How long is this bridge

maris

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Jul 13, 2011
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I may be under the wrong topic, so if I am please forgive me and direct me to the correct one if you do not mind.

My problem:
If a bridge crosses over a river that is 1520 ft wide and one bank of the river holds 1/5 of the bridge and the other holds 1/6 of the bridge how long is the bridge?

My first instinct was to take 1/5 of 1520 and then 1/6 of 1520, add them and then subtract the total from 1520, but I stopped as I realized this isn't the way to do it. I am blocked, and I know it must be simpler than I am making it.

Any suggestions?

Thanks very much for your patience and assistance.
Maris
 


maris said:
Any suggestions?

:idea: Let x = the requested unknown (i.e., the length of the entire bridge)

Since the river itself is 1,520 feet wide, the part of x over water is 1520.

The expression 1/5x is the part of x over one bank, and the expression 1/6x is the part of x over the other bank.

All three sections must add up to x.

Write an equation to model this fact, and solve this equation for x.

Please show your work, if you would like more help with this exercise.

Cheers ~ Mark 8-)

 
maris said:
I may be under the wrong topic, so if I am please forgive me and direct me to the correct one if you do not mind.

My problem:
If a bridge crosses over a river that is 1520 ft wide and one bank of the river holds 1/5 of the bridge and the other holds 1/6 of the bridge how long is the bridge?

My first instinct was to take 1/5 of 1520 and then 1/6 of 1520, add them and then subtract the total from 1520, but I stopped as I realized this isn't the way to do it. I am blocked, and I know it must be simpler than I am making it.

Any suggestions? (See below)

Thanks very much for your patience and assistance.
Maris
mmm's advice above is very good.

In general, it is an excellent method for almost all "word problems" to translate the problem into mathematical language before doing anything else. That is because mathematical language has been designed over centuries to facilitate logical thought about things like numbers, space, and time.

Almost always, the first step is to "name things mathematically," which means to assign a unique symbol to each concept that the problem concerns. Note that the first thing mmm did was say x = length of bridge. Obviously the choice of x rather than L or b is purely arbitrary. The point is that the concept "length of bridge" is now captued in a symbol that can be manipulated easily in mathematics.

The second step is to write down USING YOUR CHOSEN SYMBOLS and numbers all the information ALREADY GIVEN in the problem. See how mmm did it.
part of bridge over water = 1520
part of bridge over one bank = (1/5)x
part of bridge over other bank = (1/6)x.

The third step is to form one or more equations relevant to your problem USING YOUR SYMBOLS and the GIVENS. All the other steps in solving the problem are mechanical, but setting up the equation for a new kind of problem may take some creativity. In this case, mmm gave you a hint.

The last step is to solve your equation. At this point, it is no longer a word problem; you just apply mechanically the rules of algebra or calculus or whatever to solving your equation.

If you learn to follow this procedure, "word problems" will lose their terror. At least when I was being taught math, no one told me this procedure (or if they did I was daydreaming that day). So I thought it worthwhile to generalize on mmm's answer in case you too had never been given a general technique for solving word problems. If you can find in a library Polya's book "How To Solve It," I advise reading it.
 
1 - 1/5 - 1/6 = 19/30 : OK?

... go figure, young man :wink:
 
Denis said:
1 - 1/5 - 1/6 = 19/30 : OK?

... go figure, young man :wink:
The answer is 2400ft? I have no idea how they got that and whatever is above doesn't make sense.
 
maris said:
Denis said:
1 - 1/5 - 1/6 = 19/30 : OK?

... go figure, young man :wink:
The answer is 2400ft? I have no idea how they got that and whatever is above doesn't make sense.
mmm did steps 1 and 2 of my suggested procedure for you.

His hint for doing step 3 was that the length of the bridge must equal the sum of the lengths of its parts.
Using the symbols from steps 1 and 2 to express mathematically that the whole equals the sum of its parts gives the following equation

x = [(1/5)x] + 1520 + [(1/6)x]

Step 4 is to use the mechanical tools of algebra to solve that equation.
x = [(6/30)x] + 1520 + [(5/30)x] Lowest common denominator
x[1 - (6/30) - (5/30)] = 1520 Gather like terms on one side of the equation
x(30 - 6 - 5)/30 = 1520 Lowest common denominator
19x/30 = 1520 Simplify (Lo and behold, denis's 19/30 pops out when solving the equation)
x = 1520 * 30 / 19 Cross multiply
x = 2400 Simplify

There is nothing mystical here. Steps 1, 2, and 4 are mechanical. Step 3 is the only one that requires thinking about how to express a relationship between the desired answer (now symbolized by x) and the known data (now sympolized by x/5, x/6, and 1520). If an answer exists, it must come from what is already known.
 
OK thanks yea I need to study this answer a bit. I am a bit confused now and where the (1-) came from I have no idea? Is this a linear equation? so I can look up similar problems in the GRE math book.
 
ok I have to go away for a bit right now. Ill get back on in a couple of hours.
 
Yes this is a linear equation.

x = 1 * x ALWAYS
So x - [(6/30)x] - [(5/30)x] = [1 * x] - [(6/30)x] - [(5/30)x] But x is a common factor
So x - [(6/30)x] - [(5/30)x] = x[1 - (6/30) - (5/30)] = x[(30/30) - (6/30) - (5/30)] But (1/30) is a common factor
So x - [(6/30)x] - [(5/30)x] = x(30 - 6 - 5)(1/30) = x(19)/30 = 19x/30
 
ok great jeff. Yea I am use to using shortcuts. I would have done that problem the second way or i guess (short hand). I am just having a problem for some reason putting the question above it together completely.
 
maris said:
ok great jeff. Yea I am use to using shortcuts. I would have done that problem the second way or i guess (short hand). I am just having a problem for some reason putting the question above it together completely.
I saw my immediately preceding post did not really answer what I think was confusing you and was editing it while you responded to it. I probably should have just done a new post because now your response deals with a post of mine that is entirely changed.
 
maris said:
I am a bit confused now and where the (1-) came from I have no idea?
If you give your girlfriend 1/4 of your apple, then you have left 1 - 1/4 = 3/4 of an apple, right?

1 bridge - 1/5 of the bridge - 1/6 of the bridge ...... capish?

1 = 30/30
1/5 = 6/30
1/6 = 5/30
30/30 - 6/30 - 5/30 = (30 - 6 - 5) / 30 = 19/30 .... yo?
 
JeffM said:
maris said:
Denis said:
1 - 1/5 - 1/6 = 19/30 : OK?

... go figure, young man :wink:
The answer is 2400ft? I have no idea how they got that and whatever is above doesn't make sense.
mmm did steps 1 and 2 of my suggested procedure for you.

His hint for doing step 3 was that the length of the bridge must equal the sum of the lengths of its parts.
Using the symbols from steps 1 and 2 to express mathematically that the whole equals the sum of its parts gives the following equation

x = [(1/5)x] + 1520 + [(1/6)x]

This step becomes clear if you draw a picture

TTTTT
rrrrrrrrrrrrrrrrrrrTTTTTT

Above there are 30 spots (the LCM of 5 and 6 is 30 - that makes my drawing easier). So "T"s show the supports and the "r"s show the river. And all together show the bridge. Now you should see the equation staring at you:

x = x/6 + 1520 + x/5
 
JeffM said:
Yes this is a linear equation.

x = 1 * x ALWAYS
So x - [(6/30)x] - [(5/30)x] = [1 * x] - [(6/30)x] - [(5/30)x] But x is a common factor
So x - [(6/30)x] - [(5/30)x] = x[1 - (6/30) - (5/30)] = x[(30/30) - (6/30) - (5/30)] But (1/30) is a common factor
So x - [(6/30)x] - [(5/30)x] = x(30 - 6 - 5)(1/30) = x(19)/30 = 19x/30

Jeff still having a hard time finding the solution to this problem.
 
Maris, if you can't solve: x = x/6 + 1520 + x/5,
then you need classroom help; not much can be done here...
 
Denis said:
Maris, if you can't solve: x = x/6 + 1520 + x/5,
then you need classroom help; not much can be done here...

I figured it out. It would have been easier if you guys would have said 1/5x then converts to 5x and 1/6x converst to 6x. I worked it from there and got an answer. You guys didnt provide that step. Thanks I know it can be tedious to work with others about math over a textbox forum, but if you are on here to help then you should not tell someone they need classroom help. Thanks I was missing a step.
 
maris said:
Denis said:
Maris, if you can't solve: x = x/6 + 1520 + x/5,
then you need classroom help; not much can be done here...

I figured it out. It would have been easier if you guys would have said 1/5x then converts to 5x and 1/6x converst to 6x. I worked it from there and got an answer. You guys didnt provide that step. Thanks I know it can be tedious to work with others about math over a textbox forum, but if you are on here to help then you should not tell someone they need classroom help. Thanks I was missing a step.
BUT you were told that a few times: use LCM of 30.
1/5x does not convert to 5x: it converts to 6x: 1/5x * 30 = 6x.
You were told to multiply through by 30:
30x = 5x + 45600 + 6x
19x = 45600
x = 45600/19 = 2400
Is that what you got?
 
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