In Pearson's Square, the middle value must be intermediate between the two values on the left side.

In Pearson's Square, the middle value must be intermediate between the two values on the left side.

• A. Always true
• B. Never true
• C. Only if the left values are equal
• D. Irrelevant

Answer: (A)

In Pearson's Square, the middle value represents the target nutrient percentage you want in the final mix, and the two values on the left are the nutrient percentages of the two feeds you’re blending. The mix you can produce with those two feeds will always fall between those two ingredient percentages because the final percentage is a weighted average of them. If the target lies outside that range, you can’t reach it with just those two feeds; you’d need a third ingredient or a different approach. If the target exactly matches one of the left values, you would simply use that feed all by itself (the other proportion becomes zero), which is still a valid edge case of the same principle. So the statement is true in all cases, including when the target equals an end value. The other options don’t fit because the target is not irrelevant, and it isn’t restricted only to the equal-left-value scenario, and it isn’t never true.

In Pearson's Square, the middle value represents the target nutrient percentage you want in the final mix, and the two values on the left are the nutrient percentages of the two feeds you’re blending. The mix you can produce with those two feeds will always fall between those two ingredient percentages because the final percentage is a weighted average of them. If the target lies outside that range, you can’t reach it with just those two feeds; you’d need a third ingredient or a different approach. If the target exactly matches one of the left values, you would simply use that feed all by itself (the other proportion becomes zero), which is still a valid edge case of the same principle. So the statement is true in all cases, including when the target equals an end value. The other options don’t fit because the target is not irrelevant, and it isn’t restricted only to the equal-left-value scenario, and it isn’t never true.