Let number of 10 paisa, 20 paisa and 50 paisa coins be ‘9x’, ‘7x’ and ‘5x’, respectively. ATQ, (0.1 × 9x) + (0.2 × 7x) + (0.5 × 5x) = 48 Or, 0.9x + 1.4x + 2.5x = 48 Or, 4.8x = 48 So, x = 10 So, number of 20 paisa coins = 7x = 70 Therefore, required number of coins = 0.40 × 70 = 28
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