Question 369 Data Sufficiency 2020 GMAT Official Guide

Question 369 Data Sufficiency 2020 GMAT Official Guide

Video explanation [PQID: DS75271.01]: A pentagon with 5 sides of equal length and 5 interior…

Comments

Prashantsays

Thanks. I think eyeballing/estimating also works. The diagonal of the pentagon is smaller than the diameter of the circle. And the perimeter of the pentagon is less than the circumference of the circle. So even if we take the diameter of the circle at the max of 8, the circumference of the circle will be 8xPie, which is only ever so slightly greater than 26. So, we can safely conclude that if a diagonal of a regular pentagon is less than 8, its perimeter will be less than 26.

This is definitely an intriguing approach. Let me expand on your idea and see if there is a flaw in my thinking. If the diagonal of the pentagon is say $7.98$, then the diameter of the circle will be greater than $7.98$. That means the circumference must exceed $7.98\pi$, which is approximately equal to $25.07$. What if the circumference of the circle was $27$ and the perimeter of the pentagon was a value less than $27$ but greater than $26$, in which case the answer to the question is YES. However, if the circumference of the circle was $25$, then we know for sure that the perimeter of the pentagon will have to be less than $25$, which would answer the question as NO. So with this logic, I am not able to demonstrate sufficiency. Unless, I have made a mistake somewhere in my argument.

Thank you Mr. Dabral. Can we instead approach this question like this: for any regular polygon, the limit between the perimeter and the longest straight line possible is pi (which is when the number of sides approach infinity- i.e. a circle). The diagonal is the longest line possible in a regular pentagon. So the ratio of this diagonal to the perimeter has to be less than pie. Therefore, the perimeter has to be less than 26.

ps. I’m not a Maths major, so my line of thinking may be way off. Just trying to connect the dots between the GMAT question maker’s mindset, and the enigmatic pi.

Great, this works! This is a different argument (at least from what I can interpret) and will definitely do the job in this case. The assertion is that the ratio of the perimeter of a regular polygon to its longest diagonal is a fixed number that only depends on the number of sides of the polygons. Further for regular polygons, this ratio is always less than $\pi$. For example, in the case of a square this ratio is equal to $2\sqrt{2}$, which is close to $2.83$. In case of a regular hexagon, this ratio is exactly equal to $3$, and as the number of sides increases this ratio approaches $\pi$. With that in place, the perimeter of the given pentagon will be less than $\pi(8)=25.13$ for all values of diagonals less than 8. I like this argument a lot, it makes it a lot easier than using the similar triangle argument.

We can make one additional comment, that the ratio of perimeter to the longest diagonal for a regular pentagon must fall between that of a square, $2\sqrt{2}$, and that of a hexagon, which is $3$. This immediately tells us that in the case of a pentagon, the perimeter has to be less than 24, which would answer the question as a definite NO.

Can you please explain what was the flaw in the following reasoning?

Diagonal of the pentagon < 8
Diagonal of the pentagon < Diameter of circle that the pentagon is inscribed in

Taking the smallest value for diameter as 8, that would always satisfy the above 2, Circumference of circle is (Pi)d which is 25.13 (approx.)
So, the circumference of the pentagon will always be less than 26.

You are right that the circumference of the circle must be greater than $\pi(8) \approx 25.13$. However, that does not guarantee that the circumference must be less than $26$. It could be say, $25.5$ or it could be $26.5$, because all we know is that the circumference exceeds $25.13$.

Prashant says

Thanks. I think eyeballing/estimating also works. The diagonal of the pentagon is smaller than the diameter of the circle. And the perimeter of the pentagon is less than the circumference of the circle. So even if we take the diameter of the circle at the max of 8, the circumference of the circle will be 8xPie, which is only ever so slightly greater than 26. So, we can safely conclude that if a diagonal of a regular pentagon is less than 8, its perimeter will be less than 26.

Prashant says

Just recalculating 8xPie above – it will be less than 26, not greater than 26. This just makes our conclusion all the more valid.

GMAT Quantum says

Hi Prashant,

This is definitely an intriguing approach. Let me expand on your idea and see if there is a flaw in my thinking. If the diagonal of the pentagon is say $7.98$, then the diameter of the circle will be greater than $7.98$. That means the circumference must exceed $7.98\pi$, which is approximately equal to $25.07$. What if the circumference of the circle was $27$ and the perimeter of the pentagon was a value less than $27$ but greater than $26$, in which case the answer to the question is YES. However, if the circumference of the circle was $25$, then we know for sure that the perimeter of the pentagon will have to be less than $25$, which would answer the question as NO. So with this logic, I am not able to demonstrate sufficiency. Unless, I have made a mistake somewhere in my argument.

Dabral

Prashant says

Thank you Mr. Dabral. Can we instead approach this question like this: for any regular polygon, the limit between the perimeter and the longest straight line possible is pi (which is when the number of sides approach infinity- i.e. a circle). The diagonal is the longest line possible in a regular pentagon. So the ratio of this diagonal to the perimeter has to be less than pie. Therefore, the perimeter has to be less than 26.

ps. I’m not a Maths major, so my line of thinking may be way off. Just trying to connect the dots between the GMAT question maker’s mindset, and the enigmatic pi.

Prashant

GMAT Quantum says

Hi Prashant,

Great, this works! This is a different argument (at least from what I can interpret) and will definitely do the job in this case. The assertion is that the ratio of the perimeter of a regular polygon to its longest diagonal is a fixed number that only depends on the number of sides of the polygons. Further for regular polygons, this ratio is always less than $\pi$. For example, in the case of a square this ratio is equal to $2\sqrt{2}$, which is close to $2.83$. In case of a regular hexagon, this ratio is exactly equal to $3$, and as the number of sides increases this ratio approaches $\pi$. With that in place, the perimeter of the given pentagon will be less than $\pi(8)=25.13$ for all values of diagonals less than 8. I like this argument a lot, it makes it a lot easier than using the similar triangle argument.

We can make one additional comment, that the ratio of perimeter to the longest diagonal for a regular pentagon must fall between that of a square, $2\sqrt{2}$, and that of a hexagon, which is $3$. This immediately tells us that in the case of a pentagon, the perimeter has to be less than 24, which would answer the question as a definite NO.

Thanks,

Dabral

Aayushi Verma says

Hi,

Can you please explain what was the flaw in the following reasoning?

Diagonal of the pentagon < 8

Diagonal of the pentagon < Diameter of circle that the pentagon is inscribed in

Taking the smallest value for diameter as 8, that would always satisfy the above 2, Circumference of circle is (Pi)d which is 25.13 (approx.)

So, the circumference of the pentagon will always be less than 26.

Thanks in advance.

Aayushi

GMAT Quantum says

Aayushi,

You are right that the circumference of the circle must be greater than $\pi(8) \approx 25.13$. However, that does not guarantee that the circumference must be less than $26$. It could be say, $25.5$ or it could be $26.5$, because all we know is that the circumference exceeds $25.13$.

I hope this makes sense.