Fraction: Students not Going to Trip

WBahn

Joined Mar 31, 2012
29,978
Did you not read my post #12 and #18?
I wasn't responding to your Post #12 or #18. But now that you went out of the way to put them on the table, neither of those posts pointed out, explicitly, that Zulfi doesn't need the classic one equation per unknown because he is not solving for all of the unknowns. That point is implicit in your recommendations, and hence easily overlooked. So Zulfi could very easily end up thinking that what you've recommended is how to get that "third equation" he is looking for, when that's not what you were doing.

I also recommended that Zulfi start things off by clearly defining what he is trying to solve for. You did essentially the same thing, but in a slightly indirect faction. I chose to do it is a very direct fashion. I could have (perhaps should have) done it even more direct.

Fraction of students not going on trip = (students not going on trip) / (number of students)

That's a clear statement of what we want to find.

Then I could have illustrated the next step in the process of moving from what we want to what we have to work with:

students not going on trip = (juniors not going) + (seniors not going)

number of students = (juniors) + (seniors)

The next step would be to utilize the information given to move these equations closer to something that can be evaluated.
 
I wasn't responding to your Post #12 or #18. But now that you went out of the way to put them on the table, neither of those posts pointed out, explicitly, that Zulfi doesn't need the classic one equation per unknown because he is not solving for all of the unknowns. That point is implicit in your recommendations, and hence easily overlooked. So Zulfi could very easily end up thinking that what you've recommended is how to get that "third equation" he is looking for, when that's not what you were doing.

I also recommended that Zulfi start things off by clearly defining what he is trying to solve for. You did essentially the same thing, but in a slightly indirect faction. I chose to do it is a very direct fashion. I could have (perhaps should have) done it even more direct.

Fraction of students not going on trip = (students not going on trip) / (number of students)

That's a clear statement of what we want to find.

Then I could have illustrated the next step in the process of moving from what we want to what we have to work with:

students not going on trip = (juniors not going) + (seniors not going)

number of students = (juniors) + (seniors)

The next step would be to utilize the information given to move these equations closer to something that can be evaluated.
Part of the philosophy of the homework help forum is to guide the student, not give them the answer. So one could argue that telling them explicitly what to do rather than hinting might be going too far. I think I told him just enough for him to understand that he needed to create the fraction 1 - Trip/P. He didn't seem to catch on after post #12, so I went a little further in post #18 and made sure he understood that if he created a fraction with numerator and denominator expressed in terms of only j (which he had done in deriving his expression for Trip) he would get the numerical result needed.
 

Alec_t

Joined Sep 17, 2013
14,280
What rules did they teach in your school days
That the precedence order was exponentiation, multiplication/division, addition/subtraction. I don't recall being taught anything beyond that about precedence. "Associative" certainly was never mentioned :).
 

WBahn

Joined Mar 31, 2012
29,978
That the precedence order was exponentiation, multiplication/division, addition/subtraction. I don't recall being taught anything beyond that about precedence. "Associative" certainly was never mentioned :).
So it sounds like multiplication and division were at the same precedence (and the same for addition and subtraction). They may not have used the word "associative" in this context -- I'm pretty sure I never heard it in this context until college. But do you recall what they told you to do when you had more than one operator at the same precedence? Such as

x = 16 / 8 / 2

I was taught that you do them left to right, so this becomes

x = (16 / 8) / 2 = 1

and not

x = 16 / (8 / 2) = 4

The term "associative" was only used (in grade school, as I recall) in the context of "the associative property of multiplication" meaning that

x = (a * b) * c = a * (b * c)

and similarly, of course, for addition. Division and subtraction were neither commutative nor associative, in that sense.

But I definitely recall the left-to-right rule for operations at the same precedence being taught very early on.
 
What rules did they teach in your school days?

My understanding (and there's always variation on this kind of stuff) is that the basic rules (exponentiation before multiplication before addition) are considered natural and predate the development of algebraic symbolism in the 1600's.

The notion of associativity and equal precedence of multiplication/division and addition/subtraction dates to the late 1800's or early 1900's. A lot of the more-or-less standardization came as a result of the widespread introduction of math textbooks in the early 20th century. The driver on this particular point was the recognition that division is the multiplicative inverse and addition (should be subtraction?) is the additive inverse, and hence these two operations can be removed in any expression using higher-precedence unary operators and replaced with multiplication/addition and therefore should be considered the same as far as precedence and associativity are concerned.

x = a - b + c = a + (-b) + c
x = a / b * c = a * (b^-1) * c

By I also understand that there was a movement, which never gained much ground, that said that implied multiplication has higher precedence than explicit multiplication/division, in which case

x = a/bc

would bind the bc first because of the implied multiplication.

However, this caused problems because then

a/bc ≠ a/b·c

and that was simply more trouble than it was worth.

I don't know when the heyday of this movement was or how widespread it was. It certainly has no traction to speak of today since it would once again cause all kinds of problems in translating expressions involving both implied and explicit multiplication into computer programming expressions, which largely require only explicit operations.

I have seen people claim that the various acronyms, such as BEDMAS and PEMDAS, give an explicit order of precedence for each separate operator, but this has never been the case (as far as I know). These two, for instance, differ on whether multiplication comes before division or not. But, my understanding is that these always had the understanding (sometimes written explicitly) that multiplication/division and addition/subtraction were at the same level. So they are sometimes written as BE(DM)(AS) or PE(MD)(AS).

I don't recall ever seeing any of these acronyms when I was growing up. I think things developed pretty naturally because first we learned addition/subtraction. We learned left-to-right and then we learned parentheses. Then we learned multiplication/division. Again we learned left-to-right and parens. Then when we mixed them we were simply taught that the latter were done before the former, with parens always overriding anything. Then when we learned exponentiation it was just natural that it was yet another higher-order operation. The fine print that it is right-associative didn't come till later, IIRC.
Expand the above for a possible correction.

WBahn, how many times have you delivered this lecture/essay about the importance of using parentheses when posting mathematical expressions using ordinary text, rather than the more elegant TEX. I think you should make it a sticky in this forum.
 

WBahn

Joined Mar 31, 2012
29,978
WBahn, how many times have you delivered this lecture/essay about the importance of using parentheses when posting mathematical expressions using ordinary text, rather than the more elegant TEX. I think you should make it a sticky in this forum.
That might not be a bad idea.

Right now, though, alec_t and I are having an interesting side discussion about the history of how these things have been taught in the past. I see it as valuable not only for its own sake, but also because it helps establish the different backgrounds and perspectives that members have on this stuff depending not only on where they went to school, but when.
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
Thanks for your work. I found that deriving equation is really a difficult for me. So i have switched to using values. This is what GRE is also doing. However, i did not look at their solution but my answer is correct.

JTrip = 1/4J
STrip = 2/3S
Now its given J= (2/3)S
At this point its clear that Senior S is greater than Junior J. So lets chose 90 as the value for S.
J=(2/3 ) 90 = 60
Now JTrip = (1/4) * 60 = 15
STrip = (2/3) * 90 = 60
Total on Trip = 60 +15 = 75
Total in School = 90 + 60 = 150
Total not in Trip = 150-75=75
Fraction = 75/150 = 1/2

Thanks WBahn for telling me that i have to find the fraction out of total.
 

WBahn

Joined Mar 31, 2012
29,978
To flesh out the process I was suggesting you might consider:

First, clearly define what we are looking for:

\(
\text{Fraction of students not going on trip \; = \; \frac{(students not going on trip)}{(number of students)}}
\)

Now catalog the information given to us:

(juniors going on trip) = (1/4)(juniors)

(seniors going on trip) = (2/3)(seniors)

(juniors) = (2/3)(seniors)

Now start moving what we are looking for toward being expressed in terms of what we are given by writing each parameter on the right in terms of something that is at least closer toward the information we have available:

(students not going on trip) = (juniors not going) + (seniors not going)

(number of students) = (juniors) + (seniors)

This let's us write out final goal as:

\(
\text{
Fraction of students not going on trip = \frac{(juniors not going) + (seniors not going)}{(juniors) + (seniors)}
}
\)

Now make some more progress in the desired direction:

(juniors not going) = (juniors) - (juniors going on trip)

(seniors not going) = (seniors) - (seniors going on trip)

\(
\text{
Fraction of students not going on trip = \frac{ \[(juniors) - (juniors going on trip)\] + \[(seniors) - (seniors going on trip)\] }{(juniors) + (seniors)}
}
\)

Rearranging things in the numerator might make a little bit of sense at this point:

\(
\text{
Fraction of students not going on trip = \frac{ \[(juniors) + (seniors)\] - \[(juniors going on trip) + (seniors going on trip)\] }{(juniors) + (seniors)}
}
\)

Now we are in a position to directly apply our given information. We'll do this in two steps to make it clear:

\(
\text{Fraction of students not going on trip = }\frac{\text{ \[(juniors) + (seniors) \] - \[\(\frac{1}{4}\)(juniors) + \(\frac{2}{3}\)(seniors)\] }} {\text{(juniors) + (seniors)}}
\)

\(
\text{Fraction of students not going on trip = }\frac{ \text{\[\(\frac{2}{3}\)(seniors) + (seniors)\] - \[\(\frac{1}{4}\)\(\frac{2}{3}\)(seniors) + \(\frac{2}{3}\)(seniors)\] }} {\text{\(\frac{2}{3}\)(seniors) + (seniors)}}
\)

Now we can see that every term has (seniors) in it, so we can cancel them all out.

\(
\text{Fraction of students not going on trip = }\frac{ \[\(\frac{2}{3}\) + 1\] - \[\(\frac{1}{4}\)\(\frac{2}{3}\) + \(\frac{2}{3}\)\] } {\(\frac{2}{3}\) + 1}
\)

Now we just have a number to solve. We can make things easy by multiplying top and bottom by 6

\(
\text{Fraction of students not going on trip = }\frac{ \[4 + 6\] - \[1 + 4\] } {4 + 6} \; = \; \frac{1}{2}
\)

I think whoever came up with the problem should have chosen an answer other than 1/2, since that makes it hard to spot when students answer the wrong question.

On the other hand, I had a Physics prof that would intentionally design problems such that you could make one or more obvious mistakes and get a numerically correct answer, to which you would still lose nearly all the points with a comment, "Right answer, wrong approach. -10" or something like that. So we made him a stamp that simply said, "Right answer. -10". He got the last laugh when he used it on the next exam.
 
Top