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integer values of 'x'

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by sanju09 » Thu Aug 26, 2010 8:34 pm
If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by phillybeat » Thu Aug 26, 2010 8:55 pm
sanju09 wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]

is it A
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by sanju09 » Thu Aug 26, 2010 8:59 pm
phillybeat wrote:
sanju09 wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]

is it A
why A? few words please
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by phillybeat » Thu Aug 26, 2010 9:04 pm
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]

is it A
why A? few words please
my assumption is

acute angle triangle means all all angles less than 90º .

if one angle is 90 then

c^2 = a^2+b^2

here inserting those two values 10,12

12^2= 10^2+a^2
a^2 = 44
so a has to be less than 44

hence 1,2,3,4,5,6

the closest one is 7 ..
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by sanju09 » Thu Aug 26, 2010 9:22 pm
phillybeat wrote:
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]

is it A
why A? few words please
my assumption is

acute angle triangle means all all angles less than 90º .

if one angle is 90 then

c^2 = a^2+b^2

here inserting those two values 10,12

12^2= 10^2+a^2
a^2 = 44
so a has to be less than 44

hence 1,2,3,4,5,6

the closest one is 7 ..
sorry, that's not the best of yours

let me cheat the source to help you out:

Finding the answer to this question requires one to know two rules in geometry.

Rule 1: For an acute angled triangle, the square of the LONGEST side MUST BE LESS than the sum of squares of the other two sides.

Rule 2: For any triangle, sum of any two sides must be greater than the third side.
The mind is everything. What you think you become. -Lord Buddha



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by phillybeat » Thu Aug 26, 2010 9:35 pm
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]

is it A
why A? few words please
my assumption is

acute angle triangle means all all angles less than 90º .

if one angle is 90 then

c^2 = a^2+b^2

here inserting those two values 10,12

12^2= 10^2+a^2
a^2 = 44
so a has to be less than 44

hence 1,2,3,4,5,6

the closest one is 7 ..
sorry, that's not the best of yours

let me cheat the source to help you out:

Finding the answer to this question requires one to know two rules in geometry.

Rule 1: For an acute angled triangle, the square of the LONGEST side MUST BE LESS than the sum of squares of the other two sides.

Rule 2: For any triangle, sum of any two sides must be greater than the third side.

let me give it a shot with your rules ..

a^2 < 10^2+12^2
a^2<244.

so a can be from 1-15.

as per rule 2

a+10 > 12

we can eliminate values 1,2 for a from 15
so i am going with answer E..
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by sanju09 » Thu Aug 26, 2010 9:44 pm
phillybeat wrote:
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]

is it A
why A? few words please
my assumption is

acute angle triangle means all all angles less than 90º .

if one angle is 90 then

c^2 = a^2+b^2

here inserting those two values 10,12

12^2= 10^2+a^2
a^2 = 44
so a has to be less than 44

hence 1,2,3,4,5,6

the closest one is 7 ..
sorry, that's not the best of yours

let me cheat the source to help you out:

Finding the answer to this question requires one to know two rules in geometry.

Rule 1: For an acute angled triangle, the square of the LONGEST side MUST BE LESS than the sum of squares of the other two sides.

Rule 2: For any triangle, sum of any two sides must be greater than the third side.

let me give it a shot with your rules ..

a^2 < 10^2+12^2
a^2<244.

so a can be from 1-15.

as per rule 2

a+10 > 12

we can eliminate values 1,2 for a from 15
so i am going with answer E..
Please take rules correctly. What's 'a' in your shot? Why is that assumed to be the longest side of the triangle? Where did you use Rule 2 there?
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by sanju09 » Thu Aug 26, 2010 10:11 pm
phillybeat wrote:lets say if i assume the otheway around
12 is the longest side
then

12^2<10^2+a^2

so a >6 and a<12

means a = 7,8,9,10,11 which is 5 values.. looking at answer choices this assumption is wrong ( 12 is longest side).. that leaves one option a is longest.

then a is from 1-15..

rule 2 a is positive integer so

a+12 > 10 or a+10>12.

only 13 values satisfy two conditions..hence i went with 13
The least value of x that obeys both the rules above is 7.

10^2 + 7^2 > 12^2.

The greatest value of x that obeys both the rules above is 15.

10^2 + 12^2 > 15^2.


Hence, the values of x that satisfy both the rules are 7 through 15. [spoiler]9 values in all.

my B[/spoiler]
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by RCV » Tue Sep 07, 2010 11:19 pm
sanju09 wrote:
phillybeat wrote:lets say if i assume the otheway around
12 is the longest side
then

12^2<10^2+a^2

so a >6 and a<12

means a = 7,8,9,10,11 which is 5 values.. looking at answer choices this assumption is wrong ( 12 is longest side).. that leaves one option a is longest.

then a is from 1-15..

rule 2 a is positive integer so

a+12 > 10 or a+10>12.

only 13 values satisfy two conditions..hence i went with 13
The least value of x that obeys both the rules above is 7.

10^2 + 7^2 > 12^2.

The greatest value of x that obeys both the rules above is 15.

10^2 + 12^2 > 15^2.


Hence, the values of x that satisfy both the rules are 7 through 15. [spoiler]9 values in all.

my B[/spoiler]
great shot sanju09, very deep rooted philosophy of triangles is used on gmat then why they call it only elementary, anybody please help :?
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by Adi_Pat » Wed Sep 08, 2010 7:49 am
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:
phillybeat wrote:
sanju09 wrote:If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible?
(A) 7
(B) 9
(C) 11
(D) 12
(E) 13


[spoiler]Source: https://gmat-math.blocked/2010/02/4gmat[/spoiler]

is it A
why A? few words please
my assumption is

acute angle triangle means all all angles less than 90º .

if one angle is 90 then

c^2 = a^2+b^2

here inserting those two values 10,12

12^2= 10^2+a^2
a^2 = 44
so a has to be less than 44

hence 1,2,3,4,5,6

the closest one is 7 ..
sorry, that's not the best of yours

let me cheat the source to help you out:

Finding the answer to this question requires one to know two rules in geometry.

Rule 1: For an acute angled triangle, the square of the LONGEST side MUST BE LESS than the sum of squares of the other two sides.

Rule 2: For any triangle, sum of any two sides must be greater than the third side.


I was a little confused by this one....I was using rule 2 above to solve this...

so...if the sum of any two sides must be greater than the third side...

the smallest possible value of the third side...= 3 ( using 10 + third side > 12 )
and the greatest possible value of the third side... = 21 ( using 10 +12 > third side )

So I get that we could have any value b/w 3 and 21...which is also not an answer choice...
WHAT AM I DOING WRONG ???
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by lokesh r » Wed Sep 08, 2010 12:44 pm
According to basic rules of triangle.

12-10<x<12+10 = 2<x<22

Since question asks number of integer values that x may take, answer should be 19,which is not in answer option.

I doubt correctness of the question.
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by sanju09 » Wed Sep 08, 2010 9:56 pm
Adi_Pat wrote:I was a little confused by this one....I was using rule 2 above to solve this...

so...if the sum of any two sides must be greater than the third side...

the smallest possible value of the third side...= 3 ( using 10 + third side > 12 )
and the greatest possible value of the third side... = 21 ( using 10 +12 > third side )

So I get that we could have any value b/w 3 and 21...which is also not an answer choice...
WHAT AM I DOING WRONG ???
You are using Rule 2 only and ignoring the Rule 1 at the same time, remember, the triangle is acute angled too.
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by Arcane66 » Thu Sep 09, 2010 8:40 am
This question seems fairly easy but I don't understand your complex reasonings. I know acute is less than 90 degrees and obtuse is greater than 90 degrees. I then just figured you can't have an angle that, when added with another angle, doesn't sum to a number greater than the third angle. So, I just picked numbers. 3-11 inclusive work. So the answer would be B, 9.
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by sanju09 » Fri Sep 10, 2010 12:59 am
Arcane66 wrote:This question seems fairly easy but I don't understand your complex reasonings. I know acute is less than 90 degrees and obtuse is greater than 90 degrees. I then just figured you can't have an angle that, when added with another angle, doesn't sum to a number greater than the third angle. So, I just picked numbers. 3-11 inclusive work. So the answer would be B, 9.
I don't understand your effortless reasonings. Rule 1 is inescapable here, no matter how complex it sounds to some, but you can't confirmingly get to the right answer without suitably blending the two rules cited above.
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by msbinu » Fri Sep 10, 2010 9:14 am
Rules of triangle are
1. length of any side cannot be greater that the sum of other 2
2. Length cannot be shorter than difference of other sides

Applying rule 2
Length of third cannot be shorter than 2
. Let the length be 3 . its > 2 and <= 22 ( rule 1 )
length 4 : > 2 and < = 22
.
.
Length 10 > 2 and <=22

Length 11 >2 but now > 22 .

Hence side can be any thing from 3 to 10 .
Hence total 7 possibilities are there for the third side
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