A) \[16\frac{4}{11}\] min past 6
B) \[49\frac{1}{11}\] min past 6
C) \[50\frac{3}{11}\]min past 6
D) Both A or B
Correct Answer: D
Solution :
At 6 O'clock, the hour hand is at 6 and the min. hand is at 12 i.e. the two hands are 30 min. spaces apart. For perpendicular to each other min. hand to cover 15 min. and 45 min. space. Now, 55 min. are gained in 60 min. I. 15 min. will be gained in \[\frac{60}{55}\times 15\] min. \[=\frac{180}{11}\min .\,\,\,=\,\mathbf{16}\frac{\mathbf{4}}{\mathbf{11}}\mathbf{min}\mathbf{.}\] II. 45 mm. will be gained in \[\frac{60}{55}\times 45\] mm. \[=\frac{540}{11}\min =\mathbf{49}\frac{\mathbf{1}}{\mathbf{11}}\mathbf{min}.\] Hence, both hands are perpendicular at \[\underline{\mathbf{16}\frac{\mathbf{4}}{\mathbf{11}}\mathbf{min}\mathbf{. past 6 and 49}\frac{\mathbf{1}}{\mathbf{11}}\mathbf{min}\mathbf{. past 6}}\]
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