In a triangle ABC, PQ is drawn parallel to BC such that points P and Q lie on AB and AC respectively, If PQ ∶ BC = 2 ∶ 5, then AP ∶ PB is:  

1
\(\frac{2}{3}\)
2
\(\frac{2}{5}\)
3
\(\frac{3}{5}\)
4
\(\frac{3}{2}\)

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