Find at what time between 9 to 10 o'clock will the hands of a clock be in the same straight line but not together.

9 : 16\(\frac{4}{11}\)

Use this formula for this type of question: (5n - 30)× \(\frac{12}{11}\), where n is the minimum value of time which is given in the question.

Here, n = 9

Therefore,

The hand will be in straight line but not together after = (5 × 9 - 30) × \(\frac{12}{11}\) = 15 × \(\frac{12}{11}\) = × \(\frac{180}{11}\) = 16\(\frac{4}{11}\) min

The hand will be in straight line but not together at = 9 : 16\(\frac{4}{11}\)