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I like the solution - it’s helpful.
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I don’t quite agree with the solution. I also arrived at all 3 cases. However, for y=2x which is in the middle range, y can take only those values which exist for x being in that range, which is -10, -8, -6, -4, -2, 0, 2, 4, 6 and 8 and not 19 values
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I don’t quite agree with the solution. I also arrived at all 3 cases. However, for y=2x which is in the middle range, y can take only those values which exist for x being in that range, which is -10, -8, -6, -4, -2, 0, 2, 4, 6 and 8 and not 19 values

You are missing a point. Let’s break it down cleanly:

In the middle case, when -5 < x < 5, we have y = 2x.

  • Since x can be any real number in that interval, y can be any real number between −10 and 10.
  • That means y is not restricted to even integers only. It can hit all integer values from −9 up to 9, inclusive.
  • So in this interval, y covers 19 distinct integer values.

Your reasoning (-10, -8, ..., 8) would be correct if x were restricted to integers, but here x is any real number. That’s why the solution’s count of 19 is correct.

If still not clear you can check alternative solutions here: https://gmatclub.com/forum/if-y-x-5-x-5 ... 73626.html

Hope it helps.
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Hey. Understood the miss on my part. This explanation clears up things. Thanks a lot!
Bunuel


You are missing a point. Let’s break it down cleanly:

In the middle case, when -5 < x < 5, we have y = 2x.

  • Since x can be any real number in that interval, y can be any real number between −10 and 10.
  • That means y is not restricted to even integers only. It can hit all integer values from −9 up to 9, inclusive.
  • So in this interval, y covers 19 distinct integer values.

Your reasoning (-10, -8, ..., 8) would be correct if x were restricted to integers, but here x is any real number. That’s why the solution’s count of 19 is correct.

If still not clear you can check alternative solutions here: https://gmatclub.com/forum/if-y-x-5-x-5 ... 73626.html

Hope it helps.
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