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Difficulty:   35% (medium)

Question Stats: 72% (02:02) correct 28% (02:50) wrong based on 37 sessions

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If Ted bought n baseball cards that cost $4 each, then bought twice as many baseball cards at$7 each, and n-2 cards at $6 each, then the average (arithmetic mean) cost, in dollars per baseball card, is equal to A. 6 B. (6n - 3)/n C. (6n - 6)/(2n - 1) D. 12n/(2n - 1) E. 17/3 _________________ Intern  B Joined: 15 Aug 2017 Posts: 12 Re: If Ted bought n baseball cards that cost$4 each, then bought twice as  [#permalink]

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n no of cards each $4 cost=4n 2n no of cards each$7 cost=7X2n=14n
n-2 no of cards each$6 cost=6X(n-2)=6n-12 Mean cost= (4n+14n+6n-12)/(n+2n+n-2) =6 So A examPAL Representative P Joined: 07 Dec 2017 Posts: 1153 Re: If Ted bought n baseball cards that cost$4 each, then bought twice as  [#permalink]

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Bunuel wrote:
If Ted bought n baseball cards that cost $4 each, then bought twice as many baseball cards at$7 each, and n-2 cards at $6 each, then the average (arithmetic mean) cost, in dollars per baseball card, is equal to A. 6 B. (6n - 3)/n C. (6n - 6)/(2n - 1) D. 12n/(2n - 1) E. 17/3 Since we have variables in our question and answer, we'll pick easy numbers to work with. This is an Alternative approach. Let's say that n = 2. Then Ted bought 2 cards at$4 and 4 cards at $7 giving an average of 8+28/6 =$6 per card.
(A) could be correct, let's check the others.
(B) is 33/6 which is not 6. No!
(C) is 30/11 which is not 6. No!
(D) is 72/11 which is not 6. No!
(E) is 17/3 which is not 6. No!

Therefore (A) is our answer.
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Re: If Ted bought n baseball cards that cost $4 each, then bought twice as [#permalink] Show Tags Bunuel wrote: If Ted bought n baseball cards that cost$4 each, then bought twice as many baseball cards at $7 each, and n-2 cards at$6 each, then the average (arithmetic mean) cost, in dollars per baseball card, is equal to

A. 6
B. (6n - 3)/n
C. (6n - 6)/(2n - 1)
D. 12n/(2n - 1)
E. 17/3

$$\frac{4n + 14n +6n - 12}{(n + 2n + n - 2)}$$

= $$\frac{24n - 12}{4n - 2}$$

= $$6$$ , Answer must be (A)
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Bunuel wrote:
If Ted bought n baseball cards that cost $4 each, then bought twice as many baseball cards at$7 each, and n-2 cards at $6 each, then the average (arithmetic mean) cost, in dollars per baseball card, is equal to A. 6 B. (6n - 3)/n C. (6n - 6)/(2n - 1) D. 12n/(2n - 1) E. 17/3 We are given that the number of 4-dollar cards = n, the number of 7-dollar cards is 2n, and the number of 6-dollar cards = n - 2. Using the formula: average = sum/number we have: [4n + 7(2n) + 6(n-2)]/(n + 2n + n-2) = average (4n + 14n + 6n - 12)/(4n - 2) = average (24n - 12)/(4n - 2) = average 6(4n - 2)/(4n - 2) = 6 Answer: A _________________ Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: If Ted bought n baseball cards that cost$4 each, then bought twice as   [#permalink] 16 Jan 2018, 17:45
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