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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 =
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12 Dec 2013, 11:03

1

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 =
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12 Feb 2014, 06:19

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1

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.

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

Re: If the sides of a triangle have lengths x, y, and z, x + y =
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01 Dec 2016, 08:17

Remember the theory of the triangle, Side of the triangle will be greater than difference of the remaining two sides and less than sum of the two remaining sides. Let say in this case, X-Y<Z<X+Y. Looking at the answers easily we can eliminate one answer i.e. 28 which is anyway not following the first equation x + y = 30. The rest of the two options 36 is the correct answer.

Re: If the sides of a triangle have lengths x, y, and z, x + y =
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02 Dec 2016, 01:01

I guess the fastest and clearest approach here is to use the given choices 28 - 36 - 42. P of triangle = x+y+z 1 ) x+y+z=28 we are given that x+y=30 so first option is false since answer can not be negative

2 ) x+y+z=36 we are given that x+y=30 so 30+z=36 z=6 y= 14 x = 16 the third side must be less than sum of other two sides and more than the difference of the other two sides of the triangle. it could be P of the triangle like that: z<x+y z>x-y and the same for x and y

3 ) x+y+z=42 if we try the same property above, we will find that it couldn't be P of the triangle. So answer is D if it helps, please press kudos for me.

Re: If the sides of a triangle have lengths x, y, and z, x + y =
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16 Sep 2017, 16:26

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Top Contributor

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

First of all, the perimeter CANNOT equal 28 We know this because we're told that x + y = 30, which means the sum of two sides is 30 In order for the perimeter (x+y+z) to equal 28, side z would have to have length -2, which makes no sense. ELIMINATE A, C, D, and E

On test day, I wouldn't spend any more time on this question. However, let's keep going. . .

Next, we can show that the perimeter CANNOT equal 42 IMPORTANT RULE: If two sides of a triangle have lengths A and B, then . . . DIFFERENCE between A and B < length of third side < SUM of A and B We're told that y + z = 20, which means the sum of sides y and z is 20 The above rule tells us that the third side (side x) must be LESS THAN 20 If x is less than 20, and y+z = 20, it's impossible for the perimeter (x+y+z) to equal 42

Finally, the perimeter (x+y+z) CAN equal 36 If x = 16 y = 14, and z = 6, then all of the conditions are met, AND the perimeter is 36

Re: If the sides of a triangle have lengths x, y, and z, x + y =
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13 Sep 2019, 16:35

1

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

If we add the two equations, we have x + 2y + z = 50. Subtracting x + y + z (i.e, the perimeter of the triangle) from this, we have y = 50 - (x + y + z). Now let’s check the numbers in the given Roman numerals.

I. 28

If the perimeter is 28, then y = 50 - 28 = 22. However, it’s not possible for y + z = 20 (since z would have to be -2). Therefore, 28 can’t be the perimeter.

II. 36

If the perimeter is 36, then y = 50 - 36 = 14. In this case, x + 14 = 30 → x = 16 and 14 + z = 20 → z = 6. So we have x = 16, y = 14 and z = 6. We can see that these 3 numbers can be the side lengths of a triangle.

III. 42

If the perimeter is 42, then y = 50 - 42 = 8. In this case, x + 8 = 30 → x = 22 and 8 + z = 20 → z = 12. So we have x = 22, y = 8 and z = 12. However, these 3 numbers can’t be the side lengths of a triangle since 8 + 12 is not greater than 22.