(Is my solution correct?) Rate Word Problem: "Tom can build a bicycle in ten days. After four days,..."

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wavewavewow

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Tom can build a bicycle in ten days. After four days of doing the job, his friend, Jerry joins him, and the two of them finished building the bicycle in three and a half days. How long will it take Jerry to build the bicycle alone?

Here is my solution:
answer.jpg

Is my solution correct? Thank you in advance!
 
wavewavewow said:
Tom can build a bicycle in ten days. After four days of doing the job, his friend, Jerry joins him, and the two of them finished building the bicycle in three and a half days. How long will it take Jerry to build the bicycle alone?

Here is my solution: [huge image]
Click to expand...
Your image is huge, which makes it difficult to follow, and the writing is unclear. I *think* you have posted the following:

Tom = T
Jerry = J
T = 1/10 = 0.1
WT = 1/10 (4) = 0.4
TJ = 1/3.5 = 0.29
TJ = T + J
0.29 = 0.1 + J
J = 0.29 - 0.1
J = 0.19
TJ = 1/0.19 = 526 [??]

When you say that "Tom = T", what do you mean? Perhaps something along the lines of "T: the amount of one build that Tom can do in one day"? If so, say so. (The defining of variables is quite important, especially when communicating with others.)

For what do "WT" and "TJ" stand?

Why did you take the exact "1/3.5" and convert it to an approximate decimal value? (This will necessarily lead to errors.)

wavewavewow said:
Is my solution correct?
Click to expand...
When you plugged your answer back into the original exercise, what did you get? (You can check the answer to *any* "solving" exercise by plugging it back into the original context.)

Thank you!

Eliz.
 
How much of a job does Tom do in 1 day? How much of a job does Tom do in 7.5 days?
What part of the job did Jerry do? How many days did it take?
 
wavewavewow said:
Tom can build a bicycle in ten days. After four days of doing the job, his friend, Jerry joins him, and the two of them finished building the bicycle in three and a half days. How long will it take Jerry to build the bicycle alone?

Here is my solution:
View attachment 35676

Is my solution correct? Thank you in advance!
Click to expand...
No, your solution is not correct.

So where have you gone wrong?

Well, to start with, as has already been pointed out (AHABPO), you have introduced variables (T, J & W) that you have not defined (properly)!
We are left to infer from values subsequently assigned to these terms exactly what you mean by them; that is very bad practice which leads to confusion (and subsequent error).

I suspect (again, AHABPO) that you intended that:-

T ≡ what fraction of the whole job Tom can do in one day
​
and          (the "whole job" being to build a complete bike from start to finish.)​

J ≡ what fraction of the whole job Jerry can do in one day
(which is what you need to find in order to obtain the final answer)​

But you then also introduce W, which you do not attempt to define at all but which appears to be:-

what fraction of the whole job has been completed by someone over a particular time period?​

Though this (important) variable then subsequently disappears from your analysis altogether!

Your next "definition" appears to be (again by inference) that:-

TJ ≡ What fraction of the the whole job Tom and Jerry can do together in one day.​

Then you go on to assert that this (TJ) equates to the sum of T & J, which would be correct if (and only if) you were relating the time it took both of them to complete the whole job.

For example:
If it takes me 2 days to build a house (ie: I build ½ houses/day) and I have friend who takes 4 days to build one (ie: he builds ¼ houses/day) then how many houses will we build in 4 days?
I will build two and he will build one so that is 3 we build (together) in 4 days, ie: ¾ houses/day we build together, yes?
And ¾ = ½ + ¼ so that 'analysis' does hold good.

Unfortunately, however, you equate TJ to \(\displaystyle \frac{1}{3.5}\) (which you then 'convert' to a decimal fraction when it should have been expressed as a common fraction \(\displaystyle \left(\frac{2}{7}\right)\) as what converting it to a decimal fraction does (again AHABPO) is to introduce a rounding error)!

But...

\(\displaystyle \frac{1}{3.5}\left(\frac{2}{7}\right)\) is NOT the fraction of the whole job that they do together, it is only a fraction of part of the whole job that they do together!

What part?

Well, Tom over 4 days has already completed 0.4 or 40% of the job. Let us call that WT4 (what fraction of the whole job Tom completes in 4 days; thus reintroducing your (previously ill-defined) W term).

That means that WTJ3.5 would be 0.6 or 60% or \(\displaystyle \frac{6}{10}\) of the job
but you want to know what their (combined) rate is to complete the whole job!

If they complete \(\displaystyle \frac{6}{10}\) of the job in 3.5 days then you would need to multiply both those numbers by 1⅔ ie: \(\displaystyle \frac{5}{3}\) because \(\displaystyle \frac{6}{10} \times \large{1⅔} \normalsize{ = \frac{6}{10} \times \frac{5}{3} = \frac{30}{30} =}\large{ 1}\) which means that together they would complete the whole job in 3.5 × \(\displaystyle \frac{5}{3}\) days ie: \(\displaystyle \frac{17.5}{3}\) days.

So the (correct) TJ "rate" would be: \(\displaystyle \frac{1}{\frac{17.5}{3}} = \frac{3}{17.5}\) or \(\displaystyle \frac{6}{35}\)

and now
you can correctly say that:-

TJ = T + J \(\displaystyle \rm{\implies \frac{6}{35} = \frac{1}{10} + J \\\implies J = \frac{6}{35} - \frac{1}{10} = \frac{60}{350} - \frac{35}{350} = \frac{25}{350}}\) = What?​
​
(ie: What fraction of the whole job Jerry can do in one day.)​


Do you now see why it is important to leave everything in (common) fractional form rather than converting anything to decimal "equivalents"?

So your 'method' was correct (in principle) but failed (in practice) because you did not recognize that they would not be completing the whole job in 3½ days when working together.

Unfortunately, I suspect (as you have already demonstrated) that reaching the correct answer via the method you chose might well be beyond your capabilities, however, there is a much simpler approach which I think the question setter(s) actually intended should be used.

The simpler approach, again, AHABPO (by @Steven G) is to work out how much of the job Tom will have completed after working on it over the whole 7½ days.
ie: WT7.5 = 0.1 x 7.5 = 0.75 = 75% = ¾
So, if Ton has built ¾ of the bike, how much of it has Jerry built?
Clearly that means Jerry has built ¼ of the bike in the 3½ days he worked on it.
And if Jerry builds ¼ of a bike in 3½ days then he will build ½ of a bike in 7 days, so how long will it take Jerry to build a whole bike?
(Which is exactly what you were asked to find out!)

Hope that helps. ?
 
Last edited:
My apologies for any confusion. As I knew I was going to have to repeat it more than once I thought it would save me time, space and typing to create an acronym for the phrase "as has already been pointed out".

I thought it a fairly standard way of 'creating'/explaining an acronym was to put it in brackets immediately after the phrase itself, eg: the "International Olympics Committee" (IOC) or "mental abuse to humans" (MATH ?) but it seems not everyone is familiar with that practice.

Sorry. ?
 
I prefer using units in the equation …

let [imath]t[/imath] be the number of days it takes Jerry to build a bike alone

[imath]\dfrac{1 \text{ bike}}{10 \text{ days}} \left(4+3.5 \right) \text{ days } + \dfrac{1 \text{ bike }}{t \text{ days}} \left(3.5 \text{ days }\right) = 1 \text{ bike done}[/imath]

solve for [imath]t[/imath]
 
skeeter said:
I prefer using units in the equation …

let [imath]t[/imath] be the number of days it takes Jerry to build a bike alone

[imath]\dfrac{1 \text{ bike}}{10 \text{ days}} \left(4+3.5 \right) \text{ days } + \dfrac{1 \text{ bike }}{t \text{ days}} \left(3.5 \text{ days }\right) = 1 \text{ bike done}[/imath]

solve for [imath]t[/imath]
Click to expand...
Your "equation" holds good, of course, but I very much doubt if the OP would be able to solve it in the format you have provided; it might even confuse her/him further still! ?

If the "units" were to be removed, thus...
[math]\dfrac{1}{10}\times \left(4+3.5 \right) + \dfrac{1}{t}\times3.5 = \large{1}[/math]then s/he might be able to solve it (though, Ah ha'e ma doobts) but that wouldn't explain where s/he went wrong. ?
 
Units can and should help students. It confirms if things are definitely wrong.
While taking Physics inits were my best friends.
 
The Highlander said:
Your "equation" holds good, of course, but I very much doubt if the OP would be able to solve it in the format you have provided; it might even confuse her/him further still! ?

If the "units" were to be removed, thus...
[math]\dfrac{1}{10}\times \left(4+3.5 \right) + \dfrac{1}{t}\times3.5 = \large{1}[/math]then s/he might be able to solve it (though, Ah ha'e ma doobts) but that wouldn't explain where s/he went wrong. ?
Click to expand...
Why did you write "equation"?
 
skeeter said:
I prefer using units in the equation …

let [imath]t[/imath] be the number of days it takes Jerry to build a bike alone

[imath]\dfrac{1 \text{ bike}}{10 \text{ days}} \left(4+3.5 \right) \text{ days } + \dfrac{1 \text{ bike }}{t \text{ days}} \left(3.5 \text{ days }\right) = 1 \text{ bike done}[/imath]

solve for [imath]t[/imath]
Click to expand...
My only concern is that at the end you wrote 1 bike done instead of 1 bike.
 
Steven G said:
My only concern is that at the end you wrote 1 bike done instead of 1 bike.
Click to expand...
just a habit ...

these problems usually involve rates of individuals/machines performing a job, i.e. one job done
 
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