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

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Math Expert

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Posts: 61524

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

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14 Oct 2019, 07:29

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Difficulty:

15% (low)

Question Stats:

93% (01:12) correct

7% (01:49) wrong

based on 43 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?

A. $250
B. $500
C. $750
D. $1,000
E. $1,250

Spoiler: OA

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

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14 Oct 2019, 07:35

Bunuel wrote:

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

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

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14 Oct 2019, 08:29

Bunuel wrote:

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]

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14 Oct 2019, 09:08

Bunuel wrote:

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]

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14 Oct 2019, 10:46

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, 10:46

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