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16 Sep 2014, 00:47
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If the average (arithmetic mean) price of \(m\) items is $125 and the average (arithmetic mean) price of another \(n\) items is $205, what is the average (arithmetic mean) price of the combined \(m+n\) items?
(1) \(m + n = 30\)
(2) \(n = 2m\)
(1) \(m + n = 30\)
(2) \(n = 2m\)
16 Sep 2014, 00:47
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Official Solution:
If the average (arithmetic mean) price of \(m\) items is $125 and the average (arithmetic mean) price of another \(n\) items is $205, what is the average (arithmetic mean) price of the combined \(m+n\) items?
To find the average price of the \(m+n\) items, we use the formula: \(\frac{\text{total price of all items}}{\text{total number of items}}\). This translates to \(\frac{125m+205n}{m+n}\).
(1) \(m + n = 30\).
Not sufficient.
(2) \(n = 2m\).
Substituting this into our formula we get \(\frac{125m+205n}{m+n}=\frac{125m+205(2m)}{m+2m}\). This simplifies to \(\frac{535m}{3m}\). With the term \(m\) eliminated, we can determine the exact numerical value for the average price. Sufficient.
Answer: B
If the average (arithmetic mean) price of \(m\) items is $125 and the average (arithmetic mean) price of another \(n\) items is $205, what is the average (arithmetic mean) price of the combined \(m+n\) items?
To find the average price of the \(m+n\) items, we use the formula: \(\frac{\text{total price of all items}}{\text{total number of items}}\). This translates to \(\frac{125m+205n}{m+n}\).
(1) \(m + n = 30\).
Not sufficient.
(2) \(n = 2m\).
Substituting this into our formula we get \(\frac{125m+205n}{m+n}=\frac{125m+205(2m)}{m+2m}\). This simplifies to \(\frac{535m}{3m}\). With the term \(m\) eliminated, we can determine the exact numerical value for the average price. Sufficient.
Answer: B
23 Oct 2023, 00:19
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I've revised the question and solution, incorporating additional details for improved clarity. I trust this makes it more comprehensible.
