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14 Oct 2019, 08:29
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B
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Difficulty:
5%
(low)
Question Stats:
89% (01:15) correct
11%
(01:20)
wrong
based on 294
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A school purchased 2 computers whose prices were $1,000 and $2,000, respectively. The school is to purchase 2 more computers from a list of computers whose prices range from $1,000 to $2,000, inclusive. The greatest possible average (arithmetic mean) price of the 4 computers is how much greater than the least possible average price of the 4 computers?
A. $250
B. $500
C. $750
D. $1,000
E. $1,250
A. $250
B. $500
C. $750
D. $1,000
E. $1,250
14 Oct 2019, 08:35
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Bunuel
Given:
1. A school purchased 2 computers whose prices were $1,000 and $2,000, respectively.
2. The school is to purchase 2 more computers from a list of computers whose prices range from $1,000 to $2,000, inclusive.
Asked: The greatest possible average (arithmetic mean) price of the 4 computers is how much greater than the least possible average price of the 4 computers?
Greatest average of 4 computers = ($2000*3 + $1000)/4 = $7000/4 = $1750
Least average of 4 computers = ($2000 + 3*$1000)/4 = $5000/4 = $1250
Difference = $1750 - $1250 = $500
IMO B
14 Oct 2019, 09:29
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Bunuel
Max Average \(= \frac{1000 + 2000 + 2000 + 2000}{4} = 1750\)
Min Average \(= \frac{1000 + 2000 + 1000 + 1000}{4} = 1250\)
So, the difference is $ 500 , Answer must be (B)
