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## Interior dimensions of a rectangular box

This topic has 2 expert replies and 8 member replies
faraz_jeddah Master | Next Rank: 500 Posts
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#### Interior dimensions of a rectangular box

Sun Sep 22, 2013 12:32 am
The measurements obtained for the interior dimensions of a rectangular box are 200 cm by 200cm 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

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Sun Sep 22, 2013 4:54 am
faraz_jeddah wrote:
The measurements obtained for the interior dimensions of a rectangular box are 200 cm by 200cm 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
Let L = 200, W = 200, and H = 300.
When each dimension changes by 1cm, the result is the following:

The LENGTH changes by 1cm, implying that the product of the OTHER TWO DIMENSIONS -- W*H -- will change by 1cm:
1 * 200 * 300 = 60000.

The WIDTH changes by 1cm, implying that the product of the OTHER TWO DIMENSIONS -- L*H -- will change by 1cm:
1 * 200 * 300 = 60000.

The HEIGHT changes by 1cm, implying that the product of the OTHER TWO DIMENSIONS -- L*W -- will change by 1cm:
1 * 200 * 200 = 40000.

Note:
Because each dimension is included in 2 of the 3 products above, there is some OVERLAP among the 3 changes in volume.
Thus:
Approximate total change in volume = 60000 + 60000 + 40000 = 160000.

The correct answer is C.

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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Sun Sep 22, 2013 12:11 pm
Hi All,

Mitch's approach is the most efficient way to solve this problem (without doing lots of calculations). The process of "estimating" is the key here. Look to take advantage of this option whenever:

1) The answer choices are 'spread out'
2) The word "approximation" (or similar) is used in the question

In this question, the phrase "closest maximum....difference" is a wordy way of say "approximate."

GMAT assassins aren't born, they're made,
Rich

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vinay1983 Legendary Member
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Sun Sep 22, 2013 2:48 am
Let us take the maximum measurements to be 301*201*201 (considering 1 cm error), and the lowest possible measurement to 199*199*299(1 cm error)

So the maximum possible difference is (301*301*201)-(199*199*299)

12160701-11840699 = 320002

Option E

I hope I am right!

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faraz_jeddah Master | Next Rank: 500 Posts
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Sun Sep 22, 2013 3:18 am
vinay1983 wrote:
Let us take the maximum measurements to be 301*201*201 (considering 1 cm error), and the lowest possible measurement to 199*199*299(1 cm error)

So the maximum possible difference is (301*301*201)-(199*199*299)

12160701-11840699 = 320002

Option E

I hope I am right!
Good try Vinay. But the answer is incorrect.

The actual volume = 200x200x300

The calculation should be = (200x200x300) - (199x199x299)
OR
= (201x201x301) - (200x200x300)

Also, your approach is not feasible in the real gmat.

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vinay1983 Legendary Member
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Sun Sep 22, 2013 3:24 am
Yes Yes Yes got it, Actual volume is 200*200*300 the other configuration can change as you have mentioned. My bad!I hope I don't do this silly mistake on the actual exam.Feeling sad for this.

Option C is correct then!

But my answer would have been correct if there was no condition as such.Right?

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faraz_jeddah Master | Next Rank: 500 Posts
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Sun Sep 22, 2013 3:27 am
vinay1983 wrote:
Yes Yes Yes got it, Actual volume is 200*200*300 the other configuration can change as you have mentioned. My bad!I hope I don't do this silly mistake on the actual exam.Feeling sad for this.

Option C is correct then!

But my answer would have been correct if there was no condition as such.Right?
Which condition are you talking about?

Regarding your approach either You used a calculator or you have a super computer for a brain.

How would you do it in the test?

_________________
A good question also deserves a Thanks.

Messenger Boy: The Thesselonian you're fighting... he's the biggest man i've ever seen. I wouldn't want to fight him.
Achilles: That's why no-one will remember your name.

vinay1983 Legendary Member
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Sun Sep 22, 2013 3:49 am
faraz_jeddah wrote:
vinay1983 wrote:
Yes Yes Yes got it, Actual volume is 200*200*300 the other configuration can change as you have mentioned. My bad!I hope I don't do this silly mistake on the actual exam.Feeling sad for this.

Option C is correct then!

But my answer would have been correct if there was no condition as such.Right?
Which condition are you talking about?

Regarding your approach either You used a calculator or you have a super computer for a brain.

How would you do it in the test?
I would use brute force, though right now I am thinking of an alternate way to do this.I mentioned those figures to get an idea(so used calci to illustrate). But you are right, can't use a calci in actual exam. Thinking!

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theCodeToGMAT Legendary Member
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Sun Sep 22, 2013 4:18 am
Observed Observations : 200, 300, 200
Original Values could have been: 201,201,301 or 199,199,299

Difference in Volume= 201*201*301-200*200*300 or 200*200*300 - 199*199*299
Calculation for 201*201*301-200*200*300 is comparatively simpler = 12160701-12000000 = 160701

So, [C]

Not sure of the best solution.... but even this calculation dint took more than 1 minute..

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rakeshd347 Master | Next Rank: 500 Posts
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Wed Oct 16, 2013 2:36 am
GMATGuruNY wrote:
faraz_jeddah wrote:
The measurements obtained for the interior dimensions of a rectangular box are 200 cm by 200cm 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
Let L = 200, W = 200, and H = 300.
When each dimension changes by 1cm, the result is the following:

The LENGTH changes by 1cm, implying that the product of the OTHER TWO DIMENSIONS -- W*H -- will change by 1cm:
1 * 200 * 300 = 60000.

The WIDTH changes by 1cm, implying that the product of the OTHER TWO DIMENSIONS -- L*H -- will change by 1cm:
1 * 200 * 300 = 60000.

The HEIGHT changes by 1cm, implying that the product of the OTHER TWO DIMENSIONS -- L*W -- will change by 1cm:
1 * 200 * 200 = 40000.

Note:
Because each dimension is included in 2 of the 3 products above, there is some OVERLAP among the 3 changes in volume.
Thus:
Approximate total change in volume = 60000 + 60000 + 40000 = 160000.

The correct answer is C.
Can someone please explain me the red part above. I thought we have to do this. 200*200*300-201*201*301=answer.

Correct me if I am wrong. But I can't get my head around how did mitch calculated the red part.

gocoder Master | Next Rank: 500 Posts
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Fri Sep 22, 2017 4:19 am
I have used algebraic approach to arrive at the same result.
Let x=200,y=300, and z=200 be the sides of the rectangular box.
The difference in volume = (x+1)(y+1)(z+1)-xyz
(x+1)(y+1)(z+1)-xyz
x(y+1)(z+1)+1(y+1)(z+1)-xyz
x(yz+y+z+1)+(yz+y+z+1)-xyz
xyz+xy+xz+x+yz+y+z+1-xyz

Finding xy+xz+yz+x+y+z-1 will give the answer
we aren't looking for the exact value, only the one that's approx.=xy+xz+yz

200*300+200*200+200*300=160,000
10000(6+4+6)=16*10000=160,000

Hence choice C is the answer

ii)if we had to find (x+2)(y+2)(z+2)-xyz
x(y+2)(z+2)+2(y+2)(z+2)-xyz
x(yz+4+2(y+z))+2(yz+4+2(y+z))-xyz
xyz+4x+2xy+2xz+2yz+8+4y+4z-xyz
2(xy+xz+yz)+4(x+y+z)+8

Likewise, for (x+3)(y+3)(z+3)-xyz, we get the difference as 3(xy+xz+yz)+9(x+y+z)+27 and the approx difference will be 3(xy+xz+yz)

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