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)

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.

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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)
Please clarify.

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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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.

Great explanation Bunuel. K+1 from me.

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

Bunuel, can you please suggest some problems of this type, if any?

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

Answer : C

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.

Would you please help here? I know that this can be helpful for other similar questions.

Thank you so much,

Bunuel , Any similar questions on CUBE ?

Thanks

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?

So the answer is 201*201*301-199*199*299?

Thanks,

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.
Thus, Can someone please help out.

Regards.

Check this: https://gmatclub.com/forum/the-measureme ... l#p1270826

is not it kind of overlapping? each dimension is included in 2 of the 3 products above.

Can you clarify please?