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05 Feb 2019, 12:03
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
49% (02:45) correct
51%
(03:24)
wrong
based on 59
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Not Attempted Yet
The water tank shown is a prism with 3 rectangular faces and 2 equilateral triangular faces. Each triangular face has height 6, and the water surface is parallel to the face ABCD. If the volume occupied by the water is half the volume of the water tank, which of the following is closest to the length of x ?
(A) 1.73
(B) 2.83
(C) 3.46
(D) 4.24
(E) 5.20
GMATH practice exercise (Quant Class 17)
Attachment:
05-Feb19-5z.gif [ 25.11 KiB | Viewed 2614 times ]
05 Feb 2019, 15:14
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volume of the prism tank = area of side triangle * length of tank
volume of water = \(\frac{1}{2}\) volume of whole tank
(area of triangle with height x)*(length of tank) = \(\frac{1}{2}\) (area of triangle with height 6)*(length of tank)
(area of triangle with height x) = \(\frac{1}{2}\) (area of triangle with height 6)
because the water surface is parallel to the ABCD face of the tank, the two triangles are similar (AAA)
similar triangles have side ratio = \(\frac{x}{y}\) and area ratio of \(\frac{x^2}{y^2}\)
in our case, the ratio of areas = \(\frac{(area-of-triangle-with-height-x)}{(area-of-triangle-with-height-6)}\) = \(\frac{1}{2}\) = \(\frac{x^2}{y^2}\)
so the ratio of sides = \(\frac{x}{y}\) = \(\frac{1}{\sqrt{2}}\)
knowing that the bigger triangle has height 6, then \(x\) = \(\frac{6}{\sqrt{2}}\) = \(\frac{6\sqrt[]{2}}{2}\) = \(3\sqrt[]{2}\) = \(3*1.4\) \(\approx\) \(4.2\)
So D
volume of water = \(\frac{1}{2}\) volume of whole tank
(area of triangle with height x)*(length of tank) = \(\frac{1}{2}\) (area of triangle with height 6)*(length of tank)
(area of triangle with height x) = \(\frac{1}{2}\) (area of triangle with height 6)
because the water surface is parallel to the ABCD face of the tank, the two triangles are similar (AAA)
similar triangles have side ratio = \(\frac{x}{y}\) and area ratio of \(\frac{x^2}{y^2}\)
in our case, the ratio of areas = \(\frac{(area-of-triangle-with-height-x)}{(area-of-triangle-with-height-6)}\) = \(\frac{1}{2}\) = \(\frac{x^2}{y^2}\)
so the ratio of sides = \(\frac{x}{y}\) = \(\frac{1}{\sqrt{2}}\)
knowing that the bigger triangle has height 6, then \(x\) = \(\frac{6}{\sqrt{2}}\) = \(\frac{6\sqrt[]{2}}{2}\) = \(3\sqrt[]{2}\) = \(3*1.4\) \(\approx\) \(4.2\)
So D
06 Feb 2019, 06:07
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fskilnik
Very good, Mahmoudfawzy83 ! Congrats (kudos!) and thank you for your contribution!
Let me offer our "official solution": (It is only different of yours in the "wording".)
\(?\,\, \cong \,\,x\)
\(\frac{1}{2}\,\, = \,\,\frac{{{V_{{\text{water}}}}}}{{{V_{{\text{tank}}}}}}\,\,\mathop = \limits^{\left( * \right)} \,\,\frac{{{S_{\Delta EFG}} \cdot FH}}{{{S_{\Delta EAD}} \cdot DC}}\,\,\,\mathop = \limits^{FH = DC} \,\,\,\frac{{{S_{\Delta EFG}}}}{{{S_{\Delta EAD}}}}\,\,\,\mathop = \limits^{\left( {**} \right)} \,\,\,{\left( {\frac{x}{6}} \right)^2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{x}{6} = \frac{{\sqrt 2 }}{2}\)
\(\left( * \right)\,\,{\text{formula}}\,\,{\text{given}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {**} \right)\,\,{\text{similarity}}\,\,{\text{property}}\)
\(? = x\,\, = \,\,3\sqrt 2 \,\, \cong \,\,3 \cdot 1.41 = 4.23\)
The correct answer is therefore (D).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
