Interior dimensions of a rectangular box

[faraz_jeddah]

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

[vinay1983]

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!

[faraz_jeddah]

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.

[vinay1983]

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?

[faraz_jeddah]

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?

[vinay1983]

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!

[theCodeToGMAT]

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..

[GMATGuruNY]

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.

[[email protected]]

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

[rakeshd347]

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]

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)

[Scott@TargetTestPrep]

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

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