A) \[5\frac{5}{11}min.past4/4\]
B) \[54\frac{6}{11}min.past4/4\]
C) \[8\frac{2}{11}min.past4/4\]
D) \[10\frac{10}{11}min.past4/4\]
Correct Answer: B
Solution :
At 4 o' clock, the hour hand is at 4 and the min hand is at 12 i.e., the two hands are 20 min. spaces apart. To be in the 180° angle they have to make 30 min. space. So, the min. hand will have 20 gain 20 + 30 = 50 min. spaces over the hour hand Now, 55 min. are gained in 60 min. 50 min. will be gained in \[\left[ \frac{60}{55}\times 50 \right]mn.\] = \[\frac{600}{11}=54\frac{6}{11}\min .\] The hands will make 180° angle at \[\mathbf{54}\frac{\mathbf{6}}{\mathbf{11}}\mathbf{min}\mathbf{.}\,\,\mathbf{past}\,\mathbf{4}\]
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