how do you get from \(2t^2+89t\) to \(t^2 = 64*25\)

got it! that helps a lot, thank you!

hi mam

one simple question
ratio of time taken by A : B = 64 : t

how this is that it takes 'A' 64 minutes extra to complete some portion of the work formerly done by 'B' in 't' minutes...? isn't it such that it takes "A + B" to complete the work in "t" minutes....?

thanks in advance, mam

hi mam
I have read the explanation you have provided in another post, and perhaps, there is nothing more to explain, thanks.

but my little doubt still centers on a simple concept: "whatever work B did in 't' minutes is now done by A in 64 minutes". yes, A has to expend some extra minutes on the work previously done by B, but how much time B actually needs to do that extra job..? when A and B work together, they need 't' minutes to complete the job..

if supposed that, when A and B work together, they need 5 minutes to finish the job, and A alone needs 8 minutes to complete the job. Here, A has to spend some extra 3 minute on completion the job. So far so good.

So, according to you, time taken by A : time taken by B = 3 : 5

here it is assumed that, to do the extra job done by A, B needs 5 minutes, but according to the fact, given, 5-minute is required by A and B when they work together

Sorry to knock you again, mam

Ok, think of the job as assembling toys. Say we have 45 toys to assemble.
A is assembling toys and B is assembling toys. Together they start at 12 noon and are done in 2 ( = t) hours i.e. at 2 pm.
So in 2 hrs, say A assembled 20 toys and B assembled 25 toys. They got done by 2 pm since they were working together.

Now think - Say only A started working at 12 o clock. By 2 o clock, A assembled 20 toys (since that is A's speed). But we still have 25 toys left. Why? Because B did not work. Otherwise B would have assembled these 25 toys by 2 pm. Now A works alone till 4:30 pm to assemble these 25 toys i.e. for another 2.5 hrs.

So can we say that whatever work was done by B in 2 (= t) hrs was done by A in the additional time?
If B took 2 hrs to assemble 25 toys and A took 2.5 hrs to do the same work,
So their ratio of time take which is 2:2.5 (= 4:5) is also their ratio of speeds since they did the same work in this time?

Does this help?
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\({\rm{times}}\,\,{\rm{A,}}\,B\,\,\,\,\,\, \to \,\,\,\,\,\min\)

\(? = {A \over B}\)


\(\left( * \right)\,\,\,\left\{ \matrix{
A = t + 64 \hfill \cr
B = t + 25 \hfill \cr} \right.\)

\({1 \over A} + {1 \over B} = {1 \over t}\,\,\,\,\left( {{\rm{stem}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{1 \cdot B} \over {A \cdot B}} + {{1 \cdot A} \over {B \cdot A}} = {1 \over t}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{{2t + 89} \over {\left( {t + 64} \right)\left( {t + 25} \right)}} = {1 \over t}\)

\(2{t^2} + 89t = {t^2} + 89t + 25 \cdot 64\,\,\,\,\,\,\mathop \Rightarrow \limits^{t\,\, > \,\,0} \,\,\,\,\,t = 5 \cdot 8 = 40\)


\(?\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,{{40 + 64} \over {40 + 25}}\,\, = \,\,{{104:13} \over {65:13}} = {8 \over 5}\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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