| At what time between 9 O'clock and 10 O'clock, will the hands of a clock be in the same straight line but not together? |
A) \[16\frac{2}{11}\text{min}\,\,\text{past}\]9
B) \[16\frac{2}{11}\text{min}\,\,\text{past}\,\,\text{10}\]
C) \[16\frac{4}{11}\text{min}\,\,\text{past}\,\,9\]
D) \[16\frac{4}{11}\text{min}\,\,\text{past}\,\,10\]
Correct Answer: C
Solution :
| At 9 O'clock, the hour hand is at 9 and the minute hand is at 12. It means that the two hands are 15 min spaces apart. To be in the same straight line (but not together), they will be'30 min space apart. |
| \[\therefore \]The minute hand will have to gain\[(30-15)=15\min \] spaces over the hour hand. |
| As we know, 55 min spaces are gained in 60 min. |
| \[\therefore \]15 min will be gained in |
| \[\left( \frac{60}{55}\times 15 \right)\min =\frac{180}{11}=16\frac{4}{11}\min \] |
| Hence, the hands will be in the same straight line but not together at \[16\frac{4}{11}\]min past 9. |
| Alternate Method |
| Here, \[n=9\]and \[n+1=10\,\,(n>6)\] |
| The hands will be in the same straight line at |
| \[(5n-30)\times \frac{2}{11}\min \]past n |
| \[=(5\times 9-30)\frac{12}{11}\min \text{past}\,\,9\] |
| \[=\frac{15\times 12}{11}\min \,\,\text{past}\,\,9\] |
| \[=\frac{180}{11}\min \,\,\text{past 9}\] |
| \[=16\frac{4}{11}\min \text{past}\,\,9\] |
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