Binary logic,Clock And Calenders | Reasoning | Discussion Page

Find the tim between 2 and 3 O'clock at which the minute hand and the hour hand are perpendicular to each other.

A) $10\frac{27}{11}$min past 2

B) $27\frac{10}{11}$min past 2

C) $27\frac{3}{11}$min past 2

D) $3\frac{10}{11}$min past 2

Option C

When the two hands are perpendicular $\theta = 90^{o}$ and h=2

$\therefore \theta=$$(\frac{11}{2}m-30h)$ or $(30h-\frac{11}{2}m)$

$90=\frac{11}{2}m-30\times 2$

$\frac{11}{2}m=150$

$m=\frac{300}{11}$

$27\frac{3}{11}$ min past $2$.

or $90=30 \times 2-\frac{11}{2}$

$\frac{11}{2}=-30$

As m cannot be negative,this case is not possible.

so, the hands are perpendicular to each other only once ie, at $27\frac{3}{11}$ min past $2$.

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