how to create formular for this problem

[Abuufauzan]

I HAVE A BOOK OF 604 PAGES EACH PAGE CONTAIN 15 LINES. EACH STUDENT SUPPOSED TO READ THIS BOOK FOR 10 MONTHS WHERE 1 MONTH HAS 24 DAYS. BUT IF STUDENT CANT FINISH THIS YEAR HE/SHE WILL CONTINUE TO READ THIS BOOK NEXT YEAR SO THAT

CREATE A FORMULAR HOW MANY DAYS RMAIN TO FINISH READING THIS BOOK

 

[Agent Smith]

R = Lines read per day.
Number of days = D
Lines read in D days = DR
Lines to read = 604 × 15 = 9060
Lines left to read (after D days of reading) = 9060 - DR
Days remaining to read = [imath]\frac{9060 - DR}{R} = \frac{9060}{R} - D[/imath]

[imath]\frac{9060}{R}[/imath] = Number of days it'll take to finish reading at a rate of R lines per day.
[imath]D[/imath] = Number of days read

 

[Steven G]

I don't think that we need to know how many lines are on each page--just that they all have the same number of lines..
10 months = 10*24days = 240 days.
You need to read 604 pages in 240 days. So on average, how many pages must Rmain read per day to finish before the 10 months is over.?

If it turns out that Rmain must average X complete pages plus a portion of a page, then we can resort to lines. I just feel that if you have to read 5 1/3 pages per day, you can figure out what a third of a page is w/o knowing how many lines it has. Just for the record, it can equally happen that you have to read 234 complete lines plus 1/3 of a line. Either way, you still have this problem.

For the record, I think that the answer should be an equality as Rmain can finish before the 10 month have past.

 

[The Highlander]

R = Lines read per day.
Number of days = D
Lines read in D days = DR
Lines to read = 604 × 15 = 9060
Lines left to read (after D days of reading) = 9060 - DR
Days remaining to read = [imath]\frac{9060 - DR}{R} = \frac{9060}{R} - D[/imath]

[imath]\frac{9060}{R}[/imath] = Number of days it'll take to finish reading at a rate of R lines per day.
[imath]D[/imath] = Number of days read

I don't think that we need to know how many lines are on each page--just that they all have the same number of lines..
10 months = 10*24days = 240 days.
You need to read 604 pages in 240 days. So on average, how many pages must Rmain read per day to finish before the 10 months is over.?

If it turns out that Rmain must average X complete pages plus a portion of a page, then we can resort to lines. I just feel that if you have to read 5 1/3 pages per day, you can figure out what a third of a page is w/o knowing how many lines it has. Just for the record, it can equally happen that you have to read 234 complete lines plus 1/3 of a line. Either way, you still have this problem.

For the record, I think that the answer should be an equality as Rmain can finish before the 10 month have past.

I don't think that this a problem that can be properly addressed with the information provided by the OP.

Firstly, it looks to me like English is not the OP's first language and I strongly suspect that "RMAIN" was meant to be "REMAIN" (just a spelling mistake and not someone's name). I may be wrong, ofc, but that's how it looks to me.

I also suspect that the OP is actually looking for some way to plan out how much of the book to read (on a daily basis) so that it can be finished within the current academic year and doesn't have to be carried forward into a second "YEAR"?

It is unclear whether the reader has already begun reading the text and is part way through reading it or is just about to commence reading it now and wants to know at what rate it needs to be read (on a daily basis) in order to finish it within the available time (eg: how many pages (or lines) need to be read in each of the available days left in the current year.)

It is pointless trying to create any kind of "FORMULAR" (sic) without full information on several aspects of the situation that simply have not been provided.

Hence my reply to the OP (below).

 

[The Highlander]

I HAVE A BOOK OF 604 PAGES EACH PAGE CONTAIN 15 LINES. EACH STUDENT SUPPOSED TO READ THIS BOOK FOR 10 MONTHS WHERE 1 MONTH HAS 24 DAYS. BUT IF STUDENT CANT FINISH THIS YEAR HE/SHE WILL CONTINUE TO READ THIS BOOK NEXT YEAR SO THAT

CREATE A FORMULAR HOW MANY DAYS RMAIN TO FINISH READING THIS BOOK

Hi @Abuufauzan.

We really need more information in order to help you properly.

This book you mention has 640 pages and 15 lines in each page.

That amounts to 640 × 15 = 9,600 lines in total.

You say that (ideally) it has to be read in 10 months with each month having 24 days available for reading it.

That means there are 24 × 10 = 240 reading days available.

Is the reader about to begin reading the book at the start of that 240 day period or is the reader already part way through the 240 day time period?

Has the "STUDENT" already read any of the book or is s/he just about to start reading it?

Do you actually want to know how many pages (or lines) need to be read each day over the full 10 month period?

Or are you asking for a formula to calculate the rate (how fast) at which it has to be read (to finish it within the available period) if reading it has already begun and the reader is already part way into the 10 month period?

If the reader is at the very beginning of the 10 months and is just about to start reading the book then s/he would have to read (an average of) 2⅔ (640 ÷ 240) pages per day throughout the 10 months. (That would be the same as 40 (9600 ÷ 240) lines per day.)

If that is all you want to know then please submit another post clarifying that and confirming you now have all that you need.

If that doesn't meet your requirements then please provide clearer information about the situation you want a formula for...

What do you want the formula to provide?
You (seemed to) have asked for it to calculate how many days remain(?) to finish reading the book.

The simple answer to that is: 240 - D where D is the number of (complete) days that have passed since the beginning of the 10 month period (ie: the academic year?)

That seems far too simple so I suspect that what you really want is to know how much of the book needs to be read each day to finish it in the 10 months available (so it doesn't have to be continued into the following year) but that depends on what I have said above and below...

Has the "STUDENT" already started reading the book (and, if so, how much of it has s/he read so far)?

And, has the 10 month period already begun (and, if so, how many days/months have already passed)?

Simple formulae could be created to take account of these conditions if we knew that was what had to be dealt with...

Let the 240 days in the academic (10 month) year be numbered 1 to 240.

Let D be the current day's number; if the 10 month period has not started yet, let D = 0.

The number of days that remain to finish reading the book is simply: 240 - D.

Alternatively, if one wishes to calculate how much of the book should be read (on average) each day in order to finish it within the specified 10 months, then...

Let P be the number of (complete) pages of the book that a student has already read and/or L be the number of lines that a student has already read.

Let R_P be the number of pages a student must read (on average) each day in order to finish reading this book within the 10 months.

Let R_L be the number of lines a student must read (on average) each day in order to finish reading this book within the 10 months.

Then: \(\displaystyle \sf R_P = \frac{640-P}{240-D}\)

and: \(\displaystyle \sf R_L = \frac{9600-L}{240-D}\)

Please now tell us whether you have all you need or provide clearer (and fuller) information about the situation any "STUDENT" may be in and exactly what it is a formula needs to predict.

Thank you.

 

[Abuufauzan]

First of all I Thank you for your contribution

 

[Abuufauzan]

I deally im creating a PHP reading Progress System on the way i stucked because of many ideas of creating a good formula to get reading progress from the studen.

NOW I solved in this way

$total_lines_reading_by_student = "";
$lines_per_page = 15
$lines_per_chapter = 300;
$current_line = 0;
$current_chapter = 0;
$chapter_pages = 20;

// Calculate the number of pages
$number_of_pages = ceil($total_lines / $lines_per_page);
if($current_chapter == 30){
$number_of_pages = ($number_of_pages);
}else{
$number_of_pages = $number_of_pages;
}

$all_lines = 9060;
$lines_per_day = 30;
$total_lines_reading_by_student = 66 ;

// Calculate days left
$days_left = ceil(($all_lines - $lines_read) / $lines_per_day);

THANK YOU ALL

R = Lines read per day.
Number of days = D
Lines read in D days = DR
Lines to read = 604 × 15 = 9060
Lines left to read (after D days of reading) = 9060 - DR
Days remaining to read = 9060−𝐷𝑅𝑅=9060𝑅−𝐷R9060−DR=R9060−D

9060𝑅R9060 = Number of days it'll take to finish reading at a rate of R lines per day.
𝐷D = Number of days read

 

[The Highlander]

I deally im creating a PHP reading Progress System on the way i stucked because of many ideas of creating a good formula to get reading progress from the studen.

NOW I solved in this way

$total_lines_reading_by_student = "";
$lines_per_page = 15
$lines_per_chapter = 300;
$current_line = 0;
$current_chapter = 0;
$chapter_pages = 20;

// Calculate the number of pages
$number_of_pages = ceil($total_lines / $lines_per_page);
if($current_chapter == 30){
$number_of_pages = ($number_of_pages);
}else{
$number_of_pages = $number_of_pages;
}

$all_lines = 9060;
$lines_per_day = 30;
$total_lines_reading_by_student = 66 ;

// Calculate days left
$days_left = ceil(($all_lines - $lines_read) / $lines_per_day);

THANK YOU ALL

R = Lines read per day.
Number of days = D
Lines read in D days = DR
Lines to read = 604 × 15 = 9060
Lines left to read (after D days of reading) = 9060 - DR
Days remaining to read = 9060−𝐷𝑅𝑅=9060𝑅−𝐷R9060−DR=R9060−D

9060𝑅R9060 = Number of days it'll take to finish reading at a rate of R lines per day.
𝐷D = Number of days read

I'm not going to try to decipher your 'code' but I'm pleased that you appear to have solved the problem to your own satisfaction.

You're welcome to any help or insight that may have been provided by the forum.

Good luck with your project.

 

[Agent Smith]

[imath]\text{Rate} \geq \frac{\text{Total Pages}}{\text{Academic Term}}[/imath]