How to Solve a Tricky Algebra Problem: Clock Gains and Time Differences

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    Algebra
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Homework Help Overview

The problem involves a clock that gains time and is currently slow. The original poster seeks to determine how long it will take for the clock to show the correct time, given its rate of gain and current delay.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the clock's gain (x minutes every 24 hours) and its current delay (y minutes). There are attempts to formulate equations connecting these variables, with some participants questioning how to properly express the relationship.

Discussion Status

Several participants are exploring different formulations of the problem, with some expressing confusion about the connections between the variables. There is an ongoing dialogue about the correct approach to relate the gain and the delay, and hints of productive reasoning are emerging as participants clarify their thoughts.

Contextual Notes

Some participants express uncertainty about their algebra skills and the assumptions underlying their equations. There is mention of specific numerical examples to illustrate the problem, but no definitive conclusions have been reached.

lionely
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Homework Statement


A clock gains x min. every 24 hr. If it is y min. slow at present, in how many hours will it show the correct time?

Homework Equations


N/A

The Attempt at a Solution


I called the current time Tc so
I have:

Tc = Tc- y

I know x will get added to the above equation every 24 hours but I don't know how to connect them. So I get stuck.
I tried something like (Tc = Tc -y + x)/24 but I don't think this make sense.

Could someone give me a hint? Help is greatly appreciated.

(P.S. I know this is simple but do you guys ever buck up on a simple problem that you just can't seem to get out? or do I need to re-study algebra?)
 
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If it catches up x minutes every 24 hours, how many minutes does it catch up after h hours?
 
It would catch by hx\24 minutes every h hour
 
lionely said:
It would catch by hx\24 minutes every h hour
So how many hours h does it take to catch up y minutes?
 
hx/24y ?
 
lionely said:
hx/24y ?
##\frac{hx}{24}=y##

h=?

This is a rate x time = distance problem

Chet
 
Okay that's one part I don't get I didn't have (hx)/24 = y I had (hx)/(24y). I wondering about how to connect x to y , I mean what would make x = y . I'm still a bit confused about the reasoning behind the answer sorry.
 
lionely said:
Okay that's one part I don't get I didn't have (hx)/24 = y I had (hx)/(24y). I wondering about how to connect x to y , I mean what would make x = y . I'm still a bit confused about the reasoning behind the answer sorry.
If a clock gains 2 minutes every hour and it starts out behind by 10 minutes, how long does it take to catch up?

If a clock gains x/24 minutes every hour and it starts out behind by y minutes, how long does it take to catch up?
 
For the first question it would take 5 hours, but the second one eh I still can't see it. I did the first like so

if the time was actual time was 12 then the time shown would be 11:50 so after 1 hour(h1) the time would be 12:52 , then 1:54(h2) then 2:56(h3) then 3:58(h4) then 4:00(h5)
so yeah 5 hours.
 
  • #10
Ahhhhhhhhhhhhhhh I see it now I am such an idiot. Thank you so much!
 

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