I. y/16 = 4/y
II. x3= (2 ÷ 50) × (2500 ÷ 50) × 42× (192 ÷ 12)
I. y/16 = 4/y y 2 = 16 × 4 y 2 = 64 y = ± 8 II. x3 = (2 ÷ 50) × (2500 ÷ 50) × 42 × (192 ÷ 12) x3 = 2/50 × 2500/50 × 42 × 192/12 x3 = 512 x = 8 Hence, x ≥ y
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