# Oxford PAT 2011, Question 25

## 5 thoughts on “Oxford PAT 2011, Question 25”

1. How did u get equation A?
Wouldnt it be 8ssHsL =mmHmL?

1. Yes that’s right. But if 8ssHsL = mmHmL then, cancelling out the H and L, we get 8sss = mmm, i.e. 8s³ = m³. So, taking cube roots of both sides, we get 2s = m, or s = m/2.

2. James says:

I see, thanks for the explanation- this site has been very helpful.

3. James says:

Hi, why have you crossed out smaller?
Thanks

1. Hi James, thanks for your question.

I crossed out the ‘smaller’ and replaced with ‘larger’ because I don’t think there is a solution otherwise. The word ‘smaller’ is just an error in the question.

The question says that, for all the boxes, length/width is a constant (call it L), and height/length is a constant (call it H/L). Then height/width = (height/length)(length/width) = H. So if the widths of the small, medium and large boxes are s, m and l respectively, the volumes are LHs^3, LHm^3 and LHl^3 respectively.

But if we look at C, we can see that the base area of the small box is Ls^2, and the base area of the large box is Ll^2. But l must be bigger than s, so the base area of the large box must be larger than the base area of the small box, not larger.