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A pack of baseball cards consists of 12 outfielder cards and 8 infield

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A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post 23 Sep 2015, 22:44
2
4
00:00
A
B
C
D
E

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  65% (hard)

Question Stats:

59% (01:39) correct 41% (01:28) wrong based on 184 sessions

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A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 40 percent of the pack would be outfielder cards?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


Kudos for a correct solution.

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Re: A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post 24 Sep 2015, 01:35
2
Bunuel wrote:
A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 40 percent of the pack would be outfielder cards?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


Kudos for a correct solution.


Solution: Let x be the number of outfielder cards remaining of removal.
x <= 0.4(8+x) ===> 0.6x <= 3.2 ==> x <= 5.33
S0, x = 5
So no. of cards to be removed is 12-5 = 7.

Option D
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Re: A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post 24 Sep 2015, 06:46
2
1
Bunuel wrote:
A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 40 percent of the pack would be outfielder cards?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


Kudos for a correct solution.


Let x be the number of outfirlder cards that must be removed.
Then,
\(\frac{12-x}{20-x}*100\leq{40}\)

\(\frac{12-x}{20-x}*5\leq{2}\)

\(60-5x\leq{40-2x}\)

\(20\leq{3x}\)

\(\frac{20}{3}\leq{x}\)

\(6.66\leq{x}\)

So, x should at least be 7

Answer:- D
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Re: A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post 08 Mar 2016, 01:52
Will try substituting.

Try choice B) 5
(12-5)/(8+12-5)=7/15 ~7/14 -> so just over 50% NOT GODD

Try choice C) 7
(12-7)/(8+12-7)=5/13=0.38 something or 38% NO MORE than 40 > C is the answer!
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A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post Updated on: 12 Jul 2017, 18:51
1
Bunuel wrote:
A pack of baseball cards consists of 12 outfielder cards and 8 infielder cards. What is the lowest number of outfielder cards that would have to be removed from the pack so that no more than 40 percent of the pack would be outfielder cards?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8


Kudos for a correct solution.




Let x no. of O cards to be removed .
so no. of O cards in pack now =12-x and it should represent 40 % of toal no. of cards in pack (which is now 20-x)
12-x= .4(20-x)
x=6.67
so x must be 7
Ans D

Originally posted by rohit8865 on 10 Apr 2016, 02:11.
Last edited by rohit8865 on 12 Jul 2017, 18:51, edited 1 time in total.
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Re: A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post 12 Jul 2017, 12:49
Let X be the players in 40% deck
Hence
40% (8+x) =>x
x<=5.33

Therefore x = 5

Answer is D
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Re: A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post 14 Jul 2017, 08:44
Since the number of infielder cards remain the same , the number of outfielder cards removed = x
=> Total cards in the pack after removal = 20-x
=>least number of outfielder cards removed means maximum % possible in the pack of outfileder card = 40%
=> Infielder cards = 60% => 0.6 * (20-x) = 8
=> x = 6.66 ..rounding up to 7
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A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post 14 Jul 2017, 10:11
Let's try with (C) since it's the mid-value.

\(\frac{12-6}{20-6} = \frac{3}{7} = 42\% > 40\%.\)

(D): \(\frac{12-7}{20-7} = \frac{5}{13} = 38\% < 40\%.\) Ans - D.
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Re: A pack of baseball cards consists of 12 outfielder cards and 8 infield  [#permalink]

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New post 29 Sep 2018, 04:05
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Re: A pack of baseball cards consists of 12 outfielder cards and 8 infield &nbs [#permalink] 29 Sep 2018, 04:05
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