The following models of CASIO calculator may work with these methods: fx570ES, fx570ES Plus, fx115ES, fx115ES Plus, fx991ES, and fx991ES Plus.
Before we go to Calculator technique, let us first understand the movements of the hands of our continuously driven clock.
For simplicity, let "dial" be the unit of one hand movement and there are 60 dials in the complete circle as shown in the figure.
 When the minutehand moves 60 dials, the hour hand moves 5 dials. The ratio of the two movements, hourhand over minutehand, is 5/60 = 1/12. Thus, if the minutehand will move xminutes, the hourhand moves by x/12 minutes.
 When the secondhand moves 60 dials, the minutehand moves 1 dial. The ratio of the two movements, minutehand over secondhand, is 1/60. Thus, if the secondhand will move xseconds, the minutehand moves by x/60 seconds, and the hourhand also moves by 1/12 of x/60 or x/720 seconds.
 The relationship of handmovements can also be translated in terms of degree unit which I found handy in calculator technique for board exam problems. We know that a complete circle is equal to 360° and equal to 60 dials. Thus, 1 dial is equivalent to 360°/60 = 6° and five dials is equivalent to 5(6°) = 30°. Note that 1 dial move of the minutehand is equivalent to 1 minute of time, and five dials move of the hourhand is equivalent to 1 hour of time.
Knowing all of the above, we can now develop the calculator technique for solving clockrelated problem. We will solve some example here in order to apply this time saving technique.
Problem
What time after 3:00 o'clock will the minutehand and the hourhand of the clock be (a) together for the first time, (b) perpendicular for the first time, and (c) in straight line for the first time?
Traditional Solution
Solution by Calculator
The following calculator keys will be used.
Name 
Key 
Operation 

Shift 

SHIFT 
Mode 

MODE 

Name 
Key 
Operation 

Stat 

SHIFT → 1[STAT] 
AC 

AC 

The relationship between the movements of the clock hands is linear. We can therefore use the Linear Regression in STAT mode.
Approach No. 1
Take 3:00 pm as reference point. Initially, the minutehand of the clock is at 0 dial and the hourhand of the clock is advance by 15 dials, thus, coordinates (0, 15). After 1 hour (4:00 pm), the minutehand advanced by 60 dials leaving the hourhand 40 dials, thus, coordinates (60, 40).
MODE → 3:STAT → 2:A+BX
(a) Together for the first time: The distance between the hands of the clock is zero. We will therefore find X when Y is zero in our table.
AC → 0 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
0xcaret
= 16.36
Thus, time = 3:16.36 pm
(b) Perpendicular for the first time: The hourhand is behind by 15 dials by the minute hand. Let us find X when Y is 15.
AC → 15 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
15xcaret
= 32.73
Thus, time = 3:32.73 pm
(c) Straight line for the first time: The hourhand is behind by 30 dials by the minute hand, thus find X when Y is 30.
AC → 30 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
30xcaret
= 49.09
Thus, time = 3:49.09 pm
The above approach works fine but you need to mentally visualize the hands of the clock to get the proper sign (positive or negative) and value of the coordinates. See for example if the given time is 10:00 pm, the hands will be in straight line for the first time with the hour hand advancing the minute hand by 30 dials. Thus Y is +30 and not 30. This mental visualization takes the same effort as the traditional solution; the only difference is the absence of drawing. Without the drawing is good already but we can do better than that. The next approach will be more consistent, the only catch is that you need to memorize the numbers 30 and 330. I think it is not hard to memorize that numbers.
Approach No. 2
In the first approach, both X and Y are in dial units. In this second approach, the X coordinate will be in dial and Y coordinates in degrees. Recall that in 1 hour, the hourhand will move 5 dials equivalent to 30° and the minutehand moves for 60 dials or 360°. The 1 hour difference is therefore 360°  30° = 330° for the hour and minutehands of the clock. At 3:00 pm, the minutehand is at 90° in reference with the hourhand, thus coordinates (0, 90). After 1 hour, that is at 4:00 pm, the minute hand advanced the right hand by 330°  90° = 240°, thus coordinates (60, 240)
MODE → 3:STAT → 2:A+BX
X 
Y 
Explanation 

0 
90 
← 3 × 30 
60 
240 
← 330  90 
(a) Together for the first time: The angle between the hands of the clock is zero. Find X when Y is zero in our table.
AC → 0 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
0xcaret
= 16.36
Thus, time = 3:16.36 pm
(b) Perpendicular for the first time: The angle between the hourhand and minutehand is 90°. Let us find X when Y is 90.
AC → 90 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
90xcaret
= 32.73
Thus, time = 3:32.73 pm
(c) Straight line for the first time: The angle between the hourhand and minutehand is 180°, thus find X when Y is 180.
AC → 180 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
180xcaret
= 49.09
Thus, time = 3:49.09 pm
For me, the second approach is more rapid and easy to implement. I recommend you master just one and be good at it.
Problem
How soon after 5:00 o'clock will the hands of the clock form a (a) 60degree angle for the first time, (b) 60degree angle for the second time, and (c) 150degree angle?
Solution by Calculator Technique
MODE → 3:STAT → 2:A+BX
X 
Y 
Explanation 

0 
150 
← 5 × 30 
60 
180 
← 330  150 
(a) 60degree angle for the first time
AC → 60 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
60xcaret
= 16.36 minutes answer
(b) 60degree angle for the second time
AC → 60 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
60xcaret
= 38.18 minutes answer
(c) 150degree angle
AC → 150 SHIFT → 1[STAT] → 7:Reg → 4:xcaret
150xcaret
= 54.54 minutes answer
You may also like the calculator technique for arithmetic progression, geometric progression, and harmonic progression.
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