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

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Difficulty:   55% (hard)

Question Stats: 61% (02:04) correct 39% (02:20) wrong based on 92 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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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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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
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...
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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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
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...
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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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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dcummins wrote:
The source of this is Target Test Prep. Please update, mods. Bunuel

Done. Thank you. Could you please use Report a Problem button for such kind of issues? Thank you.
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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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If the height of the box is 0.5 inches.The ball would not fit.
Therefore i marked "E"
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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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ayusharora96 wrote:
If the height of the box is 0.5 inches.The ball would not fit.
Therefore i marked "E"

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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Veritas Prep GMAT Instructor 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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