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(1) \(a-b=15\). Must know for the GMAT: 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. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

(1) \(a-b=15\). Must know for the GMAT: 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. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

Answer: D

Could you please explain statement 2 in detail. I am unable to understand the solution for statement 2.
_________________

You have to have the darkness for the dawn to come.

(1) \(a-b=15\). Must know for the GMAT: 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. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

Answer: D

Could you please explain statement 2 in detail. I am unable to understand the solution for statement 2.

From (2) we have that if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30.
_________________

Statement 1 can be easily shown as sufficient as the difference between the two sides must be less than the third side. So the third side would be at least 15 and as given in statement the difference between the two sides is 15 so their sum would obviously be greater than 15. So perimeter will be greater than 30.

Statement 2: Area of a triangle is given to be 50. As it is a fact that for a given perimeter the equilateral triangle will have maximum area. So let us assume the side of a triangle is 10 then (root3/4)*(10)2 then we will get the area as 43.25 which less than 50. hence the perimeter has to be more than 30. Sufficient.

So answer is D
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I hope this helped. If this was indeed helpful, then you may say Thank You by giving a Kudos!

(1) \(a-b=15\). Must know for the GMAT: 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. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

Answer: D

HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

Can it be proved by a theorem?

Thanks
_________________

Thanks and Regards,

Honneeey.

In former years,Used to run for "Likes", nowadays, craving for "Kudos". :D

(1) \(a-b=15\). Must know for the GMAT: 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. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

Answer: D

HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

(1) \(a-b=15\). Must know for the GMAT: 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. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

Answer: D

HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

(1) \(a-b=15\). Must know for the GMAT: 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. So, \(a+b \gt c \gt 15\), which means that \(a+b+c \gt 30\). Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral. Let's find what this area would be: \(Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50\). Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.