Calculator Technique for Clock Problems in Algebra

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Cal cul ator Techni que for Cl ock Pr obl em s i n Al gebr a

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Calculator Technique for Clock Problems in Algebra View

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Submitted by Romel Verterra on January 30, 2013 - 12:58am

The following models of CASIO calculator may work with these methods: fx-570ES, fx-570ES Plus, fx-115ES, fx-115ES Plus, fx-991ES, and fx-991ES 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.

1. When the minute-hand moves 60 dials, the hour hand moves 5 dials. The ratio of the two movements, hour-hand over minute-hand, is 5/60 = 1/12. Thus, if the minute-hand minute-ha nd will move move x-minutes, the hour-hand moves by x/12 minutes. 2. When the the second-hand second-hand moves 60 dials, the minute-hand moves 1 dial. The ratio of the two two movements, movements, minute-ha minute-hand nd over second-hand, second-hand, is i s 1/60. Thus, Thus, if the secondsecondhand will will move x-seconds, the minute-hand minute-h and moves by x/60 seconds, and the hourhand also moves also moves by 1/12 of x/60 or x/720 x/72 0 seconds. 3. The relationship of hand-movements hand-movements can also be translated in terms of degree unit which 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 minute-hand is equivalent to 1 minute of time, and five dials move of the hour-hand is equivalent to 1 hour of time.

Knowing all of the above, we can now develop the calculator technique for solving clock-related 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 minute-hand and the hour-hand of the clock be (a) together for the first time, (b) perpendicularr for the first time, and (c) in straight line for the first time? perpendicula

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Calculator Technique for Clock Problems in Algebra

Traditional Solution

Click here to show or hide the solution

Solution by Calculator The following calculator keys will be used. Name

Key

Operation

Name

Key

Operation

Shift

SHIFT

Stat

SHIFT → 1[STAT]

Mode

MODE

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 minute-hand of the clock is at 0 dial and the hour-hand of the clock is advance by 15 dials, thus, coordinates (0, 15). After 1 hour (4:00 pm), the minute-hand advanced by 60 dials leaving the hour-hand 40 dials, thus, coordinates (60, -40).

MODE → 3:STAT → 2:A+BX X

Y

0

15

60

-40

(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:x-caret 0x-caret = 16.36

Thus, time = 3:16.36 pm

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(b) Perpendicular for the first time: The hour-hand 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:x-caret -15x-caret = 32.73


(c) Straight line for the first time: The hour-hand is behind by 30 dials by the minute hand, thus find X when Y is -30.

AC → -30 SHIFT → 1[STAT] → 7:Reg → 4:x-caret -30x-caret = 49.09


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 hour-hand will move 5 dials equivalent to 30° and the minute-hand moves for 60 dials or 360°. The 1 hour difference is therefore 360° - 30° = 330° for the hour and minute-hands of the clock. At 3:00 pm, the minutehand is at -90° in reference with the hour-hand, 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)


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.



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(b) Perpendicular for the first time: The angle between the hour-hand and minute-hand is 90°. Let us find X when Y is 90.



(c) Straight line for the first time: The angle between the hour-hand and minute-hand is 180°, thus find X when Y is 180.



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) 60-degree angle for the first time, (b) 60-degree angle for the second time, and (c) 150-degree angle?

Solution by Calculator Technique


Y

Explanation

0

-150

← -5 × 30

60

180

← 330 - 150

(a) 60-degree angle for the first time

AC → -60 SHIFT → 1[STAT] → 7:Reg → 4:x-caret -60x-caret = 16.36 minutes

answer

(b) 60-degree angle for the second time

AC → 60 SHIFT → 1[STAT] → 7:Reg → 4:x-caret 60x-caret = 38.18 minutes

answer

(c) 150-degree angle

AC → 150 SHIFT → 1[STAT] → 7:Reg → 4:x-caret http://www.mathalino.com/blog/romel-verterra/calculator-technique-clock-problems-algebra

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Calculator Technique for Clock Problems in Algebra

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