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5:36. the 28 minute mark is between 5 and 6; there the hour is 5. there are 5 minutes between the 5 (25 minutes) and 6 (30 minutes). every minute represents 12 minutes (60 minutes/5 minutes). 28 minutes is the 3rd minute b/t 5 and 6. 3 x 12=36. 5:36. on a clock, the small hand (hour hand) moves to the next minute marker every 12 minutes.

Q: A clock has only one hand the small one and it is on the 28 minute mark so what time is it?

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twice an hour, at the 15 minute mark and 45 minute mark.

The minute hand and hour hand both move clockwise to mark time and are in the same position at 12 noon or midnight.

144 degrees. Each minute mark around the clock face is 6 degrees.

0 degreesEach hour mark is worth 30 degrees. The hour hand will have moved 1/4 of 30 (7.5 degrees) from the 3 o'clock mark. The minute hand is on the 3 o'clock mark. That puts the hour hand 7.5 degrees ahead of the minute hand.

A Better Approach (with reasoning)There are 2 cases depending on the working of the watch.Case 1: the movement of the second, minute and hour hands are continuous (not step-wise or click-based)Answer is: the hour and minute hands overlap every hour.Case 2 (very unusual): the hour hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the minute hand crosses (or reaches 12) and the minute hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the second hand crosses (or reaches 12).Answer: every 65 minutes. Interviewers often expect this answer cos they do not think accurately. The exact times are:0000 (12:00 AM)0105 (01:05 AM)0211 (02:11 AM)0316 (03:16 AM)0422 (04:22 AM)0527 (05:27 AM)0633 (06:33 AM)0738 (07:38 AM)0844 (08:44 AM)0949 (09:49 AM)1055 (10:55 AM)1200 (12:00 PM)1305 (01:05 PM)1411 (02:11 PM)1516 (03:16 PM)1622 (04:22 PM)1727 (05:27 PM)1833 (06:33 PM)1938 (07:38 PM)2044 (08:44 PM)2149 (09:49 PM)2255 (10:55 PM)Reasoning for Case 1:-----------------------------When do they overlap? At every (n + (n/11)) hours where n = 0, 1, 2, 3, ..., 24.How did I find this out? The following is the reasoning i used:At 0000, the hour and minute hands overlap. So number of overlaps now is 1. The minute hand races away and never again overlaps during the next one hour.Now, the minute hand moves at 360o/hour and the hour hand moves at 30o/hour.At 0100, the hour hand would be the 1 mark (or 30o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0100), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as:30o + (30o)x(T) = (360o)x(T) How did I get to this equation? Note that, at 0100, when the minute hand starts moving from the 12 mark, the hour hand is already ahead of the minute hand by 30o. If the minute hand moves at a speed of 360o/hour, then in some time (T), it would cover (360o)(T) degrees. If the hour hand hand moves at a speed of 30o/hour, then in the same time (T), it would cover (30o)(T) degrees. Since the hour hand is already 30o ahead from the 12 mark, the total degrees covered by the hour hand from the 12 mark would then be (30o + the number of degrees covered in time T) which is (30o + (30o)x(T)).Now the condition when the two hands will overlap is that they should have covered the same number of degrees at a moment (or)No. of degrees covered by minute hand = No. of degrees covered by hour hand(or)30o + (30o)x(T) = (360o)x(T)If you solve this equation to find the value of T, you would get30o = (360o)x(T) - (30o)x(T)30o = (360o - 30o) x T30o = 330o x T(or)T = 30o/330oT = 1/11At 0200, the hour hand would be the 2 mark (or 60o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0200), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as:60o + (30o)x(T) = (360o)x(T)Solving this equation, you will the value of T = 2/11At 0300, using the same reasoning (the hour hand at 90o past the 12 mark) and modifying the equation accordingly (90o + (30o)x(T) = (360o)x(T)), you would get the answer for T = 3/11.In general, the value for T for every hour is T = n/11 where n = 0, 1, 2, 3, ..., 24. So the exact time when the two hands overlap can be written as:the hour (n) + the time taken during that hour (T)(or)n + n/11where n = 0, 1, 2, 3, ..., 24.AM12:001:052:113:164:225:276:337:388:449:4910:55 PM12:001:052:113:164:225:276:337:388:449:4910:5522 is correct.The hands overlap about every 65 minutes, not every 60 minutes. In a day, the hands would only overlap 22 times, as illustrated in the table above.I would propose that the hands always overlap, as they're both attached at the center of the dial.If you didn't want to be facetious (or, at least, less facetious), you would still have to ask how many hands were on the clock. It may have a second hand, for example, or be digital (no hands at all).

Related questions

If both started at 12, in forty minutes, the minute hand would reach the 8 mark on the clock. The 8 mark symbolizes 8 hours past 12. So it would take 8 hours for the hour hand to travel as far as the minute hand travels in 40 minutes.

twice an hour, at the 15 minute mark and 45 minute mark.

The minute hand and hour hand both move clockwise to mark time and are in the same position at 12 noon or midnight.

144 degrees. Each minute mark around the clock face is 6 degrees.

0 degreesEach hour mark is worth 30 degrees. The hour hand will have moved 1/4 of 30 (7.5 degrees) from the 3 o'clock mark. The minute hand is on the 3 o'clock mark. That puts the hour hand 7.5 degrees ahead of the minute hand.

The hour hand is at 22.5 degrees (clockwise) from the 12 mark, and the minute hand points to the 9.

An acute angle is any angle that is between 0° and 90°.At the exact hour mark, the minute hand is always at the 12.And so, the hours where the clock form an acute angle are:1 o'clock2 o'clock10 o'clock11 o'clockThus, there are 4 hours.

Each interval between numbers on a clock is 30 degrees. When the minute hand is on the 6 (which is the half mark), the hour hand has already moved 15 degrees from 12, because half of 30 is 15. To make calculating the angle easier, we can apply what we know. We know that a straight line is 180 degrees. Now subtract 15 from 180, and you get 165 degrees.

It doesn't matter where it is on the clock. If the clock is working properly, the speed of the hand is constant.The hand's angular speed is 360 degrees per minute = 6 degrees per second.For the linear speed, the tip of the second-hand revolves in a circle whose circumference is(2 pi) times (length of the hand) = 4 pi centimeters.It revolves once per minute. So the speed of the tip is (4 pi) cm/minute, or (240 pi) cm/hour.In numbers, the speed at the tip is:12.6 cm/minute2.09 mm/sec7.54 meters/hour0.000469 mile/hour593.7 feet/day12.593 furlongs/fortnight.Notice that this is the speed at the second-hand's tip. Other points on it travel slower.The closer the point is to the center, the slower its speed is. At the center, it spins, butthe linear speed is zero.

"The bawdy hand of the dial is now upon the prick of noon" (2,4) Is it the hand of a clock and is the "prick" the mark that marks twelve on a clock face, or is it another kind of hand and another kind of prick? Depends how dirty your mind is. Mercutio's is always filthy.

There are two buttons labeled 'H' and 'M' to the lower left of your tachometer. Hold the 'H' down to change the hour, and 'M' for minute. The clock on your radio, and driver console should change at the same time.

There is a small bump on the back of the timing belt backing plate that correlates with the mark on the camshaft. The crankshaft mark lines up with a small slit in the case in about the 4 o clock position.