Question: Round $5187$ to 2 significant figures, and $0.004506$ to 2 significant figures.

Step 1. Significant figures count from the first non-zero digit. In $5187$ the first two are $5$ and $1$.

Step 2. Look at the digit to the right of the last one kept — here $8$, which is $\ge 5$, so round up:

$$5187 \approx 5200 \quad (2 \text{ s.f.})$$

The zeros are place-holders that keep the number its correct size — $52$ would be a different number entirely.

Step 3. For $0.004506$: leading zeros are never significant, so counting starts at the $4$. The first two significant figures are $4$ and $5$; next digit is $0$ ($< 5$, round down):

$$0.004506 \approx 0.0045 \quad (2 \text{ s.f.})$$

Step 4. Trailing-zero rule going the other way: $9.9999$ to 3 s.f. is $10.0$ — keep the final zero so that three figures still show.

⚠ Watch out: $0.037914$ to 2 s.f. is $0.038$, not $0.04$ — front zeros don't count as figures. And $7289$ to 2 s.f. is $7300$, never $73$: rounding must not change the size of the number.

Question: Without a calculator, estimate the value of $2013 \times 39$.

Step 1. Round each number to 1 significant figure — the whole point is numbers easy enough for mental arithmetic:

$$2013 \approx 2000, \qquad 39 \approx 40$$

Step 2. Compute with the rounded values:

$$2000 \times 40 = 80\,000$$

Step 3. Present with the approximation sign:

$$2013 \times 39 \approx 80\,000$$

(The exact value is $78\,507$ — the estimate is close, and that is all an estimate promises.)

Step 4. The same technique handles roots by rounding to a nearby square or cube rather than strictly 1 s.f.: $\sqrt{98.5} \approx \sqrt{100} = 10$, $\;\sqrt[3]{26.8} \approx \sqrt[3]{27} = 3$ — pick the friendly value, not the blindly rounded one.

⚠ Watch out: Write $\approx$, never $=$ — an estimation question loses its mark on the equals sign. And round before computing (that is the method being tested); computing exactly and then rounding answers a different question, and examiners can tell.