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# A school purchased 2 computers whose prices were $1,000 and$2,000, re

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Re: A school purchased 2 computers whose prices were $1,000 and$2,000, re [#permalink]
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 Thus, the correct answer is option B. Intern Joined: 31 Oct 2018 Posts: 2 Own Kudos [?]: 1 [0] Given Kudos: 204 Re: A school purchased 2 computers whose prices were$1,000 and $2,000, re [#permalink] 1000+2000+2000+2000=7000 1000+2000+1000+1000=5000 (7000/4)-(5000/4) =2000/4 =500 Posted from my mobile device Re: A school purchased 2 computers whose prices were$1,000 and \$2,000, re [#permalink]
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