Agreed. I suspect the problem is deliberately written to generate confusion.
48÷2(9+3)
- Thread starter Splash
- Start date
Agreed. I suspect the problem is deliberately written to generate confusion.
LOL, you're batting .000:
I'm embarrassed to have graduated from the same school as you. Please don't represent us.
Motorcycle Mike
Well-known member
Why? Because you disapprove of him giving the finger, or you are another one who, despite every online calculator as well as Sushi's real calculator, thinks that the answer is not 288?
Oh... maybe we are the ones that should be ashamed to have you as a fellow Waterloo grad... afterall, you are the one who thinks you can make a perpetual motion machine with a motorcycle and hydrogen generator... how is that coming by the way? Ever run your EX500 on hydrogen?
By the way, this already has been beaten to death in more places that here: http://www.reddit.com/search?q=48÷2(9%2B3)
The overall consensus with most online calculators, most real calculators, and most international forums would agree that 288 is the correct answer, but that it is an intentionally misleading question.
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Splash
Well-known member
Regardless what my opinion of the answer is (I'm not afraid to say I disagree with the answer 288 ), he should not use the fact that he's a Waterloo grad to imply that his answer is correct.
Lets look on the flipside, "IF" he is wrong, he somewhat tarnishes the reputation of our school, which reflects on me as a grad of the same school. Not a HUGE hit, but a hit nonetheless. If hes right, makes us look great! Sure. I prefer not to have him take that risk for us and let him just say "I'm a graduate of a recognized post secondary institution".
No where do you see that I boast that I'm from a school because I do not represent the rest of our graduates.
What does my interest in alternative fuel technologies have to do with this? Have you read any works from Stan Meyers or Ravi's hydrogen cell?
Just because YOU dont think its viable, why put a negative spin on it?
Again irrelevant. I'm discussing on THIS forum. Have I read all of the internetz? My post was to address his post.
As sushi's attached picture shows, different calculators show different answers, so that answer is not definitive. You cannot say that because the majority of calculators say its 288, the answer must be 288. Just like wikipedia, because more people believe it, does it make it right?
Try this equation, and tell me what you get.
48 / 2 (9x +3) = 288
Simplify this equation and solve for x. If this answer is correct, X should = 1 correct?
Now try this equation.
48 / 2(9x + 3) = 2
Simply this equation and solve for x. If this answer is correct, X should = 1 also correct?
Please tell me what you get.
Motorcycle Mike
Well-known member
It is quite obvious that you think that 2 is the correct answer, as that is what you voted for in the poll.
I brought up your beliefs in perpetual hydrogen machines as a reason to be ashamed that you are a fellow Waterloo grad (as you were implying with lightcycle), which you previously bragged about and implied gave you some credibility. I have not read any of the works you cited, but based on the laws of thermodynamics I find highly unlikely that one could put water in a tank and have a machine generate enough energy from that hydrogen to both move the vehicle as well as separate the hydrogen from the oxygen. I do remember you assuring that it is possible though, based on your qualifications as a biochemistry grad from Waterloo - and I remember feeling shame that a fellow Waterloo grad was using the name of our alma mater to justify junk science.
Contrary to your statement that you never use your degree to attempt to establish credibility, I saw you do just that on ex-500.com. How else would I know what you studied and from where as I don't know you?
Finally, I solved your equations:
48 / 2 (9x +3) = 288 : x=1 : http://www.wolframalpha.com/input/?i=48+%2F+2+(9x+%2B3)+%3D+288
48 / 2(9x + 3) = 2 : x = -0.3241 : http://www.wolframalpha.com/input/?i=48+/+2(9x+++3)+=+2
I brought up your beliefs in perpetual hydrogen machines as a reason to be ashamed that you are a fellow Waterloo grad (as you were implying with lightcycle), which you previously bragged about and implied gave you some credibility. I have not read any of the works you cited, but based on the laws of thermodynamics I find highly unlikely that one could put water in a tank and have a machine generate enough energy from that hydrogen to both move the vehicle as well as separate the hydrogen from the oxygen. I do remember you assuring that it is possible though, based on your qualifications as a biochemistry grad from Waterloo - and I remember feeling shame that a fellow Waterloo grad was using the name of our alma mater to justify junk science.
Contrary to your statement that you never use your degree to attempt to establish credibility, I saw you do just that on ex-500.com. How else would I know what you studied and from where as I don't know you?
Finally, I solved your equations:
48 / 2 (9x +3) = 288 : x=1 : http://www.wolframalpha.com/input/?i=48+%2F+2+(9x+%2B3)+%3D+288
48 / 2(9x + 3) = 2 : x = -0.3241 : http://www.wolframalpha.com/input/?i=48+/+2(9x+++3)+=+2
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jolomatic
Well-known member
Would it be crazy to suggest that there is more than one accepted convention?
And that the correct answer changes, depending on the convention you are using?
As long as you are consistent, and the question is phrased according to the convention being used to solve it, it really should not be an issue.
And that the correct answer changes, depending on the convention you are using?
As long as you are consistent, and the question is phrased according to the convention being used to solve it, it really should not be an issue.
The correct answer is of course both 2 and 288 depending on if the original author wanted you to use multiplication or the division operator first. This is of course the one and only correct answer and puts the onus back on the author to clarify which result he intended to achieve.
Kinda like quantum mechanics.
Kinda like quantum mechanics.
Could you explain your motives for putting x anywhere and trying to solve for it?
You can put an X anywhere because as long as X is a 1 when you solve it it will be correct.
E.G
5-1 = 4
5-1X = 4, what is X? 1. If x does not = 1 then the answer is not 4
(1*5)/5= 5
(1*5)/5X=5, what is X? 1. If x does not = 1 then the answer is not 5
Basically anything multiplied by one or divided by one will equal the initial number. Its just a way to check an answer.
Something similar to this is always imagining a 1 infront of a variable.
E.G.
4-X can also be written as 4-(1)X
or
5X^2-X+7 can also be written as 5(1)X^2-(1)X+7
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It is the easiest way to check your answer. If you enter x for any variable and put = "your answer" at the end, then when you solve for x you will get x = "the variable it replaced".
Yes the 2 ways at looking at this question are to have 48÷2 represented as a fraction in that case as shown here: http://www.wolframalpha.com/input/?i=48+%2F+2+(9x+%2B3)+%3D+288
and the other way is since there is no multiplication sign between the 2 and the parentheses 2(9+3) is to look at it as the 2 is part of the parentheses.
The question it self is horribly written and misleading (and would never be seen written this way in any Math text book regardless of the school)
Lets just say (9+3) = x
Is the question 48 divided by 2 times x or is it 48 divided by 2x?
for the people that got the answer 288 would have to agree that: 48(9+3)÷2 = 48÷2(9+3)
El Zilcho, Please explain how this is wrong?
I would love to hear it?
Here is one that hopefully you might be able to answer: 2+2=4 Put an x anywhere in that equation and solve for it and see what answer you get?
I guess you have not taken math that will require you to substitute in order to solve...
Good, you managed to state the obvious. I never denied this so there is no point in mentioning it. I see that you like investigating.
I don't recall bragging about my credentials. Nor did I say that electrolysis is the definitive answer to our fuel problems. Whatever thread that was, it was purely discussion and I didn't use my BSc to say you're wrong and I'm right. It was a discussion thread asking for opinions.
Did you even read that post? On that forum, I never stated where I graduated from. I merely stated I have credentials from the field of science to stop people from immediately hating on my thread. People who have no background on the subject like to immediately jump in and say "That idea is stupid". This is where my gripe with yours is different because the problem is with Light cycle's post stating where he's accredited from PLUS doing the question wrong PLUS showing an incorrect proof. I don't need a BSc to say that this elementary math problem has been done incorrectly.
And finally, you did not solve the equation, you typed it into a calculator that we've already established that cannot provide a definitive answer.
Do it by hand and give it a try.
Click on the images i have provided. It will show step by step about solving it.
Found out the attached image is small so im going to upload it.
http://peterchen.me/stuff/Capture.PNG
http://peterchen.me/stuff/Capture1.PNG
http://peterchen.me/stuff/Capture2.PNG
http://peterchen.me/stuff/Capture3.PNG
As you can see, image #1 and #2 shows what you get if you do multiplication first, while image #3 and #4 shows division first.
So the answer varies depending on how you were taught in school. I was taught that multiplication was always done before division while some others were taught that they are equal so which ever comes first.
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I already explained that. The original problem is in the ordering of arithmetic operations. In order to evaluate the expression you need to follow a certain order. In order to solve an equation based on the original expression you also need to follow a certain order. As long as the order you follow in both cases is the same, you are going to get consistent answers from both (even if the order of operations is wrong). Hence, you cannot use one to prove the other.
Doing so is like trying to "prove" Ohm's law with a multimeter. In reality, a multimeter only measures one quantity and calculates the others internally using Ohm's law.
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