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# The measurements obtained for the interior dimensions of a

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Based on (a+b)^2 formula:

Calculated volume = 200 * 200 * 300 = 200 ^ 2 * 300

Actual volume = 201 * 201 * 301
= (200 + 1)^2 * (300 + 1)
= (200^2 + 1 + 400) * (300 + 1)
= 200^2 * 300 + 300 + 400 * 300 + 200^2 + 1 + 400

Actual volume - calculated volume = 300 + 120000+ 40000 + 1 + 400 = 160,701 ~ 160,000 (C)
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Hello Bunuel,
Sorry, but I dont understand the solution.
I thought,since it is given that the dimensions have at most an error of 1 cm. So maximum possible difference in volume would be: (201* 201 * 301) - (200 *200*300). Pls suggest what i am doing wrong.

Posted from my mobile device
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imhimanshu
Hello Bunuel,
Sorry, but I dont understand the solution.
I thought,since it is given that the dimensions have at most an error of 1 cm. So maximum possible difference in volume would be: (201* 201 * 301) - (200 *200*300). Pls suggest what i am doing wrong.

Posted from my mobile device

Yes, that's correct. But the way I suggested gives and approximate answer which is much easier to calculate than (201*201*301) - (200*200*300)= 160,701.
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Bunuel
Yes, that's correct. But the way I suggested gives and approximate answer which is much easier to calculate than (201*201*301) - (200*200*300)= 160,701.

Thanks Bunuel for the clarification. However, my solution assumes that the error of 1 cm would lead to increase in the dimensions of the boxes.
Can it be possible that the resultant volume would be bigger had I taken the values(200*200*300) - (199*199*299)?
I mean is there any number property that justifies that the difference would be always be bigger in the following case

If x<y<z , where x,y and z are positive numbers
then y-x< z-y .

according to this question, we can assume that :
x = (199*199*299)
y = (200*200*300)
z= (201*201*301)

Is it always treue -

(200*200*300) - (199*199*299) < (201*201*301) - (200*200*300)
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I had the same doubt initially as well, but then I considered an easier problem and the relationship became clear; you can prove it to yourself by considering the behavior of the area of squares, and then extrapolate that result to cuboids. Consider the area of squares with sides 1, 2, and 3; the areas woukd be 1, 4, 9. As you increase the side by one unit, the /absolute/ difference becomes increasingly large (note, however, that the /percentage/ increase becomes increasingly small).

Algebraically, and letting the length of the side be x, you can represent this difference as (x + 1)^2 - x^2 = 2x + 1.

In other words, the difference is 1 greater than twice the initial length (x), ie, it is a function /of the initial length/.

You can extrapolate this result to n dimensions, including three.

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Bunuel
Yes, that's correct. But the way I suggested gives and approximate answer which is much easier to calculate than (201*201*301) - (200*200*300)= 160,701.

Thanks Bunuel for the clarification. However, my solution assumes that the error of 1 cm would lead to increase in the dimensions of the boxes.
Can it be possible that the resultant volume would be bigger had I taken the values(200*200*300) - (199*199*299)?
I mean is there any number property that justifies that the difference would be always be bigger in the following case

If x<y<z , where x,y and z are positive numbers
then y-x< z-y .

according to this question, we can assume that :
x = (199*199*299)
y = (200*200*300)
z= (201*201*301)

Is it always treue -

(200*200*300) - (199*199*299) < (201*201*301) - (200*200*300)

The red part is not always true:
Consider this: x=1, y=100, and z=101.

As for the other point:
Yes, (200*200*300) - (199*199*299) < (201*201*301) - (200*200*300) but $$(200*200*300) - (199*199*299) \approx{ (201*201*301) - (200*200*300)}$$, so you can get the answer no matter which case you consider.

My solution deals with this problem conceptually, to get approximate maximum difference, which allows to avoid calculation of (200*200*300) - (199*199*299) or (201*201*301) - (200*200*300).

Hope it helps.
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Bunuel
imhimanshu
The measurements obtained for the interior dimensions of a rectangular box are 200 cm by 200 cm by 300cm. If each of the three measurements has an error of at most 1 centimeter, which of the following is the closes maximum possible difference, in cubic centimeters, between the actual capacity of the box and the capacity computed using these measurements?

A. 100,000
B. 120,000
C. 160,000
D. 200,000
E. 320,000

Changing the length by 1 cm results in change of the volume by 1*200*300 = 60,000 cubic centimeters;
Changing the width by 1 cm results in change of the volume by 200*1*300 = 60,000 cubic centimeters;
Changing the height by 1 cm results in change of the volume by 200*200*1 = 40,000 cubic centimeters.

So, approximate maximum possible difference is 60,000 + 60,000 + 40,000 = 160,000 cubic centimeters.

Great explanation Bunuel. K+1 from me.
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Consider X , Y , Z as original sides of the rectangular box .
Now X-1 , Y -1 , Z -1 or X + 1 , Y + 1 , Z + 1 are the new sides .

Difference in volume = ( X-1 ) ( Y - 1 ) ( Z - 1 ) - XYZ .
= XYZ + XY + YZ + ZX + 1 - X - Y - Z - XYZ .
= XY + YZ + ZX + 1 - ( X + Y + Z )
= 200*200 + 200 * 300 + 200 * 300 + 1 - ( 200 + 300 + 200 )
= 160001 - 700
= 159301
Hence C . or 16701 if taken otherwise
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Re: The measurements obtained for the interior dimensions of a [#permalink]
Bunuel, can you please suggest some problems of this type, if any?
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For those of you familiar with derivatives from calculus can use it to arrive at the same conclusion as Bunuel stated at the beginning.
Lets assume the following for the rectangular box,
length(l) = 200, breadth(b) = 200, height(h) = 300

Since Volume (V) = lxbxh,
taking its derivative we get the following,

dV = bxhxdl + lxhxdb + lxbxdh (here dl, db,dh refers to the difference in measurement of the length, breadth and height)

Hence, dV (which is the difference in the volume) = 200x300x1 + 200x300x1 + 200x200x1 = 160000

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Bunuel
imhimanshu
The measurements obtained for the interior dimensions of a rectangular box are 200 cm by 200 cm by 300cm. If each of the three measurements has an error of at most 1 centimeter, which of the following is the closes maximum possible difference, in cubic centimeters, between the actual capacity of the box and the capacity computed using these measurements?

A. 100,000
B. 120,000
C. 160,000
D. 200,000
E. 320,000

The options are well spread so we can approximate.

Changing the length by 1 cm results in change of the volume by 1*200*300 = 60,000 cubic centimeters;
Changing the width by 1 cm results in change of the volume by 200*1*300 = 60,000 cubic centimeters;
Changing the height by 1 cm results in change of the volume by 200*200*1 = 40,000 cubic centimeters.

So, approximate maximum possible difference is 60,000 + 60,000 + 40,000 = 160,000 cubic centimeters.

Hi Bunuel,

Thanks for this explanation. I had set an algebraic approach and I wanted to take out common factor but at some point I didn´t know how to go on.

Algebraic approach
(301) (201) (201) - (300) (200) (200)
(200 + 101) (200 + 1) (200 + 1) - (200 + 100) (200) (200)

I know that the 200 of both the computed capacity and the actual capacity can be factored out, but I don´t know how to go on.

Thank you so much,
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Re: The measurements obtained for the interior dimensions of a [#permalink]
Bunuel , Any similar questions on CUBE ?

Thanks
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Re: The measurements obtained for the interior dimensions of a [#permalink]
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Hi there,

The questions states that there is a 1 cm error to each of the measurements.

So can we assume that the error is +-1cm?

Thanks,
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imhimanshu
Hello Bunuel,
Sorry, but I dont understand the solution.
I thought,since it is given that the dimensions have at most an error of 1 cm. So maximum possible difference in volume would be: (201* 201 * 301) - (200 *200*300). Pls suggest what i am doing wrong.

Posted from my mobile device

Hello All,
I still have a doubt here. The error can be atmost 1 cm i.e +- 1cm .
Therefore to maximize the difference between the Actual and the faulty volume can be (201 x 201 x 301) - (199 x 199 x 299).
Then why in the above post have we assumed the (200 x 200 x 300), when we can further maximize the difference and get the value. Unfortunately this is not yielding the correct answer either.

Regards.
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Re: The measurements obtained for the interior dimensions of a [#permalink]
grsm
imhimanshu
Hello Bunuel,
Sorry, but I dont understand the solution.
I thought,since it is given that the dimensions have at most an error of 1 cm. So maximum possible difference in volume would be: (201* 201 * 301) - (200 *200*300). Pls suggest what i am doing wrong.

Posted from my mobile device

Hello All,
I still have a doubt here. The error can be atmost 1 cm i.e +- 1cm .
Therefore to maximize the difference between the Actual and the faulty volume can be (201 x 201 x 301) - (199 x 199 x 299).
Then why in the above post have we assumed the (200 x 200 x 300), when we can further maximize the difference and get the value. Unfortunately this is not yielding the correct answer either.

Regards.

Check this: https://gmatclub.com/forum/the-measureme ... l#p1270826
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Re: The measurements obtained for the interior dimensions of a [#permalink]
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Bunuel
imhimanshu
The measurements obtained for the interior dimensions of a rectangular box are 200 cm by 200 cm by 300cm. If each of the three measurements has an error of at most 1 centimeter, which of the following is the closes maximum possible difference, in cubic centimeters, between the actual capacity of the box and the capacity computed using these measurements?

A. 100,000
B. 120,000
C. 160,000
D. 200,000
E. 320,000

The options are well spread so we can approximate.

Changing the length by 1 cm results in change of the volume by 1*200*300 = 60,000 cubic centimeters;
Changing the width by 1 cm results in change of the volume by 200*1*300 = 60,000 cubic centimeters;
Changing the height by 1 cm results in change of the volume by 200*200*1 = 40,000 cubic centimeters.

So, approximate maximum possible difference is 60,000 + 60,000 + 40,000 = 160,000 cubic centimeters.