A student is experimenting with lightning rods and places one into the ground. She finds that the approximate length of rod

above ground is 5/7 of a yard
and the approximate length
below ground is 1/3 of a yard.
How many times farther above the ground is the lightning rod than below the ground?

1 answer

To find how many times farther above the ground the lightning rod is compared to below ground, we need to compare the two lengths.

The length of the rod above ground is \( \frac{5}{7} \) yards, and the length of the rod below ground is \( \frac{1}{3} \) yards.

We want to find the ratio of the length above ground to the length below ground:

\[ \text{Ratio} = \frac{\text{Length above ground}}{\text{Length below ground}} = \frac{\frac{5}{7}}{\frac{1}{3}} \]

To divide fractions, we multiply by the reciprocal of the denominator:

\[ \frac{\frac{5}{7}}{\frac{1}{3}} = \frac{5}{7} \times \frac{3}{1} = \frac{5 \times 3}{7 \times 1} = \frac{15}{7} \]

Thus, the ratio of the length above ground to the length below ground is \( \frac{15}{7} \).

Interpreting this result, it means that the lightning rod is \( \frac{15}{7} \) times farther above the ground than below the ground.

To express \( \frac{15}{7} \) as a mixed number, we can divide:

\[ 15 \div 7 = 2 \text{ remainder } 1 \]

So, \( \frac{15}{7} = 2 \frac{1}{7} \).

Therefore, the lightning rod is approximately \( 2.14 \) times farther above the ground than below the ground.

The final answer is:

\[ \frac{15}{7} \text{ or } 2 \frac{1}{7} \text{ times.} \]