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Clock word problem

paulmdrdo

Active member
May 13, 2013
386
1. how soon after noon will the hands of the clock extend in opposite direction?

2. what time after 11 o'clock will the hands of the clock be at right angle for the 2nd time?

can you please give hints on this, so i can start solving it. thanks! i have no idea where to start honestly that's why i can't show any preliminary work.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
1. how soon after noon will the hands of the clock extend in opposite direction?

2. what time after 11 o'clock will the hands of the clock be at right angle for the 2nd time?

can you please give hints on this, so i can start solving it. thanks! i have no idea where to start honestly that's why i can't show any preliminary work.
Each hour, the minutes hand rotates through 360 degrees, and the hour hand rotates through 360/12 = 30 degrees. So after $x$ minutes the minute hand will have rotated through $\dfrac x{60}\times 360$ degrees, and the hour hand will have rotated through $\dfrac x{60}\times 30$ degrees. Can you take it from there?
 

paulmdrdo

Active member
May 13, 2013
386
still no good. there's nothing i can get from question 1.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
still no good. there's nothing i can get from question 1.
What initial and final angles do you require to be subtended by the hands?
 

paulmdrdo

Active member
May 13, 2013
386
What initial and final angles do you require to be subtended by the hands?
for them to be in opposite direction the hands should form a straight line which has 180 deg. is that correct? i still dont know how to set it up.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
for them to be in opposite direction the hands should form a straight line which has 180 deg. is that correct? i still dont know how to set it up.
You are correct about the final angle, which is $\pi$ radians. The initial angle is $0$ radians. So, you want to find out how many minutes after noon it will take for the difference between the angular position of the minute hand and the hour hand is $pi$ radians. You can use degrees if you would rather. This is essentially a distance/rate/time problem.

What is the angular position of the two hands at time $t$ as measured from their initial position?
 

paulmdrdo

Active member
May 13, 2013
386
let
$x=$ distance traveled by the minute hand(in minutes).
$x/12 = $distance traveled by the hour hand(in minutes).

now, $x-\frac{x}{12}=30min(180\deg)$

then,

$12x-x=360=11x=360=x=32\frac{8}{11}$

the answer is $12:32:43$ is this correct?

p.s why do we still need to mention the initial angle here?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
let
$x=$ distance traveled by the minute hand(in minutes).
$x/12 = $distance traveled by the hour hand(in minutes).

now, $x-\frac{x}{12}=30min(180\deg)$

then,

$12x-x=360=11x=360=x=32\frac{8}{11}$

the answer is $12:32:43$ is this correct?
Yes, that is correct, and it is a neat method. But I have a couple of comments to improve the presentation.

let $\color{blue}{x=}$ distance traveled by the minute hand (in minutes). You can't measure distances in minutes. In any case, what you are measuring here is not a distance but an angle. What you actually mean here is "Let $x\:=$ time taken for the minute hand to reach the required position." Your equation then says that, after that time, the difference in position between where the minute hand has reached and where the hour hand has reached is equal to the time taken for the minute hand to rotate through 180 degrees. That is the correct equation for the time!

$\color{blue}{12x-x=360=11x=360}\color{red}{=}\color{blue}{x=32\frac{8}{11}}$ Avoid writing a mathematical argument by using a string of equals signs unless the things on each side really are equal. That red equals sign looks horrible. What you should say here is
"$12x-x=360=11x$ and therefore $x=32\frac{8}{11}$."

p.s why do we still need to mention the initial angle here?
That is because initially, at noon, the angle between the hands is $0$. Your solution takes that for granted, but it might be better to state it explicitly.
 

paulmdrdo

Active member
May 13, 2013
386
that's enlightening!

now, i thought of other way of solving it i just want to confirm if it is valid.

since the minute hand is moving 12 times as fast as the hour hand

i thought of this,

let
$x =$ time taken for the hour hand to reach required position.
$12x=$time taken for the minute hand to reach required position.

then, $12x-x=30=11x=30$ so, $x=\frac{30}{11}=2.72...$

can you tell where's the mistake in my equation?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
that's enlightening!

now, i thought of other way of solving it i just want to confirm if it is valid.

since the minute hand is moving 12 times as fast as the hour hand

i thought of this,

let
$x =$ time taken for the hour hand to reach required position.
$12x=$time taken for the minute hand to reach required position. If the minute hand moves 12 times as fast, it will only take (1/12)th the time taken by the hour hand! That will give you the same equation as before.

then, $12x-x=30=11x=30$ so, $x=\frac{30}{11}=2.72...$

can you tell where's the mistake in my equation?
. . .
 

paulmdrdo

Active member
May 13, 2013
386
do you mean that we can interchange these two

as

$x = $time taken for the HH to reach the required position
$\frac{x}{12}=$ time taken for the MH to reach the required position?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
do you mean that we can interchange these two

as

$x = $time taken for the HH to reach the required position
$\frac{x}{12}=$ time taken for the MH to reach the required position?
Yes, because they both reach their required positions at the same time!