Question

The water level of a river is below normal levels. Its water level is at 1 1/2 meters below sea level. During the summer, the river drops another 2 1/3 meters below sea level. What is the water level at the end of summer? Remember, the elevation is below sea level. Account for this in your answer.

Answer 1

To find the water level of the river at the end of summer, we need to add the two negative elevations:

1. Initial Water Level: \( 1 \frac{1}{2} \) meters below sea level

   • Converting this to an improper fraction: \[ 1 \frac{1}{2} = \frac{3}{2} \text{ meters below sea level} \]

2. Summer Drop: \( 2 \frac{1}{3} \) meters below sea level

   • Converting this to an improper fraction: \[ 2 \frac{1}{3} = \frac{7}{3} \text{ meters below sea level} \]

Next, we need to add these two values together. Since both are below sea level, we will combine them, keeping in mind that they are both negative values.

We need a common denominator to perform the addition. The least common multiple of 2 and 3 is 6.

3. Convert both fractions to have a denominator of 6:

   • For \(\frac{3}{2}\): \[ \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \]
   • For \(\frac{7}{3}\): \[ \frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6} \]

4. Now, we can add the fractions: \[ \frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6} \]

This result, \(\frac{23}{6}\) meters, means that at the end of summer, the river is \( \frac{23}{6} \) meters below sea level.

5. Converting \(\frac{23}{6}\) to a mixed number: \[ 23 \div 6 = 3 \text{ remainder } 5 \] Therefore, \[ \frac{23}{6} = 3 \frac{5}{6} \text{ meters below sea level} \]

Thus, the water level at the end of summer is \( 3 \frac{5}{6} \) meters below sea level.

Answer 2

it says its wrong

Answer 3

Let’s carefully re-evaluate the calculations to make sure we arrive at the correct answer step-by-step.

1. Initial Water Level: The initial water level is given as \( 1 \frac{1}{2} \) meters below sea level, which can be expressed as: \[ 1 \frac{1}{2} = \frac{3}{2} \text{ meters below sea level} \]

2. Summer Drop: The river drops another \( 2 \frac{1}{3} \) meters below sea level, which can be converted as: \[ 2 \frac{1}{3} = \frac{7}{3} \text{ meters below sea level} \]

3. Combining the Values: To find the total drop from sea level, we need to add these two quantities: \[ Total = \frac{3}{2} + \frac{7}{3} \]

4. Finding a Common Denominator: The least common multiple of 2 and 3 is 6.

   • Convert \(\frac{3}{2}\) to a denominator of 6: \[ \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \]

   • Convert \(\frac{7}{3}\) to a denominator of 6: \[ \frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6} \]

5. Adding the Converted Fractions: \[ \frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6} \]

6. Final Calculation:

The water level at the end of summer is at \(\frac{23}{6}\) meters below sea level.

To express \(\frac{23}{6}\) as a mixed number:

• Divide 23 by 6, which gives 3 with a remainder of 5: \[ 23 \div 6 = 3 \quad \text{(whole part)}, \quad \text{remainder } 5 \] Thus, \(\frac{23}{6} = 3 \frac{5}{6}\).

Therefore, the water level at the end of summer is 3 \(\frac{5}{6}\) meters below sea level.

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