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If Ted bought n baseball cards that cost $4 each, then bought twice as

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If Ted bought n baseball cards that cost $4 each, then bought twice as  [#permalink]

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New post 09 Jan 2018, 23:48
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Question Stats:

71% (01:16) correct 29% (01:59) wrong based on 35 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

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Re: If Ted bought n baseball cards that cost $4 each, then bought twice as  [#permalink]

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New post 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
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Re: If Ted bought n baseball cards that cost $4 each, then bought twice as  [#permalink]

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New post 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!

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]

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New post 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)
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Re: If Ted bought n baseball cards that cost $4 each, then bought twice as  [#permalink]

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New post 10 Jan 2018, 11:19
set n = 5

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

Answer = A

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]

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New post 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

Answer: A
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Re: If Ted bought n baseball cards that cost $4 each, then bought twice as &nbs [#permalink] 16 Jan 2018, 17:45
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