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Sum of marks scored by 'A', 'B', 'C', 'D' and 'E' = 58 X 5 = 290 Sum of marks scored by 'A'. 'C' and 'D' = 55 X 3 = 165 So, sum of marks scored by 'B' and 'E' = 290 - 165 = 125 Let the marks scored by 'B' = 'x' Then, marks scored by 'E' = (x + 33) We have, x + x + 33 = 2x + 33 = 125 So, x = (125 - 33) ÷ 2 = 46 Hence, option d.
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