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

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Math Expert
Joined: 02 Sep 2009
Posts: 60496
A school purchased 2 computers whose prices were $1,000 and$2,000, re  [#permalink]

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14 Oct 2019, 08:29
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Difficulty:

15% (low)

Question Stats:

91% (01:09) correct 9% (01:33) 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 _________________ SVP Joined: 03 Jun 2019 Posts: 1938 Location: India GMAT 1: 690 Q50 V34 Re: A school purchased 2 computers whose prices were$1,000 and $2,000, re [#permalink] ### Show Tags 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 Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4859 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: A school purchased 2 computers whose prices were$1,000 and $2,000, re [#permalink] ### Show Tags 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) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Senior Manager Joined: 10 Apr 2018 Posts: 309 Location: India Concentration: General Management, Operations GMAT 1: 680 Q48 V34 GPA: 3.3 Re: A school purchased 2 computers whose prices were$1,000 and $2,000, re [#permalink] ### Show Tags 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.
Intern
Joined: 31 Oct 2018
Posts: 1
Re: A school purchased 2 computers whose prices were $1,000 and$2,000, re  [#permalink]

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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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