A) \[\text{5}\frac{\text{5}}{\text{11}}\text{,}\,\,\text{38}\frac{\text{2}}{\text{11}}\,\,\text{min}\,\,\text{past}\,\,\text{4}\]
B) \[\frac{\text{60}}{\text{11}}\text{,}\,\,\frac{\text{210}}{\text{11}}\,\,\text{min}\,\,\text{past}\,\,\text{4}\]
C) \[\frac{\text{70}}{\text{11}}\text{,}\,\,\frac{\text{240}}{\text{11}}\text{min}\,\,\text{past}\,\,\text{4}\]
D) \[\frac{\text{80}}{\text{11}}\text{,}\,\,\frac{\text{250}}{\text{11}}\text{min}\,\,\text{past}\,\,\text{4}\]
Correct Answer: A
Solution :
| At 40'clock, there is 20 min of space between minute hand and hour hand. |
| So, angle difference \[=20\times 6{}^\circ =120{}^\circ \] |
| To make it\[90{}^\circ \], minute hand must move\[30{}^\circ \]. |
| Relative angle \[=5\frac{1{}^\circ }{2}\] |
| \[\therefore \]\[\frac{30}{5.5}=\frac{300}{55}=\frac{60}{11}=5\frac{5}{11}\,\,\text{min}\,\,\text{past}\,\,\text{4}\] |
| Between 4 and 5 O' clock, minute hand makes another \[90{}^\circ \] with hour hand. |
| So, total angle to cover \[=90{}^\circ +120{}^\circ =210{}^\circ \] |
| Time to cover this distance \[=\frac{210}{5.5}\] |
| \[=\frac{210\times 10}{55}=\frac{420}{11}=38\frac{2}{11}\min \] |
| Between 4 and 5 O' clock, right angle between hour hand and minute hand is measured twice one at \[5\frac{5}{11}\min \] past 4 and another at \[38\frac{2}{11}\min \]past 4. |
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