Clock

What the exam asks. A time such as 4:20 or 7:24 is given and the angle between the two hands is asked. The same idea also comes as a movement question (through how many degrees does a hand turn in an interval?), as a change of the angle over a few minutes, and with a second hand.

Why it works. Both hands are measured from the 12. The minute hand stands at 6M degrees and the hour hand at 30H + M/2 degrees, so the angle between them is simply the difference.

The method. Find the two positions, subtract, and fold anything above 180° with 360° − θ. The three sub-cases at the end only change what goes into M, or ask for a turn instead of a position.

At H hours M minutes the hour hand is at 30H + M/2 degrees and the minute hand at 6M degrees. Their difference is

θ = |30H + M/2 − 6M| = |30H − 5.5M|.

[[fig:0]]

The angle between the hands is the smaller one, so if θ comes out above 180°, take 360° − θ. The bigger one is the reflex angle, and it belongs to the same instant. At 2:48, θ = |60 − 264| = 204°, so the angle between the hands is 360° − 204° = 156° and the reflex angle is 204°. If the question says "larger angle" or "reflex", give 204°.

Three quick cases.

• Minutes zero. The angle is 30 × H, folded to 180° or less. 4:00 gives 120°, 6:00 gives 180°, 9:00 gives 270°, so the angle is 90°.
• Seconds given. Put them in M as a fraction. At 6:30:30, M = 30.5, so θ = |180 − 167.75| = 12.25°. The larger angle is 360° − 12.25° = 347.75°.
• 24-hour time. Convert first. At 17:36 the dial reads 5:36, so H = 5.

[[fig:1]]

Never forget that the hour hand drifts. At 5:30 it is not on the 5, it is halfway to the 6, so the angle is 15°, not 0 or 30°. Mixed numbers also need care: for 7:46411, write M = 510/11, so the angle is |210 − 5.5 × 510/11| = |210 − 255| = 45°. Angle = |30H − 5.5M|; if it is above 180°, subtract it from 360°.

Sub-case A: the angle turned by a hand in an interval

A hand's turn is its speed multiplied by the time it runs. The minute hand turns 6° per minute and the hour hand turns 0.5° per minute, so the only job is to find the length of the interval in minutes.

[[fig:2]]

Take the minute hand from 8:00 to 8:40. The interval is 40 minutes, so it turns 40 × 6 = 240°. The picture shows this: the hand starts at the 12 and ends at the 8, and 8 × 30° = 240°. The hour hand in the same 40 minutes turns 40 ÷ 2 = 20°.

A minute hand turns exactly 12 times as much as the hour hand in the same interval, because 6 ÷ 0.5 = 12. So if the hour hand has turned 60°, the minute hand has turned 720°, two complete rounds. Convert the interval to minutes first. Hour hand: minutes ÷ 2. Minute hand: minutes × 6.

Sub-case B: the change of the angle over a short interval

The minute hand gains on the hour hand at 5.5° per minute. So over any interval, the lead of the minute hand changes by

Change = 5.5 × (minutes).

From 12:00 to 12:16 the minute hand turns 16 × 6 = 96° and the hour hand turns 16 × 0.5 = 8°. At 12:00 the hands are together, so at 12:16 the angle between them is 96 − 8 = 88°, which is also 16 × 5.5. In the same way, from 7:45 to 7:47 the angle changes by 2 × 5.5 = 11°.

One catch. The "angle between the hands" is the smaller angle, and it folds back at 0° and 180°. The multiplication 5.5 × minutes gives the change in the lead, which is the same as the change in the angle only when the hands do not become together or opposite during the interval. From 7:30 to 7:50 the lead goes from −45° to +65°, a change of 20 × 5.5 = 110°. But the angle between the hands goes from 45° to 65°, because the hands met (at 7:38211) in between and the minute hand then went ahead. If a question asks "by how many degrees does the minute hand gain on the hour hand", answer 110°. If it asks about the angle between the hands, check for a fold first. Change in the lead = 5.5° × minutes.

Sub-case C: the second hand and times with seconds

The second hand goes round the whole dial in one minute, so it turns 360° per minute, that is 6° per second. Compare the three hands:

To connect the second hand with the other hands, change its degrees into minutes: time = degrees ÷ 360 (in minutes). Then the minute hand turns 6° for every minute of that time. For example, while the second hand turns through 480°, the time is 480 ÷ 360 = 4/3 minutes, and the minute hand turns 6 × 4/3 = 8°.

When a time includes seconds, put the seconds into M as a fraction (36 seconds is 0.6 minute). At 9:15:36 the three hands are at: hour hand = 270 + 7.8 = 277.8°, minute hand = 6 × 15.6 = 93.6°, second hand = 6 × 36 = 216°. The angle between the hour and minute hands is 277.8 − 93.6 = 184.2°, so the angle between them is 360 − 184.2 = 175.8°.

And for the second hand against the minute hand at 4:20:30: second hand at 6 × 30 = 180°, minute hand at 6 × 20.5 = 123°, so the angle is 57°. Second hand: 6° per second. Minute hand in the same time: 0.1° per second.

What the exam asks. "At what time between 5 and 6 o'clock are the hands 30° apart?" or "for the first time after 2:00 PM the hands make 30°". This is the reverse of the previous type: the angle is known and the time is the unknown.

Why it works. The minute hand gains 5.5° every minute, so the angle between the hands changes at a steady rate. Inside one hour block, "the hands are θ apart" simply means that the minute hand's lead equals θ, or −θ, or the same angle measured the long way round. One equation then gives the time.

The method. Use M = (2/11)(30H + r) and keep 0 ≤ M < 60. The sub-cases that follow are the same idea: a gap given in minute-spaces, and two [New Pattern] forms (hands equally far from a numeral, hands exchanged) which have their own short formulas.

The minute hand gains 5.5° every minute, so the angle between the hands changes steadily. In the block between H and H + 1 o'clock, the minute hand's lead is d = 5.5M − 30H. The hands make the angle θ when d equals

• +θ: the minute hand is θ ahead of the hour hand,
• −θ: the minute hand is θ behind,
• ±(360° − θ): the same angle measured the long way round.

Solving 5.5M − 30H = r gives one formula for all of them:

M = (2/11) × (30H + r), where r is any of +θ, −θ, +(360 − θ), −(360 − θ).

Keep only the values with 0 ≤ M < 60. A value outside that range belongs to a neighbouring hour block, so we throw it away for this block.

[[fig:0]]

The graph is for the 2 o'clock block (H = 2, so 30H = 60). The lead d starts at −60° and climbs to +270°. It crosses the +30° line at M = 180/11 = 16411, and the −30° line at M = 60/11 = 5511. So at 2:05511 and 2:16411 the hands are 30° apart, as the two dials below show.

[[fig:1]]

How many answers can you expect? The angle 0° or 180° usually happens once in an hour block. There is no overlap between 11 and 12 (the next one is exactly 12:00) and no opposite position between 5 and 6 (it is exactly 6:00). Any other angle happens usually twice, sometimes once, because the lead sweeps only 330° in an hour.

A window question. "When are the hands 80° apart between 4:30 and 5:30?" Work out each block that touches the window. Block 4: M = (2/11)(120 ± 80) = 7311 or 36411; only 4:36411 is after 4:30. Block 5: M = (2/11)(150 ± 80) = 12811 or 41911; only 5:12811 is before 5:30. So the times are 4:36411 and 5:12811.

A reflex angle of 315° is solved by the same times as 45°, because it is the same instant measured the long way. M = (2/11)(30H + r); keep 0 ≤ M < 60.

Sub-case A: the gap is given in minute-spaces

Some questions measure the gap not in degrees but in minute-spaces: "the minute hand is 4 minutes behind the hour hand", "the hands are 10 minute-spaces apart". One minute-space is one minute mark, which is 6°. So the gap of k minute-spaces is 6k degrees, and nothing new is needed: this is the main method above with θ = 6k.

• "k minutes ahead" means r = +6k.
• "k minutes behind" means r = −6k.

Put r = ±6k into M = (2/11)(30H + r) and it simplifies to a neat pair:

Ahead: M = (12/11)(5H + k). Behind: M = (12/11)(5H − k).

[[fig:2]]

Look at the picture, for the hour block 3 to 4 (H = 3, so 5H = 15). The minute hand is 4 minute-spaces behind at M = (12/11)(15 − 4) = 12, so at 3:12. It is 7 minute-spaces ahead at M = (12/11)(15 + 7) = 24, so at 3:24. You can check the first: at 3:12 the hour hand is at 96° and the minute hand at 72°, and 96 − 72 = 24° = 4 × 6°.

When the question says "k minute-spaces apart" without "ahead" or "behind", both branches are possible, and you keep the ones inside the hour block. When it says "behind", use only the minus; when it says "ahead", only the plus. 1 minute-space = 6°. Ahead: +6k. Behind: −6k.

Sub-case B [New Pattern]: hands equally far from a numeral

A newer pattern says: "between 4 and 5 o'clock, at what time are the two hands the same distance from the 5?" There are two families of answers.

1. The hands coincide, so their distance from the 5 is trivially equal. At 4:21911 they are together.
2. The hands are on opposite sides of the 5, each at the same distance from it. Here the line from the pivot to the 5 acts like a mirror, and the two hands are reflections of each other in it.

[[fig:3]]

Let P be the angle of the numeral measured from the 12 (the 5 is at 150°). The hour hand is at 30H + M/2 and the minute hand is at 6M. They are equally far from P, on opposite sides, when 6M − P = P − (30H + M/2). Rearranging:

M = (2/13) × (2P − 30H).

For H = 4 and the numeral 5 (P = 150): M = (2/13)(300 − 120) = 360/13 = 27913. The time is 4:27913. Check: the hour hand is at 120 + 13.85 = 133.85°, the minute hand at 6 × 27.69 = 166.15°, and each is 16.15° from 150°.

Always check 0 ≤ M < 60, and read the question: if it says "on opposite sides", only the second family counts; if it says only "equidistant", the coincidence time also qualifies. M = (2/13)(2P − 30H), where P = 30 × (numeral).

Sub-case C [New Pattern]: hands exchange places

Another newer pattern: a person leaves home between 2 and 3 o'clock and returns between 5 and 6 o'clock. On returning she notices that the hour hand and the minute hand have exchanged places. How long was she away?

Let her leave at a o'clock x minutes and return at b o'clock y minutes (a < b in the same 12-hour cycle, 12 treated as 0).

Exchange means the hour hand of the first moment is where the minute hand is at the second moment, and vice versa:

30a + x/2 = 6y and 6x = 30b + y/2.

x = (60/143)(12b + a), y = (60/143)(12a + b).

[[fig:4]]

For a = 2 and b = 5: x = 3720/143 = 262143 minutes, so she leaves at 2:262143, and y = 1740/143 = 1224143 minutes, so she returns at 5:1224143. The picture shows the two dials: the hour hand and the minute hand have simply swapped their positions.

Time away = (720/13) × (b − a) minutes.

What the exam asks. Two families. First, the direct time question: "At what time between 4 and 5 o'clock are the hands exactly opposite?" (also coincide, or at right angles). Second, the counting question: "How many times do the hands overlap in 12 hours, in a day, or between two clock times?"

Why it works. Coincide, opposite and right angle are the three special angles 0°, 180° and 90° of the previous type. The minute hand gains a full 360° on the hour hand every 65511 minutes, so each event happens once or twice in an hour block, with a few exceptions on exact hours. That is why both the time and the count have short formulas.

The method. Use the short formulas for the time, and "hours × rate − landmarks" for the count.

Overlap, opposite hands and the right angle are the three special cases of the previous type, with θ = 0°, 180° and 90°. They are so common that their formulas are worth knowing by heart.

• Overlap between H and H + 1: M = 60H/11.
• Opposite: M = 60(H + 6)/11 for H below 6, and M = 60(H − 6)/11 for H from 6 onwards.
• Right angle: M = (2/11)(30H + r), where r is one of +90, −90, +270 or −270.

Check that 0 ≤ M < 60 each time. The table in the foundation lists all the overlap and opposite times.

Why 60H/11 for overlap? At H o'clock the minute hand is 30H° behind the hour hand, and it gains 5.5° per minute. So the time to catch up is 30H ÷ 5.5 = 60H/11 minutes. At 5 o'clock this is 300/11 = 27311, so the hands meet at 5:27311.

[[fig:0]]

There is a beautiful pattern. The hands meet at only 11 places on the dial, and these places cut the dial into 11 equal parts of 360/11 degrees each. The label in each dot is the hour block in which that meeting happens. This also explains the gap of 65511 minutes between two meetings.

For a right angle, the hour block gives two values, because both +90 and −270 describe a right angle, and so do −90 and +270. At 10 o'clock: r = −270 gives M = (2/11)(300 − 270) = 5511, and r = −90 gives M = (2/11)(210) = 38211. The other two are negative or above 60. So the first right angle after 10 o'clock is at 10:05511, a time that students often miss because they try only ±90.

For the opposite check, at 3:49111 the hour hand is at 90 + 24.5 = 114.5° and the minute hand at 6 × 49111 = 294.5°, exactly 180° apart. Overlap M = 60H/11; opposite M = 60(H ± 6)/11; right angle M = (2/11)(30H + r), r = ±90, ±270.

Counting the events in a window, a day or a week

In every full hour block, the hands overlap once, are opposite once and make a right angle twice. The only exceptions come from the four special times where an event falls exactly on the hour:

• the overlap at 12:00 (so the 11-to-12 block has no overlap of its own),
• the opposite position at 6:00 (the 5-to-6 block has none),
• the right angles at 3:00 and 9:00 (the 2-to-3 and 8-to-9 blocks have one each).

These are the landmark times. So for a window made of whole hours:

Count = (hours in the window) × (1 for overlap or opposite, 2 for right angle) − (landmarks inside the window).

[[fig:1]]

Look at the strip for 8 a.m. to 2 p.m. The window has 6 hours, so we expect 6 overlaps. The landmark 12:00 lies inside it, so we subtract 1 and get 5: 8:43711, 9:49111, 10:54611, 12:00 and 1:05511. For right angles from 1 p.m. to 7 p.m.: 6 × 2 = 12, and only the landmark 3:00 is inside (9:00 is not), so the count is 12 − 1 = 11.

For long spans use the day totals: in 24 hours the hands overlap 22 times, are opposite 22 times and make 44 right angles. In a week the overlaps are 22 × 7 = 154. A "straight line" can mean 0° and 180° together. Then there are 22 + 22 = 44 in a day. For example, from Sunday 5 p.m. to Monday 7 p.m. (26 hours) the overlaps are 26 − 2 = 24 (midnight and noon), the opposites are 26 − 3 = 23 (6 p.m., 6 a.m., 6 p.m.), and the total is 47. Exams differ on whether "straight line" includes the overlap, so always look at the option set.

When a window starts or ends exactly on a landmark, the shortcut can be off by one. Then list the actual times and follow the wording: "between" leaves out the end points. Count = hours × rate − landmarks inside. Landmarks: 12 for overlap, 6 for opposite, 3 and 9 for right angles.

What the exam asks. "A clock is a minutes slow (or fast) at one time and b minutes fast (or slow) at a later time. At what time does it show the correct time?" Related forms ask when two clocks, or one clock, will show the right time again.

Why it works. A clock that gains or loses steadily has an error that changes along a straight line with time. It shows the correct time at the moment the error is zero, so we only need to find where the straight line crosses zero.

The method. Opposite signs: use similar triangles. Same sign: find the rate from the two errors and extend the line.

A clock that gains or loses steadily has an error that changes in a straight line with time. It shows the correct time at the moment when the error is zero.

[[fig:0]]

Suppose the clock is behind by a minutes at the start and ahead by b minutes after T hours. The error line goes from −a to +b. Where does it cross zero? By similar triangles, the crossing is at

T × a / (a + b) after the start.

The same formula works when the clock is fast first and slow later: it was ahead by a and is behind by b, so the zero-crossing is again T × a/(a + b) after the start. Remember that the numerator is the first error, and the denominator is the sum of the two.

Take "4 minutes slow at 6 a.m. on Monday and 5 minutes fast at 6 a.m. on Wednesday". T = 48 hours, a = 4, b = 5. Crossing = 48 × 4/9 = 2113 h = 21 h 20 min after Monday 6 a.m. That is Tuesday 3:20 a.m.

When both readings are on the same side, the line has not crossed zero between the two readings, so extend it. If the clock is 6 minutes slow at 2:20 and 12 minutes slow six hours later, it loses 1 minute per hour, so it was correct 6 hours before 2:20, that is, at 8:20. In the same way, "3 minutes fast at 11:30 a.m. and 9 minutes fast at 11:30 p.m." means a gain of 6 minutes in 12 hours, so it was correct 6 hours before 11:30 a.m., at 5:30 a.m.

Finally, a clock that gains or loses x minutes in a day will be correct again only when the total error reaches 12 hours (720 minutes), after 720/x days. Correct at T × a/(a + b) after the start (opposite signs); extend the line backwards when both errors are on the same side.

Sub-case: when do two clocks, or one clock, show the same time again?

Two clocks are set right together. One gains, the other loses. They begin to drift apart, and they will show the same time again only when the gap between them has grown to a full turn of the dial: 12 hours = 720 minutes for a 12-hour dial (1440 minutes for a 24-hour dial).

[[fig:1]]

Think of two runners on a circular track, running in opposite directions. They meet again when their distances add up to one full round. The drift per day is therefore

• gain + loss when one gains and the other loses (opposite directions),
• the difference of the two rates when both gain, or both lose (same direction).

Days = 720 ÷ (drift per day in minutes).

Take Radhika's wall clock that gains 2 minutes per day and her kitchen clock that loses 1 minute per day, both set right at noon on 1 March 2023. Drift = 2 + 1 = 3 minutes per day, so days = 720 ÷ 3 = 240. Counting from 1 March: the days up to 1 October are 31 + 30 + 31 + 30 + 31 + 31 + 30 = 214, and 26 more days make 240. So they show the same time again on 27 October 2023, at noon.

Check the date with care. Month lengths decide it: 31 days for January, March, May, July, August, October, December; 30 for April, June, September, November; February has 28, or 29 in a leap year. A good habit is to count whole months first and then the days that are left. Same time again when the total drift is 12 hours: days = 720 ÷ (gain + loss).

What the exam asks. A clock gains or loses time at a stated rate and is set right at some moment. Later it shows a time, and the real time is asked (or the other way round: the real time is given and the reading is asked). A rarer form says that the hands meet every G minutes of correct time and asks the gain or loss in a day.

Why it works. A steady clock keeps a fixed ratio between the time it shows and the real time that has passed. Once that ratio is known, every question is a simple proportion.

The method. Write what the clock shows in one real period, make the ratio R/D = real ÷ shown, and add R (not D) to the moment it was set right.

A clock that gains or loses time at a fixed rate shows a different time from the real one. Let D be the time shown (elapsed on the clock since it was set right) and R the real time that has passed. The two are in a fixed ratio.

[[fig:0]]

If the clock gains x hours in every 24 real hours, then while 24 real hours pass the clock shows 24 + x hours. So

R = D × 24 / (24 + x) for a clock that gains, R = D × 24 / (24 − x) for a clock that loses.

The same idea works for other periods: a clock that gains 2 hours in 12 real hours shows 14 hours in 12, so R = D × 12/14. Per hour too: a clock that gains 4 minutes every hour shows 64 minutes for every 60 real minutes, so R = D × 60/64.

Take Radhika's wall clock, which loses 1 hour in 24. She sets it right at 8 p.m. on Friday, and on Sunday it shows 6 p.m. The clock has shown 46 hours. R = 46 × 24/23 = 48 hours. Real time = Friday 8 p.m. + 48 h = Sunday 8 p.m. Notice that we add R, not D, to the setting moment.

The question can run the other way: "A clock set right at 4 a.m. loses 20 seconds every hour; what does it show at 4 a.m. next day?" Now the real time (24 h) is given, so find the total loss: 24 × 20 s = 480 s = 8 minutes. The clock shows 4 a.m. − 8 min = 3:52 a.m. Shown time D, real time R. Gaining: R = D × 24/(24 + x). Losing: R = D × 24/(24 − x). Add R, not D.

Sub-case [New Pattern]: the hands meet every G minutes of correct time

A correct clock has its hands together every 65511 minutes of true time. Suppose a clock has its hands together every G minutes of true time, measured by a correct watch. The dial movement between two meetings is the same (the hands must travel the same circle), so the faulty clock does that movement in G real minutes instead of 65511.

• If G < 65511, the clock does the same job faster: it runs fast.
• If G > 65511, it takes longer: it runs slow.

[[fig:1]]

How much? In G real minutes the faulty clock shows 65511 minutes, so it shows an extra (65511 − G) in every G real minutes. Over T real minutes:

Gain or loss = T × (65511 − G) / G.

A positive value is a gain (fast); a negative value is a loss (slow). The base of the division is G, because G is the true time that the faulty clock takes between meetings.

For G = 64 and one day (1440 minutes): (65511 − 64)/64 × 1440 = (16/11)/64 × 1440 = 32811 minutes gained. For G = 72: (65511 − 72)/72 × 1440 = −1440/11 = 1301011 minutes lost.

Two points to take care of. First, G has to be the time measured by a correct watch; the question says "of correct time". If it says that the clock's own dial shows G minutes between meetings, then G is just 65511 and there is no error from this data. Second, the answer comes out in the unit of T, so convert an hour to 60 minutes where needed. Error = T × (65511 − G) / G; positive means fast.

What the exam asks. "A clock shows T. What time will be seen in a plane mirror (or in water)?" or the reverse: "Seen through a mirror, a clock shows T; the correct time is …". Some questions give a picture of a dial, and a rarer form replaces the numerals by letters.

Why it works. A mirror in front of the clock folds the dial along the 12–6 line, so a hand at angle θ goes to 360° − θ. A mirror below the clock folds it along the 3–9 line, so a hand at θ goes to 180° − θ. Each fold turns into a one-line subtraction on the time.

The method. Mirror: 11:60 − t. Water: 6:30 − t. Then check the minutes with the sum test in the tricks below. The picture side of mirror and water images is treated fully in the Mirror and Water Image chapter, so only the clock-time side is built here.

Hold a clock in front of a vertical mirror. The mirror swaps left and right, so the 12–6 line is the fold line. A hand that points to 3 now points to 9, and a hand at 2 moves to 10. A hand at angle θ moves to 360° − θ. Since the whole clock turns into 12:00 minus the time, the picture is again a proper clock reading.

[[fig:0]]

In the picture the left clock shows 5:35. The right one is its mirror image: the hour hand has gone from 167.5° to 192.5° and the minute hand from 210° to 150°. Read in the ordinary way, the image shows 6:25.

You do not need to draw anything:

Mirror time = 11:60 − actual time.

Here 11:60 is just 12:00 written so that you can subtract minutes. For 5:35: minutes 60 − 35 = 25 and hours 11 − 5 = 6, so 6:25. The hour pairs are 1 ↔ 11, 2 ↔ 10, 3 ↔ 9, 4 ↔ 8, 5 ↔ 7, while 6 and 12 stay, and the minutes become 60 − m. More examples: 8:20 → 3:40; 11:55 → 12:05; 3:10 → 8:50; 9:00 → 3:00.

If the time shows 12 o'clock something, treat the 12 as 0: the mirror of 12:30 is 11:60 − 0:30 = 11:30, and the mirror of 11:30 is 12:30.

The mirror applied twice gives back the original time, so the same method works backwards: actual time = 11:60 − the time seen in the mirror. If the time is in 24-hour form, convert it to a 12-hour time first.

When the options are pictures of dials, flip each hand about the 12–6 line and compare. This chapter treats the time-formula side; the picture side is in the Mirror and Water Image chapter, and both use the same idea.

Sub-case A: the water image

Now lay the mirror flat below the clock. It swaps top and bottom, so the horizontal line through the 3 and the 9 is the fold line. A hand pointing to 12 now points to 6, and a hand at the 3 or the 9 stays where it is. A hand at angle θ moves to 180° − θ.

[[fig:1]]

In the picture the clock shows 2:20. The minute hand goes from the 4 to the 2 (that is, from the 20-minute mark to the 10-minute mark), and the hour hand from 70° to 110°.

There is a difficulty you must know about. After the flip, the hour hand and the minute hand do not agree with each other as they do on a real clock: the reflected hour hand lies exactly half-way between two readings that are consistent with the minute hand. So time questions follow an exam convention, which takes the later reading. This is the exam convention for the water image time:

Water time = 6:30 − actual time (add 12 hours if the result is negative or zero).

For 2:20: 6:30 − 2:20 = 4:10. Here the reflected hour hand is at 110°, midway between 3:10 (hour hand at 95°) and 4:10 (125°). For 8:20: 6:30 − 8:20 is −1:50, and adding 12 hours gives 10:10. Other examples: 3:00 → 3:30; 4:20 → 2:10; 2:47 → 3:43 (write 6:30 as 5:90 and subtract).

Always subtract the full time, borrowing from the hour when the minutes are not enough (2:47 → 6:30 − 2:47 = 3:43). Do not map hours and minutes separately, because that ignores the borrow. The method is its own reverse: 4:10 → 2:20.

Two warnings. Do not use 6:00 − t; it gets the minute hand wrong. And 3:00 is not unchanged under a water image: the hour hand stays at 3, but the minute hand goes from 12 to 6, so the picture reads 3:30. When the options are pictures of dials, flip each hand about the 3–9 line instead of using the formula.

Sub-case B [rare]: lettered and re-labelled dials

Some questions replace the numerals of the dial with letters, symbols or a different numbering. The hands still move by position, exactly as before. Only the labels have changed. So the whole method is: convert every label to its position on an ordinary dial, then read the time.

[[fig:2]]

In the first dial the numerals 1 to 12 are replaced by A to L, with A at the place of 1 and L at the place of 12. Convert each letter to its number (A = 1 … L = 12). The minute hand points at the letter C, which stands at position 3, so it is at the 15-minute mark (3 × 5). The hour hand is just past F, which stands at 6. The time is 6:15. In the figure the hands are drawn at 6:15, with the hour hand just past F and the minute hand on C.

Three checks keep you safe.

• Minutes always come from the position of the minute hand (position × 5). They never come from the letter's own number.
• The hour comes from the hour hand's position, and it sits just past the numeral of that hour (the hour hand moves slowly, 0.5° per minute).
• Check the start. A re-labelled dial may start its letters at the top (A at the 12 position) or at the 1 position. Identify where the A is before reading anything.

[[fig:3]]

The second dial starts with A at the top (the 12 position), so B is at 1, C at 2 and so on, and L is at 11. Now the hour hand between H (position 7) and I (position 8), nearer I, and the minute hand on K (position 10 → 50 minutes) give 7:50. Convert labels to positions first. Minutes = position × 5.

A reversed dial (the numerals run anticlockwise) needs the mirror-image idea from earlier in this chapter rather than a formula: read the hands' positions and mirror them. Every such question can be solved by position, so do not try to remember a new formula for each design.

What the exam asks. A direction is given for one hand at some time ("the minute hand points South at 12:00 a.m."), and the question asks for the direction of the same hand at a later time, or of the other hand.

Why it works. The dial is treated as a compass that may be turned as a whole. When it turns, both hands turn together by the same angle, so the angle between them does not change. The minute hand also makes one full circle every hour, so it returns to its direction after every whole hour.

The method. Find how many degrees the hand turns in the given time, add it to the given direction, and read the compass in steps of 45°.

Here the clock is treated as a compass. On an ordinary dial laid flat with the 12 pointing north, the 3 points east, the 6 south and the 9 west. At 3:00 the minute hand is on the 12, so it points north, and the hour hand is on the 3, so it points east.

[[fig:0]]

The question usually says that the dial is rotated: "at 3:00 the minute hand points north-east". Nothing is wrong with the clock; the whole dial has simply been turned, here by 45° clockwise. When the dial turns, both hands turn together by the same angle, and the angle between the hands does not change. So:

1. Find the angle of each hand from the 12 of the dial: hour hand 30H + M/2, minute hand 6M.
2. Find the rotation of the dial from the given direction. If the minute hand (at dial angle 0°) points north-east (45°), the rotation is +45°.
3. Add the rotation to the other hand's dial angle and read the direction.

[[fig:1]]

In the picture the dial is turned 45° clockwise. The minute hand points north-east and the hour hand, which was 90° from it, is now at 135°, which is south-east.

Some useful states of an unrotated dial: 3:00 (minute N, hour E); 9:00 (minute N, hour W); 4:30 (minute S, hour at 135°, south-east); 10:30 (minute S, hour at 315°, north-west); 1:30 (minute S, hour at 45°, north-east). The directions are measured clockwise from north: N 0°, NE 45°, E 90°, SE 135°, S 180°, SW 225°, W 270°, NW 315°.

If the problem asks about a later time, remember that the minute hand turns a full circle every hour. In 2½ hours it completes 2 circles and half a circle more, so it ends up pointing opposite to where it began. Rotate both hands together. Find the shift from the hand you are told about.

What the exam asks. Two quiet word-problem families in which the hands play no part. In the striking-clock family you are asked how many times a clock strikes in a day, or how long it takes to strike n times. In the timetable family you are given arrivals, delays, buses at fixed intervals or two work shifts, and you find a time or a common period.

Why it works. Both families are careful counting. A striking clock's time is made of the gaps between strikes, and a timetable is arithmetic with minutes in base 60.

The method. Count gaps, not strikes (n strikes have n − 1 gaps). Draw a time line, convert to the 24-hour clock, and subtract in base 60.

A striking clock rings h times at h o'clock. Two ideas solve almost every question.

Idea 1: strikes per period. In 12 hours the clock strikes 1 + 2 + 3 + … + 12 = 78 times, and in a day 156 times. If it also strikes once at every half hour, add 1 for each half hour: 24 more in a day, so 156 + 24 = 180.

Idea 2: n strikes have n − 1 gaps. The time taken to strike n times is the time of the gaps between strikes, so it is (n − 1) × gap. Unless the question says otherwise, treat each strike as instantaneous.

[[fig:0]]

The picture shows the point. Four strikes take 3 gaps, and ten strikes take 9 gaps. If 4 strikes take 9 seconds, each gap is 3 s, and 10 strikes take 9 × 3 = 27 seconds. Students who divide 9 ÷ 4 and then multiply by 10 get 22.5 s, which is the usual wrong option.

Other typical facts: 12 strikes in 33 seconds means 11 gaps of 3 seconds. A clock striking 6 takes 5 gaps, which is 15 s at 3 s per gap. A 12-hour clock strikes once at 1:00 and also once at 13:00; it does not strike 13 times. And at midnight (12:00) it strikes 12, not 24 times. n strikes = (n − 1) gaps. Strikes per day = 156.

Sub-case: elapsed time and timetable problems

Some "clock" questions need no hands at all. They ask about arrivals, delays, bus intervals and shifts. Everything is addition and subtraction of times, with one thing to remember: minutes work in base 60, not 100.

A few habits make these safe.

• Use the 24-hour clock to subtract. 6:50 p.m. is 18:50. From 18:50 take 40 minutes: 18:10, which is 6:10 p.m.
• Early and late. Scheduled time = arrival + minutes early, or arrival − minutes late. If the train is due at 5:48 p.m. and arrives 22 minutes early, it arrives at 5:26 p.m. If Radhika reached at 5:48 p.m., 22 minutes before the schedule, the schedule is 5:48 + 22 = 6:10 p.m.
• A service every f minutes. If the next one is at T, the last one was at T − f.
• "Left k minutes ago." The speaker's moment is the last departure + k.
• Overlapping shifts. The common time of two shifts is the intersection of their intervals: from the later start to the earlier end.

[[fig:1]]

The picture shows a bus problem. Buses leave every 25 minutes, and the next one is at 4:05 p.m. Mohan arrives and is told that the last bus left 10 minutes ago. The last bus was at 4:05 − 25 min = 3:40, so he arrived 10 minutes after 3:40, that is at 3:50 p.m. (the most common slip is to subtract 10 minutes from the next bus, which gives 3:55).

For shifts: one worker is in from 7:45 a.m. to 4:20 p.m. and another from 10:30 a.m. to 7:05 p.m. Both are present from the later start, 10:30 a.m., to the earlier end, 4:20 p.m., which is 5 h 50 min.

Problems of time that involve speed or work (a train catching up, a monkey climbing a pole) belong to the Arithmetic chapters and are not part of the clock types. Base 60. Convert to 24-hour. Common time = later start to earlier end.