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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

@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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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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My dad once said to me: Son, nothing succeeds like success.

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.

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

Re: If the sides of a triangle have lengths x, y, and z, x + y = [#permalink]

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12 Dec 2013, 11:03

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 __________________( - ) 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

Re: If the sides of a triangle have lengths x, y, and z, x + y = [#permalink]

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12 Feb 2014, 06:19

1

This post received KUDOS

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.

Re: If the sides of a triangle have lengths x, y, and z, x + y = [#permalink]

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15 Mar 2015, 16:08

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Re: If the sides of a triangle have lengths x, y, and z, x + y = [#permalink]

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13 May 2016, 09:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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If the sides of a triangle have lengths x, y, and z, x + y = [#permalink]

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16 May 2016, 19:45

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

my approach... x+y=30, +z will be >30. so I is out right away. A, C, D, and E are eliminated. less than 30 seconds needed to figure it out. answer choices should be given more "confusing"...

gmatclubot

If the sides of a triangle have lengths x, y, and z, x + y =
[#permalink]
16 May 2016, 19:45

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