Challenge Level

**1.** **Yes** - As $ 1 \textrm{m} = 100 \textrm{cm} $ a crate with dimensions 100cm by 50cm by 50cm will have volume $ V = 250,000 \textrm{cm}^3 $. Hence, the number of water bottles required to fill this up is

$$ N = \frac{250,000}{500} = 500 $$

**2. Yes** - If the diameter of the tree is almost half a meter, its radius is almost 25 cm. Now, if the tree's radius grows by 0.6 cm in each year, in 42 years the radius should be about $ 42 \cdot 0.6 = 25.2 \textrm{cm} $, and this is about right. However, to make a safer estimate (as it is unlikely that the tree will be growing by exactly 0.6 cm
each year) a range of the form 38 - 45 years would be better.

**3. No** - While $365 \cdot 0.3 = 10.95 \textrm{kg}$, which is close to the estimate, perhaps it would be sensible to consider that the scientist only has crisps on the *working days* of the year (since she is having a pack in her morning break). Now, a person is on average working 44 weeks each year, so they have $44 \cdot 5 = 220 $ working
days (and perhaps slightly less, if we take into account bank holidays).

Therefore, the scientist is more likely to be consuming $220 \cdot 0.3 = 6.6 \textrm{kg} $ of crisps each year.

**4. No** - A teaspoon of sugar weighs about 5 grams. So, the scientist and her friend consume together 8 cups of coffee each day, with 16 teaspoons of sugar. So, each day they consume $16 \cdot 5 = 80 \textrm{g}$ of sugar. Thus, in a two - month period, they would need $ 60 \cdot 80 = 4800 \textrm{g} $ of sugar, which is much more than a packet!

**5. No** - Once again, the calculation $365 \cdot 22 = 8030 $ is correct, but 365 is not the right number to use, as she is not working every day of the year. Using our previous estimate for the number of working days in a year, we see that the scientist is actually going to claim around $220 \cdot 22 = 4840 $ miles from work.

**6. Yes** - In this case, it is highly likely that the lab will be working all year round, so the estimate $560 \cdot 12 = 6720 $ is about right.

**7. Yes** - We can model the rubble by a large cone, whose height is 2m and base radius is 1.5m. The volume of such a cone is $$ V = \frac{1}{3}\cdot \pi \cdot r^2 \cdot h = \frac{1}{3} \cdot \pi \cdot 1.5^2 \cdot 2 \approx 4.7 m^3 $$ So the estimation is correct, and a 5 cubic meter skip should be large enough.

**8. No** - $124 \ cdot 170 = 21080 m^3 $ mean that each minute there are 21080 cubic meters of water flowing. Hence, over the course of an hour, there will be $21080 \cdot 60 = 1,264,800 m^3$ flowing, which is equivalent to 1,264.8 metric tonnes.

**9. Yes** - The area of the vegetable plot is $ A = 9.5 \cdot 11 = 104.5 m^2 $. Now, with the test data as described, we expect that *on average* there are $\frac {42 + 53}{2} = 47.5 $ worms per square meter.

Hence, a good estimate for the number of worms in the whole plot is $ N = 47.5 \cdot 104.5 = 4963.8 $, so the scientist is correct in this case.

**Well done to the students in Class 5C of the Brooklands Primary School, for sending us some very well-reasoned and clear answers to most of these short problems!**