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Danny is sitting on a rectangular box. The area of the front [#permalink]
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06 Nov 2012, 18:03
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Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face, and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box? A. 3 B. 6 C. 8 D. 9 E. 12 You grossly underestimated the time this question took you. You actually solved it in 10 minutes and 33 seconds.
Correct.
Applying the relationship of the surface areas of the three sides, you can get the values of w and h in terms of l:
Side face = w x h
Top face = w x l = 1.5 x w x h which implies l = 1.5 h > h = 2/3 l
Front face = l x h = 1.5/2 x w x h which implies l = 1.5/2 w > w =4/3 l
Thus, the sides of the box: l, h and w, have the relationship of 1, 2/3 and 4/3 respectively such that the sides of the box are factors of 24. Expand the ratio times 3 to get rid of the fraction: l:h:w have the ratio of 3:2:4. It just so happens that 3×2×4 = 24, so these values also satisfy the question' requirements.
Hence, the area of side face is w x h = 4 x 2 = 8.
Wow, so I understand all the concepts here but I have to say, to be able to understand everything fully, and to do all the calculations required in this question would take a LOT of org and concentration skills not to make an easy error.
Anyone have any special insight on this on how to break it down in a more easier manner?
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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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06 Nov 2012, 20:17
anon1 wrote: Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face, and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box? 3 6 8 9 12 You grossly underestimated the time this question took you. You actually solved it in 10 minutes and 33 seconds. Correct. Applying the relationship of the surface areas of the three sides, you can get the values of w and h in terms of l: Side face = w x h Top face = w x l = 1.5 x w x h which implies l = 1.5 h > h = 2/3 l Front face = l x h = 1.5/2 x w x h which implies l = 1.5/2 w > w =4/3 l Thus, the sides of the box: l, h and w, have the relationship of 1, 2/3 and 4/3 respectively such that the sides of the box are factors of 24. Expand the ratio times 3 to get rid of the fraction: l:h:w have the ratio of 3:2:4. It just so happens that 3×2×4 = 24, so these values also satisfy the question' requirements. Hence, the area of side face is w x h = 4 x 2 = 8. Wow, so I understand all the concepts here but I have to say, to be able to understand everything fully, and to do all the calculations required in this question would take a LOT of org and concentration skills not to make an easy error. Anyone have any special insight on this on how to break it down in a more easier manner? Lets say, front face is made of l ,b top face is made of l,w and side face is made of b,w question says, front face of the box is half the area of the top face, these faces will share one common side, lb =lw/2 => b=w/2 also that, area of the top face is 1.5 times the area of the side face, lw = 1.5 bw => l=3/2w We want area of side face, A=bw We know Volume =lbw or V = A*l Thus all we need to know is l. Now, Volume =24 lbw=24 l*2/3l*2*2/3l =24 l^3 = 24*9/2^3 l = 3 Also, volume = Area of side face * l =>A = Volume/l = 24/3 = 8
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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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17 Dec 2012, 05:37
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My Approach
Area of the sides Top = a Front = a/2 Side = a/1.5
Let x,y,z be the different sides then volume = xyz = 24 areas = xy, xz, yz ie xy×xz×yz = (xyz)² = (a³/3) = 24²
calculate a, a = 12 area of the side is a/1.5 = 8



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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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anon1 wrote: Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face, and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box?
A. 3 B. 6 C. 8 D. 9 E. 12
If you want to minimize the use of too many variables, you can look at it another way: Attachment:
Ques3.jpg [ 7.6 KiB  Viewed 7047 times ]
"The area of the front face of the box is half the area of the top face"  Front and top faces share the width so length must be half the depth. L = (1/2)D "the area of the top face is 1.5 times the area of the side face."  Top and side faces share the depth so width must be 1.5 times the length. W = (3/2)L = (3/4)D Volume = 24 = D*(1/2)D*(3/4)D = (3/8)D D = 4 Side face area = D*L = 4*(1/2)*4 = 8
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Danny is sitting on a rectangular box. The area of the front face of t [#permalink]
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29 Jul 2015, 10:51
Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face, and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box? A. 3 B. 6 C. 8 D. 9 E. 12 Looking forward to your solutions. To solve this in 2 minutes, naaaw, rather difficult for me
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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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29 Jul 2015, 10:55



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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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29 Jul 2015, 11:23
reto wrote: Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face, and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box? A. 3 B. 6 C. 8 D. 9 E. 12 Looking forward to your solutions. To solve this in 2 minutes, naaaw, rather difficult for me <2 minutes, only if you can draw a cuboid and then it becomes a matter of keeping a tab of your variables. Solutions above are straightforward. But to help you> Let l,b,h be the dimensions of the box ( in the grand scheme of things, it will not matter what face do you take the "side" face etc but you need to be consistent throughout!!). Given information: lh = lb/2 > b = 2h lb = 1.5bh > l = 1.5h Also, volume = lbh = 24 > h =2 , b =4 and side face area = 4*2 = 8 units.



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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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07 Oct 2015, 12:19
OE: Applying the relationship of the surface areas of the three sides, you can get the values of w and h in terms of l: Side face = l × h Top face = w × l = 1.5 × l × h which implies w = 1.5 h Front face = w × h = (w × l)/2 which implies h = l/2 > l = 2h Thus, the sides of the box: l, h and w, have the relationship of 2, 1 and 1.5 respectively such that the sides of the box are factors of 24. Expand the ratio by 2 to get rid of the decimal: the ratio of l:h:w is now 4:2:3. It just so happens that 4×2×3 = 24, so these values also satisfy the question' requirements. Hence, the area of side face is l × h = 4 × 2 = 8.
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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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22 May 2016, 09:33
anon1 wrote: Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face, and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 24, what is the area of the side face of the box?
A. 3 B. 6 C. 8 D. 9 E. 12
Lets suppose length= l, breadth= b, depth= d Front face area= l*w = 1/2 w*d (l=1/2 d or d=2l) top face area= w*d side face area= w*d = 1.5 d*l (w=1.5l) Volume = l*w*d= 24 l*1.5l*2l= 24 l=2 Side face area= l*d= l*2l= 2*2*2=8 C is the answer
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Re: Danny is sitting on a rectangular box. The area of the front [#permalink]
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