The question is:
A $120$-page book has $p$ lines to a page. If the number of lines were reduced by three on each page, the number of pages would need to be increased by $20$ to give the same amount of writing space. How many lines were there on each page originally?
If anyone could help me to solve this and show the working out it'd be greatly appreciated.
Let $T$ be the number of total lines in the book.
A $120$-page book has $p$ lines to a page.
This tells us that $120 p = T$.
If the number of lines where reduced by three on each page, the number of pages would need to be increased by $20$ to given the same amount of writing space.
This tells us that $(120+20)(p-3) = T$, where $$ (120+20)(p-3) = 140(p-3) = 140p - 420. $$
How many lines where on each page originally?
Putting the above together we have $$ 120p = T = 140p - 420 \implies 120p = 140p - 420 \implies 20p = 420 \implies p = 21. $$ So there were originally $21$ lines on each page.
See let lines on each page be $x$ so $120$ has $120x$ now according to new condtion we require $140$ pages fir $(x-3)$ lines . Now its given that writing space is same implies $120x=140(x-3)$ solving you get $x=21$ thus original lines are $21$
