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- Fri Jun 03, 2016 11:27 am
#26045
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=107&t=7219)
The correct answer choice is (D)
This Global question asks us to identify a combination of subzones that would be disallowed from any single zone. The process of elimination is likely to prove useful: four of the answer choices will contain subzone designations that do not violate any of the rules. To maximize the chances of identifying the correct answer choice more quickly, begin by examining the answer choices that contain the largest number of designated subzones. This approach is likely to prove advantageous, because the greater the number of variables within a group, the higher the chance that they would violate one of the rules requiring their separation. The optimal way to begin, then, would be to analyze answer choices (D) and (E) first.
Answer choice (A): The designation of two subzones—one H and one R—within the same zone does not violate any of the rules, as long as the zone in question is Z2 or Z3. This answer choice is incorrect.
Answer choice (B): The designation of two subzones—one I and one R—within the same zone does not violate any of the rules, as long as the zone in question is Z2 or Z3. This answer choice is incorrect.
Answer choice (C): The designation of three subzones, all of them R, within the same zone does not violate any of the rules, as long as the zone in question is Z2 or Z3. This answer choice is incorrect.
Answer choice (D): This is the correct answer choice. If four subzones are designated within a single zone, and none of these subzones is an I, we would need to designate a mix of H and R subzones within that zone. However, recall that if an H subzone is designated in any zone, that zone can contain a maximum of 3 subzones (H, H, R). This is because no zone contains more than 2H’s, no zone contains both an H and an I, and no zone with an H contains more than 1R. In other words, designating four subzones without the use of I subzones would be impossible, proving that answer choice (D) cannot be true.
Answer choice (E): It is possible to designate four subzones—two R’s and two I’s—within the same zone, as long as the zone in question is Z2 or Z3. This answer choice is incorrect.
(The complete setup for this game can be found here: lsat/viewtopic.php?f=107&t=7219)
The correct answer choice is (D)
This Global question asks us to identify a combination of subzones that would be disallowed from any single zone. The process of elimination is likely to prove useful: four of the answer choices will contain subzone designations that do not violate any of the rules. To maximize the chances of identifying the correct answer choice more quickly, begin by examining the answer choices that contain the largest number of designated subzones. This approach is likely to prove advantageous, because the greater the number of variables within a group, the higher the chance that they would violate one of the rules requiring their separation. The optimal way to begin, then, would be to analyze answer choices (D) and (E) first.
Answer choice (A): The designation of two subzones—one H and one R—within the same zone does not violate any of the rules, as long as the zone in question is Z2 or Z3. This answer choice is incorrect.
Answer choice (B): The designation of two subzones—one I and one R—within the same zone does not violate any of the rules, as long as the zone in question is Z2 or Z3. This answer choice is incorrect.
Answer choice (C): The designation of three subzones, all of them R, within the same zone does not violate any of the rules, as long as the zone in question is Z2 or Z3. This answer choice is incorrect.
Answer choice (D): This is the correct answer choice. If four subzones are designated within a single zone, and none of these subzones is an I, we would need to designate a mix of H and R subzones within that zone. However, recall that if an H subzone is designated in any zone, that zone can contain a maximum of 3 subzones (H, H, R). This is because no zone contains more than 2H’s, no zone contains both an H and an I, and no zone with an H contains more than 1R. In other words, designating four subzones without the use of I subzones would be impossible, proving that answer choice (D) cannot be true.
Answer choice (E): It is possible to designate four subzones—two R’s and two I’s—within the same zone, as long as the zone in question is Z2 or Z3. This answer choice is incorrect.
