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Difficulty:   25% (medium)

Question Stats: 86% (01:10) correct 14% (01:30) wrong based on 20 sessions

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If Ted brought n baseball cards costs $3 each, then brought twice as many cards baseball cards at$9 each, then the average (arithmetic mean) cost, in dollars per baseball card, is equal to

A. 4n
B. 9
C. 7n
D. 6/n
E. 7

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Intern  B
Joined: 28 Dec 2017
Posts: 30

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Bunuel wrote:
If Ted brought n baseball cards costs $3 each, then brought twice as many cards baseball cards at$9 each, then the average (arithmetic mean) cost, in dollars per baseball card, is equal to

A. 4n
B. 9
C. 7n
D. 6/n
E. 7

Average cost in dollars = $$\frac{TotalCost}{TotalQty}$$

Total Cost = (Qty$$_1$$)(Cost$$_1$$) + (Qty$$_2$$)(Cost$$_2$$)

(Qty$$_1$$)(Cost$$_1$$): $$n$$ baseball cards that cost $3 each = n($3)
(Qty$$_2$$)(Cost$$_2$$): twice as many cards (as $$n$$) that cost $9 each = 2n($9)

Total Cost in dollars: n(3) + 2n(9)
Total Qty: (n + 2n) = 3n

Average cost in dollars =
$$\frac{n(3) + 2n(9)}{3n}=\frac{3n+18n}{3n}=\frac{21n}{3n}=7$$

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Manager  G
Joined: 22 May 2015
Posts: 126

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Bunuel wrote:
If Ted brought n baseball cards costs $3 each, then brought twice as many cards baseball cards at$9 each, then the average (arithmetic mean) cost, in dollars per baseball card, is equal to

A. 4n
B. 9
C. 7n
D. 6/n
E. 7

Avr cost = (Total cost)/(Number of items)=(3*n+9*2n)/(n+2n)=21n/3n=7
Therefore, E is correct answer Re: If Ted brought n baseball cards costs $3 each, then brought twice as [#permalink] 03 Jan 2018, 19:52 Display posts from previous: Sort by If Ted brought n baseball cards costs$3 each, then brought twice as

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