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Patrick from Woodbridge School gave a really good solution to this problem. There is much to consider in this problem, so we have added a few editorial notes of interest to Patrick's solution.

1. This is a reasonable assumption, given that an apple is mostly made up of water with a density of $1$g cm$^{-3}$. [This is an over-estimate: since apples float, their density will be a bit less than that of water.]

2. This is clearly an under-estimate since spheres do not tessellate at all well (there is empty space between each sphere). It will be out by some considerable amount. It would be much easier to test the volume by filling it with water and measuring that. [Roughly speaking you can fill up to about $74\%$ of space using spheres. So the volume will be about $850/0.74 =1150$ cm$^3$.]

3. $0.0894 \times 7.3 = 0.65$ gives the rough volume of the trunk (this will be inaccurate as the trunk tapers, so it will be an overestimate, but it will not be a very big difference). However, the branches make up a lot of the tree's volume so it will be considerably under the true value. A fir tree has fewer branches, so it will still be an underestimate, but not so far out. [it is a good point that the branches are an important factor in the weight of the tree, not just the shape of the trunk.]

4. $27000\times 17=459000$ so this would be a reasonable assumption for the first $20$cm of soil. However, this fails to take into account the possibility that there is much more soil underneath the tested area, so the estimate is probably a very long way under the true value. [This assumes that soil is uniform. Some areas might be more or less fertile than this. Also we would need to know how far down the worms are likely to live.]

5. The cheetah's top speed is 100kmh$^{-1}$, but it cannot sustain this for three minutes - it is only a peak measurement. Therefore this distance is an overestimate (but not by very much). [$3$ mins at $100$kph$^{-1} = 100\times 3/60 =5$. Can a cheetah maintain top speed for 3 mins?]

6. This is a perfectly reasonable estimate. It is likely to be almost exactly right.[Assume salt concentration and water density is uniform: this is very likely. If $10$kg of water gives $0.35$kg of salt then $10\times 100/0.35 =2857$ kg of water gives $100$ kg of salt.]

7. We need more information to solve this question - I have no idea what the density of the salts is. [Four barrels will contain up to $2800$l of fluid. $3000$kg of sea water would fit providing that the density is greater than $3000/2800 = 1.07$ kg/l. The actual figure is about $1027$ kg m$^{-3}$, so the required amount of salt water will easily fit.]

8. Apparently every cell divides in two every $35$ minutes. $12$ hrs = $720$ minutes; $7200/35 = 20.57$. So the cells increase by a factor of $2^{20.57}$, which gives $43.6$ million. [This figure is probably an upper bound, as it does not take into account issues concerning possible death rate, increasing scarcity of food or other changing environmental factors.]

9. The reasoning is that $100/12 = 8.33$, so the estimate for fish will be $100\times 8.33 = 833$. I would therefore say that the estimate is a little under the correct value. [This proportional reasoning is a good first estimate. More advanced statistical considerations would give rise to a likely range of values. Of course, the estimate does not take into account environmental factors such as fish being eaten, moving away from the test area, tagged fish being possibly more likely to be captured etc.]

10. I think this sounds like it should be correct. [This is likely to be true for the first part, but not necessarily true for solutions which absorb $100\%$ of the light. Imagine a jug of black ink. Diluting this ink might still yield a completely opaque jug of liquid until quite a large volume of water has been added.]