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

You are missing the fact that \(t^2+89t+64*25 = 2t^2+89t\) is 1 equation entirely. You are considering that you get \(2t^2+89t\) from \(t^2+89t+64*25\)

This is not true.

You need to look at the complete equation, \(t^2+89t+64*25 = 2t^2+89t\) ---> cancelling \(89t\) from both sides and transferring \(t^2\) from LHS to RHS you get,

\(64*25=t^2\) ---> \(t^2=64*25\)--->\(t=8*5 = 40\)

Hope this helps.
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got it! that helps a lot, thank you!

rate of A: 1/(t+64)
rate of B: 1/(t+25)
rate together: 1/t

1/t = 1/(t+64) +1/(t+25)
1/t = t+25+t+64/(t+64)(t+25)
cross multiply:
t(2t+89) = (t+64)(t+25)
extend
t^2 + 89t = t^2 + 25t+64t +64*25
we can either leave 64*25 or use the technique doubling and halving: 64*25 = 32*50 = 16*100 = 1600.
now, we can get rid of one t^2, 89t
and we have:
t^2 = 1600
well, that's 40^2.
we can also see that 64 is 8^2 and 25 is 5^2.
thus, t must be 8*5 = 40.

to get back to the problem..
t=40.
time for A: 40+64 = 104
time for B: 40+25 = 65

rate: 104/65
A, C, D - out right away.
between B and E
there is no way from 104 to get 64 and from 65 to get 25...
answer must be B

alternatively, divide both 104 and 65 by 13 and get 8/5

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

Check this post: https://gmatclub.com/forum/if-dev-works ... l#p1957822
I have explained this concept in detail. Let me know if something is still unclear.
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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

Why do you say: "So machine B used to do this work in t mins when both were working together."? Is this true? Machines A and B used to do this work in t mins not machine B.

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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The approach below is detailed for the understanding purposes, however once you know the logic behind it, you'd be able to solve the question faster.

Consider giving Kudos if it helps :D

Posted from my mobile device

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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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VeritasKarishma is the following correct? I had to draw it out to understand your explanation:



We are looking for A time alone / over B time alone to do the whole work

ratio of A time alone doing the entire work = t (A's part of work) / 64 (A doing B's part of work) = t/64
ratio of B time alone = 25 (B doing A's part) / t (B's part) = 25/t

(t/64) = (25/t) because they both do the same work
t² = 1600
t = 40
A's time alone (64+40) / B's time alone (25+40) = 104/65 = 8/5

How do you immediately get 40/25 = 8/5 without the last step? Just plugging in t would give me 5/8. Is there a reason the fraction is set up that way in your original problem? I realized it's the same either result either way but this took me way too long to figure out!

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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Solution:

The rate of machine A is 1/(t + 64), and the rate of machine B is 1/(t + 25), and their combined rate is 1/t. Therefore, we can create the equation:

1/(t + 64) + 1/(t + 25) = 1/t

(t + 25 + t + 64) / [(t + 64)(t + 25)] = 1/t

t(2t + 89) = (t + 64)(t + 25)

2t^2 + 89t = t^2 + 89t + 1600

t^2 = 1600

t = 40 or -40

However, since t can’t be negative, t = 40. Therefore, the rates of A and B are 1/104 and 1/65, respectively. Since time is the inverse of rate, the minutes it takes A and B individually to complete the job are 104 and 65, respectively. Therefore, the ratio is 104/65 = 8/5.

Answer: C


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Cant we solve this question through allegation formula?