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A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]
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17 Oct 2012, 19:21
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A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper? (1 yard = 3 Ft) A. 8 B. 9 C. 18 D. 27 E. 36
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]
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17 Oct 2012, 20:59
aeros232 wrote: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper? (1 yard = 3 Ft) A. 8 B. 9 C. 18 D. 27 E. 36 Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet. Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)). So, it must be true that \(6a^2\leq{63}\). If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube. If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet. Answer: D. If we won't limit the value of a side to integers only, then: \(6a^2={63}\) > \(a=\sqrt{10.5}\) > \(volume=a^3=(\sqrt{10.5})^3\). Hope it's clear.
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]
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17 Oct 2012, 21:11
Bunuel wrote: aeros232 wrote: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper? (1 yard = 3 Ft) A. 8 B. 9 C. 18 D. 27 E. 36 Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet. Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)). So, it must be true that \(6a^2\leq{63}\). If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube. If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet. Answer: D. If we won't limit the value of a side to integers only, then: \(6a^2={63}\) > \(a=\sqrt{10.5}\) > \(volume=a^3=(\sqrt{10.5})^3\). Hope it's clear. Great, thanks Bunuel
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]
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19 Oct 2012, 09:42
Bunuel wrote: aeros232 wrote: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper? (1 yard = 3 Ft) A. 8 B. 9 C. 18 D. 27 E. 36 Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet. Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)). So, it must be true that \(6a^2\leq{63}\). If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube. If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet. Answer: D. If we won't limit the value of a side to integers only, then: \(6a^2={63}\) > \(a=\sqrt{10.5}\) > \(volume=a^3=(\sqrt{10.5})^3\). Hope it's clear. Bunnel Can you please let me know Why you didn't consider a=2? then the volume would be 8
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]
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19 Oct 2012, 09:45
mydreammba wrote: Bunuel wrote: aeros232 wrote: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper? (1 yard = 3 Ft) A. 8 B. 9 C. 18 D. 27 E. 36 Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet. Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)). So, it must be true that \(6a^2\leq{63}\). If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube. If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet. Answer: D. If we won't limit the value of a side to integers only, then: \(6a^2={63}\) > \(a=\sqrt{10.5}\) > \(volume=a^3=(\sqrt{10.5})^3\). Hope it's clear. Bunnel Can you please let me know Why you didn't consider a=2? then the volume would be 8 We need the largest cube that could be packed using this paper. Now, the larger the side of a cube larger the volume and since a can be 3, no need to consider smaller lengths. Hope it's clear.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]
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19 Oct 2012, 09:53
Thanks Bunnel i din't read the Q properly and can you please respond the private message on statistics which i have sent you yesterday? and for the post the link is here clarificationonstatisticssd140907.html
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]
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19 Oct 2012, 09:54
mydreammba wrote: Bunuel wrote: aeros232 wrote: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume in Ft3 of the largest cube that could be packed using this paper? (1 yard = 3 Ft) A. 8 B. 9 C. 18 D. 27 E. 36 Note that it should be mentioned that the side of the largest cube must be an integer (or dimensions should be 1 by 6, not 1 by 7).The wrapping paper has dimensions \(1*3=3\) feet by \(7*3=21\) feet. Thus, its area is \(3*21=63\) square feet. Now, the surface area of a cube is \(6a^2\), where \(a\) is the length of a side (a cube has 6 faces and the area of each face is \(a^2\)). So, it must be true that \(6a^2\leq{63}\). If \(a=4\), then \(6a^2=96\), which means that we won't have enough paper for such cube. If \(a=3\), then \(6a^2=54\). The volume of such cube is \(a^3=27\) cubic feet. Answer: D. If we won't limit the value of a side to integers only, then: \(6a^2={63}\) > \(a=\sqrt{10.5}\) > \(volume=a^3=(\sqrt{10.5})^3\). Hope it's clear. Thanks Bunnel i din't read the Q properly and can you please respond the private message on statistics which i have sent you yesterday? and for the post the link is here clarificationonstatisticssd140907.html
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume [#permalink]
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Re: A wrapping paper has dimensions 1 yard by 7 yards. What is the volume
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