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# Three cars were purchased. Is the least cost greater than

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Three cars were purchased. Is the least cost greater than [#permalink]

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05 May 2006, 22:52
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Three cars were purchased. Is the least cost greater than $12000? 1) The average price of these cars were$16,000
2) The median price is $2000 more than average price Senior Manager Joined: 29 Jun 2005 Posts: 403 Followers: 2 Kudos [?]: 19 [0], given: 0 [#permalink] ### Show Tags 05 May 2006, 23:44 Three cars were purchased. Is the least cost greater than$12000?

1) The average price of these cars were $16,000 2) The median price is$2000 more than average price

It is C
St. A is insuff. Consider: 1000, 14000 and 33000; 16000, 16000, 16000
St. B is insuff. this st. gives us nothing about the prices
Combining both st also insuff.:
Suppose X<=Y<=Z
(X+Y+Z)/3=16000
X+Y+Z=48000
Y=16000+2000=18000
X+Z=30000
The least value of Z is also 18000(Y<=Z), X=30000-18000=12000
Thus X will not be greater than 12000

Last edited by Dilshod on 06 May 2006, 00:52, edited 1 time in total.
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05 May 2006, 23:58
Dilshod wrote:
X+Z=30000
The least value of Z is also 18000(Y<=Z), X=30000-18000=12000
Thus X may be greater than or equal to 12000.

Thus Max value of X = (X+Z) - min value of Z
Min Value of Z is 18000

So Shudnt Max value of X be = 30000 - 18000 = 12000?

Thus X is never greater than 12000

Hence, C?
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06 May 2006, 00:50
sm176811 wrote:
Dilshod wrote:
X+Z=30000
The least value of Z is also 18000(Y<=Z), X=30000-18000=12000
Thus X may be greater than or equal to 12000.

Thus Max value of X = (X+Z) - min value of Z
Min Value of Z is 18000

So Shudnt Max value of X be = 30000 - 18000 = 12000?

Thus X is never greater than 12000

Hence, C?

Hi,
Yes, my mistake. thanks for correcting me. I mistook X for Z
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06 May 2006, 18:03
Agree with C
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