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Wouldn't combining both A and B give us the required info to solve this question.
It is given that (y-x) > (b-a) and x = 3a
-> y-3a > b-a
-> y> b+2a

for (y+x)/2 > (b+2a+3a)/2
-> (y+x)/2 > (b+5a)/2 > (b+a)/2


Hi,

The highlighted portion says that the range of x & y is greater than the range of a & b.
Consider x as 6. y as 8. a as 17 and b as 18.
Now y-x = 2 > b-a = 1.

But Y>b+2a = 8>18+34. This is not true.

Hope it helps. :)
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the example that you have mentioned does not satisfy the constraint that x=3a. I was menntioning what would happen if we combine both the statements. Hope you can help.
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Hi, [email protected] maybe this helps you understand why both statements aren't sufficient.
From statements one and two, we know that x = 3a and that the range of x and y > range of a and b. We can first determine the potential values for x and a. Because x, y, a, and b are all between zero and twenty, there are only a finite number of values for x and a.

x = 3a a =1, x = 3 a = 2, x = 6 a = 3, x = 9 a = 4, x = 12 a = 5, x =15 a = 6, x =18
Using the values for a and x and the information from statement two, we can still create contradictory scenarios:
Let a =1 and b = 5. The range will be 4, and the average will be 3.
Let x = 3 and y = 8. The range will be 5, and the average will be 5.5. Average of a and b < Average of x and y Now let a = 6 and b = 19. The range will be 13, and the average will be 12.5. Let x = 18 and y = 1. The range will be 17, and the average will be 9.5. Average of a and b > Average of x and y Statements one and two together are not sufficient.
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