The table shows marks got by a group of pupils from a test. a) How many pupils were in that group? b) Calculate the average mark score. c) Find the median mark.

There were $\boxed{\text{8 pupils}}$ in that group.

The average mark score is $\boxed{\text{30}}$.

The marks are: 20, 20, 30, 30, 30, 30, 40, 40.
There are 8 marks in total.

Since there is an even number of marks (8), the median is the average of the two middle values (the 4th and 5th values).
The 4th value is 30.
The 5th value is 30.

The median mark is $\boxed{\text{30}}$.

4. Subtract 140kg, 8Hg and 9726mg from 160kg 8Hg and 8692mg.

We need to calculate: $(160 \text{ kg } 8 \text{ Hg } 8692 \text{ mg}) - (140 \text{ kg } 8 \text{ Hg } 9726 \text{ mg})$.

Since $8692 \text{ mg} < 9726 \text{ mg}$, we need to borrow from the hectograms (Hg).
$1 \text{ Hg} = 100 \text{ g} = 100,000 \text{ mg}$.
Borrow 1 Hg from 8 Hg, leaving 7 Hg. Add 100,000 mg to 8692 mg.
$8692 \text{ mg} + 100,000 \text{ mg} = 108692 \text{ mg}$.
Now, $108692 \text{ mg} - 9726 \text{ mg} = 98966 \text{ mg}$.

We now have 7 Hg in the first quantity. Since $7 \text{ Hg} < 8 \text{ Hg}$, we need to borrow from kilograms (kg).
$1 \text{ kg} = 10 \text{ Hg}$.
Borrow 1 kg from 160 kg, leaving 159 kg. Add 10 Hg to 7 Hg.
$7 \text{ Hg} + 10 \text{ Hg} = 17 \text{ Hg}$.
Now, $17 \text{ Hg} - 8 \text{ Hg} = 9 \text{ Hg}$.

We now have 159 kg in the first quantity.
$159 \text{ kg} - 140 \text{ kg} = 19 \text{ kg}$.

The final result is $\boxed{\text{19 kg 9 Hg 98966 mg}}$.

5. Use the above line graph (Collected eggs for five days) to answer the questions.
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