melissadh
Badges: 12
Rep:
?
#1
Report Thread starter 2 years ago
#1
The Tickton town hall clock is rather a curiosity. It shows the correct time every hour on the hour, but the minute hand travels three times as fast down to 6 as it does back up to 12, though within each half of its revolution its speed is constant.

What is the correct time when the Tickton town hall clock shows quarter to the hour?

A. 7.5 minutes to the hour

B. 10 minutes to the hour

C. 20 minutes to the hour

D. 22.5 minutes to the hour

E. 25 minutes to the hour

Please could someone really break down and explain to me why, in the first 15 minutes, will the clock reach the half way point. I don't understand the algebraic equation at all. I really think the answer is 25, but I don't understand why my first line is wrong, but I understand the rest of the working out. Thanks for the help!
0
reply
mqb2766
Badges: 18
Rep:
?
#2
Report 2 years ago
#2
(Original post by melissadh)
The Tickton town hall clock is rather a curiosity. It shows the correct time every hour on the hour, but the minute hand travels three times as fast down to 6 as it does back up to 12, though within each half of its revolution its speed is constant.

What is the correct time when the Tickton town hall clock shows quarter to the hour?

A. 7.5 minutes to the hour

B. 10 minutes to the hour

C. 20 minutes to the hour

D. 22.5 minutes to the hour

E. 25 minutes to the hour

Please could someone really break down and explain to me why, in the first 15 minutes, will the clock reach the half way point. I don't understand the algebraic equation at all. I really think the answer is 25, but I don't understand why my first line is wrong, but I understand the rest of the working out. Thanks for the help!
The speeds are in the ratio 3:1.
As the (angle) distance travelled is the same the time taken are in the ratio 1:3.
Each part is therefore 15 minutes and therefore the time taken to reach the halfway point is 15 minutes.
0
reply
melissadh
Badges: 12
Rep:
?
#3
Report Thread starter 2 years ago
#3
(Original post by mqb2766)
The speeds are in the ratio 3:1.
As the (angle) distance travelled is the same the time taken are in the ratio 1:3.
Each part is therefore 15 minutes and therefore the time taken to reach the halfway point is 15 minutes.
Hi thank you so much for responding! Just another question to add, if thats alright, if there are four parts on a clock, then I don't understand how the half way point (2 part) is equal to 1 part? I think I'm just not understanding something really small, so sorry if I look like a douche right now.
0
reply
mqb2766
Badges: 18
Rep:
?
#4
Report 2 years ago
#4
(Original post by melissadh)
Hi thank you so much for responding! Just another question to add, if thats alright, if there are four parts on a clock, then I don't understand how the half way point (2 part) is equal to 1 part? I think I'm just not understanding something really small, so sorry if I look like a doosh right now.
1 part of time. Not angular distance.
It goes 3 times slower in the 2nd half, so spends 1 part of time going to 30 min, then 1 part going to 40 min, another to 50 & finally another to 60.
0
reply
melissadh
Badges: 12
Rep:
?
#5
Report Thread starter 2 years ago
#5
Thanks again.

I'm still slightly confused as to how the half way point is 15 minutes. I thought one part meant the time to go around the clock. I created a formula which was 4m + m = 60 and M came out as 15 and to get the half way point, I would do 15/2. Which part is wrong and why? Thank you! I've spent hours reading this question and I just don't understand it. Is this a part of a particular maths topic?
0
reply
melissadh
Badges: 12
Rep:
?
#6
Report Thread starter 2 years ago
#6
(Original post by melissadh)
Thanks again.

I'm still slightly confused as to how the half way point is 15 minutes. I thought one part meant the time to go around the clock. I created a formula which was 4m + m = 60 and M came out as 15 and to get the half way point, I would do 15/2. Which part is wrong and why? Thank you! I've spent hours reading this question and I just don't understand it. Is this a part of a particular maths topic?
It might seem really simple, but I'm just so lost and I'm worried a similar question will come up on my exam.
0
reply
mqb2766
Badges: 18
Rep:
?
#7
Report 2 years ago
#7
(Original post by melissadh)
Thanks again.

I'm still slightly confused as to how the half way point is 15 minutes. I thought one part meant the time to go around the clock. I created a formula which was 4m + m = 60 and M came out as 15 and to get the half way point, I would do 15/2. Which part is wrong and why? Thank you! I've spent hours reading this question and I just don't understand it. Is this a part of a particular maths topic?
What is m and can you explain the formula?
0
reply
melissadh
Badges: 12
Rep:
?
#8
Report Thread starter 2 years ago
#8
(Original post by mqb2766)
What is m and can you explain the formula?
I literally tried to, but I saw something similar somewhere else. If I didn't see something like that, I would have just done 30/3 = 10 min to get to the half way point then I would have been lost from there in changing it from the fake and real time.
0
reply
melissadh
Badges: 12
Rep:
?
#9
Report Thread starter 2 years ago
#9
So I really can't explain it... Thank you for being really responsive again by the way
0
reply
mqb2766
Badges: 18
Rep:
?
#10
Report 2 years ago
#10
(Original post by melissadh)
I literally tried to, but I saw something similar somewhere else. If I didn't see something like that, I would have just done 30/3 = 10 min to get to the half way point then I would have been lost from there in changing it from the fake and real time.
Do you understand your formula? I'm not being funny, but if you get that wrong in the first place the rest of the question will be wrong.
What is m and why the 4m + m = 60?
0
reply
melissadh
Badges: 12
Rep:
?
#11
Report Thread starter 2 years ago
#11
I thought m was the time, if the speed was constant, to get 60minutes (in the real time). Sorry! I meant 3m + m = 60. 3 being the speed. M being the time. M being the time comes up again because I thought it would be the time assuming there was no increase (x3) in the speed, so its the speed between 30-60 if that makes sense. The answer would be 15, but I calculated m to be the time to get to 60 mins, not 30. If I didn't do this I would have just done 3m = 30 then m=10, m being the time it takes to get to the half way point. Hence I'd have 50 mins left to fill up on the second half of the clock and I'd divide it by two to get to the quarter to the hour point. The last method is what I did initially, but it was wrong, then I tried to think of it as having 4 parts and by looking at the clock as a whole.
0
reply
melissadh
Badges: 12
Rep:
?
#12
Report Thread starter 2 years ago
#12
However, I get the first method makes no sense because m has a range of meanings, none of which apply to the question. I just made a sloppy guess because I couldn't think of any other method
0
reply
ThatGuy107
Badges: 13
Rep:
?
#13
Report 2 years ago
#13
22.5 mins to hour:

3m + m = 60
m = 15
Therefore to half past the hour it is 15 mins.
So the remaining half an hour is 45 mins, therefore half of this 45 mins is 22.5 mins.
15 + 22.5 = 37.5
60-37.5 = 22.5 minutes in reality.
0
reply
mqb2766
Badges: 18
Rep:
?
#14
Report 2 years ago
#14
(Original post by mqb2766)
(Do you understand your formula? I'm not being funny, but if you get that wrong in the first place the rest of the question will be wrong.
What is m and why the 4m + m = 60?
Sorry ... posts overlapped. To try and break it down.

You're told in the first half of the clock (0-30 angular minutes) the speed is 3 times faster than the second half (30-60 angular minutes). Note, angular minutes are distances on the clockface. The speeds are constant in those two intervals.

So either set up an equation for the speed or the time. Let's set up an equation for velocity
* Obviously the sum of the two parts is 60 minutes.
* Let the speed in the second half be "v", so the speed in the first half is "3v".
speed = distance/time or
time = distance/speed
The time taken in the first half is 30/(3v) and the time taken in the seond half is 30/v. So
60 = 30/(3v) + 30/v
Can you solve for v from there and relate back to the first and second part?

Note that if the velocity is 3 times faster for the same distance, the time is 1/3 of the value in the first region, or 3 times larger in the second region. So we could solve
60 = 30*t + 30*3*t
where t is the time taken to cover 1 angular minute in the first half of the clock face (0-30 angular minutes). Again, solve for t.

If you're confident, its easier to use ratios as suggested in the first response.


1
reply
melissadh
Badges: 12
Rep:
?
#15
Report Thread starter 2 years ago
#15
(Original post by Thomas_Grimes_17)
22.5 mins to hour:

3m + m = 60
m = 15
Therefore to half past the hour it is 15 mins.
So the remaining half an hour is 45 mins, therefore half of this 45 mins is 22.5 mins.
15 + 22.5 = 37.5
60-37.5 = 22.5 minutes in reality.
Hi Thomas, thank you!

Do you mind explaining why m=15 for only half the past hour? I keep thinking it's for the whole hour. Thanks!
0
reply
ThatGuy107
Badges: 13
Rep:
?
#16
Report 2 years ago
#16
(Original post by melissadh)
Hi Thomas, thank you!

Do you mind explaining why m=15 for only half the past hour? I keep thinking it's for the whole hour. Thanks!
No problem:
So the ratio of going down to 6 vs up to twelve is 1:3 meaning it is 3x as long to get back up to 12 as it takes to get down to 3.
Therefore splitting the whole hour in this ratio, 60/4 = 15 minutes for the 1st half hour
15:45 is the actual ratio of 1:3
1
reply
ThatGuy107
Badges: 13
Rep:
?
#17
Report 2 years ago
#17
(Original post by Thomas_Grimes_17)
takes to get down to 3.
Should read down to 6.
0
reply
melissadh
Badges: 12
Rep:
?
#18
Report Thread starter 2 years ago
#18
(Original post by mqb2766)
Sorry ... posts overlapped. To try and break it down.

You're told in the first half of the clock (0-30 angular minutes) the speed is 3 times faster than the second half (30-60 angular minutes). Note, angular minutes are distances on the clockface. The speeds are constant in those two intervals.

So either set up an equation for the speed or the time. Let's set up an equation for velocity
* Obviously the sum of the two parts is 60 minutes.
* Let the speed in the second half be "v", so the speed in the first half is "3v".
speed = distance/time or
time = distance/speed
The time taken in the first half is 30/(3v) and the time taken in the seond half is 30/v. So
60 = 30/(3v) + 30/v
Can you solve for v from there and relate back to the first and second part?

Note that if the velocity is 3 times faster for the same distance, the time is 1/3 of the value in the first region, or 3 times larger in the second region. So we could solve
60 = 30*t + 30*3*t
where t is the time taken to cover 1 angular minute in the first half of the clock face (0-30 angular minutes). Again, solve for t.

If you're confident, its easier to use ratios as suggested in the first response.

My! That makes a lot of sense! So t = 1/2. 1/2 x 30 = 15 mins!
1
reply
melissadh
Badges: 12
Rep:
?
#19
Report Thread starter 2 years ago
#19
(Original post by Thomas_Grimes_17)
No problem:
So the ratio of going down to 6 vs up to twelve is 1:3 meaning it is 3x as long to get back up to 12 as it takes to get down to 3.
Therefore splitting the whole hour in this ratio, 60/4 = 15 minutes for the 1st half hour
15:45 is the actual ratio of 1:3


This makes so much sense. It's crazy - thank you so so much! Both of you guys^
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (103)
13.24%
I'm not sure (33)
4.24%
No, I'm going to stick it out for now (241)
30.98%
I have already dropped out (19)
2.44%
I'm not a current university student (382)
49.1%

Watched Threads

View All