The measurements obtained for the interior dimensions of a

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Hope it helps.

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. :-D :-D

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

Bunuel

i took this measurement as 21 21 31 and 20 20 30 and i computed to get 1671 ,is this approximation okay??

Here it worked because again the options are well spread but your way has one flaw: 1 centimetre for 20 is 5% while 1 centimetre for 200 is 0.5%, so for other similar questions it could give much more skewed result.

Solution:

The maximum possible difference, in cubic centimeters, between the actual capacity of the box and the capacity computed using these measurements is:

201 x 201 x 301 - 200 x 200 x 300

However, this can be approximated by the total volume of the 6 sides of the box with each side having a thickness of 0.5 cm. Therefore, the approximation is:

2(200 x 300 x 0.5) + 2(200 x 300 x 0.5) + 2(200 x 200 x 0.5)

60,000 + 60,000 + 40,000

160,000

Answer: C

Video solution from Quant Reasoning starts at 5:20
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I have a doubt:
Why it's +1, why it can't be 200-1, 200-1 & 300-1 as the question says the error is almost 1 cm? if we assume 199 199 & 299, the answer will be around 240,000, which is an option (wrong) on the Official GMAT Mock?

Iotaa

200*200*300 = 12,000,000
199*199*200 = 11,840,699
Difference = 159,301

I like the percent difference approach.
1 is 0.5% of 200 and 0.333% of 300.
We are increasing (or decreasing!) two dimensions by 0.5% and one dimension by 0.333%. The only reason the difference between increasing and decreasing would be large is if the compounding effects of these changes (that is, where two or more of them overlap) is large compared to the overall. We are talking about very small overlaps compared to the overall, so lets ignore that compounding effect.
0.5% + 0.5% + 0.333% = 1.333%
1% of 12,000,000 = 120,000
0.333% of 12,000,000 = 40,000
1.333% of 12,000,000 = 160,000
Answer choice C.