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A circle has 10 points on its circumference.
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04 Aug 2023, 08:26
04 Aug 2023, 08:26
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63% (00:49) correct
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A circle has 10 points on its circumference.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Quantity A |
Quantity B |
Ratio of number of Quadrilaterals to the Number of Hexagons that can formed using these 10 points |
1 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE Math Essentials - A most comprehensive handout!!
\(\Longrightarrow\) ALL the GRE Quant and Verbal Practice Directories in ONE place (2023)
\(\Longrightarrow\) Best GRE Test Books of 2023
Re: A circle has 10 points on its circumference.
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08 Aug 2023, 02:28
08 Aug 2023, 02:28
Expert Reply
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OE
A circle has 10 points (A to J) on its circumference. We need to find the ratio of total
number of quadrilaterals to total number of hexagons that can be formed.
Quadrilaterals: Has four sides. Therefore, we need 4 points to form it. Total points are
10. Since order in which points are selected does not matter.
Since if any four points are selected only one quadrilateral can be formed
For example, if we consider 4 points say A, C, E and H then only one quadrilateral ACEH
can be formed.
Hence, the total number of quadrilaterals formed would be equal to the number of
groups of 4 points that can be selected.
Hence, we will apply combination.
The total number of combinations of 4 points or number of ways to select 4 points out
of 10 \(= C^{10}_4=\frac{10!}{4!6!}\)
Hexagons: Has six sides.
Similarly, if any six points are selected only one hexagon can be formed
For example, if we consider 6 points say A, B, C, D, E and H then only one quadrilateral
ABCDEH can be formed.
Hence the total number of hexagons formed would be equal to the number of groups of
6 points that can be selected.
Number of ways to choose that \(= C^{10}_6=\frac{10!}{6!4!}\)
The ratio is
\(\dfrac{\frac{10!}{4!6!}}{\frac{10!}{6!4!}}=1\)
C is the answer
A circle has 10 points (A to J) on its circumference. We need to find the ratio of total
number of quadrilaterals to total number of hexagons that can be formed.
Quadrilaterals: Has four sides. Therefore, we need 4 points to form it. Total points are
10. Since order in which points are selected does not matter.
Since if any four points are selected only one quadrilateral can be formed
For example, if we consider 4 points say A, C, E and H then only one quadrilateral ACEH
can be formed.
Hence, the total number of quadrilaterals formed would be equal to the number of
groups of 4 points that can be selected.
Hence, we will apply combination.
The total number of combinations of 4 points or number of ways to select 4 points out
of 10 \(= C^{10}_4=\frac{10!}{4!6!}\)
Hexagons: Has six sides.
Similarly, if any six points are selected only one hexagon can be formed
For example, if we consider 6 points say A, B, C, D, E and H then only one quadrilateral
ABCDEH can be formed.
Hence the total number of hexagons formed would be equal to the number of groups of
6 points that can be selected.
Number of ways to choose that \(= C^{10}_6=\frac{10!}{6!4!}\)
The ratio is
\(\dfrac{\frac{10!}{4!6!}}{\frac{10!}{6!4!}}=1\)
C is the answer
Re: A circle has 10 points on its circumference.
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15 Dec 2024, 02:45
15 Dec 2024, 02:45
1
why can this not be treated like a round table problem? for the hexagons, we fix one point and the remaining 5 points can be arranged in the following ways: 1*9*8*7*6*5. For the quadrilaterals, we have 1*9*8*7. So the value of the ratio will be a fraction and B will be greater
