Cedar Court and Maple Court have the same number of occupied apartment

Question

Cedar Court and Maple Court have the same number of apartments. Was the average (arithmetic mean) monthly electricity usage per apartment in Cedar Court greater than the average (arithmetic mean) monthly electricity usage per apartment in Maple Court?

(1) The median monthly electricity usage per apartment in Cedar Court was greater than the median monthly electricity usage per apartment in Maple Court.
(2) The range of the monthly electricity usage per apartment in Cedar Court was the same as the range of the monthly electricity usage per apartment in Maple Court.

Bunuel · Accepted Answer

GMAT Club Official Solution:

Cedar Court and Maple Court have the same number of apartments. Was the average (arithmetic mean) monthly electricity usage per apartment in Cedar Court greater than the average (arithmetic mean) monthly electricity usage per apartment in Maple Court?

We need to determine whether the average for Cedar Court was greater than the average for Maple Court. Since both had the same number of apartments, this is equivalent to determining whether the total usage for Cedar Court was greater than the total usage for Maple Court.

(1) The median monthly electricity usage per apartment in Cedar Court was greater than the median monthly electricity usage per apartment in Maple Court.

A greater median does not guarantee a greater average. One building could have a higher median but a lower average if the data are distributed differently. Not sufficient.

(2) The range of the monthly electricity usage per apartment in Cedar Court was the same as the range of the monthly electricity usage per apartment in Maple Court.

Having the same range also does not determine the average. Two groups can have the same range but different averages. Not sufficient.

(1)+(2) Even together, a greater median and the same range still do not determine which building had the greater average.

Example 1:
Cedar Court: 4, 5, 6
Maple Court: 1, 2, 3

Here, Cedar Court has the greater median, and the two groups have the same range. The total usage for Cedar Court is  greater  than the total usage for Maple Court.

Example 2:
Cedar Court: 1, 4, 5
Maple Court: 2, 3, 6

Here, Cedar Court has the greater median, and the two groups have the same range. The total usage for Cedar Court is  less  than the total usage for Maple Court.

Not sufficient.

Answer: E.

Adit_ · Answer

Statement 1:
Clearly INSUFFICIENT.
There is no AP that is mentioned so we can choose whatever values before and after median.

Statement 2:
Clearly INSUFFICIENT.
Range tells us nothing at all regarding the mean.

Combining the two:
Key thing to note is we are given same number of apartments but we can choose how many ever we want.
Even if the range is same and median of one is higher than the other, the values in between can range between anything.
The gap between the medians or any values can easily be bridged with more number of apartments if median of one is way higher than the other.
INSUFFICIENT.

Answer:  Option E

thisisakr · Answer

Statement 1 The median being greater in Cedar Court does not tell us about the mean. A distribution can have a higher median but a lower mean depending on how the values are spread. For example, Cedar Court could have a few very low outliers that drag the mean below Maple Court's mean even though the median is higher. So the median alone cannot determine which average is greater. ——  insufficient .
Statement 2: Knowing that the range is the same in both courts tells us nothing about where the values are concentrated or what the average is. Two sets can have identical ranges but very different means ——  insufficient .

Combining both statements: Even together, knowing that Cedar Court has a higher median and the same range still does not pin down the mean. The mean depends on the entire distribution of values, not just the median and range. We can construct examples where Cedar Court's mean is higher and examples where it is lower, both consistent with the given conditions. So the two statements together are still ——  insufficient .

E

BongBideshini · Answer

(1)  Even though the median monthly electricity cost of Cedar is more than Maple's, the average of the cost can change based on whether the data sets are right-skewed or left-skewed.  Insufficient

(2)  The same goes for range. Even if the range is the same, Cedar Court can have a greater or lesser average, depending on the dataset.  Insufficient

(1)+(2)

Case 1:

Cedar: {2, 2, 3, 5, 5} Average = 3.4
Maple:{2, 2, 2, 2, 5} Average = 2.6
 Avg C> Avg M

Case 2:

Cedar: {2, 2, 3, 3, 5} Average = 3.0
Maple: {2, 2, 2, 5, 5} Average = 3.2
 Avg C< Avg M

Insufficient

Ans E

NidhiYad · Answer

Let:

Both Cedar and Maple have n apartments (same number given)
We are asked: Is mean(Cedar) > mean(Maple)?
Statement (1): Median(Cedar) > Median(Maple)

Median only tells us about the middle value, not the overall distribution.

You can easily construct cases:
Cedar has higher median but some very low values → lower mean
Or Cedar has higher median and higher mean

❌ Insufficient

Statement (2): Range(Cedar) = Range(Maple)
Range = max − min

This tells us spread, not where the values lie.

Same range, but Cedar values could be mostly lower
Or mostly higher

❌ Insufficient

Combine (1) and (2)

Now we know:

Cedar has a higher median
Both have same spread (range)
Same number of apartments

Still not enough.Why?
Because you can shift distributions while keeping:

Same range
Higher median

...but still manipulate extremes to change the mean.

Answer : E

DropletMaverick · Answer

When i read the question, i calculated using the number of apartments as the parameter. My mistake was assuming that the mean equals the median for consecutive numbers. However, the question required us to consider consumption as the parameter. I chose option A. Its about the parameter consideration more than actual calculation.

Cedar Court and Maple Court have the same number of occupied apartment

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Cedar Court and Maple Court have the same number of occupied apartment

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