An electrical company decides to replace the lightbulbs in all of the nonworking street lights on two specific streets. The company learns that 2/3

Question

of the lights do not work on the first street and 1/6
of the lights do not work on the second street. What is the total fractional portion of the nonworking lights on these two streets?

Answer 1

Let's denote the total number of street lights on the first street as \( A \) and on the second street as \( B \).

According to the problem:

• On the first street, \( \frac{2}{3} \) of the lights do not work.
• On the second street, \( \frac{1}{6} \) of the lights do not work.

To find the total fractional portion of the nonworking lights on both streets combined, we can compute the nonworking lights as follows:

1. Nonworking lights on the first street: \[ \frac{2}{3} \text{ of } A = \frac{2}{3}A \]

2. Nonworking lights on the second street: \[ \frac{1}{6} \text{ of } B = \frac{1}{6}B \]

Now, to find the total nonworking lights from both streets, we add the nonworking fractions together.

To do this, we need to find a common denominator for the fractions \( \frac{2}{3} \) and \( \frac{1}{6} \).

The least common multiple (LCM) of 3 and 6 is 6. We will convert \( \frac{2}{3} \) to have a denominator of 6:

\[ \frac{2}{3} = \frac{2 \times 2}{3 \times 2} = \frac{4}{6} \]

Now, we can add the fractions:

\[ \frac{4}{6} + \frac{1}{6} = \frac{4 + 1}{6} = \frac{5}{6} \]

Thus, the total fractional portion of the nonworking lights on these two streets is

\[ \frac{5}{6} \]