After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
Question:
After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
Options
Answer: 25
Explanation:
Step 1: Set up equation (1 + x/100)(1 + 2x/100) = 1.875 Expand: 1 + 3x/100 + 2x²/10000 = 1.875 3x/100 + 2x²/10000 = 0.875 Multiply both sides by 10000: 300x + 2x² = 8750 Divide by 2: x² + 150x − 4375 = 0 Step 2: Solve quadratic x = [−150 ± √(150² − 4 × 1 × (−4375))]/2 x = [−150 ± √(22500 + 17500)]/2 x = [−150 ± √40000]/2 x = [−150 ± 200]/2 Positive solution: x = (−150 + 200)/2 = 25
Explanation:
Step 1: Set up equation (1 + x/100)(1 + 2x/100) = 1.875 Expand: 1 + 3x/100 + 2x²/10000 = 1.875 3x/100 + 2x²/10000 = 0.875 Multiply both sides by 10000: 300x + 2x² = 8750 Divide by 2: x² + 150x − 4375 = 0 Step 2: Solve quadratic x = [−150 ± √(150² − 4 × 1 × (−4375))]/2 x = [−150 ± √(22500 + 17500)]/2 x = [−150 ± √40000]/2 x = [−150 ± 200]/2 Positive solution: x = (−150 + 200)/2 = 25
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