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Difficulty:
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64%
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Is the perimeter of triangle ABC greater than 20?
(1) BC - AC = 10.
(2) The area of the triangle is 20.
(1) BC - AC = 10.
(2) The area of the triangle is 20.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
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Bunuel
I pick 'A'
(1) Sides AB, BC, AC
Given BC-AC=10
BC=AC+10
BC>10
And BC-AC< AB <BC+AC
so AB >10
so whatever is the value of AC
AB+BC+AC >20 Sufficient
(2) Since type of triangle is not given, it is not possible to find only one set of lengths of sides. Not Sufficient
Answer A
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
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
(A) is the straightforward answer and it's not correct. The question is a bit trickier than this.
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The answer should be C.
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.
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.
OA?
(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.
OA?
GMAT 1: 750 Q50 V42
Posts: 124
Question Stem : Is AB + BC + AC > 20?
St. (1) : BC = AC + 10
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).
_________________
St. (1) : BC = AC + 10
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!
https://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 : https://gmatclub.com/forum/word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : https://gmatclub.com/forum/work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : https://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html
https://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 : https://gmatclub.com/forum/word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : https://gmatclub.com/forum/work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : https://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html
Posts: 1,012
Bunuel
AB+AC+BC = 20?
Use third-side rule: The third side cannot be > sum of the rest two sides and smaller than the difference of the two sides:
traingle-sides-85784.html#p643924
Also: The smallest perimeter is of the equilateral triangle for a given area or perimeter.
1: BC-AC=10.
BC>10
AC>0
AB>BC-AC
Add up these all:
AB+AC+BC > 10+0+BC-AC
AB+AC+BC > 10+0+(10+x) where x>0
AB+AC+BC > 20+x. Suff.
2: The area of the triangle is 20.
The smallest perimeter of this triangle is s if the triangle is equilateral. If so, s is:
20 = s^2 (Sqrt3/4)
s^2 = 80/(Sqrt 3)
s = 6.8
Perimeter = 3s = 3(6.8) = 20.40>20. SUFF.
D.
sriharimurthy
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.
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GMAT 1: 750 Q50 V42
Posts: 124
Thanks Bunuel.
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?
_________________
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!
https://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 : https://gmatclub.com/forum/word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : https://gmatclub.com/forum/work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : https://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html
https://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 : https://gmatclub.com/forum/word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : https://gmatclub.com/forum/work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : https://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html
sriharimurthy
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.
\(Area=s^2*\frac{\sqrt{3}}{4}=(\frac{20}{3})^2*\frac{\sqrt{3}}{4}=\frac{100*\sqrt{3}}{9}=~\frac{173}{9}<20\)
Think this way is easier. \(\sqrt{3}\approx{1.73}\).
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Posts: 99
for statement 2 u r using properties of equilateral triangle but no where in the q it is mentioned it is equilateral triangle
anilnandyala
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.
Hope it's clear.
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Bunuel
Bunuel,
is it necessary to solve it? what if you just recognized that you could solve it and the answer is either > 20 or < 20? I'm just thinking in terms of timing strategy...
psirus
You have to solve it. If we get that the minimum perimeter possible for a triangle with an area of 20 is less than 20 then we won't be able to answer the question. Similarly if we get that the maximum area possible for a triangle with a perimeter of 20 is more than 20 (for example 25) then knowing that area is 20 won't mean that perimeter must be more than 20.
Hope it's clear.
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Solution:
Q-stem rephrased- Is AB+AC+BC >20
St(1):- BC-AC=10
=> BC = AC + 10
Since one of the sides is greater than 10, sum of the other sides is also greater than 10 and the perimeter greater than 20.
This is because the sum of two sides is greater than the third side. (Sufficient)
St(2)Area of the triangle is 20
For a given area equilateral triangle has the smallest perimeter
Now if the triangle was equilateral with area 20 then its side would have been 20/3 and thus its area would have been
=√3/4 * (20/3) * (20/3)
= √3 * (100/9) which is
√3 * 11.11 and this is LESS than 20
= >Thus p=20 is NOT enough to produce area=20 even for the best case (for equilateral triangle that has minimum perimeter)
=>The Perimeter must be more than 20. (Sufficient) (option d)
The second statement was indeed testing on the Maximum/Minimization model and first statement was testing the basic properties of triangles.
Devmitra Sen
GMAT SME
Q-stem rephrased- Is AB+AC+BC >20
St(1):- BC-AC=10
=> BC = AC + 10
Since one of the sides is greater than 10, sum of the other sides is also greater than 10 and the perimeter greater than 20.
This is because the sum of two sides is greater than the third side. (Sufficient)
St(2)Area of the triangle is 20
For a given area equilateral triangle has the smallest perimeter
Now if the triangle was equilateral with area 20 then its side would have been 20/3 and thus its area would have been
=√3/4 * (20/3) * (20/3)
= √3 * (100/9) which is
√3 * 11.11 and this is LESS than 20
= >Thus p=20 is NOT enough to produce area=20 even for the best case (for equilateral triangle that has minimum perimeter)
=>The Perimeter must be more than 20. (Sufficient) (option d)
The second statement was indeed testing on the Maximum/Minimization model and first statement was testing the basic properties of triangles.
Devmitra Sen
GMAT SME
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.
\(Area=s^2*\frac{\sqrt{3}}{4}=(\frac{20}{3})^2*\frac{\sqrt{3}}{4}=\frac{100*\sqrt{3}}{9}=~\frac{173}{9}<20\)
Think this way is easier. \(\sqrt{3}\approx{1.73}\).[/quoteHi, for given parameter the largest area will be of equilateral triangle but why are you saying that area should be > 20 ? why this comparison..the area is largest amongst various possible triangles right?
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