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28 Apr 2025, 09:33
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On a shipment, the weights (in kilograms) of a group of boxes represented a set of consecutive even integers. If the average weight was 42 kilograms, what was the lightest box?
(1) The standard deviation of the box weights was 4 kilograms.
(2) The sum of the weights of the lightest and heaviest boxes was 84 kilograms.
(1) The standard deviation of the box weights was 4 kilograms.
(2) The sum of the weights of the lightest and heaviest boxes was 84 kilograms.
28 Apr 2025, 09:33
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Official Solution:
On a shipment, the weights (in kilograms) of a group of boxes represented a set of consecutive even integers. If the average weight was 42 kilograms, what was the lightest box?
First of all, in evenly spaced sets, such as a set of consecutive even integers, the average always equals the median. Next, the fact that the median of a set of consecutive even integers is 42, an even integer, indicates that the number of weights (boxes) is odd, with the median being the middle weight. Thus, there would be an equal number of weights below and above the median (the middle term). So we can have the following cases for the distribution:
• {42} - if there were one box;
• {40, 42, 44} - if there were three boxes;
• {38, 40, 42, 44, 46} - if there were five boxes;
and so on.
(1) The standard deviation of the box weights was 4 kilograms.
The standard deviation of a set shows how much variation there is from the mean, how widespread a given set is. So, a low standard deviation indicates that the data points tend to be very close to the mean, whereas a high standard deviation indicates that the data are spread out over a larger range of values. Notice that as we increase the number of weights (boxes), the set becomes more widespread, and the standard deviation increases (with one box, {42}, it is 0; with three boxes, {40, 42, 44}, it is approximately 1.6; with five boxes, {38, 40, 42, 44, 46}, it is approximately 2.8; and so on). Therefore, there would be only one possible distribution where the standard deviation is exactly 4 (just for reference, it would be {36, 38, 40, 42, 44, 46, 48}). Thus, we can determine the full set and the weight of the lightest box. Sufficient.
(2) The sum of the weights of the lightest and heaviest boxes was 84 kilograms.
Since the average (median) is 42, the sum of the weights of the lightest and heaviest boxes would be 84 kilograms for any odd number of boxes (check the examples above). Thus, this statement alone is not sufficient.
Answer: A
On a shipment, the weights (in kilograms) of a group of boxes represented a set of consecutive even integers. If the average weight was 42 kilograms, what was the lightest box?
First of all, in evenly spaced sets, such as a set of consecutive even integers, the average always equals the median. Next, the fact that the median of a set of consecutive even integers is 42, an even integer, indicates that the number of weights (boxes) is odd, with the median being the middle weight. Thus, there would be an equal number of weights below and above the median (the middle term). So we can have the following cases for the distribution:
• {42} - if there were one box;
• {40, 42, 44} - if there were three boxes;
• {38, 40, 42, 44, 46} - if there were five boxes;
and so on.
(1) The standard deviation of the box weights was 4 kilograms.
The standard deviation of a set shows how much variation there is from the mean, how widespread a given set is. So, a low standard deviation indicates that the data points tend to be very close to the mean, whereas a high standard deviation indicates that the data are spread out over a larger range of values. Notice that as we increase the number of weights (boxes), the set becomes more widespread, and the standard deviation increases (with one box, {42}, it is 0; with three boxes, {40, 42, 44}, it is approximately 1.6; with five boxes, {38, 40, 42, 44, 46}, it is approximately 2.8; and so on). Therefore, there would be only one possible distribution where the standard deviation is exactly 4 (just for reference, it would be {36, 38, 40, 42, 44, 46, 48}). Thus, we can determine the full set and the weight of the lightest box. Sufficient.
(2) The sum of the weights of the lightest and heaviest boxes was 84 kilograms.
Since the average (median) is 42, the sum of the weights of the lightest and heaviest boxes would be 84 kilograms for any odd number of boxes (check the examples above). Thus, this statement alone is not sufficient.
Answer: A
20 May 2025, 16:07
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I like the solution - it’s helpful. Really good question
