A Level Chemistry

Any help will be much appreciated for the attached question:

You would expect the rate equation to take the form
rate = k [X]^p [Y]^q [Z]^r
They’ve told you that the reaction is zero order with respect to Y (e.g q = 0, so [Y]^q = [Y]^0 = 1), so we can say the rate equation reduces to
rate = k [X]^p [Z]^r
We can see that in experiments 1 and 2 that when [Z] is held constant and [X] is doubled that the rate quadruples. That suggests that it must be second order with respect to X (e.g p = 2). This implies
rate = k [X]^2 [Z]^r
Now comparing experiments 1 and 3 (where [X] is halved and [Z] is doubled), we can see that the relative rate halves. If we call the concentrations of X and Z in experiment 1 x and z, respectively, we have
rate in experiment 1 = k x^2 z^r
rate in experiment 3 = k (0.5x)^2 (2z)^r
= k (0.25x^2) (2^r z^r)
= (0.25 * 2^r) k x^2 z^r
So experiment 3 has a rate that is 0.25 * 2^r times that of experiment 1. But we know that this factor is 0.5 as stated in the table, so
0.5 = 0.25 * 2^r
2 = 2^r
2^1 = 2^r
1 = r
So the rate equation is
rate = k [X]^2 [Z]
And the overall order is 2 + 0 + 1 = 3