directly proportional to π‘₯Β². (a) When π‘₯ = 6, 𝑧 = 6. Calculate the value of 𝑧 when π‘₯ = 7. 𝑧 is inversely proportional to 𝑦. (b) When 𝑦 = 5, 𝑧 = 1/55. Calculate the value of 𝑧 when 𝑦 = 9. (c) Circle the letter of the graph which shows the relationship between π‘₯ and 𝑦.

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Video Transcript

𝑧 is directly proportional to π‘₯ squared. Part a) When π‘₯ equals six, 𝑧 equals six. Calculate the value of 𝑧 when π‘₯ equals seven. 𝑧 is inversely proportional to 𝑦. Part b) When 𝑦 equals five, 𝑧 equals one over 55. Calculate the value of 𝑧 when 𝑦 equals nine. There is also a part c) to this question which we’ll look at in a moment.

In part a) of the question, we’re told that 𝑧 is directly proportional to π‘₯ squared. So we can express this using the proportionality symbol. We need though an equation that relates 𝑧 and π‘₯ squared to each other.

If 𝑧 is directly proportional to π‘₯ squared, then it means that 𝑧 is some multiple of π‘₯ squared. So we can write 𝑧 equals π‘˜π‘₯ squared. π‘˜ here is just a number. It’s a constant which we call the constant of proportionality. And we need to work out its value.

To do so, we use the information given in the question, which is that when π‘₯ is equal to six, 𝑧 is also equal to six. So we have a pair of values for π‘₯ and 𝑧, which we can substitute into this equation. We have then that six, that 𝑧, is equal to π‘˜ multiplied by six squared. That’s π‘₯ squared. Six squared is 36. So we have six equals 36π‘˜.

To solve for π‘˜, we need to divide both sides of this equation by 36, as this will cancel out the factor 36 on the right, leaving just π‘˜. On the left, we have the fraction six over 36. But this can be simplified by dividing both the numerator and denominator by six to give one-sixth. We found that the value of π‘˜ is equal to one-sixth.

Next, we substitute this value of π‘˜ back into the equation relating 𝑧 and π‘₯. So we have that 𝑧 is equal to one-sixth of π‘₯ squared. The question asked us to calculate the value of 𝑧 when π‘₯ equals seven. So finally, we need to substitute seven for π‘₯ in this equation. We have that 𝑧 is equal to one-sixth multiplied by seven squared. Seven squared is equal to 49. So multiplying the two values together, we have that 𝑧 is equal to 49 over six. We’ll leave our answer as a fraction rather than converting to a decimal.

Now let’s look at part b. In this part, we’re told that 𝑧 is inversely proportional to 𝑦. If two quantities are in inverse proportion, then it means that as one quantity increases, the other decreases at the same rate, whereas in direct proportion, the two quantities increase at the same rate together. If 𝑧 is inversely proportional to 𝑦, then it, in fact, means that 𝑧 is proportional to the reciprocal of 𝑦, one over 𝑦.

We can express this using the proportionality symbol. And then we can write it as an equation. 𝑧 is equal to some multiple of one over 𝑦. In this case, I’ll use the letter 𝑐 for the constant of proportionality to distinguish it from the π‘˜ we used in part a.

As in part a, we’re given a pair of values. When 𝑦 is equal to five, 𝑧 is equal to one over 55. So we can use these values to find the value of our constant of proportionality. Substituting one over 55 for 𝑧 and five for 𝑦, we have the equation one over 55 equals 𝑐 over five.

To find the value of 𝑐, we need to multiply both sides of this equation by five. We have that five over 55 equals 𝑐. But this fraction can be simplified by dividing both the numerator and denominator by five. The simplified fraction for 𝑐 then is one over 11, one eleventh.

Substituting this value for 𝑐 back into our equation relating 𝑧 and 𝑦, we have that 𝑧 is equal to one over 11 multiplied by one over 𝑦, which we can simplify as one over 11𝑦 by multiplying the two fractions together.

Finally, we need to calculate the value of 𝑧 when 𝑦 equals nine. So we’re going to substitute nine for 𝑦 in this equation. We have that 𝑧 is equal to one over 11 multiplied by nine. And as 11 multiplied by nine is 99, the fraction simplifies to one over 99. So we found that when 𝑦 is equal to nine, 𝑧 is equal to one over 99.

Part c) of the question says, β€œCircle the letter of the graph which shows the relationship between π‘₯ and 𝑦.” We’ve been given four possible graphs that could describe the relationship between π‘₯ and 𝑦. Now if you recall from the earlier parts of the question, we were told how π‘₯ was related to 𝑧 and how 𝑦 was related to 𝑧, but not how π‘₯ and 𝑦 were related to each other. We were told that 𝑧 was directly proportional to π‘₯ squared. And using the information given in the question, we found that 𝑧 was equal to one-sixth of π‘₯ squared. We were also told that 𝑧 was inversely proportional to 𝑦. And using the second piece of information we were given, we found that 𝑧 was equal to one over 11𝑦. We can use these equations to find an equation linking π‘₯ squared and 𝑦.

We have that one-sixth of π‘₯ squared is equal to one over 11𝑦. In order to work out which graph belongs with this equation, let’s rearrange it to make 𝑦 the subject. Our first step is to multiply both sides of the equation by 𝑦, as 𝑦 is currently in the denominator on the right-hand side. Doing so gives 𝑦 over six π‘₯ squared equals one over 11. We can then multiply both sides of the equation by six and then divide both sides of the equation by π‘₯ squared to give 𝑦 equals six over 11π‘₯ squared.

In fact, it doesn’t really matter what the value of the constant is. So we can replace six over 11 with the letter π‘˜. And we found that 𝑦 equals π‘˜ over π‘₯ squared. This means that 𝑦 is inversely proportional to π‘₯ squared. So we need to decide which of the four graphs matches with this.

We can rule out graph D straight away because this is a straight line passing through the origin, which means that 𝑦 is directly proportional to π‘₯, with the equation 𝑦 equals π‘˜π‘₯. So it’s not graph D. We can also rule out graph C. This is a quadratic curve passing through the origin. So this would have the form 𝑦 equals π‘˜π‘₯ squared, meaning that 𝑦 is directly proportional to π‘₯ squared rather than inversely proportional to π‘₯ squared.

We’re left with graphs A and B, both of which are reciprocal curves. So they both look plausible. The key difference is in what happens for negative values of π‘₯. In graph A, we find that the 𝑦-values are positive, whereas in graph B, the 𝑦-values will be negative. Whenever we square an π‘₯-value, whether it’s positive or negative, we always get a positive answer. And as our constant π‘˜ is the same for positive and negative π‘₯-values, the sign of 𝑦 will therefore be the same for both positive and negative π‘₯-values.

This means that we have to rule out graph B because here the signs of 𝑦 are different for positive and negative π‘₯-values. In fact, graph B does represent an inversely proportional relationship. But it’s the relationship that 𝑦 is inversely proportional to π‘₯ β€” 𝑦 equals π‘˜ over π‘₯ β€” rather than inversely proportional to π‘₯ squared.

Graph A has the correct shape for a curve of the form 𝑦 equals π‘˜ over π‘₯ squared. It’s a reciprocal graph and the 𝑦-values are always positive. So graph A shows the relationship between π‘₯ and 𝑦.

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