Let the first number be ‘5x’ So, second number = (7/5) × 5x = ‘7x’ And, third number = (9/7) × 7x = ‘9x’ ATQ, (5x + 7x + 9x) = 294 Or, 21x = 294 Or, x = (294/21) = 14 Now, the first number = 5x = 5 × 14 = 70 And, third number = 9x = 9 × 14 = 126 So, the required difference = (126 – 70) = 56
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