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What is the sum of the predecessor of the number 1320 and the successor of the number 42906?

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44226

44227

44229

44224

answer is C.

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Detailed Solution

Given numbers = 1320 and 42906

∴ Predecessor of 1320 = 1320 − 1 = 1319

Successor of 42906 = 42906 + 1 = 42907

Thus, required sum = 1319 + 42907 = 44226.

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