A contractor's insurance policy has a $2 million per occurrence limit with a $5 million aggregate limit. Three separate incidents occur with damages of $1.8 million, $1.2 million, and $2.5 million respectively. How much will the insurance company pay in total?
Correct Answer
B) $5.0 million
Per occurrence payments: $1.8M + $1.2M + $2.0M (limited) = $5.0M, which equals the aggregate limit.
Why This Is the Correct Answer
Option B ($5.0 million) is correct. Each claim is capped at the $2M per-occurrence limit, then the aggregate acts as a second ceiling. Incident 1: $1.8M paid in full (under $2M limit). Incident 2: $1.2M paid in full (under $2M limit). Running total: $3.0M. Incident 3: $2.5M would be limited to $2.0M per-occurrence, but only $2.0M remains under the $5M aggregate, so exactly $2.0M is paid. Total: $1.8M + $1.2M + $2.0M = $5.0M, which equals the aggregate limit.
Why the Other Options Are Wrong
Option A: $3.0 million
$3.0 million is incorrect. This would only account for the first two incidents ($1.8M + $1.2M) and ignore the third incident entirely, or it represents the first incident capped incorrectly. The aggregate has not yet been exhausted after two incidents.
Option C: $5.5 million
$5.5 million is incorrect. This adds all three incidents at face value ($1.8M + $1.2M + $2.5M = $5.5M) without applying either the per-occurrence limit to the third incident or the $5M aggregate cap. Both limits must be applied.
Option D: $4.5 million
$4.5 million is incorrect. This may result from incorrectly limiting the third incident to the per-occurrence cap ($2.0M) but then miscalculating the total ($1.8 + $1.2 + $1.5 = $4.5M), or from some other error. The aggregate cap of $5M governs the final answer.
Memory Technique
Think of the per-occurrence limit as a 'bucket' per incident and the aggregate as the 'tank' that holds all buckets. Each bucket can hold at most $2M. The tank holds at most $5M total. Fill each bucket (up to $2M), pour into the tank, stop when the tank is full.
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