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If the sides of a triangle have lengths x, y, and z, x + y =

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If the sides of a triangle have lengths x, y, and z, x + y = [#permalink] New post 04 May 2011, 18:35
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If the sides of a triangle have lengths x, y, and z, x + y = 30, and y + z = 20, then which of the following could be the perimeter of the triangle?

I. 28
II. 36
III. 42

A I only
B II only
C I and II only
D I and III only
E I, II, and III

[Reveal] Spoiler:
OA is B.

I don't have the solution but this is how I think it is done. Pls verify the reasoning.

The question is basically asking us to determine the limits on x.

x + y = 30 ---- (1)
y + z = 20 ---- (2)
x - z = 10. This means y > 10 [Axiom : The third side is greater than the difference of the two sides.]

x + y = 30
y > 10
From this we get x < 20.
y + z = 20
x < 20
Adding we get x + y + z < 40 -----> I think this step is correct

From (2) we have y < 20. Since side z is non-negative. From (1) we have x > 10.
y + z = 20
x > 10
Adding we get x + y + z > 30 ------> I think this step is correct

Hence 30 < x + y + z < 40. Hence B
[Reveal] Spoiler: OA
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Re: Triangle [#permalink] New post 04 May 2011, 18:52
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(I) is out as x+y = 30 > 28 (perimeter can't be < sum of two sides)

And all answers excepy B contain I as option

So Answer - B
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Re: Triangle [#permalink] New post 04 May 2011, 18:56
Brilliant !!! I was assuming there will be a solution like this. Thanks so much :-D

There is one more thing - can you also verify the explanation in the spoiler ? Cheers

subhashghosh wrote:
(I) is out as x+y = 30 > 28 (perimeter can't be < sum of two sides)

And all answers excepy B contain I as option

So Answer - B
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Re: Triangle [#permalink] New post 04 May 2011, 19:07
This took me more than 2 min's though :)

Using POE since x+y = 30, means 1 can be nullified altogether.

But a better approach will take a STAB at this,

I am focusing on Z (min) and Z (max) values. x+y = 30, y+z = 20 means x-z = 10

Z(min) = 1, means X = 11 and Y = 19 thus Perimeter (Z min) = 31

Z(max) = 9, since Y >10, means Y = 11 and X = 19 thus Perimeter (Zmax) = 37. Thus B fits in.

Likewise, one can try for either X(min) or Y (min) and max values too. Keeping the limits X<20,Z<30 and Y>10.
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Re: Triangle [#permalink] New post 05 May 2011, 03:48
@gmat1220, I think you're right. I also deduced x + y + z < 40 initially (by using the length of 3rd side < sum of two sides), and then I spotted the odd man out in the answer choices.
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Re: Triangle [#permalink] New post 05 May 2011, 03:49
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gmat1220 wrote:
If the sides of a triangle have lengths x, y, and z, x + y = 30, and y + z = 20, then which of the following could be the perimeter of the triangle?
I. 28
II. 36
III. 42

A I only
B II only
C I and II only
D I and III only
E I, II, and III


perimeter is x+y+z = ?


Apply POE
1) Clearly, x+y=30 then how can x+y+z = 28?. OUT

2) x+y+z=36
x+2y+z=50
Subtracting, x+2y+z-(x+y+z) = 50-36,
y=14
x+y=30 (given), hence x=16
y+z=20 (given) hence z=6
x+y+z => 16+14+6 = 36.
Also, (14-6)<16<(14+6). Same can be tested for other sides as well.

3) x+y+z=42
x+2y+z=50
Subtracting, y=8
x+y=30 (given), hence x=22
y+z=20 (given) hence z=12
x+y+z => 22+8+12=42
BUT X(22) IS NOT LESS THAN SUM OF OTHER TWO SIDES (8+12=20).
It doesn't satisfy triangle inequality theorem.
Hence, OUT.

OA. B
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Re: Triangle [#permalink] New post 05 May 2011, 08:18
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The POE approach above works fast. The algebraic approach is:

First, establish the equation we are looking or x + y + z = ? and name it A

if we add both given equations we can get x + y + z + y = 50. Isolate A and you get A + y = 50

Now we know from triangle inequality theorem that x - z < y < x +z. We can get x - z by substracting both equation we are given and use the other for x + z. So we get 10 < y < 20 so:

so A = 50 - GT (10) so A = LT (40)
and A = 50 - LT (20) so A = GT (30)

30 < A < 40

Only II (36) meets this criteria.

I think you can solve under 2mn with this or even better by recognizing the trick subhashghosh explained.

Hope this is helpful.
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Re: If the sides of a triangle have lengths x, y, and z, x + y = [#permalink] New post 29 Nov 2013, 20:29
20------10-------10--------40-------N
19------11-------9---------39-------Y
18------12-------8---------38-------Y
17------13-------7---------37-------Y
16------14-------6---------36-------Y
x-------y----------z------per.----triangle?
15------15-------5---------35------Y
14------16-------4---------34------Y
13------17-------3---------33------N
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Re: If the sides of a triangle have lengths x, y, and z, x + y = [#permalink] New post 30 Nov 2013, 02:58
Expert's post
gmat1220 wrote:
PS : What do you think we must guess. Or is there a more intuitive approach which guarantees the result in less than 2 mins?

If the sides of a triangle have lengths x, y, and z, x + y = 30, and y + z = 20, then which of the following could be the perimeter of the triangle?

I. 28
II. 36
III. 42

A I only
B II only
C I and II only
D I and III only
E I, II, and III

[Reveal] Spoiler:
OA is B.

I don't have the solution but this is how I think it is done. Pls verify the reasoning.

The question is basically asking us to determine the limits on x.

x + y = 30 ---- (1)
y + z = 20 ---- (2)
x - z = 10. This means y > 10 [Axiom : The third side is greater than the difference of the two sides.]

x + y = 30
y > 10
From this we get x < 20.
y + z = 20
x < 20
Adding we get x + y + z < 40 -----> I think this step is correct

From (2) we have y < 20. Since side z is non-negative. From (1) we have x > 10.
y + z = 20
x > 10
Adding we get x + y + z > 30 ------> I think this step is correct

Hence 30 < x + y + z < 40. Hence B


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Re: If the sides of a triangle have lengths x, y, and z, x + y = [#permalink] New post 12 Dec 2013, 10:03
PS : What do you think we must guess. Or is there a more intuitive approach which guarantees the result in less than 2 mins?

If the sides of a triangle have lengths x, y, and z, x + y = 30, and y + z = 20, then which of the following could be the perimeter of the triangle?

As others have pointed out, we can rule out I.) because it indicates that all three sides add up to 28 when the question says that just two sides add up to 30.

x + y = 30
y + z = 20

x + z + 2y = 50
We can solve by ruling out answer choices, so let's say that we assume x + y + z = 36

x + z + 2y = 50
x + y + z = 36
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y = 14

x + y = 30
x + (14) = 30
x = 16

x + y = 30
(16) + y = 30
y = 14

We don't even need to test III.) because it is always lumped in with I.) which we know is not possible.

B.)


I. 28
II. 36
III. 42

A I only
B II only
C I and II only
D I and III only
E I, II, and III
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Re: If the sides of a triangle have lengths x, y, and z, x + y = [#permalink] New post 12 Feb 2014, 05:19
x+y=30 & y+z=20 so x+2y+z=50

x+y+z=50-y

If perimeter is 28 then y=50-28=22, and y+z=20 z cannot be negative. I is out.
If perimeter is 36 then y=50-36 = 14. z=6, x=16. no problem here.
If perimeter is 42 then y=50-42 =8. x=22, z=12. x cannot be greater than sum of y & z. III is out.

B is answer.
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Re: If the sides of a triangle have lengths x, y, and z, x + y =   [#permalink] 12 Feb 2014, 05:19
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