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Math Expert V
Joined: 02 Sep 2009
Posts: 59039
A school purchased 2 computers whose prices were $1,000 and$2,000, re  [#permalink]

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Difficulty:   25% (medium)

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?

A. $250 B.$500
C. $750 D.$1,000
E. $1,250 _________________ SVP  P Joined: 03 Jun 2019 Posts: 1839 Location: India Re: A school purchased 2 computers whose prices were$1,000 and $2,000, re [#permalink] ### Show Tags 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 _________________ "Success is not final; failure is not fatal: It is the courage to continue that counts." Please provide kudos if you like my post. Kudos encourage active discussions. My GMAT Resources: - Efficient Learning All you need to know about GMAT quant Tele: +91-11-40396815 Mobile : +91-9910661622 E-mail : kinshook.chaturvedi@gmail.com Board of Directors D Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4831 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 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  P Joined: 10 Apr 2018 Posts: 279 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 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  B
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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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] 14 Oct 2019, 11:46
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