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integer values of 'x'

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]
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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?
The mind is everything. What you think you become. -Lord Buddha



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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.
The mind is everything. What you think you become. -Lord Buddha



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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.
The mind is everything. What you think you become. -Lord Buddha



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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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