CSAT Solved Papers/ 2021/Q57

2021 CSAT — Q57

Quant Logical & quantitative reasoning 2.5 marks Medium

At which one of the following times, do the hour hand and the minute hand of the clock make an angle of $180^\circ$ with each other?

A At 7:00 hours
B Between 7:00 hours and 7:05 hours
C At 7:05 hours
D Between 7:05 hours and 7:10 hours Answer

Worked rationale

At $t$ minutes past $7{:}00$ : the hour hand is at $210^\circ + 0.5t$ (it starts at $7\times 30 = 210^\circ$ ), the minute hand at $6t$ . The angle between them is

|\,(210 + 0.5t) - 6t\,| = |210 - 5.5t|.

Set this to $180^\circ$ :

210 - 5.5t = 180 \;\Rightarrow\; 5.5t = 30 \;\Rightarrow\; t = \tfrac{60}{11} \approx 5.45\ \text{min}.

(The other root $210 - 5.5t = -180$ gives $t\approx 70.9$ min, outside this hour.) So the hands are $180^\circ$ apart at about $7{:}05{:}27$ , i.e. between $7{:}05$ and $7{:}10$ .

Answer: (d) Between 7:05 hours and 7:10 hours.

Why the other options miss

A

the wrong assumption: thinks the hands are opposite at the top of the hour; at $7{:}00$ the angle is $210^\circ$ , not $180^\circ$ .
B

an arithmetic slip: solves $5.5t = 30$ as $t < 5$ (e.g. uses $6t$ relative speed instead of $5.5t$ ).
C

rounds too early: rounds $t = 60/11 \approx 5.45$ down to exactly $5$ minutes.

Specialist insight

The minute hand gains on the hour hand at $6 - 0.5 = 5.5^\circ$ per minute, and at $7{:}00$ it trails by $210^\circ$ . To reach a straight line ( $180^\circ$ apart) it must close $210 - 180 = 30^\circ$ , taking $30/5.5 = 60/11 \approx 5.45$ min. The decisive detail is the relative speed $5.5^\circ/\text{min}$ (not $6$ ) and recognising $60/11$ lands just past $7{:}05$ . Use the gain-rate, not absolute positions, and the band falls out immediately.

The trap, in one line

Close the $210^\circ$ lead to $180^\circ$ at $5.5^\circ$ /min: $t = 30/5.5 = 60/11 \approx 5.45$ min $\Rightarrow$ between 7:05 and 7:10 $\Rightarrow$ (d).

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