Each time Kim pours water from a jug into a glass, exactly $10 \%$ of the water remaining in the jug is used. What is the minimum number of times that she must pour water into a glass so that less than half the water remains in the jug?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Solution 1

We can suppose that the jug contains 1 litre of water at the start. The following table shows the quantity of water poured in each glass and the quantity of water remaining in each glass after each pouring, stopping when the quantity of water remaining is less than $0.5 \mathrm{~L}$.

| Number of glasses | Number of litres poured | Number of litres remaining |
| :---: | :--- | :--- |
| 1 | $10 \%$ of $1=0.1$ | $1-0.1=0.9$ |
| 2 | $10 \%$ of $0.9=0.09$ | $0.9-0.09=0.81$ |
| 3 | $10 \%$ of $0.81=0.081$ | $0.81-0.081=0.729$ |
| 4 | $10 \%$ of $0.729=0.0729$ | $0.729-0.0729=0.6561$ |
| 5 | $10 \%$ of $0.6561=0.06561$ | $0.6561-0.06561=0.59049$ |
| 6 | $10 \%$ of $0.59049=0.059049$ | $0.59049-0.059049=0.531441$ |
| 7 | $10 \%$ of $0.531441=0.0531441$ | $0.531441-0.0531441=0.4782969$ |

We can see from the table that the minimum number of glasses that Kim must pour so that less than half of the water remains in the jug is 7 .

Solution 2

Removing $10 \%$ of the water from the jug is equivalent to leaving $90 \%$ of the water in the jug.

Thus, to find the total fraction remaining in the jug after a given pour, we multiply the previous total by 0.9 .

We make the following table; stopping when the fraction of water remaining in the glass is first less than 0.5 (one half).

| Number of glasses poured | Fraction of water remaining |
| :---: | :--- |
| 1 | $0.9 \times 1=0.9$ |
| 2 | $0.9 \times 0.9=0.81$ |
| 3 | $0.9 \times 0.81=0.729$ |
| 4 | $0.9 \times 0.729=0.6561$ |
| 5 | $0.9 \times 0.6561=0.59049$ |
| 6 | $0.9 \times 0.59049=0.531441$ |
| 7 | $0.9 \times 0.531441=0.4782969$ |

We can see from the table that the minimum number of glasses that Kim must pour so that less than half of the water remains in the jug is 7 .

ANswer: (C)