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A school purchased 2 computers whose prices were $1,000 and $2,000, re
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14 Oct 2019, 08:29
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Question Stats:
91% (01:08) correct 9% (01:00) wrong based on 33 sessions
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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?
Re: A school purchased 2 computers whose prices were $1,000 and $2,000, re
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14 Oct 2019, 08:35
Bunuel wrote:
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
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
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Re: A school purchased 2 computers whose prices were $1,000 and $2,000, re
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14 Oct 2019, 09:29
Bunuel wrote:
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
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)
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Re: A school purchased 2 computers whose prices were $1,000 and $2,000, re
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14 Oct 2019, 10:08
Bunuel wrote:
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
Maximum average=(1000+2000+2000+2000)/4=1750 Minimum average =(1000+2000+1000+1000)/4=1250 So, the difference in the two averages is=1750-1250=500
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91% (01:08) correct 
