Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Triangle Property : The sum of any two sides of a triangle is always greater than the third.

Since we are given that one of the sides is greater than 10, the sum of the other two sides must also be greater than 10. Hence the perimeter will always be greater than 20. Statement is sufficient.

St. (2) : A = 40

Triangle Property : For triangles with same area, the perimeter is smallest for an equilateral triangle.

Area of equilateral triangle with side x = \(\frac{\sqrt{3}}{4}x^2\)

Therefore, \(\frac{\sqrt{3}}{4}x^2\) = 40

\(x^2 = \frac{160}{\sqrt{3}}\)

Now, in order to speed up calculations, I will assume \(\sqrt{3}\) to be equal to 2.

If the condition is satisfied with \(\sqrt{3}\) equal to 2 then it will definitely be satisfied with the actual value of \(\sqrt{3}\) which is less than 2.

Therefore, \(x^2 = \frac{160}{2}\) = 80

This tells us that x is almost 9. More importantly, it tells us that x is greater than 8. Thus perimeter will be 3*x = 24.

Since this is the minimum perimeter possible (actually it is still less than what the actual minimum would be due to our approximations), we can conclude that the question stem will always be true.

Hence Sufficient.

Answer : D

Another interesting triangle property :For triangles with same perimeter, the area is maximum for an equilateral triangle. (If you think about it, this property goes hand in hand with the one we used in St. 2).

_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Though I just realized something. I considered the area given to be 40 by mistake. It is actually 20.

Although the logic will be same, the calculations will be harder since we will have to use the real value of \(\sqrt{3}\). (no margin for approximations!)

Sorry for the mistake guys. However, as long as you understand the logic, it shouldn't matter. At least you'll know how to go about approaching such questions in the future!

Cheers.

Ps. Any trick for the calculations Bunuel?

I would go backward.

Let's assume the perimeter is 20. The largest area with given perimeter will have the equilateral triangle, so side=20/3. Let's calculate the area and if the area will be less than 20 it'll mean that perimeter must be more than 20.

will go with A 1. BC = 10+AC. If BC is the longest side then going by the rule that sum of the 2 smaller sides of triangle will be more than the larger side we will get perimeter more than 20. suff 2. we can get sides as 8,5,4 and 10,4,7 satisying that area is 20. for 8,5,4 perimeter is 17 and for 10,4,7 perimeter is 21. hence insuff

basing S1 on the rule that "the length of a triangle is always greater than the absolute difference of the lengths of other two sides" we know that other side must be > 10 and from S1 we know that atleast one other side is 10, hence sufficient.

S2) not sufficient. ab=40, a,b can be 5,8 or 2, 20
_________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Triangle Property : The sum of any two sides of a triangle is always greater than the third.

Since we are given that one of the sides is greater than 10, the sum of the other two sides must also be greater than 10. Hence the perimeter will always be greater than 20. Statement is sufficient.

St. (2) : A = 40

Triangle Property : For triangles with same area, the perimeter is smallest for an equilateral triangle.

Area of equilateral triangle with side x = \(\frac{\sqrt{3}}{4}x^2\)

Therefore, \(\frac{\sqrt{3}}{4}x^2\) = 40

\(x^2 = \frac{160}{\sqrt{3}}\)

Now, in order to speed up calculations, I will assume \(\sqrt{3}\) to be equal to 2.

If the condition is satisfied with \(\sqrt{3}\) equal to 2 then it will definitely be satisfied with the actual value of \(\sqrt{3}\) which is less than 2.

Therefore, \(x^2 = \frac{160}{2}\) = 80

This tells us that x is almost 9. More importantly, it tells us that x is greater than 8. Thus perimeter will be 3*x = 24.

Since this is the minimum perimeter possible (actually it is still less than what the actual minimum would be due to our approximations), we can conclude that the question stem will always be true.

Hence Sufficient.

Answer : D

Another interesting triangle property :For triangles with same perimeter, the area is maximum for an equilateral triangle. (If you think about it, this property goes hand in hand with the one we used in St. 2).

Yes, the OA is D. It was the hard one.

This problem could be solved knowing the properties sriharimurthy mentioned. +1.

For (1): The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.

For (2): A. For a given perimeter equilateral triangle has the largest area. B. For a given area equilateral triangle has the smallest perimeter. _________________

Geometry Is the perimeter of triangle ABC greater than 20? (1) BC-AC=10. (2) The area of the triangle is 20.

Please help to solve this best

Stmt1 says: BC-AC=10, which implies BC>10, and by property of triangles AB>10. This means AB+BC > 20, So Perimeter>20 ie. stmt1 is suff. Stmt2 says: Area = 20, Area > Perimeter, implies perimter <20, i.e stmt2 is suff.

OA is C.
_________________

The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done!

for statement 2 u r using properties of equilateral triangle but no where in the q it is mentioned it is equilateral triangle

We are not told that ABC is an equilateral triangle.

Let's assume the perimeter is 20. The largest area with given perimeter will have the equilateral triangle, so side=20/3. Then we calculated the area of this hypothetical equilateral triangle and get that its area<20 but statement (2) says area=20 so as p=20 is not enough to produce area=20 even for the best case (for equilateral triangle) then perimeter must be more than 20.

The explanation for why Statement 1 is sufficient is already given above

For Statement 2 it is stated that Area = 20

=> 1/2*B*H = 20 or B*H = 40.

If we look at the possible twin factors that could yield 40 we have {1,40}, {2,20}...{5,8}..

If we consider the base to be any one of the numbers occuring in the twin factors, and the height to be the other number, both the sides other than the base will always have a length that is greater than the height.

This implies that the perimeter will always be greater than 20 if the area is 20.

I go for D each statement alone is suff. (1) is evident (2) is more tricky but going for one extrem (base, height) (40,1) to the other (1, 40) and the mid-couple (20, 2), the perimeter will seemingly always be greater than 20.

Though I just realized something. I considered the area given to be 40 by mistake. It is actually 20.

Although the logic will be same, the calculations will be harder since we will have to use the real value of \(\sqrt{3}\). (no margin for approximations!)

Sorry for the mistake guys. However, as long as you understand the logic, it shouldn't matter. At least you'll know how to go about approaching such questions in the future!

Cheers.

Ps. Any trick for the calculations Bunuel?
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Let's assume the perimeter is 20. The largest are with given perimeter would have the equilateral triangle, so side=20/3. Let's calculate the area and if the area will be less than 20 it'll mean that perimeter must be more than 20.

Think this way is easier.

Yes. You are right. It will be easier. Thanks!
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Though I just realized something. I considered the area given to be 40 by mistake. It is actually 20.

Although the logic will be same, the calculations will be harder since we will have to use the real value of \(\sqrt{3}\). (no margin for approximations!)

Sorry for the mistake guys. However, as long as you understand the logic, it shouldn't matter. At least you'll know how to go about approaching such questions in the future!

Cheers.

Ps. Any trick for the calculations Bunuel?

I would go backward.

Let's assume the perimeter is 20. The largest area with given perimeter will have the equilateral triangle, so side=20/3. Let's calculate the area and if the area will be less than 20 it'll mean that perimeter must be more than 20.

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...