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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The concept is a bit difficult to understand but once you do, it's quite rewarding. But you need to understand ratios and how speed, time and distance are connected.

Let's take some easier numbers:

Say the work consists of making 30 identical widgets.
Now, machine A makes 1 widget in 1 minute.
Machine B makes 2 widgets in 1 minute.
So when they both work together, they make 3 widgets in 1 min and complete the entire work in 10 mins ( say = t). In these 10 mins, machine A makes 10 widgets and machine B makes 20 widgets.

If only machine A works, it takes 30 mins to complete the work (20 mins more than t)
Now think: why does it take 20 mins more? because it now needs to make extra 20 widgets which machine B was making in 10 mins. So for the same work - 20 widgets - machine A takes 20 mins while machine B takes 10 mins.
Ratio of time taken by A : B = 20/10 (for same amount of work)
This means ratio of speed A : B = 10:20 = 1/2 (this stays constant because their speeds are constant)

If only machine B works, it takes 15 mins to complete the work (5 mins more than t)
Now think: why does it take 5 mins more? because it now needs to make extra 10 widgets which machine A was making in 10 mins. So for the same work - 10 widgets - machine A takes 10 mins while machine B takes 5 mins.
Ratio of time taken by A : B = 10/5 (for same amount of work)
This means ratio of speed A : B = 5:10 = 1/2 (this stays constant because their speeds are constant)
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