Hi,
Base of the triangle is distance between (-1,0) and (4,0) i.e. 5 units
Height of triangle is|A|
Area of triangle is (1/2).5.(|A|) = 5|A|/2
From (1) : A<3
If A = 2, area = 5units <15
If A = -8, area = 20units >15
Insufficient
From(2) : right angled triangle and hypotenuse is 5 units
So, (1^2+A^2)+(4^2+A^2) = 5^2 => 2A^2 = 8 =>|A| = 2
So area is 5 units.
Sufficient
Hence, B
Area of Triangle > 15?
This topic has expert replies
Source: Beat The GMAT — Data Sufficiency |
-
Frankenstein
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
-
SoCan
- Senior | Next Rank: 100 Posts
- Posts: 97
- Joined: Sun May 15, 2011 9:19 am
- Thanked: 18 times
- Followed by:1 members
With regards to the second statement:
Frankenstein's solution is great, but this is one of those questions where it's probably best to know that you can solve it and not spend the time going through the calculations - at least when you're taking the real test.
If you draw a quick sketch, you should be able to quickly see there's only one way to draw a right triangle with a positive A and only one way with a negative A. Since the hypotenuse lies on the x-axis, you know the height is the same with you have A or -A. Therefore, you know there's only one area possible. That means it's sufficient. You can go through all the calculations if you want, but this approach will save you time.
Frankenstein's solution is great, but this is one of those questions where it's probably best to know that you can solve it and not spend the time going through the calculations - at least when you're taking the real test.
If you draw a quick sketch, you should be able to quickly see there's only one way to draw a right triangle with a positive A and only one way with a negative A. Since the hypotenuse lies on the x-axis, you know the height is the same with you have A or -A. Therefore, you know there's only one area possible. That means it's sufficient. You can go through all the calculations if you want, but this approach will save you time.
-
Frankenstein
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
Hi,SoCan wrote:With regards to the second statement:
Frankenstein's solution is great, but this is one of those questions where it's probably best to know that you can solve it and not spend the time going through the calculations - at least when you're taking the real test.
If you draw a quick sketch, you should be able to quickly see there's only one way to draw a right triangle with a positive A and only one way with a negative A. Since the hypotenuse lies on the x-axis, you know the height is the same with you have A or -A. Therefore, you know there's only one area possible. That means it's sufficient. You can go through all the calculations if you want, but this approach will save you time.
Agree with you. But, even if we come to the point that it can be solved, we still need to check whether the area is less than 15 or not. So, we still need to do a bit of calculation. Of course, this calculation can be done simpler by using slope approach or even logic that the height of the triangle, with hypotenuse as base is less than any of the sides of the right triangle.
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
-
SoCan
- Senior | Next Rank: 100 Posts
- Posts: 97
- Joined: Sun May 15, 2011 9:19 am
- Thanked: 18 times
- Followed by:1 members
You just need to know whether the statement is sufficient to definitively answer the question yes or no. If the area is not less than 15, the statement is still sufficient. You just need to make sure the answer isn't 'maybe'. If you know there can only be one value for the area, you know it's either yes or no, but it doesn't matter which one.Frankenstein wrote:Hi,SoCan wrote:With regards to the second statement:
Frankenstein's solution is great, but this is one of those questions where it's probably best to know that you can solve it and not spend the time going through the calculations - at least when you're taking the real test.
If you draw a quick sketch, you should be able to quickly see there's only one way to draw a right triangle with a positive A and only one way with a negative A. Since the hypotenuse lies on the x-axis, you know the height is the same with you have A or -A. Therefore, you know there's only one area possible. That means it's sufficient. You can go through all the calculations if you want, but this approach will save you time.
Agree with you. But, even if we come to the point that it can be solved, we still need to check whether the area is less than 15 or not.
-
Frankenstein
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
Hi,SoCan wrote:You just need to know whether the statement is sufficient to definitively answer the question yes or no. If the area is not less than 15, the statement is still sufficient. You just need to make sure the answer isn't 'maybe'. If you know there can only be one value for the area, you know it's either yes or no, but it doesn't matter which one.Frankenstein wrote:Hi,SoCan wrote:With regards to the second statement:
Frankenstein's solution is great, but this is one of those questions where it's probably best to know that you can solve it and not spend the time going through the calculations - at least when you're taking the real test.
If you draw a quick sketch, you should be able to quickly see there's only one way to draw a right triangle with a positive A and only one way with a negative A. Since the hypotenuse lies on the x-axis, you know the height is the same with you have A or -A. Therefore, you know there's only one area possible. That means it's sufficient. You can go through all the calculations if you want, but this approach will save you time.
Agree with you. But, even if we come to the point that it can be solved, we still need to check whether the area is less than 15 or not.
Agreed!
Overlooked the 2nd paragraph of your post in which you mentioned about only one area possible. Nice work!
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
-
smackmartine
- Legendary Member
- Posts: 516
- Joined: Fri Jul 31, 2009 3:22 pm
- Thanked: 112 times
- Followed by:13 members
IMO B
One alternative way to solve this question :
we know that base of the triangle with base (-1,0) and (4,0) has length 5
1) A>3
at A= 4 , area of the triangle is (1/2)*5*A = (1/2)*5*4= 10 < 15, but
at A= 8 , area of the triangle is (1/2)*5*A = (1/2)*5*8= 20 > 15.
Insufficient.
2) The triangle is right.
As soon as you know the triangle is right angled and one of its side is "5", two Pythagorean triplets come in mind.
3-4-5 and 5-12-13 are the only possible options.However, we cannot consider 5-12-13 in this case, because side with length 5 must be hypotenuse and in case of 5-12-13 , 13 is the hypotenuse(the angle between straight lines where (-1,0) and (4,0) converge at point (A,0) should be 90 degrees).
So only option is 3-4-5
As we know that we can calculate the value of "A" or area of the triangle using 3-4-5 triplets so , we should select [spoiler]"B"[/spoiler] i.e sufficient and move on.
We can get A as follows: (no need to do this on test)
(1/2)*5*A = (1/2)*4*3
=> A = 12/5
One alternative way to solve this question :
we know that base of the triangle with base (-1,0) and (4,0) has length 5
1) A>3
at A= 4 , area of the triangle is (1/2)*5*A = (1/2)*5*4= 10 < 15, but
at A= 8 , area of the triangle is (1/2)*5*A = (1/2)*5*8= 20 > 15.
Insufficient.
2) The triangle is right.
As soon as you know the triangle is right angled and one of its side is "5", two Pythagorean triplets come in mind.
3-4-5 and 5-12-13 are the only possible options.However, we cannot consider 5-12-13 in this case, because side with length 5 must be hypotenuse and in case of 5-12-13 , 13 is the hypotenuse(the angle between straight lines where (-1,0) and (4,0) converge at point (A,0) should be 90 degrees).
So only option is 3-4-5
As we know that we can calculate the value of "A" or area of the triangle using 3-4-5 triplets so , we should select [spoiler]"B"[/spoiler] i.e sufficient and move on.
We can get A as follows: (no need to do this on test)
(1/2)*5*A = (1/2)*4*3
=> A = 12/5
Last edited by smackmartine on Wed Jun 01, 2011 11:34 pm, edited 1 time in total.
-
cans
- Legendary Member
- Posts: 1309
- Joined: Mon Apr 04, 2011 5:34 am
- Location: India
- Thanked: 310 times
- Followed by:123 members
- GMAT Score:750
Some corrections in your solutionsmackmartine wrote:IMO B
One alternative way to solve this question :
we know that base of the triangle with base (-1,0) and (4,0) has length 5
1) A>3
at A= 4 , area of the triangle is (1/2)*5*A = (1/2)*5*4= 10 < 15, but
at A= 8 , area of the triangle is (1/2)*5*A = (1/2)*5*8= 20 > 15.
Insufficient.
2) The triangle is right.
As soon as you know the triangle is right angled and one of its side is "5", two Pythagorean triplets come in mind.
3-4-5 and 5-12-13 are the only possible options.However, we cannot consider 5-12-13 in this case, because side with length 5 must be hypotenuse and in case of 5-12-13 , 13 is the hypotenuse(the angle between straight lines where (-1,0) and (4,0) converge at point (A,0) should be 90 degrees).
So only option is 3-4-15
As we know that we can calculate the value of "A" or area of the triangle using 3-4-5 triplets so , we should select [spoiler]"B"[/spoiler] i.e sufficient and move on.
We can get A as follows: (no need to do this on test)
(1/2)*5*A = (1/2)*4*3
=> A = 12/5
1) a) option is A<3 and not A>32
2) b) 3-4-5 is not the only right angled triangle with hypotenuse 5.
and if you solve the question here, the sides will be root(5), 2*root(5) and 5. Thus value of A=12/5 is incorrect.
-
smackmartine
- Legendary Member
- Posts: 516
- Joined: Fri Jul 31, 2009 3:22 pm
- Thanked: 112 times
- Followed by:13 members
Nowhere,I used 32 , but yes I used the wrong sign ">" instead of "<" Sorry about that.cans wrote:Some corrections in your solutionsmackmartine wrote:IMO B
One alternative way to solve this question :
we know that base of the triangle with base (-1,0) and (4,0) has length 5
1) A>3
at A= 4 , area of the triangle is (1/2)*5*A = (1/2)*5*4= 10 < 15, but
at A= 8 , area of the triangle is (1/2)*5*A = (1/2)*5*8= 20 > 15.
Insufficient.
2) The triangle is right.
As soon as you know the triangle is right angled and one of its side is "5", two Pythagorean triplets come in mind.
3-4-5 and 5-12-13 are the only possible options.However, we cannot consider 5-12-13 in this case, because side with length 5 must be hypotenuse and in case of 5-12-13 , 13 is the hypotenuse(the angle between straight lines where (-1,0) and (4,0) converge at point (A,0) should be 90 degrees).
So only option is 3-4-5
As we know that we can calculate the value of "A" or area of the triangle using 3-4-5 triplets so , we should select [spoiler]"B"[/spoiler] i.e sufficient and move on.
We can get A as follows: (no need to do this on test)
(1/2)*5*A = (1/2)*4*3
=> A = 12/5
1) a) option is A<3 and not A>32
2) b) 3-4-5 is not the only right angled triangle with hypotenuse 5.
and if you solve the question here, the sides will be root(5), 2*root(5) and 5. Thus value of A=12/5 is incorrect.
So if A< 3 , value could be positive such as 1 or 2 or could be negative such as -10 or -20 . Because we are dealing with absolute values of Y coordinate to calculate area, area could be (1/2)*5*2 =5 <15 (at A=2) or (1/2)*5*|-20| =50 >15 (at A=-20) ,So insufficient.
Typo "So only option is 3-4-15" (I corrected it in the original post)
@cans, I did mention that 5-12-13 is also one triplet but we cannot consider that. Please see the explanation again.
Now, please refer the diagram.
The reason why I used (1/2)*5*A = (1/2)*4*3 because
Area of a triangle = 1/2 * base * height
We can calulate area in two ways :
Considering PQ as base and PR as height ,area = (1/2)* 3* 4
and
Considering QR as base and A as height ,area = (1/2)* 5 * A
Because both are same triangles, the area will also be equal.
Let me know if you still have some doubts.
-
Frankenstein
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
Hi,smackmartine wrote:Nowhere,I used 32 , but yes I used the wrong sign ">" instead of "<" Sorry about that.cans wrote:Some corrections in your solutionsmackmartine wrote:IMO B
One alternative way to solve this question :
we know that base of the triangle with base (-1,0) and (4,0) has length 5
1) A>3
at A= 4 , area of the triangle is (1/2)*5*A = (1/2)*5*4= 10 < 15, but
at A= 8 , area of the triangle is (1/2)*5*A = (1/2)*5*8= 20 > 15.
Insufficient.
2) The triangle is right.
As soon as you know the triangle is right angled and one of its side is "5", two Pythagorean triplets come in mind.
3-4-5 and 5-12-13 are the only possible options.However, we cannot consider 5-12-13 in this case, because side with length 5 must be hypotenuse and in case of 5-12-13 , 13 is the hypotenuse(the angle between straight lines where (-1,0) and (4,0) converge at point (A,0) should be 90 degrees).
So only option is 3-4-5
As we know that we can calculate the value of "A" or area of the triangle using 3-4-5 triplets so , we should select [spoiler]"B"[/spoiler] i.e sufficient and move on.
We can get A as follows: (no need to do this on test)
(1/2)*5*A = (1/2)*4*3
=> A = 12/5
1) a) option is A<3 and not A>32
2) b) 3-4-5 is not the only right angled triangle with hypotenuse 5.
and if you solve the question here, the sides will be root(5), 2*root(5) and 5. Thus value of A=12/5 is incorrect.
So if A< 3 , value could be positive such as 1 or 2 or could be negative such as -10 or -20 . Because we are dealing with absolute values of Y coordinate to calculate area, area could be (1/2)*5*2 =5 <15 (at A=2) or (1/2)*5*|-20| =50 >15 (at A=-20) ,So insufficient.
Typo "So only option is 3-4-15" (I corrected it in the original post)
@cans, I did mention that 5-12-13 is also one triplet but we cannot consider that. Please see the explanation again.
Now, please refer the diagram.
The reason why I used (1/2)*5*A = (1/2)*4*3 because
Area of a triangle = 1/2 * base * height
We can calulate area in two ways :
Considering PQ as base and PR as height ,area = (1/2)* 3* 4
and
Considering QR as base and A as height ,area = (1/2)* 5 * A
Because both are same triangles, the area will also be equal.
Let me know if you still have some doubts.
The triplets you mentioned are integers ones. But, the side lengths need not be integers here. And the triplet you mentioned is not the solution here because with the triplet you mentioned the third vertex will not be on the Y-axis. You get the third vertex as (0,2) or (0,-2).
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
-
cans
- Legendary Member
- Posts: 1309
- Joined: Mon Apr 04, 2011 5:34 am
- Location: India
- Thanked: 310 times
- Followed by:123 members
- GMAT Score:750
about 32, that was my typo error. i wanted to write 3 actually...smackmartine wrote:Nowhere,I used 32 , but yes I used the wrong sign ">" instead of "<" Sorry about that.cans wrote:Some corrections in your solutionsmackmartine wrote:IMO B
One alternative way to solve this question :
we know that base of the triangle with base (-1,0) and (4,0) has length 5
1) A>3
at A= 4 , area of the triangle is (1/2)*5*A = (1/2)*5*4= 10 < 15, but
at A= 8 , area of the triangle is (1/2)*5*A = (1/2)*5*8= 20 > 15.
Insufficient.
2) The triangle is right.
As soon as you know the triangle is right angled and one of its side is "5", two Pythagorean triplets come in mind.
3-4-5 and 5-12-13 are the only possible options.However, we cannot consider 5-12-13 in this case, because side with length 5 must be hypotenuse and in case of 5-12-13 , 13 is the hypotenuse(the angle between straight lines where (-1,0) and (4,0) converge at point (A,0) should be 90 degrees).
So only option is 3-4-5
As we know that we can calculate the value of "A" or area of the triangle using 3-4-5 triplets so , we should select [spoiler]"B"[/spoiler] i.e sufficient and move on.
We can get A as follows: (no need to do this on test)
(1/2)*5*A = (1/2)*4*3
=> A = 12/5
1) a) option is A<3 and not A>32
2) b) 3-4-5 is not the only right angled triangle with hypotenuse 5.
and if you solve the question here, the sides will be root(5), 2*root(5) and 5. Thus value of A=12/5 is incorrect.
So if A< 3 , value could be positive such as 1 or 2 or could be negative such as -10 or -20 . Because we are dealing with absolute values of Y coordinate to calculate area, area could be (1/2)*5*2 =5 <15 (at A=2) or (1/2)*5*|-20| =50 >15 (at A=-20) ,So insufficient.
Typo "So only option is 3-4-15" (I corrected it in the original post)
@cans, I did mention that 5-12-13 is also one triplet but we cannot consider that. Please see the explanation again.
Now, please refer the diagram.
The reason why I used (1/2)*5*A = (1/2)*4*3 because
Area of a triangle = 1/2 * base * height
We can calulate area in two ways :
Considering PQ as base and PR as height ,area = (1/2)* 3* 4
and
Considering QR as base and A as height ,area = (1/2)* 5 * A
Because both are same triangles, the area will also be equal.
Let me know if you still have some doubts.
and about error b) Frank has explained why your solution has an error. you are considering only integer values for sides where as for (-1,0), (4,0) and (0,A) to form a right angled triangle, 2 sides are not an integer. and that's why i said the sides will be root(5), 2*root(5) and 5. And thus value of A will be 2.
Hope it's clear.
-
smackmartine
- Legendary Member
- Posts: 516
- Joined: Fri Jul 31, 2009 3:22 pm
- Thanked: 112 times
- Followed by:13 members
Ahh! got it! Thankscans wrote:
about 32, that was my typo error. i wanted to write 3 actually...
and about error b) Frank has explained why your solution has an error. you are considering only integer values for sides where as for (-1,0), (4,0) and (0,A) to form a right angled triangle, 2 sides are not an integer. and that's why i said the sides will be root(5), 2*root(5) and 5. And thus value of A will be 2.
Hope it's clear.
