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A number of equal sized baseballs are stored in a box with a width of

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A number of equal sized baseballs are stored in a box with a width of  [#permalink]

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New post 21 Oct 2018, 14:06
1
4
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

60% (02:06) correct 40% (02:19) wrong based on 95 sessions

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A number of equal sized baseballs are stored in a box with a width of 18 inches and a length of 24 inches. If each ball is the same size, and they are lined up tangent to each other in the box, what is the maximum number of balls that can fit in the bottom layer of the box?

(1) The surface area of each ball is \(64\pi\) inches^2
(2) If the diameter of each ball were 4 inches less, the maximum number of balls that could fit into the box would be 4 times the number of balls that currently fit into the box.

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Re: A number of equal sized baseballs are stored in a box with a width of  [#permalink]

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New post 21 Oct 2018, 14:41
stmt 1

surface area of each ball =4 pi r^2=64pi
=> r=4 inches

Length of the box is 24 inches , so maximum 3 balls (length wise) and 2 balls (width wise)
total 3* 2 = 6 balls in bottom layer can fit

sufficient

stmt 2:
if diameter is 4 inches less, that means now diameter is 8-4 = 4 inches , we can calculate the number of balls on bottom layer that is 6 balls (length wise) and 4 balls widh wise
so total 6 *4 =24 balls in bottom layer can fit

sufficient


Ans D

As height is not given, question only asks about bottom layer .
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Re: A number of equal sized baseballs are stored in a box with a width of  [#permalink]

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New post 21 Oct 2018, 19:19
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Abhi077 wrote:
A number of equal sized baseballs are stored in a box with a width of 18 inches and a length of 24 inches. If each ball is the same size, and they are lined up tangent to each other in the box, what is the maximum number of balls that can fit in the bottom layer of the box?
1) The surface area of each ball is \(64\pi inches^2\)
2) If the diameter of each ball were 4 inches less, the maximum number of balls that could fit into the box would be 4 times the number of balls that currently fit into the box.



We are looking for the radius of each ball..

1) The surface area of each ball is \(64\pi inches^2\)
So \(4*π*r^2=64*π......r=4\)
So dia is 4*2=8 and number of the ball that will fit in is 24/8=3 along one side and 18/8=2.zyz so 2 along other side
Answer 3*2=6
Sufficient

2) If the diameter of each ball were 4 inches less, the maximum number of balls that could fit into the box would be 4 times the number of balls that currently fit into the box.
Let the dia be x, new dia is x-4...
\(\frac{18*24}{(x-4)^2}\)=\(4*\frac{18*24}{x^2}\)
So (x-4)^2*4=x^2.....\(4x^2-32x+64=x^2.......3x^2-32x+64=0.......\)
\(3x^2-24x-8x+64=0.....(3x(x-8)-8(x-8)=0.....(3x-8)(x-8)=0\)

Now X can be 8/3 or 8 but 8/3-4 will be negative so not possible...
Answer 8..
Again sufficient

D

sunita123, you are reading info of statement I in solving II by taking initial DIA as 8
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Re: A number of equal sized baseballs are stored in a box with a width of  [#permalink]

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New post 22 Oct 2018, 22:09
Thank you Chetan2u for pointing out the mistake ..

Can you please explain how did you arrive at this -

18∗24(x−4)218∗24(x−4)2= 4∗18∗24x2

Thank you !

chetan2u wrote:
Abhi077 wrote:
A number of equal sized baseballs are stored in a box with a width of 18 inches and a length of 24 inches. If each ball is the same size, and they are lined up tangent to each other in the box, what is the maximum number of balls that can fit in the bottom layer of the box?
1) The surface area of each ball is \(64\pi inches^2\)
2) If the diameter of each ball were 4 inches less, the maximum number of balls that could fit into the box would be 4 times the number of balls that currently fit into the box.



We are looking for the radius of each ball..

1) The surface area of each ball is \(64\pi inches^2\)
So \(4*π*r^2=64*π......r=4\)
So dia is 4*2=8 and number of the ball that will fit in is 24/8=3 along one side and 18/8=2.zyz so 2 along other side
Answer 3*2=6
Sufficient

2) If the diameter of each ball were 4 inches less, the maximum number of balls that could fit into the box would be 4 times the number of balls that currently fit into the box.
Let the dia be x, new dia is x-4...
\(\frac{18*24}{(x-4)^2}\)=\(4*\frac{18*24}{x^2}\)
So (x-4)^2*4=x^2.....\(4x^2-32x+64=x^2.......3x^2-32x+64=0.......\)
\(3x^2-24x-8x+64=0.....(3x(x-8)-8(x-8)=0.....(3x-8)(x-8)=0\)

Now X can be 8/3 or 8 but 8/3-4 will be negative so not possible...
Answer 8..
Again sufficient

D

sunita123, you are reading info of statement I in solving II by taking initial DIA as 8
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Re: A number of equal sized baseballs are stored in a box with a width of  [#permalink]

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New post 16 May 2019, 02:43
The source of this is Target Test Prep. Please update, mods. Bunuel
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Re: A number of equal sized baseballs are stored in a box with a width of  [#permalink]

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New post 16 May 2019, 03:03
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Re: A number of equal sized baseballs are stored in a box with a width of  [#permalink]

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New post 24 May 2019, 13:55
Bunuel VeritasKarishma
If the height of the box is 0.5 inches.The ball would not fit.
Therefore i marked "E"
Please explain
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Re: A number of equal sized baseballs are stored in a box with a width of  [#permalink]

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New post 25 May 2019, 01:44
ayusharora96 wrote:
Bunuel VeritasKarishma
If the height of the box is 0.5 inches.The ball would not fit.
Therefore i marked "E"
Please explain


It is apparent in the question that the height of the box is irrelevant. The words "bottom layer" clarify that one could stack balls on top of each other in the box.
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Re: A number of equal sized baseballs are stored in a box with a width of   [#permalink] 25 May 2019, 01:44
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