# Fraction: Students not Going to Trip

#### zulfi100

Joined Jun 7, 2012
630
Hi,

I have following question:

1/4 of all the juniors and 2/3 of all the seniors are going on a trip. If there are 2/3 as many juniors as seniors, what
fraction of the students are not going on the trip?

Trip = 1/4Juniors(J) + 2/3Seniors(S)

I cant understand what is meant by "2/3 as many juniors as seniors". I think that J=2/3S.

Zulfi.

Last edited:

#### Alec_t

Joined Sep 17, 2013
10,708
I think that J=2/3S.
Think about that. So if the number of seniors (S) is 15, the number of juniors (J) is 2/(3*15) = 2/45 ???
I think you need to be more careful how you express fractions, or you will be marked down .

#### zulfi100

Joined Jun 7, 2012
630
Hi,
<I think you need to be more careful how you express fractions, or you will be marked down >
Thats why i am asking help from this forum.

I cant understand:
there are 2/3 as many juniors as seniors

How the above can be represented mathematically?

Zulfi.

#### dl324

Joined Mar 30, 2015
9,577
How the above can be represented mathematically?
Is this homework?

#### zulfi100

Joined Jun 7, 2012
630
No.
I think my equation j=2/3S is right. If wrong tell me the correct one.

Zulfi.

#### Alec_t

Joined Sep 17, 2013
10,708
I think you intended J=(2/3)*S or, to put it another way, J=2S/3.

#### zulfi100

Joined Jun 7, 2012
630
Hi.
After replacing the S with value in terms of J in the first equation i got:

Trip= 1/J + 2/3*3/2J

Trip= 5J/4

I dont know what to do now.

Zulfi.

#### Alec_t

Joined Sep 17, 2013
10,708
In post #1 you say:
1/4 of all the juniors and 2/3 of all the seniors are going on a trip.
Then you say:
Not Going to Trip = 1/4Juniors(J) + 2/3Seniors(S)
Do you see a problem there?

#### zulfi100

Joined Jun 7, 2012
630
Hi,
Yes its a prob but then i corrected ("Not Trip" to Trip)in post#7
Trip= 5J/4---eq1
J=(2/3)*S----eq2
Now i have 3 unknows: Trip, S & J and two equations. How to create one more equation?

Okay 3rd eq would contain students not going to trip but it would create one additional unknown, how to solve this problem?
Zulfi.

Last edited:

#### Alec_t

Joined Sep 17, 2013
10,708
How to create one more equation?
Write an equation for the total number of students.

#### zulfi100

Joined Jun 7, 2012
630
Total=TotalNumOfStudents=Trip + NotToTrip
NotToTrip = 3J/4 + 1*S/3
Total=5J/4 + 3J/4 + 1*S/3
Again S=3J/2

Total=5J/4 + 3J/4 + J/2
Total=5J/2
Zulfi.

#### The Electrician

Joined Oct 9, 2007
2,751
If P is the total number of students, that can be expressed as P = s+j. Express P as an expression only in j.

You have derived an expression for Trip in terms only involving j.

The fraction of students going on the trip is Trip/P, so the fraction of students not going is 1 - Trip/P.

#### WBahn

Joined Mar 31, 2012
25,068
Think about that. So if the number of seniors (S) is 15, the number of juniors (J) is 2/(3*15) = 2/45 ???
I think you need to be more careful how you express fractions, or you will be marked down .
The expression 2/3S is the same as (2/3)S and 2S/3. This is because, by the pretty much universal rules of basic arithmetic precedence, multiplication and division are both of the same precedence and are left-associative. Thus when both appear together they are evaluated left-to-right. If you want the expression 2/(3S) then you must explicitly override the normal rules using parentheses.

#### Alec_t

Joined Sep 17, 2013
10,708
@ WBahn
I guess the rules have changed since my school days . Ah well, we live and learn. Thanks for that.

#### WBahn

Joined Mar 31, 2012
25,068
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 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.

#### zulfi100

Joined Jun 7, 2012
630
Hi
Thanks.
P = s+j
Trip= 5J/4
Fraction of Student going to Trip= (5j/4)/P
Fraction of students not going to trip = 1-(5j/4)/P
= (P-(5j/4))/P

P = s+j
=5J/2

If P is different kindly show me.

Zulfi.

#### WBahn

Joined Mar 31, 2012
25,068
What do you get for the question that was asked?

Namely, what fraction of all the students are NOT going on the trip?

Your answer should be a simple fraction; no S's, no J's, no P's. It should be something like 3/4 or 1/2 or 71/93.

The information given gives you no way to find out how many students there are, how many seniors there are, how many junior there are, how many students are going on the trip, nor how many aren't.

But you are not asked to find any of those. You are asked to find the fraction of all students that are not going on the trip, and you are given enough information to give a specific, concrete answer to that question.

#### The Electrician

Joined Oct 9, 2007
2,751
Hi
Thanks.
P = s+j
Trip= 5J/4
Fraction of Student going to Trip= (5j/4)/P
Fraction of students not going to trip = 1-(5j/4)/P
= (P-(5j/4))/P

P = s+j
=5J/2

If P is different kindly show me.

Zulfi.
You didn't follow the procedure I suggested. I said "Express P as an expression only in j." You did this to find an expression for Trip only in terms of j.

Then the fraction on the right side of Trip = (5j/4)/P will become a pure numerical fraction, not containing j any more.

#### WBahn

Joined Mar 31, 2012
25,068
Hi,
Yes its a prob but then i corrected ("Not Trip" to Trip)in post#7
Trip= 5J/4---eq1
J=(2/3)*S----eq2
Now i have 3 unknows: Trip, S & J and two equations. How to create one more equation?

Okay 3rd eq would contain students not going to trip but it would create one additional unknown, how to solve this problem?
Zulfi.
You don't need the classic three equations, three unknowns because you aren't asked to solve for S or J or even the number going on the trip. You are asked to solve for a ratio involving these quantities, which cuts down the number of equations you need.

First and foremost thing you need to try to do -- set up an equation that will answer the question being asked:

Fraction of students not going on trip = [(juniors not going) + (seniors not going)] / (juniors + seniors)

Now try to get expressions for each of the pieces.

#### The Electrician

Joined Oct 9, 2007
2,751
You don't need the classic three equations, three unknowns because you aren't asked to solve for S or J or even the number going on the trip. You are asked to solve for a ratio involving these quantities, which cuts down the number of equations you need.

First and foremost thing you need to try to do -- set up an equation that will answer the question being asked:

Fraction of students not going on trip = [(juniors not going) + (seniors not going)] / (juniors + seniors)

Now try to get expressions for each of the pieces.
Did you not read my post #12 and #18?