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Re: Is the area of the right angled triangle ABC>25?
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05 Mar 2016, 03:54
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1
seesharp wrote:
Is the area of the right angled triangle ABC>25?
1) AC = 6 2) AB = 10
Hi, whenever we see a restrictions, <,>, atleast etc, more often than not, the answer is possible through a statement..
1) AC = 6 Since, one of the sides is 6, the other can take any value Insuff..
2) AB = 10 It is given that hyp is 10.. An isosceles right angle triangle will have max area.. so with Hyp as 10 , the max area possible is when the other two sides are equal.. so each side = Hyp/\(\sqrt{2}\)=10/\(\sqrt{2}\).. area = 1/2 * 10/\(\sqrt{2}\)*10/\(\sqrt{2}\) =>100/4=25.. so max area possible is 25.. so teh area of ABC can never be >25.. Suff
Re: Is the area of the right angled triangle ABC>25?
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05 Mar 2016, 04:10
chetan2u wrote:
seesharp wrote:
Is the area of the right angled triangle ABC>25?
1) AC = 6 2) AB = 10
Hi, whenever we see a restrictions, <,>, atleast etc, more often than not, the answer is possible through a statement..
1) AC = 6 Since, one of the sides is 6, the other can take any value Insuff..
2) AB = 10 It is given that hyp is 10.. An isosceles right angle triangle will have max area.. so with Hyp as 10 , the max area possible is when the other two sides are equal.. so each side = Hyp/\(\sqrt{2}\)=10/\(\sqrt{2}\).. area = 1/2 * 10/\(\sqrt{2}\)*10/\(\sqrt{2}\) =>100/4=25.. so max area possible is 25.. so teh area of ABC can never be >25.. Suff
ANS B
Thanks, so you assumed the triangle to be an isosceles to give the maximum possible area. That is great. How about if statement 2 was AB=11? What would the answer be then? C or E?
Re: Is the area of the right angled triangle ABC>25?
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05 Mar 2016, 04:13
seesharp wrote:
chetan2u wrote:
seesharp wrote:
Is the area of the right angled triangle ABC>25?
1) AC = 6 2) AB = 10
Hi, whenever we see a restrictions, <,>, atleast etc, more often than not, the answer is possible through a statement..
1) AC = 6 Since, one of the sides is 6, the other can take any value Insuff..
2) AB = 10 It is given that hyp is 10.. An isosceles right angle triangle will have max area.. so with Hyp as 10 , the max area possible is when the other two sides are equal.. so each side = Hyp/\(\sqrt{2}\)=10/\(\sqrt{2}\).. area = 1/2 * 10/\(\sqrt{2}\)*10/\(\sqrt{2}\) =>100/4=25.. so max area possible is 25.. so teh area of ABC can never be >25.. Suff
ANS B
Thanks, so you assumed the triangle to be an isosceles to give the maximum possible area. That is great. How about if statement 2 was AB=11? What would the answer be then? C or E?
The answer will be Statement 2 will be INSUFF and answer will be C.. But the trap lies in assuming C to be the answer without trying for MAX possible AREA..
_________________
Re: Is the area of the right angled triangle ABC>25?
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05 Mar 2016, 04:19
seesharp wrote:
chetan2u wrote:
seesharp wrote:
Is the area of the right angled triangle ABC>25?
1) AC = 6 2) AB = 10
Hi, whenever we see a restrictions, <,>, atleast etc, more often than not, the answer is possible through a statement..
1) AC = 6 Since, one of the sides is 6, the other can take any value Insuff..
2) AB = 10 It is given that hyp is 10.. An isosceles right angle triangle will have max area.. so with Hyp as 10 , the max area possible is when the other two sides are equal.. so each side = Hyp/\(\sqrt{2}\)=10/\(\sqrt{2}\).. area = 1/2 * 10/\(\sqrt{2}\)*10/\(\sqrt{2}\) =>100/4=25.. so max area possible is 25.. so teh area of ABC can never be >25.. Suff
ANS B
Thanks, so you assumed the triangle to be an isosceles to give the maximum possible area. That is great. How about if statement 2 was AB=11? What would the answer be then? C or E?
the LOGIC, which may help you in some other Qs too, is.. Given sum of any two numbers, the product will be MAX when the two numbers are same.. Example SUM=a+b =10.. max product ab = 10/2 * 10/2= 25.. that is a aand b are 5 each.. try any combination of a and b, product will not >25..
_________________
Is the area of the right angled triangle ABC>25?
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05 Mar 2016, 12:08
chetan2u wrote:
The answer will be Statement 2 will be INSUFF and answer will be C.. But the trap lies in assuming C to be the answer without trying for MAX possible AREA..
Yes, its a C-trap question, Its creator knows that students will not try max area. But why B would be insuff if AB=11? why looking at value 10 of AB makes us look for max area. if its about Pythagorean triplets then we can consider the same for value 6 of AC and choice A would be correct too? Please explain chetan2u _________________
Re: Is the area of the right angled triangle ABC>25?
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05 Mar 2016, 12:55
HKD1710 wrote:
chetan2u wrote:
The answer will be Statement 2 will be INSUFF and answer will be C.. But the trap lies in assuming C to be the answer without trying for MAX possible AREA..
Yes, its a C-trap question, Its creator knows that students will not try max area. But why B would be insuff if AB=11? why looking at value 10 of AB makes us look for max area. if its about Pythagorean triplets then we can consider the same for value 6 of AC and choice A would be correct too? Please explain chetan2u
Its better if you look at a few cases when AB=11.
Case 1: it is an iscosceles right triangle, giving you \(AC=BC=11/\sqrt{2}\) ---> \(Area = 0.5*11/\sqrt{2}*11/\sqrt{2} = 121/4 > 25\) but if
Case 2: \(AC=\sqrt{120}\) and BC =1, you still get AB=11 but in this case, \(Area = 0.5*\sqrt{120}*1\) <25
This is the reason why statement 2 of this hypothetical question will not be sufficient.
I was wondering could you please explain where I made a mistake in my solution for statement B?
AB=10, BC=6, AC=5. Area= 15 = We get our "no" value. AB=10, BC=9, AC=8. Area= 36 = We get our "yes" value.
Hence, insufficient.
I feel like I am making a very foolish mistake somewhere in my solution that I am not able to see for some reason. If you could please tell me where I am making a mistake, I would greatly appreciate it!
I was wondering could you please explain where I made a mistake in my solution for statement B?
AB=10, BC=6, AC=5. Area= 15 = We get our "no" value. AB=10, BC=9, AC=8. Area= 36 = We get our "yes" value.
Hence, insufficient.
I feel like I am making a very foolish mistake somewhere in my solution that I am not able to see for some reason. If you could please tell me where I am making a mistake, I would greatly appreciate it!
Hi.. You are looking at right angled triangle with hypotenuse as 10, so the other two sides cannot be (6,5) or (9,8).. Since it is right angled triangle, the square of hypotenuse,10, should be equal to sum of squares of other two sides.. But 6^2+5^2=36+21... This is not equal to 10^2.. Similarly for 9^2+8^2.. The are will be max when both sides are equal , so \(a^2+a^2=10^2....2a^2=100....a=√50\) This max area is (1/2)*√50*√50=25.. So area will never be >25
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46% (01:43) correct 
