Eliminating Alcohols: Why Do (A) and (B) Give Different Products?

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baldbrain
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Homework Statement
I just have a problem with (A) & (B)
Matching table of E2 elimination reactions with alc.KOH and heat: four alkyl bromide structures (A) to (D) matched to Column II product counts (p) 0, (q) 1, (r) 2, (s) 3.

The attempt at a solution
Analysing all possible products;
(A) can give 1-pentene (minor) + 2-pentene (major). But you also have the E-Z diastereomers of 2-pentene. So that's total 3 possible products.
(B) will give only the diastereomers E & Z 2-pentene. So 2 possible products
Hence, (A)→(s) & (B)→(r)
But they've given (A)→(r) & (B)→(s)
How can it be?
 
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  • Matching table of E2 elimination reactions with alc.KOH and heat: four alkyl bromide structures (A) to (D) matched to Column II product counts (p) 0, (q) 1, (r) 2, (s) 3.
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Are you sure? No hitch? We aren't missing anything?
 
Nope. S, R, Q, P.
 
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