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18 Jul 2019, 23:44
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In the figure, is a < b < c ?
(1) x = (y + z)/2
(2) y = (z − x)/2
Attachment:
2019-07-19_1043.png [ 14.68 KiB | Viewed 1624 times ]
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20 Jul 2019, 22:52
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Bunuel
In the figure, is a < b < c ?
(1) x = (y + z)/2
(2) y = (z − x)/2
Attachment:
2019-07-19_1043.png
For a<b<c to be true, we need x<y<z to be true.
Note that x,y,z can only be positive integers
Statement 1
Since x is the average of y and z, it lies between y and z. Therefore x is not the smallest. So, x<y<z is not true.
Sufficient
Statement 2 - I am confused if this is a correct question because of the following reason.
y = (z-x)/2
As per the equation, z is definitely greater than x. We need to know if y is definitely greater than x or not.
Taking cases ( keep in mind that the sum of 2 sides of a triangle is always greater than the third side)
If z = 10, x =8, then y =2.
This set of sides is not possible since x+y is not greater than z. Let's try another
z = 18, x = 10, y = 4.
This set is also not possible for the same reason. Let's try another
z=6, x =3, y = 1.5. This set is also not possible for the same reason
=> Statement 2 is not giving us a valid set of side values for a triangle.
I would choose A as the answer because of this reason.
08 Feb 2022, 00:21
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Bunuel
In the figure, is a < b < c ?
(1) x = (y + z)/2
(2) y = (z − x)/2
If an angle is the smallest in a triangle, the side opposite to it will be the smallest.
a is smallest means, the side opposite x should be the smallest.
(1) \(x = \frac{(y + z)}{2}\)
This means x is the average of y and z. So, either x is between y and z OR all are equal.
In both cases, answer is NO for a<b<c.
Sufficient
(2) \(y = \frac{(z − x)}{2}........z=2y+x\)
Surely z is more than both x and y.
If y>x, the answer is yes, and if x>y, the answer is no.
However, the statement is not valid as any side has to be more than the difference of other two sides.
So, y>z-x, but z-x=2y...........y>2y Not possible for a positive number.
A.
