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If the height of each box is 2.8^2/[square_root]349[/square_root] mtrs

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If the height of each box is 2.8^2/[square_root]349[/square_root] mtrs  [#permalink]

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New post 01 Sep 2018, 05:08
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Question Stats:

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If the height of each box is \(\frac{2.8^2}{\sqrt{349}}\) meters, then how many more boxes could be placed on top of the pile?
A) 0
B) 1
C) 2
D) 6
E) 7

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If the height of each box is 2.8^2/[square_root]349[/square_root] mtrs  [#permalink]

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New post 01 Sep 2018, 05:21
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souvonik2k wrote:
If the height of each box is \(\frac{2.8^2}{\sqrt{349}}\) meters, then how many more boxes could be placed on top of the pile?
A) 0
B) 1
C) 2
D) 6
E) 7


Since \(2.8^2\) =7.84,i.e. approx 8 and \(\sqrt{349}\) is approx 19, so 8/19 is 0.4 approx .
And since 6 box are already there =6*0.4=2.4 meter
and one more box can be entered to total approx=2.8 ,if we add one more it goes above 3 m.
Ans:B
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Re: If the height of each box is 2.8^2/[square_root]349[/square_root] mtrs  [#permalink]

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New post Updated on: 01 Sep 2018, 06:26
souvonik2k wrote:
If the height of each box is \(\frac{2.8^2}{\sqrt{349}}\) meters, then how many more boxes could be placed on top of the pile?
A) 0
B) 1
C) 2
D) 6
E) 7
Given expression roughly turns to 1/2. Something
So total box in 3 meter will be
Total height divided by height of individual boxes that is 9 boxes

6 are already present. It can accommodate approximately 1 more.

Imho, answer is B

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Originally posted by sumit411 on 01 Sep 2018, 05:26.
Last edited by sumit411 on 01 Sep 2018, 06:26, edited 1 time in total.
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Re: If the height of each box is 2.8^2/[square_root]349[/square_root] mtrs  [#permalink]

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New post 01 Sep 2018, 05:28
souvonik2k wrote:
If the height of each box is \(\frac{2.8^2}{\sqrt{349}}\) meters, then how many more boxes could be placed on top of the pile?
A) 0
B) 1
C) 2
D) 6
E) 7
Given expression roughly turns to 1/3
So total box in 3 meter will be
Total height divided by height of individual boxes. This comes out to be 9 boxes

6 are already present. It can accommodate approximately 3 more. Note, 2 is the nearest number

Imho, answer is C

Thank you = Kudos
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Re: If the height of each box is 2.8^2/[square_root]349[/square_root] mtrs  [#permalink]

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New post 01 Sep 2018, 05:33
sumit411 wrote:
souvonik2k wrote:
If the height of each box is \(\frac{2.8^2}{\sqrt{349}}\) meters, then how many more boxes could be placed on top of the pile?
A) 0
B) 1
C) 2
D) 6
E) 7
Given expression roughly turns to 1/3
So total box in 3 meter will be
Total height divided by height of individual boxes that is 9 boxes

6 are already present. It can accommodate approximately 3 more. Note, 2 is the nearest number

Imho, answer is C

Thank you = Kudos


Hi sumit411,
I did not understand the first line of your reply.Can you tell me how you reached 1/3
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Re: If the height of each box is 2.8^2/[square_root]349[/square_root] mtrs  [#permalink]

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New post 01 Sep 2018, 06:28
AnkitOrYadav wrote:
sumit411 wrote:
souvonik2k wrote:
If the height of each box is \(\frac{2.8^2}{\sqrt{349}}\) meters, then how many more boxes could be placed on top of the pile?
A) 0
B) 1
C) 2
D) 6
E) 7
Given expression roughly turns to 1/3
So total box in 3 meter will be
Total height divided by height of individual boxes that is 9 boxes

6 are already present. It can accommodate approximately 3 more. Note, 2 is the nearest number

Imho, answer is C

Thank you = Kudos


Hi sumit411,
I did not understand the first line of your reply.Can you tell me how you reached 1/3
Hey ankit,
I did a mistake in approximating things. I was half asleep I guess

Anyway, I have edited the post

Thank you = Kudos
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Re: If the height of each box is 2.8^2/[square_root]349[/square_root] mtrs   [#permalink] 01 Sep 2018, 06:28
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