Standard Form: Question 3

Worked solution

Setting up

The total mass is the number of beads multiplied by the mass of one bead:

$\text{Total mass} = \left(6 \times 10^{5}\right) \times \left(8 \times 10^{-9}\right) \text{ g}$

Method 1: multiply coefficients, add indices

Separate the number parts from the powers of ten:

$\left(6 \times 8\right) \times \left(10^{5} \times 10^{-9}\right)$

Multiply the coefficients:

$6 \times 8 = 48$

Add the indices (when multiplying powers with the same base, you add the exponents; you do not multiply them):

$10^{5} \times 10^{-9} = 10^{\,5 + (-9)} = 10^{-4}$

So far:

$48 \times 10^{-4}$

This is not yet in standard form, because the coefficient $48$ is not between $1$ and $10$. Rewrite $48$ in standard form and combine the powers:

$48 = 4.8 \times 10^{1}$

$48 \times 10^{-4} = 4.8 \times 10^{1} \times 10^{-4} = 4.8 \times 10^{\,1 + (-4)} = 4.8 \times 10^{-3}$

Method 2: convert to ordinary numbers first

Write each value as an ordinary number:

$6 \times 10^{5} = 600\,000, \qquad 8 \times 10^{-9} = 0.000\,000\,008$

Multiply:

$600\,000 \times 0.000\,000\,008 = 0.0048$

Now convert back to standard form by moving the decimal point $3$ places to the right:

$0.0048 = 4.8 \times 10^{-3}$

Both methods agree.

Check

A number is in correct standard form $a \times 10^{n}$ when $1 \le a < 10$ and $n$ is an integer. Here $a = 4.8$ (valid) and $n = -3$ (negative, as expected for a small total mass). ✓

• Distractor C $\left(48 \times 10^{-4}\right)$ has $a = 48$, which is not in the range $1 \le a < 10$.
• Distractor A $\left(4.8 \times 10^{-4}\right)$ comes from forgetting to add $1$ to the index when rewriting $48$ as $4.8 \times 10^{1}$.
• Distractor D $\left(4.8 \times 10^{3}\right)$ comes from a sign error on the final exponent.

Final answer

$\boxed{4.8 \times 10^{-3}\ \text{g}}$