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

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Bunuel
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A school purchased 2 computers whose prices were $1,000 and $2,000, re [#permalink] New post   14 Oct 2019, 08:29
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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
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B
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Re: A school purchased 2 computers whose prices were $1,000 and $2,000, re [#permalink] New post   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 [#permalink] New post   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 [#permalink] New post   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

Thus, the correct answer is option B.
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Re: A school purchased 2 computers whose prices were $1,000 and $2,000, re [#permalink] New post   14 Oct 2019, 11:46
1000+2000+2000+2000=7000
1000+2000+1000+1000=5000
(7000/4)-(5000/4)
=2000/4
=500

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Re: A school purchased 2 computers whose prices were $1,000 and $2,000, re [#permalink]
14 Oct 2019, 11:46
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