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ATQ, Let the number of Pen and Marker bought be ‘p’ and ‘b’, respectively According to the question, 6p + 5b = 100 Since, 5b and 100 both are multiples of 5 therefore, ‘p’ must be a multiple of 5 Therefore, possible values of p = 0, 5, 10, 15 And, corresponding values of b = 20, 14, 8, 2 Therefore, he could have bought the items in 4 ways.
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