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In the figure above, if x, y and z are integers such that x < y < zIn

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In the figure above, if x, y and z are integers such that x < y < zIn [#permalink]

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New post 15 Aug 2017, 02:25
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In the figure above, if x, y and z are integers such that x < y < z, then the least and the greatest possible values of x + z are

(A) 59 and 91
(B) 59 and 135
(C) 91 and 178
(D) 120 and 135
(E) 120 and 178

Attachment:
2017-08-15_1321_001.png
2017-08-15_1321_001.png [ 5.38 KiB | Viewed 779 times ]

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Re: In the figure above, if x, y and z are integers such that x < y < zIn [#permalink]

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New post 15 Aug 2017, 03:40
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In the figure above, if x, y and z are integers such that x < y < z, then the least and the greatest possible values of x + z are

(A) 59 and 91
(B) 59 and 135
(C) 91 and 178
(D) 120 and 135
(E) 120 and 178

For max value:
x=1 and y=2; Thus, z=177 => x+z=178

For min value:
x=1 and y=89; Thus, z=90 => x+z=91

Answer "C"
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Re: In the figure above, if x, y and z are integers such that x < y < zIn [#permalink]

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New post 15 Aug 2017, 04:15
Bunuel wrote:
Image
In the figure above, if x, y and z are integers such that x < y < z, then the least and the greatest possible values of x + z are

(A) 59 and 91
(B) 59 and 135
(C) 91 and 178
(D) 120 and 135
(E) 120 and 178

Attachment:
2017-08-15_1321_001.png



Hi..

x<y<z....
x+y+z=180..
LEAST value of x+z...
Make y the MAX. For this make y=z-1 and z the least that is 1..
So z+y=180-1=179.....z+z-1=179....2z=180.....z=90 and y=z-1=90-1=89...
\(x+z=1+90=91\)

MAX value of x+z...
Make y the least. For this make x=1 and y=2, so z=177...
\(x+z=1+177=178\)

C
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Re: In the figure above, if x, y and z are integers such that x < y < zIn [#permalink]

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New post 15 Aug 2017, 10:27
Hi chetan2u,

How did you deduce this "For this make y=z-1 and z the least that is 1"?
Can you kindly explain?

Thanks,
Abhishek
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Re: In the figure above, if x, y and z are integers such that x < y < zIn [#permalink]

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New post 15 Aug 2017, 10:41
The answer is C.

x+y+z = 180
For the greatest value of x+z, y must be minimal, but greater than 1: 1+2+177 = 180. 1+177 = 178
For least value of x+z, y must be maximal, but less than 90: 1+89+90 = 180. 1+90 = 91
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Re: In the figure above, if x, y and z are integers such that x < y < zIn [#permalink]

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New post 15 Aug 2017, 14:07
Why C?
180-91=89 which could be y
and z>y
Why not D


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Re: In the figure above, if x, y and z are integers such that x < y < zIn [#permalink]

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New post 18 Aug 2017, 09:10
Bunuel wrote:
Image
In the figure above, if x, y and z are integers such that x < y < z, then the least and the greatest possible values of x + z are

(A) 59 and 91
(B) 59 and 135
(C) 91 and 178
(D) 120 and 135
(E) 120 and 178



Since x, y, and z are the interior angles of a triangle, the relation x + y + z = 180 holds. Passing y to the right-hand side of the equality, we get x + z = 180 - y. Therefore, if we can determine the least and the greatest possible value for y, we will be able to determine the greatest and the least possible value of x + z.

Since x and y are integers and x cannot be zero, the least possible value of x is 1. Therefore, the least possible value of y is 2. Then, the greatest possible value of x + z is 180 - 2 = 178.

Let’s determine the greatest possible value of y. To do that, we need to minimize the values of x and z. We already know the smallest possible value of x is 1. Thus, we are looking for the greatest integer value of y where y < z and y + z = 179. We note that y attains its greatest value when y and z are as close to each other as possible. Since 179/2 = 89.5, we see that the greatest integer value of y is 89 (and the smallest possible integer value of z is 90). Then, the least possible value of x + z is 180 - 89 = 91.

Answer: C
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Re: In the figure above, if x, y and z are integers such that x < y < zIn   [#permalink] 18 Aug 2017, 09:10
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