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David, Suzy, and Mary each purchased a new car. The average (arithmeti
[#permalink]
02 Nov 2023, 00:46
02 Nov 2023, 00:46
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David, Suzy, and Mary each purchased a new car. The average (arithmetic mean) price of the three cars was $30,000. The price of Suzy’s car was $30,000.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The median price of the three cars |
$ 30,000 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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David, Suzy, and Mary each purchased a new car. The average (arithmeti
[#permalink]
02 Dec 2023, 08:18
02 Dec 2023, 08:18
1
Let D = cost of David's car, S = cost of Susan's car, M = cost of Mary's car.
Given that the mean of the three cars is 30k, we know that \(D+S+M = 3*30k = 90k\)
And we're also given that \(S = 30k\), so \(D+M=60k\)
Case 1: both D and M = 30k
All 3 are 30k, so 30k is the median.
Case 2: D < 30k, M > 30k
Then S is the median since it lies in between D and M, so 30k is still the median.
Case 3: D > 30k, M < 30k
S is again the median, so 30k is still the median.
In all 3 possible cases, the median = 30k, so C is the answer.
Given that the mean of the three cars is 30k, we know that \(D+S+M = 3*30k = 90k\)
And we're also given that \(S = 30k\), so \(D+M=60k\)
Case 1: both D and M = 30k
All 3 are 30k, so 30k is the median.
Case 2: D < 30k, M > 30k
Then S is the median since it lies in between D and M, so 30k is still the median.
Case 3: D > 30k, M < 30k
S is again the median, so 30k is still the median.
In all 3 possible cases, the median = 30k, so C is the answer.
gmatclubot
David, Suzy, and Mary each purchased a new car. The average (arithmeti [#permalink]
02 Dec 2023, 08:18
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