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# S98-20

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Math Expert
Joined: 02 Sep 2009
Posts: 52285

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16 Sep 2014, 00:53
00:00

Difficulty:

95% (hard)

Question Stats:

35% (02:14) correct 65% (03:35) wrong based on 75 sessions

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Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 52285

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16 Sep 2014, 00:53
1
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

The closest answer is 42%.

_________________
Intern
Joined: 07 Sep 2014
Posts: 18
Location: United States (MA)
Concentration: Finance, Economics

### Show Tags

10 Oct 2014, 15:25
Bunuel wrote:
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

The closest answer is 42%.

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?
Math Expert
Joined: 02 Sep 2009
Posts: 52285

### Show Tags

11 Oct 2014, 12:28
bsmith37 wrote:
Bunuel wrote:
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

The closest answer is 42%.

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?

The following discussion might help: approximately-what-percent-of-the-area-of-the-circle-shown-i-169744.html
_________________
Intern
Joined: 07 Sep 2014
Posts: 18
Location: United States (MA)
Concentration: Finance, Economics

### Show Tags

11 Oct 2014, 14:57
Bunuel wrote:
bsmith37 wrote:
Bunuel wrote:
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

The closest answer is 42%.

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?

The following discussion might help: approximately-what-percent-of-the-area-of-the-circle-shown-i-169744.html

that link was helpful, but now I'm just trying to tie it back in with your slightly alternative explanation above. Still, i cannot see why you say that AX = $$\frac{\sqrt{3}r}{4}$$ because we have an equilateral triangle (AOB), with sides R, and height AX. The formula states that the height AX would be $$\frac{\sqrt{3}r}{2}$$. Please advise.
Manager
Joined: 12 May 2013
Posts: 63

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23 Dec 2014, 23:42
Bunuel wrote:
Official Solution:

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24%
B. 30%
C. 36%
D. 42%
E. 48%

If polygon $$ABCDEF$$ is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to $$180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ$$, and there are 6 interior, equal angles, then each of those angles must measure $$\frac{720^\circ}{6} = 120^\circ$$.

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center $$O$$:

Consider small triangle $$AXB$$. The hypotenuse of this right triangle, $$AB$$, has length $$r$$. Angle $$ABO$$ is $$60^\circ$$, so $$AXB$$ is a 30-60-90 triangle. This means that $$XB$$ is $$\frac{r}{2}$$ in length, and $$AX$$ is $$\frac{\sqrt{3}r}{4}$$ in length. Since $$AX$$ and $$XB$$ are the same length (by symmetry), $$AC$$ is $$\sqrt{3}r$$ in length.

This means that the area of triangle $$ABC$$ is $$\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}$$. Since there are three shaded triangles in all (including $$ABC$$), the total shaded area is $$\frac{3 \sqrt{3} r^2}{4}$$. The area of the circle is , so the ratio of the shaded area to the area of the circle is given by
$$\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43$$

The closest answer is 42%.

Hey Bunuel,
I think the highlighted part above should be ax=xc.
and that is why it is creating a bit of confusion.

Intern
Joined: 17 Mar 2015
Posts: 1

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19 Feb 2016, 05:00
I think this is a high-quality question and I agree with explanation. There is a typo. AX equals (√3*r)/2, not (√3*r)/4 as it mentioned in the second line below the picture in the explanation. The rest is correct.
Manager
Joined: 31 Jan 2017
Posts: 58
Location: India
GMAT 1: 680 Q49 V34
GPA: 4
WE: Project Management (Energy and Utilities)

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17 Feb 2017, 01:18
This is a great question.

Typo errors in the solution:

AX = √3 * r/2
AX and XC are same in length
Area of triangle ABC is 1/2 * (3√r) *(r/2) = (√3 * r^2 )/4
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Intern
Joined: 20 Jun 2015
Posts: 28
Location: India
Concentration: General Management, Operations
GMAT 1: 720 Q50 V37
GPA: 2.91

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04 Jan 2018, 01:59
I think we don't have to actually calculate and waste time in exam. Rather, we can simply see that if the area of triangle is "x" the highlighted area, when wrapped inside, is also "x". Hence, the total area becomes (2*Triangle area+Remaining area). Now we see clearly that the remaining area is a smaller area as compared to the area of triangle. Hence, Area of triangle can not be less than 33% (Option A&B goes out). At the same time, the area of Triangle can not be 50% or very close to 50%, since that would mean that remaining area is hardly 2% (Option E goes out). Now, If we choose 36% that would mean that remaining area is nearly equal to area of triangle, which is clearly not the case (Option C Goes out). The only remaining and sensible choice that remains is Option D.
Re: S98-20 &nbs [#permalink] 04 Jan 2018, 01:59
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