Clocks Questions and Answers

Answer: A

The hands cross 30 times.

The second hand turns at 6° per second and the minute hand at 0.1° per second, so the second hand gains 5.9° per second. At 09:15:00 the minute hand points at 3 (90°) and the second hand at 12 (0°), so the second hand starts 90° behind. It overtakes the minute hand each time it has gained 90° + 360°k:

\[ \begin{aligned} t_{k} &= \frac{90 + 360k}{5.9} \le 1800 \\ 90 + 360k &\le 10620 \Rightarrow k \le 29.25 \end{aligned} \]

So k = 0, 1, ..., 29: 30 crossings, the first about 15 s after 09:15:00 and the last about 15 s before 09:45:00. Over 30 minutes the second hand gains 29.5 laps, and 29 is that figure rounded down, missing the crossing that the 90° starting gap adds. 31 counts one lap too many, and 15 is half the true count.

Answer: C

Rohit spent \(65\frac{5}{11}\) minutes at the restaurant.

The minute hand moves 6° per minute and the hour hand 0.5° per minute, so the minute hand gains 5.5° per minute. To coincide again it must gain a full 360°:

\[ \begin{aligned} t &= \frac{360}{5.5} = \frac{720}{11} \\ &= 65\frac{5}{11}\ \text{min} \end{aligned} \]

He arrived when the hands met just after 1 o'clock (at about 1:05) and left when they next met (at about 2:11), so the time between is one full coincidence interval.

The other mixed numbers do not equal 720/11; the ones with 13 in the denominator are not based on the hands' relative speed at all.

Answer: A

The true time will be 11 p.m.

From 5 a.m. on the first day to 10 p.m. on the fourth day, the clock shows 3 days and 17 hours, which is 89 hours. The clock loses 16 minutes a day, so it shows 1424 minutes for every 1440 true minutes.

\[ \begin{aligned} \text{true} &= 89 \times \frac{1440}{1424} \\ &= 89 \times \frac{90}{89} = 90\ \text{hours} \end{aligned} \]

90 true hours after 5 a.m. on day 1 is 11 p.m. on day 4, an hour later than the clock shows. A losing clock always shows less time than has passed, so the true time is later than 10 p.m., which rules out 8 p.m. and 9 p.m.; 12 p.m. overstates the loss.

Answer: A

Both Statement I and Statement II are true.

The hour hand moves 30° per hour, so 135° is \(135 / 30 = 4.5\) hours. From 3:00 pm that is 7:30 pm. Statement I is true.

For Statement II use the clock-angle formula \(|30H - 5.5M|\):

\[ |30 \times 4 - 5.5 \times 25| = |120 - 137.5| = 17.5^\circ \]

Statement II is true. The trap is to forget that the hour hand moves on from the 4 during the 25 minutes (by 12.5°).

Answer: C

The hands coincide at \(32\frac{8}{11}\) minutes past 6.

At 6:00 the minute hand is 30 minute-spaces behind the hour hand. It gains 55 minute-spaces every 60 minutes, so it catches up after

\[ \begin{aligned} 30 \times \tfrac{60}{55} &= \tfrac{360}{11} \\ &= 32\tfrac{8}{11}\ \text{min} \end{aligned} \]

The general rule is \(\frac{60}{11} \times 5h\) minutes past \(h\) o'clock. The other options correspond to other hours or arithmetic slips.

Answer: C

The angle is 245°, the reflex angle.

The angle between the hands at \(H\) hours \(M\) minutes is \(|30H - 5.5M|\).

\[ \begin{aligned} |30 \times 10 - 5.5 \times 10| &= |300 - 55| \\ &= 245^\circ \end{aligned} \]

The smaller angle between the hands is \(360^\circ - 245^\circ = 115^\circ\), which is not among the options, so the keyed answer is the reflex angle of 245°. 240° (or 120°) is what you get by leaving the hour hand fixed on the 10, forgetting that it moves 5° past it in 10 minutes.

Answer: A

The hands point in opposite directions at \(49\frac{1}{11}\) minutes past 3.

At 3 o'clock the hour hand is 15 minute-divisions ahead of the minute hand. In m minutes the minute hand moves m divisions and the hour hand \(\frac{m}{12}\). For the hands to be opposite, the minute hand must be 30 divisions ahead of the hour hand:

\[ \begin{aligned} m &= 15 + \frac{m}{12} + 30 \\ \frac{11m}{12} &= 45 \\ m &= \frac{540}{11} = 49\frac{1}{11} \end{aligned} \]

The minute hand gains \(5\frac{1}{2}\) degrees a minute and must gain 90 + 180 = 270 degrees, which gives the same \(\frac{270}{5.5} = 49\frac{1}{11}\) minutes. 50 minutes ignores the hour hand's movement, and the other fractions come from arithmetic slips.

Answer: C

The watch will show 8:25 p.m.

From 7:00 a.m. Monday to 7:00 p.m. the next Monday is 7 days and 12 hours, that is 7.5 days. At 12 minutes a day the watch gains 7.5 × 12 = 90 minutes.

It started 5 minutes slow, so its net error at the end is 90 - 5 = 85 minutes fast. The true time is 7:00 p.m., so the watch shows 7:00 p.m. + 1 hour 25 minutes = 8:25 p.m.

8:30 p.m. ignores the initial 5 minutes slow, and the 5 p.m. options treat the gain as a loss.

Answer: C

The hands are perpendicular 44 times in a day.

The minute hand gains 330° on the hour hand every hour (360° - 30°), so it gains a full circle on it 11 times in 12 hours. In each such lap, the hands are at right angles twice: once when the minute hand is 90° ahead and once when it is 270° ahead (90° behind).

\[ \begin{aligned} \text{in 12 h} &= 11 \times 2 = 22 \\ \text{in 24 h} &= 2 \times 22 = 44 \end{aligned} \]

22 is the count for 12 hours only. 24 and 48 come from assuming two right angles every hour, which ignores that the hands meet only 11 times, not 12, in 12 hours.

Answer: D

The smaller angle is 100°.

The angle between the hands at H hours M minutes is \(|30H - 5.5M|\):

\[ \begin{aligned} |30 \times 4 - 5.5 \times 40| &= |120 - 220| \\ &= 100^\circ \end{aligned} \]

At 4:40 the minute hand points at 8 (240°) and the hour hand has moved two-thirds of the way from 4 to 5 (140°), so they are 100° apart. Forgetting the hour hand's movement gives 120°, which is not offered.