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09 Jan 2018, 23:48
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Difficulty:

45% (medium)

Question Stats:

71% (01:14) correct 29% (01:59) wrong based on 34 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 Joined: 15 Aug 2017 Posts: 15 Re: If Ted bought n baseball cards that cost$4 each, then bought twice as [#permalink]

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10 Jan 2018, 01:28
3
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 Joined: 07 Dec 2017 Posts: 418 Re: If Ted bought n baseball cards that cost$4 each, then bought twice as [#permalink]

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10 Jan 2018, 03:26
1
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!

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10 Jan 2018, 08:37
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) _________________ 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 ) Intern Joined: 24 Nov 2016 Posts: 32 Re: If Ted bought n baseball cards that cost$4 each, then bought twice as [#permalink]

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10 Jan 2018, 11:19
set n = 5

5 * 4 + 10 * 7 + 3 * 6 / 18 = 108/18 = 6

PS: After I've chosen 5, reading the answers i realized that i could've pick 2. That would be much easier
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Re: If Ted bought n baseball cards that cost $4 each, then bought twice as [#permalink] ### Show Tags 16 Jan 2018, 17:45 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

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