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GMATH Teacher P
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The water tank shown above is a prism, consisting of 3 rectangular fac  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 39% (02:48) correct 61% (03:27) wrong based on 38 sessions

HideShow timer Statistics 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)

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Re: The water tank shown above is a prism, consisting of 3 rectangular fac  [#permalink]

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2
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
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The water tank shown above is a prism, consisting of 3 rectangular fac  [#permalink]

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fskilnik wrote: 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)

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.
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Re: The water tank shown above is a prism, consisting of 3 rectangular fac  [#permalink]

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fskilnik wrote:

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".) $$?\,\, \cong \,\,x$$

$$\frac{1}{2}\,\, = \,\,\frac{{{V_{{\text{water}}}}}}{{{V_{{\text{tank}}}}}}\,\,\mathop = \limits^{\left( * \right)} \,\,\frac{{{S_{\Delta EFG}} \cdot FH}}{{{S_{\Delta EAD}} \cdot DC}}\,\,\,\mathop = \limits^{FH = DC} \,\,\,\frac{{{S_{\Delta EFG}}}}{{{S_{\Delta EAD}}}}\,\,\,\mathop = \limits^{\left( {**} \right)} \,\,\,{\left( {\frac{x}{6}} \right)^2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{x}{6} = \frac{{\sqrt 2 }}{2}$$

$$\left( * \right)\,\,{\text{formula}}\,\,{\text{given}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {**} \right)\,\,{\text{similarity}}\,\,{\text{property}}$$

$$? = x\,\, = \,\,3\sqrt 2 \,\, \cong \,\,3 \cdot 1.41 = 4.23$$

Could you gently explain me the ** passage called "similar property"?
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Re: The water tank shown above is a prism, consisting of 3 rectangular fac  [#permalink]

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Hi paolodeppa

The triangles we are comparing between are similar to each other.
the meaning of 'similar triangles' is that they their corresponding angles are equal.
in our case both triangles has angles of 60 degree each.
If two triangles are proved to be similar, the the ratio between any side and its corresponding side in the other triangle = $$\frac{x}{y}$$
and it can be deduced the ratio between the ares of the two triangles = $$\frac{x^2}{y^2}$$

try revising this link, it is very helpful to grasp the fundamentals:
https://gmatclub.com/forum/math-triangles-87197.html

similar triangles is an important and helpful trick to master for the GMAT. the following links can help you benefit from its applications:
https://gmatclub.com/forum/determining-the-area-of-similar-triangles-on-the-gmat-193413.html
https://gmatclub.com/forum/looking-for-similar-triangles-on-the-gmat-193036.html
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Re: The water tank shown above is a prism, consisting of 3 rectangular fac  [#permalink]

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paolodeppa wrote:
Could you gently explain me the ** passage called "similar property"?

Hi, paolodeppa !

The property: the ratio of areas of two similar polygons is equal to the square of the ratio of similarity of these figures!

In our case, we have used that for the two similar triangles related to (any) one of the triangular faces, as correctly explained by Mahmoudfawzy83 !

Another interesting question/detail: where we use the EQUILATERAL condition presented in the stem?

Answer: to give (unique) meaning to the statement "Each triangular face has height 6". In equilateral triangles, ALL heights are equal!

I hope my explanations were useful.

Regards and success in your studies (to both of you)!
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net Re: The water tank shown above is a prism, consisting of 3 rectangular fac   [#permalink] 08 Feb 2019, 06:21
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