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
Interior dimensions of a rectangular box
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- faraz_jeddah
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A good question also deserves a Thanks.
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- vinay1983
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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!
So the maximum possible difference is (301*301*201)-(199*199*299)
12160701-11840699 = 320002
Option E
I hope I am right!
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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Good try Vinay. But the answer is incorrect.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!
The actual volume = 200x200x300
The calculation should be = (200x200x300) - (199x199x299)
OR
= (201x201x301) - (200x200x300)
Also, your approach is not feasible in the real gmat.
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.
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
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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?
Option C is correct then!
But my answer would have been correct if there was no condition as such.Right?
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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Which condition are you talking about?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?
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.
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
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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!faraz_jeddah wrote:Which condition are you talking about?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?
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?
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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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..
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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Let L = 200, W = 200, and H = 300.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
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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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."
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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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Can someone please explain me the red part above. I thought we have to do this. 200*200*300-201*201*301=answer.GMATGuruNY wrote:Let L = 200, W = 200, and H = 300.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
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.
Correct me if I am wrong. But I can't get my head around how did mitch calculated the red part.
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)
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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Solution:faraz_jeddah wrote: ↑Sun Sep 22, 2013 12:32 amThe 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
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
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