Question

For which two uses of electricity was the ratio of the amounts of electricity used most nearly 3 to 1?

• A. Water heater and lights/small appliances
• B. Large appliances and lights/small appliances
• C. Air conditioner and water heater
• D. Air conditioner and lights/small appliances
• E. Air conditioner and large appliances
• A. \(\frac{1}{11}\)
• B. \(\frac{1}{9}\)
• C. \(\frac{1}{8}\)
• D. \(\frac{1}{5}\)
• J. It cannot be determined from the information given.
• A. 13%
• B. 33%
• C. 50%
• D. 65%
• O. 130%

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question asks us to find a pair of electricity uses from the pie chart whose consumption percentages have a ratio of approximately 3 to 1. We will test the ratio for each pair given in the options.
Step 2: Key Formula or Approach:
\[ \text{Ratio} = \frac{\text{Percentage of first use}}{\text{Percentage of second use}} \]
We need this ratio to be close to 3.
Step 3: Detailed Explanation:
From the pie chart, the percentages are:
\begin{itemize} \item Air Conditioner: 53% \item Water Heater: 20% \item Large Appliances: 18% \item Lights/Small Appliances: 9% \end{itemize} Now, let's test the ratio for each option:
(A) Water heater to lights/small appliances: \(\frac{20}{9} \approx 2.22\). Not close to 3.
(B) Large appliances to lights/small appliances: \(\frac{18}{9} = 2\). The ratio is exactly 2 to 1. Not 3 to 1.
(C) Air conditioner to water heater: \(\frac{53}{20} = 2.65\). This is somewhat close, but let's check other options.
(D) Air conditioner to lights/small appliances: \(\frac{53}{9} \approx 5.89\). Not close to 3.
(E) Air conditioner to large appliances: \(\frac{53}{18} \approx 2.94\). This value is the closest to 3.
Step 4: Final Answer:
The ratio for air conditioner and large appliances (2.94 to 1) is the most nearly 3 to 1.

Answer 2

Step 1: Understanding the Concept:
This is a weighted average cost problem. The cost per unit (kWh) is different for the water heater than for the other appliances. We need to calculate the total cost by considering these different rates and then find what fraction of that total cost is attributable to the water heater.
Step 2: Detailed Explanation:
1. Define variables for the cost rates.
Let \(C\) be the cost per kWh for the "rest of the electricity".
Then the cost per kWh for the water heater is \(\frac{C}{2}\).
2. Calculate the amount of electricity used for each category.
Water heater usage = 20% of total usage = 0.20 \(\times\) Total.
Rest of the usage = (53% + 18% + 9%) of total usage = 80% of total usage = 0.80 \(\times\) Total.
(Note: We don't need the 800 kWh value, as it will cancel out. We can work with percentages.)
3. Calculate the cost for each category in terms of C and Total Usage.
Cost of water heater = (Water heater usage) \(\times\) (Cost rate for water heater)
\[ = (0.20 \times \text{Total}) \times \left(\frac{C}{2}\right) = 0.10 \times \text{Total} \times C \]
Cost of the rest = (Rest of usage) \(\times\) (Cost rate for the rest)
\[ = (0.80 \times \text{Total}) \times C = 0.80 \times \text{Total} \times C \]
4. Calculate the total cost.
Total Cost = Cost of water heater + Cost of the rest
\[ = (0.10 \times \text{Total} \times C) + (0.80 \times \text{Total} \times C) = 0.90 \times \text{Total} \times C \]
5. Find the required fraction.
Fraction = \(\frac{\text{Cost of water heater}}{\text{Total Cost}}\)
\[ = \frac{0.10 \times \text{Total} \times C}{0.90 \times \text{Total} \times C} = \frac{0.10}{0.90} = \frac{1}{9} \]
Step 3: Final Answer:
The cost of the electricity used by the water heater was \(\frac{1}{9}\) of the total cost.

Answer 3

Step 1: Understanding the Concept:
This is another percent increase problem. We need to calculate the amount of electricity (in kWh) used by the water heater in July and in November, and then find the percent increase from the July amount to the November amount.
Step 2: Key Formula or Approach:
\[ \text{Percent Increase} = \left( \frac{\text{November Usage} - \text{July Usage}}{\text{July Usage}} \right) \times 100% \]
Step 3: Detailed Explanation:
1. Calculate July water heater usage.
Total electricity in July = 800 kWh.
Water heater usage in July = 20% of 800 kWh = \(0.20 \times 800 = 160\) kWh.
2. Calculate November water heater usage.
Total electricity in November = 800 kWh (same as July).
Water heater usage in November = 33% of 800 kWh = \(0.33 \times 800 = 264\) kWh.
3. Apply the percent increase formula.
July Usage (Original Value) = 160 kWh.
November Usage (New Value) = 264 kWh.
Increase = 264 - 160 = 104 kWh.
\[ \text{Percent Increase} = \left( \frac{104}{160} \right) \times 100% \]
Simplify the fraction: \(\frac{104}{160} = \frac{52}{80} = \frac{26}{40} = \frac{13}{20}\).
\[ \text{Percent Increase} = \frac{13}{20} \times 100% = 0.65 \times 100% = 65% \]
Step 4: Final Answer:
The amount of electricity used by the water heater increased by 65% from July to November.