Simultaneous Equation problem
Two taps A and B fill a swimming pool together in two hours. Alone, it takes tap A three haours less than B to fill the same pool. How many hours does it take each tap to fill the pool separately?
Is it 3 and 6 hours .
They haven't given answers @Pawanyadav. How did you solve it?
this is an equation right
Yes
ok this is what i found ....... http://www.algebra.com/algebra/homework/equations/Equations.faq.question.616566.html
this should give you a better understanding then what we gave you .. this question does not seem complete
It doesn't :(
what do you need
I want to know how one would solve this question
Will i do or you will do @HWBUSTER00
well i tired to help @Catch.me can you do it
OK the main concept of these problems is to convert words to equation, so just get sentences and write the corresponding equation Two taps A and B fill a swimming pool together in two hours. means \[A + B = 2\] Alone, it takes tap A three hours less than B to fill the same pool. \[A = B - 3\] now you have 2 equations 2 unknowns substitute in the first equation using last equation
So, \(A=2-B\) and \(A=B-3\)? Therefore \[2-B=B-3\]\[-B=B-5\]\[-2B=-5\]\[B=\dfrac{5}{2}\]
\[A=2-b\]\[A=-\dfrac{1}{2}\]Which is impossible :(
@Catch.me?
OK first equation was wrong A = x to fill 1 liter B = y to fill 1 liter so in 2 hours x : 1 2 : ?? A will fill 2/x B will fill 2/y so 2/a + 2/b = 2
no equal 2 but 1
hole tank
I'm really confused now. Can you elaborate @Catch.me?
I made a mistake in understanding the first sentence Two taps A and B fill a swimming pool together in two hours means that the amount of A filling in 2 hours + the amount of B filling in 2 hours will make the tank full lets say A fills tank in 7 hours so in 2 hours it fills 2/7 of that tank. so (2/a) + (2/b) = 1
ok
now you have 2 equations 2 unknowns solve them \[\frac{ 4b-6 }{ b ^{2}-3b }= 1 \] of course when b = 3 is neglected. so \[B ^{2}-7B+6=0\] B=6 or 1 is neglected
@Ahsome ?
How did you go from (2/a)+(2/b)=1 to that @Catch.me?
put a = b-3 and make multiply them
OK so let's say that one tap takes \(x\) hours and the other one takes \(x+3\). Then the fraction of work they do in one hour is given by \(1/x + 1/(x+3)\) but since it takes two hours for them to do the work, it is also given by \(1/2\)\[\Rightarrow \dfrac{1}{x} + \dfrac{1}{x+3} = \dfrac{1}{2}\]\[\Rightarrow x = 3\]
common denominator
Ahh, I see. So x would be the slower tap, right?
@Ahsome do you understand this question now?
No, I still don't :(
B fills 1 pool per x hours ----> rate of B filling a pool = 1/x A = B - 3 hours, so A fills 1 pool per x - 3 hours ----> rate of A filling a pool = 1/(x - 3) now we know the rates of each so we have this, both taps fill 1 pool in 2 hours, 2 hours * (rate of B) + 2 hours * (rate of A) = 1 pool filled 2* (1/x) + 2 * (1/(x-3)) = 1
when you solve that for x you get 2 answers for x one of these will not make sense if you plug into this equation A = B - 3, will give a negative result, and we know this makes no sense because the number of hours it takes to fill the pool cannot be negative, so the answer that makes sense is 6 hours for B, then you just subtract 3 from that to get 3 hours for A
the method ParthKholi used is the same thing except he used B = A + 3 instead of A = B -3 , if you divide both sides of this equation by 2 2* (1/x) + 2 * (1/(x-3)) = 1 you will get a similar equation to what he had
i hope that helps
i recommend watching some of these videos https://www.khanacademy.org/math/algebra/ratio-proportion-topic/advanced-ratios/v/another-take-on-the-rate-problem
Thank you so much @billj5. I'm kinda busy now, but I will definately look at this afterwards :)
yw
The answer was right.
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