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11/24/10 12:44 PM
Math Forum - Problem of the Week
Submissions for The Box Supper
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Short
Created
Long Answer
History
Answer
04/09/2001
The amount J – amount of money Joseph paid was an odd #
view
of money that L – amount of money Leroy paid was equaled to (J * 3 – 25)
The amount of money Jay had to pay was 5 * ( L – J )
Jay had to
pay was $65. First of all, we know that Joseph is definitely going to be paying
Submitters
Age Teacher
School
Huang, Nina
[n/a]
Taipei
American
School
Shih, Jason
[n/a]
Taipei
American
School
the least amount of money, then Leroy, then Jay. Also, Jay's payment
is less than 100 dollars.
Let's plot in some random numbers that could be the amount for Joseph.
Let's start with 15.
Leroy's payment would be 15 * 3 - 25, which is 45 - 25 = 20.
Jay's payment would be 5 * (20 - 15) which is 5 * 5 = 25.
Now knowing that, the total amount of money raised that day is 12
times Jay's payment. 12 * 15 = 180, and then it also says that the
sum of the money paid by these three gentlemen was $4 more than the
amount paid by everybody else who bought boxes. Now if we add up all
three of their payments, it would be = 15 + 20 + 25 = 60
total amount of money raised - 180 subtracted by 60 is 120. Now 120
is not 4 less than 60, so Jay didn't pay 15 dollars.
Moving on, let's try 17. Leroy's payment would be 17 * 3 - 25, which
is 51 - 25 = 26. Jay's payment would be 5 * (26 - 17) which is 5 * 9
= 45. Now knowing that, the total amount of money raised that day is
12 times Jay's payment. 12 * 17 = 204, and then it also says that the
sum of the money paid by these three gentlemen was $4 more than the
amount paid by everybody else who bought boxes. Now if we add up all
three of their payments, it would be = 17 + 26 + 45 = 88
total amount of money raised - 204 subtracted by 88 is 212. Now 212
is not 4 less than 88, so Jay didn't pay 17 dollars.
Now let's try 19. Leroy's payment would be 19 * 3 - 25, which is 57 25 = 32. Jay's payment would be 5 * (32 - 19) which is 5 * 13 = 65.
Now knowing that, the total amount of money raised that day is 12
times Jay's payment. 12 * 19 = 228, and then it also says that the
sum of the money paid by these three gentlemen was $4 more than the
amount paid by everybody else who bought boxes. Now if we add up all
three of their payments, it would be = 19 + 32 + 65 = 116
total amount of money raised - 228 subtracted by 116 is 112. Now 112
IS 4 less than 116, so Joseph paid 19 dollars!
04/09/2001
view
Jay, Judy
Ann's
boyfriend,
paid a total of
65 dollars for
that meal.
Joseph paid 19 dollars, Leroy paid 32 dollars, and Jay paid 65
dollars!
OK. First I jotted all the clus down.
Clue 1: Joseph paid (odd #) dollars
Clue 2: Leroy paid 3(odd #) - 25 dollars
Clue 3: Jay paid 5(2 * (odd #) - 25) dollars
Clue 4: Total $$ raised was 12(odd #)
Clue 5: (odd #) + 3(odd #) + 5(2 * (odd #)-25)=4 more than rest of
money raised.
So, with the first clue I set the amount Joseph paid as "x" making
the amount he paid "x" then I went back to my clues and subsituted in
all the area's with (odd #) with "x". And it looked like this.
Clue
Clue
Clue
Clue
Clue
1:
2:
3:
4:
5:
Joseph paid "x"
Leroy paid 3x Jay paid 5(2x Total $$ raised
x + 3x + 5(2x -
dollars
25 dollars
25) dollars
was 12x
25) = 4 more than rest of money raised.
So I then wrote a verbal model.
Amount Joseph paid + amount Leroy paid + amount Jay paid
+ rest of money = total $$ raised.
Then I labled my variables
x = amount joseph paid
3x - 25 = amount Leroy paid
5(2x - 25) = amount Jay paid
12x = total $$ raised
x + 3x + 5(2x - 25) - 4 = rest of the money raised aside from the
amounts Joseph, Leroy, and Jay paid.
Then, I plugged in the variables into my verbal model.
x + 3x - 25 + 5(2x - 25) + x + 3x + 5(2x - 25)- 4 = 12x
Now that's one BIG equation. Gosh!
So then, I solved for x...
28x - 304 = 12x
Solved for x
28x = 12x + 304
added 304 to both sides
16x = 304
subtracted 12x from both sides
x = 19
divided both sides by 16
x = 19
"this is the amount Joseph paid"
so, I'm not done yet.
Since x, or 19, is the amount of dollars Joseph paid,
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 1 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
and since you guys are asking for the amount of money Jay paid, and
since, Jay paid 5 times the positive difference between Leroy and
Joseph, I did this.
amount Leroy paid: 3x - 25
amount Joseph paid x
3 x - 25 - x = 2x - 25
5 times 2x - 25 = 10x - 125
I then subsituted x in and got
10 * (19) - 125 = amount joseph paid.
190 - 125 = 65!!!
Therefor, 65 dollars is the amount Jay paid for that box supper
prepared by Judy Ann, his girlfriend.
04/09/2001
view
Jay paid $65 First of all, because Joseph pay an odd number of dollar, so he paid
2n + 1, which is the expression of an odd number.
for the box
supper.
Secondly, Leroy paid 25 fewer dollars than three times what Joseph
Hsu, Morris
[n/a]
Taipei
American
School
Napolitan, Andrew
[n/a]
Homeschooled
spent, so it is 3(2n + 1) - 25.
Lastly, Jay paid five times the positive difference between what
Joseph and Leroy paid, so it is 5 * | [3(2n + 1) - 25] - (2n + 1) |.
And I know that the total amount of money raised was twelve times
Joseph's payment, and the sum of the money paid by these three
gentlemen was $4 more than the amount paid by everybody else who
bought boxes, so I can then make this equation.
12 Joseph = (Joseph + Leroy + Jay) + (Joseph + Leroy + Jay) - 4
12 Joseph = 2(Joseph + Leroy + Jay) - 4
Now I'll substitude in the numbers.
12(2n + 1) = 2[ 2n + 1 + 3(2n + 1) - 25 + 5 |3(2n + 1) - 25 - (2n +
1)| ] - 4
And I can just now simply solve it.
12(2n + 1) = 2[ 4(2n + 1) - 25 + 5 |2(2n + 1) - 25| ] - 4
12(2n + 1) = 2[ 4(2n + 1) - 25 + |10(2n + 1) - 125| ] - 4
Now there might be two ways to do it, the |10(2n + 1) - 125| part
might be either positive or negative, and I'll do positive first.
12(2n
12(2n
12(2n
304 =
304 =
288 =
9 = n
+ 1) = 2[ 4(2n + 1) - 25 + 10(2n + 1) - 125 ] - 4
+ 1) = 2[ 14(2n + 1) - 150 ] - 4
+ 1) = 28(2n + 1) - 300 - 4
16(2n + 1)
32n + 16
32n
And now I'll do the negative.
12(2n + 1) = 2[ 4(2n + 1) - 25 - 10(2n + 1) + 125 ] - 4
12(2n + 1) = 2[ -6(2n + 1) + 100 ] - 4
12(2n + 1) = -12(2n + 1) + 200 - 4
-196 = -24(2n + 1)
-196 = -48n - 24
-172 = -48n
3.58 = n
Because the answer of negative is 3.58, so it can't be the answer
since it's not an odd number. Now I'll check if this fit with the
description of Jay paid five times the positive difference between
what Joseph and Leroy paid. If it fits, then I'll get the solution.
Joseph = 2n + 1 = 19
Leroy = 3(19) - 25 = 32
32 - 19 = 13
13 is positive, so this is what I want to find, and now I just need
to times that by 5.
13 * 5 = 65
04/09/2001
view
Jay paid $65
for his box
supper.
So Jay paid $65 for the box supper.
I.
X = the amount Joseph paid for his box supper
II.
3X - 25 = the amount Leroy paid for his box supper
5[(3X - 25)] = the amount Jay paid for his box supper
12X = the amount raised from the whole fundraiser
The problem says that Leroy paid 25 dollars less than three
times the amount of Joseph. Then it says that Jay paid five times as
much of the difference of Leroy and Joseph. So the difference of
Leroy and Joseph's payment multiplied by five is how much Jay paid.
Finally, the problem says that the total amount of money made is
twelve times the amount of Joseph's payment. These equations are what
will help solve the problem. After solving the equation, I can solve
all of the other equations.
III. X + (3X - 25) + 5[(3X - 25) - X] = 6X + 2
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 2 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
This is the equation that I will use to solve the problem.
The problem say that the sum the three gentlemen paid was $4 more
than the amount paid by everybody else who bought boxes. So I can
say that by removing $2 from the three gentlemen's overall total,
that they have paid for half of the fundraiser. I know this because
the problem says that the three gentlemen have paid $4 more than
everyone else. By subtracting $2 from the three gentlemen and adding
$2 to everyone else's total, then the two groups of people would have
an even amount of money. So I can say that the three gentlemen,
(X + [3X - 25] + 5{[3X - 25] - X}) plus the three gentlemen minus 4
(everyone else, X + [3X - 25] + 5{[3X - 25] - X} - 4)will equal the
whole fundraiser (12X). But to make it easier and less time
consuming, I will add the three gentlemen's total together and it
will equal half of the fundraiser plus two (6X + 2).
IV. X + (3X - 25) + 5[(3X - 25) - X]
X + 3X - 25 + 5(2X - 25)
4X - 25 + 10X - 125
14X - 150
14X - 6X
8X
X
paid.
=
=
=
=
=
=
=
6X + 2
6X + 2
6X + 2
6X + 2
150 + 2
152
19
1)
2)
3)
4)
5)
6)
3X - 25 - X
5 x 2x; 5 x 25
4X + 10X; -25 -125
-6X; + 150
14X - 6X; 150 + 2
152 divided by 8
This is the value of X. Therefore, this is how much Joseph
I can figure everthing out from here.
3X - 25 =
3(19) - 25 =
57 - 25 =
$32 =
Leroy's
Leroy's
Leroy's
Leroy's
payment
payment
payment
payment
This is how much Leroy paid for his box supper.
5(32 - 19) = Jay's payment
5(13) = Jay's payment
$65 = Jay's payment
This is how much Jay paid for his box supper.
12(19) = amount of the entire fundraiser
$228 = amount of the entire fundraiser
This is the amount of money raised by the entire fundraiser.
There is another possiblitly that Joseph could have payed more
than Leroy. The only way to figure that out is to do this:
X + 3X - 25 + 5[X - (3X - 25)]
4X - 25 + 5(-2X + 25)
4X - 25 -10X + 125
-6X + 100
-12X
X
=
=
=
=
=
=
6X + 2
6X + 2
6X + 2
6X + 2
-98
8.166
The problem states that the amount Joseph paid was an odd
number. That means that this answer doesn't agree with the problem.
If I take this number (8.166) and put it into the equation
to find the amount Leroy paid, it will come out like this:
3(8.166) - 25 = Leroy's payment
24.498 - 25 = Leroy's payment
-.502 = Leroy's payment
The amount Leroy paid would be in negative numbers, meaning that
the fundraiser paid HIM. Along with the problem saying that Joseph's
payment is an odd number, this answer can not be true because the
problem also states that Leroy paid a "very good price" for his box
supper.
V. (1) X + 3X - 25 + 5[(3X - 25) - X] =
19 + 32 + 65 =
51 + 65 =
116 =
6X + 2
6(19) + 2
114 + 2
116
(2) X + 3X - 25 + 5(2X - 25) + 6X
19 + 57 - 25 + 5(38 - 25) + 114
19 + 32 + 5(13) +
19 + 32 + 65 +
51 +
228
228
228
228
228
228
- 2
- 2
112
112
177
228
=
=
=
=
=
=
116(the three gentlmen) - 4 = 112(everyone else)
04/09/2001
view
My answer is
that Jay spent
65 dollars on
his box
supper.
There are two different ways I have decided to check my
equation. I can use the equation in which I sovled for X (1). I can
also take half of the fundraiser,(6X), subtract it by two,(6X - 2),
add on the three gentlemen, and it should come out to the total price
that the fundraiser made. The reason I subtract two and not add two
is because I added two to get the same answer as the three
gentlemen. By subtracting half of the fundraiser by two, I
essentially get the amount paid for by everyone else. To check to
see if my solution is correct, I inserted the answer for X
into all of the X's, and proceed to solve the problem.
First I set variables for each person. I made Joseph the variable a.
Leroy, therefore became 3a - 25. I knew that Leroy paid more then
Joseph did. This is because Joseph paid a MODEST amount, which means
that he paid less then everyone else did. This also can be proved
because if you make a bigger [which will make the equation a - (3a +
25), or -2a + 25], and substitute that into the equation that is
stated below, the answer will not be a whole number, but a negative
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Sokolsky, Sasha
[n/a]
Welsh Valley
Middle School
Page 3 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
decimal
with a lot of decimal places. The equation would be:
a + -2a + 25 + 5(a - 3a - 25) - 4 = 12a - [a + -2a + 25 + 5(a - 3a 25)]
This can be simplified to -7.470588 and so on.
This helped me make an equation for
Jay. This equation was 5(3a - 25 - a).
Then I made the total amount of money raised (12 times the amount
Joseph paid) 12a. Then I made an equation:
a + 3a - 25 + 5(3a - 25 - a) - 4 = 12a - [a + 3a - 25 + 5(3a - 25 a)]
In this equation the left side is the amount of money spent by the
three boys. The right side is the amount of money that everyone else
spent. Then I simplified this equation until I could solve for a:
a + 3a - 25 + 5(2a - 25) - 4 = 12a - [a + 3a - 25 + 5(2a - 25)]
a + 3a - 25 + 10a - 125 - 4 = 12a - (a + 3a - 25 + 10a - 125)
14a - 154 = 12a - a - 3a + 25 - 10a + 125
14a - 154 = -2a + 150
14a = -2a + 302
16a = 304
a = 19
Now I have the amount that Joseph has. I know this amount fits
because it's odd. Then I use this to find out how much Leroy paid:
57 - 25 = 32
Now I can find out how much Jay spent.
5(32 - 19) = 65
04/09/2001
view
Jay payed
625 dollars
for his box
supper.
And
Let
Let
Let
this is how much Jay spent.
X= the amount that Joseph paid for his box.
3x-25= the amount that Leroy paid for his box.
5(3x-25-x)= the amount that Jay paid.
Tanenbaum, Lauren
[n/a]
[n/a]
I used n as the odd number of dollars that joseph payed.
for leroy it says that it cost him 25 fewer dollars than three times
what Joseph spent. (3n-25) For jay it says that he payed five times
the positive difference between what Joseph and Leroy paid. (n-3n-25)
When you solve this problem n equals 12.50 dollars. When you
multiply that by 5 it comes out to 62.50 dollars.
Levine, Sam
[n/a]
Solomon
Schechter Day
School
x
y
z
w
Bell, Jack
[n/a]
Fontainebleau
High School
WE know that the total amount of money raised was twelve times Joe's
payment(12x.
12x= x + 3x-25 +10x-125
12x=14x-150
2x=150
x=75
75(3)=225 x 5=1125
5(25)= 125
75(5)= 375
1125-125-375=625.
Jay paid 625 dollars for his box supper.
04/09/2001
view
04/09/2001
view
Jay payed
62.50 dollars
for Judy
Ann's
delicious
meal.
Jay paid $65
for his dinner.
=
=
=
=
amount
amount
amount
amount
3x - 25 = y
3x - y = 25
Joseph spent
Leroy spent
Jay spent
spent by all the other participants
Leroy spent $25 less than Joseph
standard format
| 5x-5y | = z
Jay paid five times the positive difference between
what Joseph and Leroy paid
-5x + 5y - z = 0 Standard format
x + y + z = w + 4
Joseph's payment
x + y + z - w = 4
total amount of money raised was twelve times
Standard format
x + y + z + w = 12x
sum of the money paid by these three gentlemen
was $4 more than the amount paid by everybody else who bought boxes.
-11x + y + z + w = 0
Standard format
Next I created two determinents from my equations and multiplied the
the 4x4 matrix witch was made up by the coefficents of the varibles
(a, b, c, d) to the power of -1, then i multiplied that and the 4x1
matrix witch is made up of the answers to the equations (e).
|
3 -1 0 0 |^-1
| -5 5 -1 0 |
*
|
1 1 1 -1 |
| -11 1 1 1 |
| 25 |
| 0 |
| 4 |
| 0 |
=
| 19
| 32
| 65
| 112
|
|
|
|
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 4 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
x
y
z
w
=
=
=
=
19
32
65
112
The amount of money Jay spent to buy they meal was $65.
Well, I set up an equation for Joesph(x) then for Leroy(3x-25) and
also for Jay( 5(3x-25-x). Then I worked the equation for Jay:3x-x=2x,
so then I had 5(2x-25), so then I multiplied 5 times 2x and 5 times a
-25. that gave me 10x-125. And the extra information says that they
raised twelve times Joesphs payment or 12x. so my equation then reads
10x-125=12x, so I cancel out 10x on both sides leaving me with
-125=2x. Then I divide to get 64.5. So Joesph's amount was 64.5 so I
plugged 64.5 in Jay's equation 5(3x-25-x), then 5(2x-25) then 10x-125,
then 10(64.5)-125, then 645-125=520. So Jay paid 520 dollars.
Explanation:
04/09/2001
view
Jay paid 520
dollars for his
box supper.
04/09/2001
view
Solution: Jay
paid 65
Let J dollars be the amount paid by Jay, H dollars by Joseph, L
dollars for his dollars by Leroy, R dollars by the rest of the people, and T dollars
meal.
be the total amount raised in the Box Supper. I use the information
Chivers, Alicia
[n/a]
[n/a]
Ko, Andrew
[n/a]
[n/a]
Hillius, Cassie
Marotz, Tara
[n/a]
[n/a]
to form the following equations.
L = 3H - 25
three times what Joseph
"It costs Leroy 25 dollars less than
spent."
J = 5|L - H|
difference between what
"Jay paid five times the positive
Joseph and Leroy paid."
I use absolute
value for the
difference.
T = J + L + H + R
Leroy, Joseph and
Total is the sum of amount paid by Jay,
the rest.
T = 12H
Joseph's payment."
"Total amount raised was twelve times
L + J + H = 4 + R
$4 more than that
"Sum paid by Leroy, Jay, and Joseph was
paid by the rest"
I can combine T = 12H and T = J + L + H + R.
J + L + H + R = 12H
R = 11H - J - L
Next I substitute R = 11R - J - L into L + J + H = 4 + R.
L + J + H = 4 + 11H - J - L
2L + 2J - 10H = 4
I divide each term in the equation
by 2.
L + J - 5H = 2
Case I:
If L < H, then J = 5(H - L) = 5H - 5L.
by substituting L = 3H - 25.
J = 5H - 5L =
Next I solve J in terms of H
5H - 5(3H - 25) = 5H - 15H
+ 125 = -10H + 125
Now I substitute L and J into L + J - 5H = 2.
3H - 25 - 10H + 125 - 5H = 2
-12H + 100 = 2
-12H = -98
H = 98/12 = 49/6
This is not a solution because it should cost Joseph an odd number of
dollars.
Case II:
If H < L, then J = 5(L - H) = 5L - 5H.
by substituting L = 3H - 25.
J = 5L - 5H =
Next I solve J in terms of H
5(3H - 25) - 5H = 15H - 125 - 5H = 10H - 125
Now I substitute L and J into L + J - 5H = 2.
3H - 25 + 10h - 125 - 5H = 2
-150 + 8H = 2
8H = 152
H = 152/8 = 19 (an odd number of dollars!)
Now I use the value of H to find J.
J = 10H - 125 = 10(19) - 125 = 190 - 125 = 65
Therefore, Jay paid 65 dollars for his meal.
04/09/2001
view
In
In our
solution, we J=
L=
found that
A=
J+L+A+E=12J T=
our
the
the
The
The
solution, we let
amount Joe paid
amount Leroy paid
amount J paid
amount everyone else paid
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 5 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
So here is our solution:
04/09/2001
view
Jay paid $65
for his meal.
L=3J-25
J=5(L-J)
A=5(3J-25-J)
A=5(2J-25)
A=10J-125
J+L+A+E=12J
Let J equal the sum Joseph paid for his meal,
Let L equal the sum Leroy paid for his meal,
Let X equal the sum Jay paid for his meal.
Vats, Naalti
[n/a]
Detroit Country
Day School
We know that J should be odd
We also know that L is $25 less than three times J:
L = 3J - 25
Finally, we know that X should be five times the positive difference
between J and L.
If L is greater than J, then:
X = 5 (3J - 25 - J)
= 10J - 125
If J is greater than L, then:
X = 5(J - (3J - 25))
= -10J + 125
As long as J is greater than 12.5, L will be greater than J because
if J is equal to 12.5, X is equal to 0. We'll just have to see what
J ends up being, and then decide which of the two equations to use.
Let y equal the amount paid by everyone else who bought a box.
We know that the total amount of money raised should equal twelve
times what Joseph paid:
J + L + X + y = 12J
We also know that the amount of money paid by Joseph, Leroy, and Jay
is four more than that paid by everyone else:
J + L + X = y + 4
If we subtract four from both sides, we can find out what y is in
terms of J, L, and X.
J + L + X - 4 = y
Now, we can substitute this expression in for y in our orignal
equation for the total amount of money raised:
J + L + X + J + L + X - 4 = 12J
We know L and X in terms of J, so if we substitute those original
expressions in, we get:
J + (3J-25)+ (10J-125) + J + (3J-25) + (10J-125) - 4 = 12J
If we group together the same terms, we get:
2J + 2(3J - 25)+ 2(10J - 125)- 4 = 12J
Now we just have to simplify:
2J + 6J - 50 + 20J - 250 - 4 = 12J
28J - 304 = 12J
16J = 304
J = 19
Now we know the sum Joseph paid, and we know it should be an odd
number, which it is. To find what Jay paid, we just have to solve
the equation in which we expressed Jay's sum paid in terms of
Joseph's sum paid.
First, we have to decide which of the two equations to use. We
should use the one where we said that L is greater than J because J
is greater than 12.5.
originally, we had:
X = 10J - 125
Now we plug in the value we got for J, or Joseph's sum, to find the
amount Jay paid:
X = 190 - 125
X = 65
So we know that Jay paid $65 for his meal.
To check our work, we simply have to see if the total sum paid equals
twelve times Joseph's sum. First, we need to write out what Joseph
paid, what Leroy paid, what Jay paid, and what everyone else paid:
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 6 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
J = 19
L = 3(19) - 25
= 32
X = 65
y = J + L + X - 4
= 19 + 32 + 65 - 4
= 112
Now we find the total amount of money paid by everyone:
J + L + X + y
= 19 + 32 + 65 + 112
= 228
If our values are correct, that sum should equal twelve times
Joseph's sum, which is:
12(19)
= 228
Now we know the sums are correct!
04/10/2001
view
Jay paid $15 I put all the known facts into formulas:
for that
basket.
Joseph is an odd number
Damminga, Dave
[n/a]
Skyview High
School
Taipei
American
School
Taipei
American
School
LeRoy = Joseph * 3 - 25
Jay = |Joseph - LeRoy| * 5
Total = 12 * Jospeh
Total - (Jospeh + LeRoy + Jay) = Jospeh + LeRoy + Jay + 4
Jay = ?
After this, I tried to plan out how to solve this problem.
My method came down to an educated guess and check.
First, I would pick a resonable low number that
was odd as the amount Joseph paid.
I would use that number throughout my formulas
until I came upon Jay's price. From that, I would
adjust Joseph's price until an answer worked, which was $15.
I came upon this after 4 tries ($27, $25, $21, $17.)
After solving this, I found that Jospeh's bid was $11, LeRoy's was
$8, and Jay's was $15, and that the total was $132
Jay paid $15 for that basket.
04/10/2001
view
i dunt know
i dunt know
Fuang, Michelle
[n/a]
04/10/2001
view
Jay spent $65
to buy the
box supper
made by her
girlfriend.
The first thing I did was to find out the exact amount of money each
man paid for the box suppers that their girlfriends made.
Lin, Michael
[n/a]
Since the amount of money that Joseph (A) paid was an odd amount, so
I will represent it with x. The expression I will use for this is:
A = x
Leroy (B)paid 25 fewer dollars than three times what Joseph spent, so
this is the equation for the money he spent in buying his box supper.
B = 3x - 25
Jay (C), the man who bought the most expensive one, paid five times
the positive difference between what Joseph and Leroy paid for their
box suppers. This is the expression I came up with for the amount
that Jay paid.
C = 5[(3x - 25) - x]
= 5(2x - 25)
Now I put the extra facts into the form of an equation.
1. the total amount of money raised was twelve times Joseph's payment
1. T = 12x
2. the sum of the money paid by these three gentlemen (A + B + C) was
$4 more than the amount paid by everybody else who bought boxes. So
thi means that the amount everyone else paid was $4 less than the
total of these three gentlemen. After finding the amount everone ELSE
paid, in order to find the total amount of money all together, I
added the amount the 3 gentlement paid, which totals up to (T).
2. [(A + B + C) - 4] + (A + B + C) = T
---> [x + 3x - 25 + 5(2x - 25) - 4] + x + 3x - 25 + 5(2x - 25) = T
---> (4x - 25 + 10x - 125 - 4) + (4x - 25 + 10x - 125) = T
---> 28x - 304 = T
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Then by sticking in the equation from number 1, this is what I got.
---> T = 28x - 304
---> 12x = 28x - 304
---> 16x = 304
---> x = 19
To find out how much money Jay spent, all I did was to fill
in "Jay's" formula with the other expressions. This is what happened.
Amount of money Jay paid: 5(2x - 25)
--->
--->
--->
--->
5[2(19) - 25)
5(38 - 25)
5(13)
$65
Amount of money Joseph paid: x
---> $19
Amount of money Leroy paid: 3x - 25
---> 3(19) - 25
---> 57 - 25
---> $32
---To Discuss the possibility of Joseph paying more money than Leroy,
this is what I did. It is NOT possible since the amount Leroy paid is
three times of x (19) subtracted by 25, and Joseph paid only one x.
Each x is 35, so three times of that MINUS 25 (which is less than two
times of x) still ends up with more than one x. This excludes the
possibility that Joseph would pay more than Leroy.--Amount of moeny the rest of the people paid: (A + B + C) - 4
---> (65 + 19 + 32) - 4
---> 116 - 4
---> $112
04/10/2001
view
SHORT
ANSWER Jay
spend 65
dollars for his
box.
In conclusion, through these processes, I have found out that the the
amount of money Jay paid to buy the box supper made by his girlfriend
was $65.
EXPLANATION
The problem gives us the following information
1.
Joseph, paid odd number of dollars
2.
Leroy, it cost him 25 fewer dollars than three times what
Joseph spent
3.
Jay, paid five times the positive difference between what
Joseph and Leroy paid
4.
The total amount of money raised was twelve times Joseph's
payment
5.
The sum of the money paid by these 3 men was $4 more than
the amount paid by everybody else who bought boxes
Chang, Gary
[n/a]
Taipei
American
School
Ammar, Rebecca
[n/a]
[n/a]
Let us assume x is the number of dollars Joseph paid for his box.
Now, we construct expressions and equations based on x.
A.
Amount Joseph Paid
= x
B.
Amount Leroy Paid
= 3x - 25
C.
Amount Jay Paid
= 5[(3x - 25) - x]
= 5(2x - 25)
= 10x - 125
D.
Total Amount of Money Raised
= 12 * x
= 12x
E.
Total Amount of Money Raised by 3 Men - 4 = Total Amount of
Money Raised by Everybody Else
(x + 3x - 25 + 10x - 125) - 4 = 12x - (x + 3x - 25 + 10x 125)
14x - 150 - 4 = 12x - (14x - 150)
14x - 154 = 12x - 14x + 150
14x - 154 = -2x + 150
Now we have an equation, we use it to solve for x.
F.
04/10/2001
view
Jay paid 60$.
Solving For X
14x - 154 = -2x + 150
16x = 304
x = 19
Joseph spent 19 dollars for his box.
3x - 25 = 3(19) - 25 = 32
Leroy spend 32 dollars for his box.
(32 - 19) * 5 = 13 * 5 = 65
Jay spend 65 dollars for his box.
Joseph paid x
Leroy paid y
Jay paid z
From Sally Jo's sale : x is odd.
From Mary Lou's sale : 3x-25=y
From Judy Anns sale : 5(y-x)=z
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 8 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
From the two facts at the end of the problem :
x+y+z-4=B (what others paid)
x+y+z+B=12x
let x+y+z=A
then A+A-4=12x
2A=12x+4
A=6x+2
x+y+z=6x+2
-5x+y+z=2
But 3x-y=25 .... y=3x-25
and z=5(y-x) ..... z=5(3x-25-x)
-5x+(3x-25)+5(3x-25-x)=2
-5x+3x-25+15x-125-5x=2
8x=152
x= 19 dollars
Hence y = 3x-25 = 57-25=32dollars
and z = 5(32-19) = 65 dollars.
So Jay paid 65 dollars for his meal.
To check the answer :
x+y+z-2 = 6x
19+32+65-2 = 6x19
114 = 114
Yes!!!!!!
04/10/2001
view
Jay paid $65. First of all I wrote the algebraic equations from the given
informations. Then I solved the problem.
Antikacioglu, Arda
[n/a]
[n/a]
Let j be the Joseph’s payment
Let l be the Leroy’s payment
Let y be the Jay’s payment
1. Leroy is Mary Lou’s boy friend and he paid 25 fewer dollars than
three times what Joseph paid for Sally’s box.
l = 3j – 25
2. Jay is Judy Ann’s boy friend and he paid five tiimes the positive
differece between Joseph ans Leroy paid.
At this step we need to know which one is greate m or s. Since we
haven’t got this information we need to set two equations.
Ýf l
y
y
y
>
=
=
=
j
((3j – 25) – j) * 5
(2j – 25) * 5
10j – 125
Ýf j
y
y
y
>
=
=
=
l
(j - (3j – 25)) * 5
(-2j + 25) * 5
-10j + 125
3. the total amount of money raised was twelve times Joseph's payment
Let t be the total amount of money.
t = 12j
4. the sum of the money paid by these three gentlemen was $4 more
than the
amount paid by everybody else who bought boxes.
Let g be the sum of money paid by Joseph, Leroy and Jay
t = g + (g - 4)
g = j + l + y
if l > j
substitute (3j-25) for l, (10j- 125) for y
g = j + (3j – 25) + (10j – 125)
g = 14j – 150
t = g + (g - 4)
substitute (14j –150) for g and 12j for t
12j = (14j – 150) + (14j – 150 – 4)
12j = 28j – 304
12j – 28j = -304
-16j = -304
j = 19
if j > l
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substitute (3j - 25) for l, (-10j + 125) for y
g = j + (3j – 25) + (-10j + 125)
g = -6j + 100
t = g + (g - 4)
substitute (-6j + 100) for g and 12j for t
12j = (-6j + 100) + (-6j + 100 – 4)
12j = -12j + 196
12j + 12j = 196
24j = 196
j = 196/24
j = 8 1/6
Since Saly Jo’s box costs Joseph an odd number of dollars j can’t be
8 1/6. Hence Joseph’s payment is $19 and Leroy’s payment is more than
the Joseph’s payment. Thus Jay’s payment is 125 fewer dollars than 10
times what Joseph paid.
y = 10j – 125
Substitute 19 for j
y = 10*19 – 125
y = 190 – 125
y = 65
Jay paid $65.
Check the result :
j = 19
y = 65
l = 3*19 – 25
= 57 –25
= 32
Leroy paid $32 for Mary Lou’s box.
Find the sum of the money paid by three gentlemen.
g = j + l + y
= 19 + 32 + 65
= 116
04/10/2001
view
Jay paid $65
The
t =
t =
=
total amount of money raised was twelve times Joseph's payment
12j
12*19
228
The
the
t =
=
=
=
sum of the money paid by these three gentlemen was $4 more than
amount paid by everybody else who bought boxes.
g + (g –4)
116 + (116 –4)
116 + 112
228
Since I reach the same amount of money with two equations my result
is correct.
Joseph paid an odd number of dollars to Sally Jo.
Lets define this money as A:
Ulengin, Idil
[n/a]
[n/a]
A= 2x + 1
Le Roy paid to Mary Lou as B :
B= 3 (2x + 1)-25 = 6x + 3 -25 = 6x -22 dollars
Jay paid to Juddy Ann as C:
C= 5(6x - 22 - (2x + 1)) = 5 (6x - 22 - 2x -1) =
= 20x - 115 dollars
5(4x -23)
The three boys paid in total A+ B+ C;i.e.
A + B + C = (2x + 1) + (6x - 22) + (20x - 115) = 28x - 136 dollars
The total money(TM)is 12 times what Joseph paid;i.e.:
TM= 12*A = 12 (2x + 1) = 24x + 12
The money paid
TM-(A + B C)
by others is:
The total money paid by Joseph, Le Roy and Jay is 4 more than the
money paid by the others:
A + B+ C = ( TM-(A + B + C) ) + 4
28x - 136 = 24x + 12-(28x - 136) + 4
28x - 136= (24x + 12)-(28x - 136) + 4
2*(28x - 136) = 24x + 12 + 4
56x - 272 = 24x + 16
56x - 24x = 272 + 16
32x = 288
x= 9
Jay paid:
C = 20x - 115 =20 * 9 - 115 = 180 - 115 = $65
FINAL CHECK:
Joseph paid:
A= 2x + 1 = 2 * 9 + 1 = $19
Le Roy paid::
B = 6x - 22 = 6 * 9 - 22 = 54 - 22 = $ 32
The three boys paid:
A + B + C = $19 + $32 + $65 = $116
Total money paid = 12 * A = 12 * 19 = $228
The rest of the people paid= 228 - 116 = $112
A + B + = ( TM-(A + B + C) + 4
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The Math Forum @ Drexel University
11/24/10 12:44 PM
116 = (228-116) + 4
116 = 112 + 4
116 = 116
04/10/2001
view
Jay paid $65 Jay paid $65.
Let
Let
Let
Let
H
L
J
T
=
=
=
=
Berg, Emily
[n/a]
[n/a]
Billings, Tina
Sutton, Jessica
Jacobson, Tony
[n/a]
[n/a]
Dollars paid by Joseph. It is given that H is an odd amount.
Dollars paid by Leroy.
Dollars paid by Jay.
Total dollars raised by everyone in the event.
Leroy paid $25 less than three times what Joseph paid, so
* L = 3H - 25
Jay paid 5 times the positive difference between what Joseph and
Leroy paid.
If Joseph paid more than Leroy, then the difference represented by
H - L is a positive number already, and I can write
J = 5*(H - L)
Substituting in L = 3H - 25, I find
J = 5*(H - [3H - 25])
= 5*(H - 3H + 25)
= 5*(25 - 2H)
* J = 125 - 10H
If Leroy paid more than Joseph, then the difference represented by
L - H is a positive number already, and I can write
J = 5*(L - H)
Substituting in L = 3H - 25, I find
J = 5*([3H - 25] - H)
= 5*(3H - 25 - H)
= 5*(2H - 25)
* J = 10H - 125
Since I do not yet know which value, L or H, is larger, I will work
the problem each way and see which works.
The total dollars raised is 12 times Joseph's payment, so
* T = 12H
The sum of the money paid by Joseph, Leroy, and Jay was $4 more than
the amount paid by everyone else.
The amount paid by Joseph, Leroy, and Jay is
(H + L + J)
Everybody else paid $4 less than these gentlemen, so everybody else
paid
(H + L + J) - 4
The total paid is the amount paid by the three men plus the amount
paid by everybody else, so
T = (H + L + J) + [(H + L + J) - 4]
= (H + L + J) + (H + L + J) - 4
= 2*(H + L + J) - 4
* T = 2*(H + L + J) - 4
I substitute into this last equation for L, J, and T to solve for H.
I will use the first value of J above (assumes H > L)
12H = 2*(H + [3H - 25] + [125 - 10H]) - 4
6H = (H + 3H -25 + 125 - 10H) - 2
= H + 3H - 10H - 25 + 125 - 2
= -6H + 98
12H = 98
H = 49/6
I did the same substitutions again but using the second value of J
above (assumes L > H)
T = 2*(H + L + J) - 4
12H = 2*(H + [3H - 25] + [10H - 125]) - 4
6H = (H + 3H - 25 + 10H - 125) - 2
= H + 3H + 10H - 25 - 125 - 2
= 14H - 152
-8H = - 152
H = 19
04/10/2001
view
Jay paid $65. Let J
Let L
Let A
Let T
J = 10H - 125 was the correct equation since it gave an odd
number for the amount Jay spent.
J = 10(19) - 125
= 190 - 125
J = 65
= to the amount Joseph paid
= to the amount Leroy paid
= to the amount Jay paid
= to the amount everyone else paid
First we figured out what everyone paid in comparison to what Joseph
paid.
L = 3J - 25
A = 5 (L - J)
A = 5 (3J - 25 - J)
A = 5 (2J - 25)
A = 10J - 125
Next we found out what everyone paid
J + L + A + T = 12J
J + 3J - 25 + 10J - 125 + T = 12J
14J - 150 + T = 12J
2J - 150 + T = 0
T = 150 - 2J
T + 4 = L + J + A
150 - 2J + 4 = 3J - 25 + J + 10J - 125
150 - 2J + 4 = 14J - 150
150 + 4 = 16J - 150
304 = 16J
19 = J
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
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11/24/10 12:44 PM
Joseph paid $19
Next we plugged 19 into where all the J's were to find Jay's payment
amount
A = 10 * 19 - 125
A = $65
First, I calcuated how much Leroy and Jay paid in terms of Joseph.
Leroy paid 3 times what Joseph paid minus 25 (3Joseph-25), and Jay
paid 5 times the difference between Leroy and Joseph (5(3Joseph-25Joseph)).
In the problem, it also stated that the amount Jay, Leroy, and Jay
paid is 4 dollares more the amount paid by everyone else (the
equation of much the three paid in terms of Joseph is Joseph+(3Joseph25)+(5(3Joseph-25-Joseph)=Eveyone else+4 or 14Joseph-150=Everyone
else+4). Also, Eveyone+(14Joseph-150)=12Joseph. Subtract the two
equations and get Everyone else-150=2Joseph. Substitude Everyone else150 for 2Joseph and get 7(everyone else-150)-150=everyone else+4.
From that I could tell that Eveyone else paid 200.67 dollares and
Joseph paid 25 dollares.
i used guess and try.
04/10/2001
view
Joseph paid
25 dollares
for his box
supper.
04/10/2001
view
jay spent 10
dollars on
judy anns
box.
Jay paid $65 Here is my solution:
dollars for
I first assigned a variable to each person.
Judy Ann's
box supper. Joseph = x
04/10/2001
view
Leroy
Jay
Zhang, Fan
[n/a]
Masterman
High School
Wiley, Kirk
[n/a]
[n/a]
Bajwa, Moazzum
[n/a]
Southeast
Raleigh High
School
= y
= z
Then, I analyzed the given information, so it could be put into
equation form.
Leroy paid 25 fewer dollars than three times what Joseph spent. So,
y = 3x - 25
Jay paid 5 times the positive difference between what Joseph and
Leroy paid.
z = 5(y - x)
The total amount of money
t = 12x
raised was twelve times Joseph's payment.
The sum of the money paid by the three gentlemen (x + y + z) was four
dollars more than the amount paid by everyone else who bought boxes.
x+y+z = 4 + [t - (x+y+z)]
So, I now have four starting equations:
y = 3x - 25
z = 5(y - x)
t = 12x
x+y+z = 4 + [t - (x+y+z)]
I decided that it would be much easier if I simply solved for x, and
then used that value in the equations. To solve for x, I needed to
get it separate from the other variables.
x + y + z = 4 + [t - (x + y + z)]
x + y + z = 4 + [12x - (x + y + z)]
x + y + z = 4 + 12x - x - y - z
x + y + z = 4 + 11x - y - z
2y + 2z
= 4 + 10x
y
= 2 + 5x - z
So I now had two equations equal to y, which makes these two
equations equal to each other.
3x - 25 = 2 + 5x - z
However, there was still a z in this equation. So, I used one of the
other equations to replace z with.
z = 5(y - x)
z = 5[(3x - 25) - x]
z = 5(2x - 25)
z = 10x - 125
(10x
3x 3x 3x 3x
8x
x
- 125) could now be substitued for z in the equation:
25 = 2 + 5x - z
25 = 2 + 5x -(10x - 125)
25 = 2 + 5x - 10x + 125
= -5x + 152
= 152
= 19
I now know that x = 19. This means that Joseph paid 19 dollars for
his meal. I could substitute this value into the other equations.
y = 3x - 25
y = 3(19) - 25
y = 57 - 25
y = 32
Leroy paid 32 dollars for his meal. To find z, or how much Jay paid,
I only needed to substitute the values of x and y.
z = 5(y-x)
z = 5(32-19)
z = 5(13)
z = 65
I could check this solution by using the other equation.
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Page 12 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
x + y + z
= 4 +
19 + 32 + 65 = 4 +
116
= 4 +
116
= 4 +
116
= 116
[t - (x + y + z)]
[(12 x 19) - (19 + 32 + 65)
[228 - 116)
112
So, the solution does work. Jay paid $65 for his meal.
Sincerely,
Moazzum Bajwa
04/10/2001
view
Jay paid 65 let x = How much Joseph paid for his meal
dollars for his
let 3x - 25= How much Leroy paid for his meal
meal
Leane, Thomas
[n/a]
[n/a]
Dillard, Colin
[n/a]
Central High
School
let 5(3x - 25 - x)= How much Jay paid for his meal
The every one else paid 4 dollars less then what these three guys paid
let
[x + 5(3x - 25 - x) + 3x - 25] - 4= How much every one else paid
The total of all this money equals 12 times what Joseph paid so to
solve for x the equation is
[x + 5(3x - 25 - x) + 3x - 25 - 4] + [x + 5(3x - 25 - x) + 3x - 25]=
12x
which boils down to
(14x - 154) + (14x - 150) = 12x
which then boils down to
16x = 304
x = 304/16
x = 19
so Joseph paid 19 dollars now to find out how much Leroy paid
3(19) - 25 = 32
Leroy paid 32 dollars so To find how much Jay paid
5(32 - 19) = 65
so Joseph = $19
Leroy = $32
Jay
= $65
Proof
Joseph paid $19 which equals x
Leroy paid 3x - 25 = 32 that checks
Jay paid 5(3x - 25 - x) = 65 that also checks
so my work checks out
Leroy could not have paid more then Jay because no matter what the
direct difference between what Leroy and Joseph paid multiplied by
five will always be bigger than what Leroy paid no matter what the
variable equaled.
04/10/2001
view
Jay paid 65
dollars.
And that is how I figured out how much Jay paid for his meal.
First I set up variables and equations:
x
y
z
n
=
=
=
=
amount Joseph paid
amount Leroy paid
amount Jay paid
total amount raised
y
5
n
n
=
*
=
=
3x - 25
|x - y| = z
12x
2(x + y + z - 4) + 4
Next I started combining equations:
n = 2(x + y + z - 4) + 4
n = 2x + 2y + 2z - 4
12x = 2x + 2y + 2z -4
10x - 2y + 4 = 2z
5x - y + 2 = z
5x - (3x - 25) + 2 = z
5x - 3x + 25 + 2 = z
2x + 27 = z
2x + 27 = 5 * |x - y|
2x + 27 = 5 * |x - (3x - 25)|
2x + 27 = 5 * |-2x + 25|
Now we have two options depending on whether -2x + 25 is positive or
negative:
2x + 27 = -10x + 125
12x = 98
x = 8.1666...
or
2x + 27 = 10x - 125
-8x = -152
x = 19
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 13 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
Since we are told that Joseph paid an odd number of dollars for his
food, x = 19 is our answer. Plugging this back in:
y
y
y
y
=
=
=
=
3x - 25
3 * 19 - 25
57 - 25
32
5 * |x - y| = z
5 * |19 - 32| = z
5 * |-13| = z
5 * 13 = z
65 = z
n = 12x
n = 12 * 19
n = 228
n
n
n
n
04/10/2001
view
Jay paid 65
dollars.
=
=
=
=
2x + 2y + 2z -4
2 * 19 + 2 * 32 + 2 * 65 - 4
38 + 64 + 130 - 4
228
We don't run across any problems so our answers are correct.
paid 65 dollars.
Leroy spent 25 dollars fewer than 3 times what Joe spent.
y = 3x - 25
Jay
Nalavade, Aman
[n/a]
[n/a]
Jay paid 5 times the positive difference of Joe and Leroy.
So the equation could be z = 5(x-y) if x is greater than y or z =
5(y-x) if y is greater than x.
The total amount of money raised is 12 times the amount spent by Joe.
T = 12x
The sum of the money paid by these 3 gentlemen was 4 dollars more than
the amount paid by everyone else.
x + y + z = 4 + T- (x + y + z)
I added x + y + z on both sides.
x + y + z + x + y + z = 4 + T- (x + y + z) + x + y + z
2x + 2y + 2z = 4 + T
I substituted 12x in for T.
2x + 2y + 2z = 4 + 12x
I subtracted 2x on both sides.
2x + 2y + 2z - 2x = 4 + 12x - 2x
2y + 2z = 4 + 10x
I next would either have to substitute 5(x-y) or 5(y-x) in for z.
I first tried 5(x-y) in for z.
2y + 2(5(x-y))
= 4 + 10x
I used distributive property.
2y + 10x - 10y = 4 + 10x
-8y + 10x = 4 + 10x
I subtracted 10x from both sides.
-8y + 10x - 10x = 4 + 10x - 10x
-8y = 4
y = -1/2
Since y can't be negative, I went back and tried the other
possibility.
2y + 2(5(y-x)) = 4 + 10x
I used distributive property.
2y + 10y - 10x = 4 + 10x
12y - 10x = 4 + 10x
I added 10x to both sides.
12y - 10x + 10x = 4 + 10x + 10x
12y = 4 + 20x
I substituted 3x - 25 in for y
12(3x - 25) = 4 + 20x
36x - 300 = 4 + 20x
I added 300 to both sides
36x - 300 + 300 = 4 + 20x + 300
36x = 20x + 304
I subtracted 20x from both sides.
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16x = 304
x = 19
To find y I plugged 19 in for x in the equation y = 3x - 25.
y = 3*19 - 25
y = 57 - 25
y = 32
Now to find z I plugged 19 in for x and 32 in for y in the equation z
= 5(y-x)
z = 5(32-19)
z = 5*13
z = 65
Therefore Jay spent 65 dollars on the box supper.0
04/10/2001
view
Jay paid $65 Price Joseph paid for his meal, where n is an integer:
(sixty-five
Because Joseph's price is odd, let J be:
dollars) for his
meal.
J = 2n + 1
Schroeder, Craig
[n/a]
[n/a]
Greer, Shakur
[n/a]
Fontainebleau
High School
Leroy paid 25 fewer dollars than three times what Joseph spent, so let
L be:
L = 3J - 25 = 3 (2n + 1) - 25 = 6n - 22
Jay paid five times the positive difference between what Joseph and
Leroy paid, so let Y be:
Y = 5 |J - L| = 5 |(2n + 1) - (6n - 22)| = 5 |-4n + 23|
The total amount of money raised was twelve times Joseph's payment, so
let T be:
T = 12J = 12 (2n + 1) = 24n + 12
The sum of the money paid by these three gentlemen was $4 more than
the amount paid by everybody else who bought boxes.
Y + L + J = 4 + (T - Y - L - J)
2Y + 2L + 2J = 4 + T
2Y + 2L + 2J = 4 + (24n + 12)
Y + L + J = 12n + 8
(5 |-4n + 23|) + (6n - 22) + (2n + 1) = 12n + 8
5 |-4n + 23| = 4n + 29
Assuming the expression in absolute values to be positive produces:
5 (-4n + 23) = 4n + 29
-20n + 115 = 4n + 29
-24n = -86
n = 43/12
Because n MUST be an integer, this cannot be the desired solution.
Thus, the portion inside absolute values is negative, so that:
-5 (-4n + 23) = 4n + 29
20n - 115 = 4n + 29
16n = 144
n = 9
A simple and routine check:
5 |-4*9 + 23| = 4*9 + 29
5 |-36 + 23| = 36 + 29
5 |-13| = 65
65 = 65
Jay's payment:
Y = 5 |-4n + 23| = 5 |-4*9 + 23| = 65
Because we want Jay's payment (Y), the desired answer is 65 dollars.
04/10/2001
view
The amount
the Jay paid
for his dinner
is $34.30.
Since we can relate everything to Jose[h's amount, I called his
amount x. Leroy's amount was $25 less than 3 times Jose[hs amount.
so I sued the equation 3x-25 to represent Leroy's amount. Jay's
amount was 5 times the positive difference of what Joseph and Leroy
paid, so I used the equation: 5(3x-25-x) to represent Jay's amount.
To find what I shold set these equations equal to (what the totla
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04/10/2001
view
Jay paid $65
for his box
supper.
11/24/10 12:44 PM
amount is), I used the fact that the total amount of money raised was
12 times Joseph's pauyment and the fact that sum of the money paid by
the three men was $4 more the amount paid by everyone else. I set
the equation 12x= x+3x-25+(5(3x-25-x))+4, and solved for x. THe
answer came out to be x=73. I then set the total of the 3 mens
amount equal to 73. This equation was x+3x-25+(5(3x-25-x))=73. I
solved for x and it came out ro be $15.93. Since x was Joseph's
amount and x was equal to $15.93, I plugged in $15.93 into Jay's
equation which was 5(3x-25-x)=(5 times the difference of Joseph and
Leroy's amonts), and I solved for x. This answer can ou to be
$34.30, So Jay's amount paid for the dinner is 34.30.
I solved this problem by entering the available information into
compatible equations.
Warner, Nathan
[n/a]
[n/a]
Hawks, Joshua
[n/a]
[n/a]
Joseph: I gave the cost of Josephs supper the value of "X".
Cost=X
Leroy: Leroy paid 3 times Josephs cost minus $25.
Cost=3X-25
Jay: Jay paid 5 times the positive difference between what
Joseph and Leroy paid.
Cost=(3X-25-X)5
Cost=(2X-25)5
(simplified)
Cost=10X-125
(simplified)
Two facts about total amount of money raised:
Total $: Money raised is equal to 12 times Josephs payment.
Total=12X
Total $:
(X+3X-25+10X-125)
This represents the sum of all the amount the three
guys
paid.
(X+3X-25+10X-125)-4
This is the amount everyone else paid.
12X=2(X+3X-25+10X-125)-4
This is the total of all the people and the three guys.
Joseph:
To solve for X ,the only unknown, I
having one total equalling the same
12X=2(X+3X-25+10X-125)-4
12X=2(14X-150)-4
(simplified)
12X=28X-300-4
(simplified)
-16X=-304
(subtracted 28X
X=-304/-16
(divided by -16
X=19
Leroy: Cost=3X-25
Cost=3(19)-25
Cost=$32
Jay: (2X-25)5
(2(19)-25)5
Jay=$65
created an equation by
total written differently.
from both sides)
form both sides)
(substitution)
(simplified)
(substitution)
*****I checked the possibility of Joseph spending more than Leroy and
the answer I got for the amount that Joseph spent when I did this was
$8.16, which is even and not a whole number and the problem asks for
an odd whole number of dollars.
Joesph=X
Leroy=3X-25
Jay=(X-3X-25)5=10X-125
(Joseph spent more than Leroy)
12X=2(X+3X-25-10X+125)-4
12X=2(-6X+100)-4
12X=-12X+196
24X=196
X=$8.16
04/11/2001
view
The amount
that Jay
payed for the
meal was 65
dollars.
First I took the facts that you had in the report of this pinic. Then
I made up their formulas, but in order to solve them they had to all
be in terms of x. So then I converted them all to terms of x. And
since you said dollars in your thing I knew it had to be in whole
dollars. so then i wrote my final equation, and came up with 65
dollars for the payment made by Jay. (Work shown down below.)
joe-odd# of $
leroy-(3x-25)
jay-5(|(x-y)|)
joe=x
leroy=y
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11/24/10 12:44 PM
jay=z
people=w
w+4=x+y+z
x+(3x-25)+5(|(x-(3x-25))|)+x+(3x-25)+5(|(x-(3x-25))|)-4 = 12x
19+(57-25)+5(|(19-(57-25))|)+19+(57-25)+5(|(19-(57-25))|)-4 = 228
19+(32)+(65)+19+(32)+(65)-4 = 228
2(19+(32)+(65))-4 = 228
2(116)-4 = 228
(232)-4 = 228
(228) = 228
Jay payed 65 dollars for the meal.
04/11/2001
view
Jay paid 65
dollars
x = what Joseph paid (odd number)
Pilling, Caleb
[n/a]
[n/a]
Zimmerer, Adam
[n/a]
[n/a]
3x - 25 = what Leroy paid
5[(3x-25) - x] = what Jay paid
12x = total amount of money raised
y = everyone except Joseph, Leroy, and Jay
If the sum of money paid by the three gentlemen was four dollers more
than the amount paid by everybody else, then:
x + (3x-25) + 5[(3x-25) - x] = y + 4 and
12x = x + (3x-25) + 5[(3x-25) - x] + y so
12x = 4x - 25 + 5(2x-25) + y
12x = 4x - 25 + 10x - 125 + y
12x = 14x - 150 + y
12x - 11x = 14x - 11x - 150 - y
x = 3x - 150 + y so
y = x - 3x + 150
y = 150 - 2x
Now that I know what y equals, I'll substitute it into this equation
from above:
x + (3x-25) + 5[(3x-25) - x] = y + 4
x + 3x - 25 + 10x - 125 = y + 4
14x - 150 = 150 - 2x + 4
( the 150 - 2x is substituted for y)
14x = 300 - 2x + 4
16x = 304
x = 19
Therefore, Joseph paid 19 dollars (an odd number),
Leroy paid 3(19) - 25 or 32 dollars, and
Jay paid 5[(3*19 - 25) - 19] or 65 dollars
04/11/2001
view
For this solution, I added Joseph's price( x ) to Leroy's price
Jay payed 65
dollars for his ( 3x - 25 ) to Jay's price { 5 [ ( 3x - 25 ) - x ] }. I then
multiplied this sum by 2, and subtracted 4 from this product to equal
meal.
the total sales( 12x ), or in mathematical terms:
2 { x + ( 3x - 25 ) + 5 [ (
As this problem works out:
2 { 4x - 25 + 5 [ 2x - 25 ]
2 { 4x - 25 + 10x - 125 } 2 { 14x - 150 } - 4 = 12x
28x - 300 - 4 = 12x
28x - 304 = 12x
16x = 304
x = 19
From this, I figured out
3x - 25 ) - x ] } - 4 = 12x
} - 4 = 12x
4 = 12x
Joseph's price. Then, I plugged this in
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04/11/2001
view
Jay has paid
$65.
11/24/10 12:44 PM
to Jay's price equation{ 5 [ ( 3x - 25 ) - x ] } to find Jay's price:
5 [ ( 57 - 25 ) - 19 ]
5 [ 32 - 19 ]
5 [ 13 ] = 65
Jay payed 65 dollars.
Let us express the foregoing givens of the problem into algebraic
expressions. The first given is that it cost Joseph an odd number of
dollar. Let x be the amount paid by Joseph. The second given is that
it cost Leroy 25 fewer dollars than three times what Joseph spent.
Let's translate that into 3x-25 and let it be the amount paid by
Leroy. Thirdly, it was given that Jay paid five times the positive
difference between what Joseph and Leroy paid. In algebraic terms,
that is 5((3x-25)-x) or 5(2x-25) or 10x-125. Adding those algebraic
expressions, we have x + 3x - 25 + 10x - 125 to be equal to 14x-150
and this represents the total amount paid by Joseph, Leroy and Jay. On
the other hand, the other two givens are that the total amount of
money raised was twelve times what Joseph paid or in algebraic term it
will be 12x and that the total amount paid by these three gentlemen
was 4 dollar more than the amount paid by everybody else who also
bought boxes or the other way to say it is that the total amount paid
by everybody else other than Joseph, Leroy and Jay is 4 dollar less
than the total amount paid by the three or in algebraic term it can be
expressed as (14x-150)-4 or 14x-154.
Lansang de Luna, Efren
[n/a]
[n/a]
Okada, Asahi
[n/a]
Pacifica High
School
Mullins, Aaron
[n/a]
[n/a]
Subtracting 14x-154 from 12x will result in 12x -(14x-154)
or 12x - 14x + 154 or equal to -2x + 154. This algebraic term is also
the total amount paid by the three since it is the difference between
the total amount collected from all and the amount paid by the rest of
the gentlemen other than Joseph, Leroy and Jay. Therefore, equating
the two sets of algebraic expressions, we have 14x - 150= -2x + 154.
Transposing the expression to the left and the constant to the right
results in 14x + 2x= 154 + 150 which is equal to 16x= 304. Solving for
x, x= 304/16= 19.
Now to find how much Jay has paid, we need to substitute the value 19
for x in our algebraic expression formulated for Jay, namely
10x - 125. Substituting, we have 10(19) - 125= 190 - 125= $65.
04/11/2001
view
Jay paid 103
dollars.
To check our solution:
Substituting our value of 19 for x in every set of algebraic
expression:
1. How much did Joseph pay? It is equal to x or $19.
2. How much did Leroy pay? It is equal to 3x-25 or 3(19)-25
or 57-25= $32.
3. How much did Jay pay? It is equal to 10x-125 or 10(19)-125
or 190-125= $65. This is also equal to 5 times the difference between
32 and 19 or 5(13)= $65.
4. How much was the total amount raised? It is equal to 12x or 12(19)
= $228.
5. How much did the gentlemen other than the three contribute? It is
equal to 12x -(14x-150) or 12x - 14x + 150 or -2x + 150
or -2(19)+ 150 or 150-38= $112 which is 4 dollar less than the total
amount paid by the three equal to 19 + 32 + 65= $116. Adding $112 and
$116 will produce a total of $228, the total amount raised.
Everything in the problem is expressed in terms of what joseph
paid, so I wrote the amount paid for Sally, Mary, and Judy's lunches
below as equations.
Let s = amount Jospeh paid for Sally Jo's lunch.
"It cost him an odd number of dollars." Not helpful yet..
Sally:
s
"It cost him 25 fewer dollars than three times what Joseph spent."
Mary: 3s - 25
"He paid five times the positive difference between what Joseph and
Leroy paid."
5(3s - 25 - s)
5(2s - 25)
Judy: 10s - 125
Joseph, Leroy, and Jay raised half of the total amount of money
paid by everyone else plus four dollars. To find how much they paid,
I first subtract 4
from 12s, and divide that by two. That gives me two equal sums, and
four dollars left over. To one part I add 4, and that would be the
part these three people paid.
(12s - 4) / 2 = 6s - 2
6s - 2 + 4 = 6s + 2
Now I set the the amount the three earned equal to what they each
paid, and solve for s.
6s + 2
6s + 2
-8s
s
=
=
=
=
s + 3s - 25 + 10s - 125
14s - 150
-152
19
Now that I have s, I can plug (oops, substitute!) into 10s - 125 and
solve for the amount jospeh paid.
10(19) - 125 =
190 - 125 = 65
Hey, if this is right, that'll mean I passed the 31-point milestone!
04/11/2001
view
The awnser to The way to figure out the awnser to the problemi s to diside what odd
the question numbers you can multipliy to get a awnser higher than 25. I tryed
every numer until I got to 19. Then I multiplyed that by three and
is $19.00.
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The Math Forum @ Drexel University
is $19.00.
04/11/2001
view
11/24/10 12:44 PM
took away 25. I got 57. I multipled the dirfrence by 12 and it was 4
more than what the other people payed.
Info:
Three young ladies, Sally Jo, Mary Lou, and Judy Ann
Jay paid $65
for his supper
box.
Joseph's cost for his box = J
Yang, Jason
[n/a]
Taipei
American
School
Police, Matt
[n/a]
Perry Middle
School
Leonbruno, Greg
[n/a]
Perry Middle
School
Chen., Kimberly
[n/a]
Taipei
American
School
Leroy's cost for his box = L
Jay's cost for his box = A
Sally Jo's boyfriend, Joseph, paid a modest but significant amount
for that box. (It cost him an odd number of dollars.)
Mary Lou's boyfriend, Leroy, won that box for a very good price.
(It cost him 25 fewer dollars than three times what Joseph spent.)
L = 3J - 25
Judy Ann's boyfriend Jay, paid top dollar for that scrumptious meal.
(He paid five times the positive difference between what Joseph and
Leroy paid.)
A = 5*|L - J|
Info 2
Amount paid by everybody else who bought boxes = P
We want to know just how much Jay paid, but you need to know two more
facts:
the sum of the money paid by these three gentlemen was $4 more than
the amount paid by everybody else who bought boxes.
J + L + A = 4 + P
J + L + A -4 = P
The total amount of money raised was twelve times Joseph's payment
(J + L + A + P) = 12J
J + L + A + J + L + A - 4 = 12J
2J + 2L + 2A - 4 = 12J
2L + 2A - 4 = 10J
L + A - 2 = 5J
Solve: for J.
L + A - 2 = 5J
(equation from above)
3J - 25 + 5*|L - J| - 2 = 5J
(substitute 1st and 2nd equation from Info1: into this one)
3J - 25 + 5*|3J - 25 - J| - 2 = 5J
(I see L, so
I'll substitute again)
3J - 25 + |10J - 125| - 2 = 5J
|10J - 125| = 2J + 27
8J = 152
J = 19
or
|10J - 125| = - (2J + 27)
|10J - 125| = - 2J - 27
12J = 98
J = 8.166
(not the answer, because it's not odd.)
Solve: for L. substitute J into the following eq.
L = 3J - 25
L = 3*19 - 25
L = 32
04/11/2001
view
Jay payed
75$.
04/11/2001
view
when i typed
it in my
calculator i
got the
answer 32.46.
Jay had paid
65 dollars for
buying the
box that his
girlfriend had
made for him.
04/11/2001
view
Solve: for A. Substitute J and L into the following eq.
A = 5*|L - J|
A = 5*|32 - 19|
A = 65
When I did this problem I wasn't sure but When I typed some of the
numbers in my calculator I got 44.00. THis seems correct so i'll go
with it.
i took all the information given and put it into my calculator. This
was a very hard problem and i dont think i have the answer at all.
In the information that the problem provided we know that three
people whp had paid with different money, and now our task for this
problem is to find out how much money Jay paid.
The first boy is Joseph, the problem said that the box cost him an
odd number of dollars. I count the money that he had paid as (2x + 1).
The second boy is Leroy, the problem said that the box cost him 25
fewer dollars than three times what Joseph spent. I count the money
that he had paid as 3(2x + 1) - 25.
The third boy is Jay, the problem said he paid five times the
positive difference between what Joseph and Leroy paid. I count the
money that he had paid as 5((2x + 1) - (3(2x + 1)- 25) or
5(3(2x + 1) - 25 - (2x + 1)), there are two equations becasue you
don't
know whether which one is positive and which one is negative, it
depends on which one is bigger, to make it's difference positive. So
you have to calculate both of them, and find out the positive one.
In one of the sentence in the problem "the total amount of money
raised was twelve times Joseph's payment", I count Joseph's money as
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11/24/10 12:44 PM
x, so the total amount of payment would be 12(2x + 1).
In the other information that the problem has give is that "the sum
of the money paid by these three gentlemen was $4 more than the
amount paid by everybody else who bought boxes." So we can know that
the equation would be the three people's money added together - 4.
Now, first I will do the 5(3(2x + 1) - 25 - (2x + 1)), and if it is
positive then that means I got the solution.
12(2x + 1) is becasue that is total number of money which Jay,
Joseph, Leroy and others paid. Then, 2[(2x + 1) + 3(2x + 1) - 25 + 5(3
(2x + 1) - 25 - (2x + 1))] - 4 is because the others people's money
is the total number of Jay, Joseph, and Leroy's money minus 4. So we
know that this equation will used twice, [(2x + 1) + 3(2x + 1) - 25 +
5(3(2x + 1) - 25 - (2x + 1))] , so we put 2 in front of the equation
in order to tell people to multiply the whole thing twice, then after
multiplying twice minus 4. We minus 4 because we know that the total
number of the three of them's money are more 4 then others, so we
have to minus 4. So you know that the whole equation would be
12(2x + 1) = 2[(2x + 1) + 3(2x + 1) - 25 + 5(3(2x + 1) - 25 - (2x +
1))] - 4
Now, let's solve the equation
12(2x + 1) = 2[(2x + 1) + 3(2x + 1) - 25 + 5(3(2x + 1) - 25 - (2x +
1))] - 4
12(2x + 1) =
12(2x + 1) =
12(2x + 1) =
12(2x + 1) =
-16(2x + 1)=
16(2x + 1) =
32x + 16 =
32x =
x =
2[4(2x + 1) - 25 + 5(2(2x + 1) - 25)] - 4
2[4(2x + 1) - 25 + 10(2x + 1) - 125] - 4
2[14(2x + 1) - 150] - 4
28(2x + 1) - 300 - 4
-304
304
304
288
9
So x = 9
Now, let me check the answer
Joseph (2x + 1) = 9 * 2 + 1 = 19
Leroy 3(2x + 1) - 25 = 3(18 + 1) - 25 = 3 * 19 - 25 = 32
32 - 19 = 13
yeah, it's positive so I know the equation is right.
So now, I started to calculate out the money that Jay spend for
buying her girlfriend's box.
x = 9
5(3(2x + 1) ¡V 25 - (2x + 1))
= 5(3(19) - 25 - 19)
= 5(13)
= 65
So we know that Jay paid 65 dollars to buy that box.
Now, I'm checking to see whether my answer is right or wrong.
The total money that the three of them paid is 116 dollars.
The problem said that the sum of the money paid by these three
gentlemen was $4 more than the amount paid by everybody else who
bought boxes.
So I'll just check whether it fit the equation or not.
288 - 116 =
112
Then, I add for to 112
112 + 4 = 116
After I added 4 to 112, it is the same number that the total number
that the three of them paid to buy their girlfriends' boxes.
Yeah! I'm right. My answer is correct.
Jay paid 65 dollars to buy the box which made by his girlfriend.
04/11/2001
view
Jay paid $65. Let J, L, Y be the amount of money paid by Joseph, Leroy and Jay and
K the amount paid by everybody else who bought boxes.
Boccardo, Giacomo
[n/a]
[n/a]
"Leroy paid 25 fewer dollars than three times what Joseph spent":
L = 3*J - 25
[1]
"Jay paid five times the positive difference between what Joseph and
Leroy paid":
Y = 5 * Abs(J - L)
[2]
"The total amount of money raised was twelve times Joseph's payment":
J + L + Y + K = 12*J
[3]
"The sum of the money paid by these three gentlemen was $4 more than
the amount paid by everybody else who bought boxes":
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J + L + Y = 4 + K
[4]
I substitute [1] in [2]:
Y = 5 * Abs(J - L) = 5 * Abs(J - (3*J - 25)) = 5 * Abs(25 - 2*J)
[5]
I substitute [4] solved for K in [3]
K = J + L + Y - 4
[4]
J + L + Y + (J + L + Y - 4) = 12*J
10*J - 2*L - 2*Y + 4 = 0
[6]
Now L and Y are expressed depending on J, therefore I substitute [1]
and [5] in [6]:
10*J - 2*L - 2*Y + 4 = 0
10*J - 2*(3*J - 25) - 2*(5 * Abs(25 -2*J)) + 4 = 0
10*J - 6*J + 50 - 10*Abs(25 - 2*J) + 4 = 0
4*J - 10*Abs(25 - 2*J) + 54 = 0
2*J - 5*Abs(25 - 2*J) + 27 = 0
There are two solutions because of the absolute value:
2*J - 5*(25 - 2*J) + 27 = 0
2*J - 5*(2*J - 25) + 27 = 0
The solution of the first equation is:
2*J - 5*(25 - 2*J) + 27 = 0
2*J - 125 + 10*J + 27 = 0
12*J - 98 = 0
J = 98/12 = $49/6
The solution of the second equation is:
2*J - 5*(2*J - 25) + 27 = 0
2*J - 10*J + 125 + 27 = 0
-8*J + 152 = 0
J = 152/8 = $19
The right solution is the second.
Substituting this value in [5] I find how much Jay paid:
Y = 5 * Abs(25 - 2*J) = 5 * Abs(25 - 2*19) = 5 * Abs(25 - 38) = 5 *
Abs(-13) = 5 * 13 = $65
Jay paid $65.
04/11/2001
view
Jay paid 65 Let x = amount Joseph paid
3x-25 = amount Leroy paid : it says that Leroy paid 15 fewer
dollars for the
dollars than three time what Joseph
box supper.
paid.
Ko, Min
[n/a]
Loyola
Academy
5(( 3x-25 )-x) = amount Jay paid : it says he paid five times the
positive difference what Joseph
and Jay paid. From the
equation, you can notice that
Leroy paid more than Joseph.
12x = the total payment
3x - 25 + 10x - 25 = 14x - 150 = totla paid by three men
12x-( x+3x-25+10x-125) = 150 - 2x = the sum of money paid by
others
Then the equation becomes
14x - 150 = 150 - 2x + 4
16x = 304
x = 19
Add 4 on the side of paid by others
because the sum of money paid by three
was $4 more than the amount paid by
others.
Simplify the equation and isolate the x.
Solve for x.
Since x represents the amount of money paid by Joseph, subsititute 19
to the equation for the payment of Jay.
5(( 3x-25 )-x) = Amount paid by Jay
5(( 3(19)-25)-19)= 5(13)
= 65
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 21 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
The amount paid by Jay is $65.
04/11/2001
view
Jay paid $65
for his box.
There is posibility that Leroy paid more than Jay when Jay paid
5(Joseph - Leroy) which is 5[ x- (3x-25) ] = 5( -2x+25)
If 3x-25 is to be greater than 5[ x- (3x-25) ] you have to set an inequality.
3x-25 > 5 [ x- (3x-25) ]
3x-25 > 5( -2x+25)
3x-25 > -10x+125
13x > 150
x > 150/13
When x is greater than 150/13 there is possibility that Leroy paid more
than Jay.
All the boxes were bought at a certain price, which can be put in
terms of Joseph’s box. I used the variable x to denote the price
Joseph paid for his box.
x = Joseph’s box
3x – 25 = Leroy’s box
5((3x – 25) – x) = Jay’s box
So, the
x + (3x
To make
x + (3x
04/11/2001
view
jay paid 85
dollars.
04/11/2001
view
Joseph paid$75 with a
total of
everybodies
boxes = 900$
Leroy paid$200 Jay
paid- $625
Sullivan, Emmet
[n/a]
Loyola
Academy
Sakal, Lindsey
[n/a]
[n/a]
Haavisto, Christine
Stark, Sarah
[n/a]
Perry Middle
School
M., Jaclyn
[n/a]
[n/a]
sum of their boxes would be:
– 25) + 5((3x – 25) – x)
it easier to write, I denoted the above sum as y, so:
– 25) + 5((3x – 25) – x) = y
The total price is 12x. This number is the sum of all the boxes
combined. It says that the three boys paid 4 dollars more than all
the other people who bought boxes. So:
y – 4 is the price that everyone else spent on boxes.
12x – y is the same thing, since this means that it is the total
amount minus the amount that the four boys spent. So, you can set
them equal to each other:
y – 4 = 12x – y
If you simplify this, you get:
2y = 12x + 4
Next, you fit in the long expression for y, so the new equation is:
2(x + (3x – 25) + 5((3x – 25) – x)) = 12x + 4
2x + (6x – 50) + 10(2x – 25) = 12x + 4
8x – 50 + 20x – 250 = 12x + 4
28x – 300 = 12x + 4
16x = 304
x = 19
At this point, we have the price Joseph paid. So, if we fit it in to
the expression that symbolizes what Jay pays:
5((3x – 25) – x)
5((57 – 25) – 19)
5(32 – 19)
5(13)
65
Jay paid $65.
first for joseph i made him J. I made leroy 3J-25 and Jay was 5
(J+L). I made everyone ther equal to 4+12J.I added joseph, Leroy ANd
Jay togther and equal to 4+12J so it looked like this. 3J-25+J+5J+15J125=4+12J then i simplified it down to this 24J-150=4+12J then down
to 12J =154 and divided 154 by 12 and got 21 then from there i just
went back to the equations ihad to begin with and filled in J for 21
and solved.
First we made an equation for each of the three guys
joesph= x
leroy= 3x-25
jay= 5(3x-25)
Then to find what x equaled we found out what x equaled by setting
those into one big equation.
x+3x-25+10x-125=12x
we set it equal to 12x because the total anount was 12x what they
paid.
So once we did that we found that x=75
So then we plugged this into the equations for each guy
Joseph then =75
Leroy then = 200
jay then = 625
04/11/2001
view
then it says the total amount was twelve times what joseph paid so we
did 75 x 12 to get $900 as the total amount raised.
Let
Jay payed
$65 for Judy
Amount of $ Joseph paid = x
Ann's supper
box.
x must be odd
Then
Amount of $ Leroy paid
= 3x - 25
Then
Amount of $ Jay paid
= 5 (3x - 25 - x)
because the difference of Joseph and Leroy's amount must be
positive.
Total amount of $ raised = 12x
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 22 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
The sum of the 3 men's payments = x + 3x - 25 + 5 (3x -25 -x)
The amount of $ paid by everybody else =
[x + 3x -25 +5 (3x - 25 -x)] -4
Total= sum of money paid by everyone else + sum of the 3 men's
amount
12x = [ x + 3x - 25 + 5 (3x - 25 - x) -4] + x + 3x -25 + 5 (3x - 25x)
12x = [x + 3x - 25 + 15x - 125 - 5x -4] + x + 3x - 25 + 5 (3x - 25x)
12x = [14x - 25 - 125 - 4] + x + 3x - 25 + 5 (3x - 25 - x)
12x = 14x - 154 + x + 3x - 25 + 5 (3x - 25 -x)
12x = 14x - 154 + x + 3x - 25 + 15x - 125 - 5x
12x = 28x - 154 - 25 - 125
12x= 28x - 304
-28x -28
--------------16x/-16 = -304/-16
x=19
Joseph paid $19 for Sally Jo's Box.
---Check--Joseph's payment = x = $19
19 is odd
Leroy's payment = 3x - 25 = $32
3 (19) -25
57 - 25 = 32
Jay's payment = 5 (3x - 25 - x) = $65
5 [3 (19) - 25 - (19)]
5 [57 - 25 - (19)]
5 [ 32 - (19)]
5 [13] = $65
Total amount of $ raised = 12x = $228
12 (19) = 228
Sum of the payment of the 3 men = $116
Joseph's payment+ Leroy's + Jay's = sum
$19 + $32 + $65
$51 + $65= $116
Sum of the amount paid by everyone else = $112
$116 - $4 = $112
Total$ =Sum of the amount paid by everyone else+ amount paid by
everyone else
12x = [x + 3x - 25 + 5 (3x - 25 - x) - 4] + x + 3x - 25 + 5 (3x - 25 x)
$228 = $112 + $116
$228= $228
04/11/2001
view
checks
Jay paid $65. To solve this problem I first identified all of the varibles
x = how much Joe paid
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Ursini, Meagan
[n/a]
Loyola
Academy
Page 23 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
3x - 25 = how much Leroy paid
5(3x - 25 - x)= how much Jay paid
12x = total money rasied
x + 3x - 25 + 5(3x - 25 - x) - 4 = total paid by everyone else
y = total money paid bye Joe, Jay and Leroy
y = x + (3x - 25) + 5(3x - 25 - x)
y = x + 3x - 25 + 10x -125
y = 14x - 150
I then used this information to set up an equation. The equation is
the total money raised by everyone (12x) equals the amount Jay, Joe
and Leroy paid (y) plus the amount they paid minus 4.
12x
12x
12x
-16x
x
=
=
=
=
=
y + (y-4)
14x - 150 + 14x - 150 - 4
28x - 304
-304
19
So the amount that Joe paid is $19. To figure out how much Jay paid I
have plug 19 back into the equation that told me how much Jay paid.
5 (3x - 25 - x)
5 ( (3) (19) - 25 - 19)
5 ( 57 - 25 - 19)
5 ( 13)
65
So the amount that Jay paid is $65.
Now I will find the amount that Leroy paid.
3x - 25
3 (19) - 25
32
So Leroy paid $32. It is possible that Joe could have paid more then
Leroy.
Now I will find the total paid by all 3 of the boys andf there are 2
ways to do that. One is to add up their amounts and the second is to
plug 19 into the equation to figure out how much they all paid.
32 + 65 + 19
116
y
y
y
y
=
=
=
=
14x - 150
14 (19) - 150
266 - 150
116
The total money paid by all of the boys was $116.
Now I will find out the total amount paid by everyone else.
x + 3x - 25 + 5(3x - 25 - x) - 4
19 + 3 (19) - 25 + 5(3 (19) - 25 - 19) - 4
19 + 57 - 25 + 5(57 - 25 - 19) - 4
19 + 57 - 25 + 5(13) - 4
19 + 57 - 25 + 65 - 4
112
The total paid by everyone else is $112.
Now I will find out the amount paid by everyone. There are two ways
to do this. One is to add up the amount paid by the 3 to the amount
paid by everyone else and the other way is to plug 19 by into the
original equation.
112 + 116
228
12x
12(19)
228
So the amount paid by everyone was $228. Now I will summarize all of
my answers.
04/11/2001
view
Jay paid $605
Joe = $19
Leroy = $32
Jay = $65
amount paid by Joe, Jay and Leroy = $116
total paid by everyone else = $112
total paid by Joe, Jay, Leroy and everyone else = 228
Joseph paid x dollars
leroy paid 3x-25 dollars
Jay paid five times (Leroy minus joseph)
so Jay paid 5((3x-25)-x)
Total money raised was 12x
Three boys money combined = 12x-4
therefore
12x-4=x+3x-25+5((3x-25)-x)
12x-4=4x-25+5(2x-25)
12x-4=4x-25+10x-125
12x-4=14x-150
146=2x
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Lenarduzzi, Michael
[n/a]
Australian
Catholic
University
Page 24 of 26
The Math Forum @ Drexel University
04/11/2001
view
Jay paid
$78.125 for
his box
supper.
11/24/10 12:44 PM
x=73
Joseph paid $73
Jay paid 5((3(73)-25)-73)
Jay paid $605
First I will assign each boy a variable.
Miller, Katherine
[n/a]
[n/a]
Creanza, Christine
[n/a]
[n/a]
J=Joseph
L=Leroy
Y=Jay
Now I will assign the variable "E" to the amount that
everyone else spent.
The three main equations that I will use for this problem
are:
#1: Y=5(25-J)
#2: E+25+J+Y=12J
#3: E+4=25+J+Y
Now I will simplify equation #2 to get an equation for
the variable "E".
E+25+J+Y=12J
E=12J-J-Y-25
E=11J-Y-25
Now I will simplify equation #3 to get another equation
for the variable "E".
E+4=25+J+Y
E=25-4+J+Y
E=21+J+Y
So since E=11J-Y-25 and E=21+J+Y, I know that 11J-Y-25=21+J+Y.
I will use this to make an equation for the variable "J".
11J-Y-25=21+J+Y
11J-J=21+25+Y+Y
10J=2Y+46
J=0.2Y+4.6
I will now substitute the above equation into equation #1 to
find out what the variable "Y" represents.
y=5(25-(0.2Y+4.6)
Y=5(20.4-0.2Y)
Y=102-Y
Y+Y=102
2Y=102
Y=51
So, Jay spent $51 for his box supper.
04/11/2001
view
The price Jay With the given information I know:
paid for his
I will let J= Joseph's price
box was 65
dollars.
Joseph= odd amount, (I will use J to represent the amount Joseph
paid)
Since Leroy spent 25 dollars less that three times what Joseph paid:
Leroy= 3J- 25
Jay paid 5 times the difference between what Joseph and Leroy spent.
(3J- 25- J) represents the difference between what Joseph and Leroy
paid.
That can be simplified and then multiplied by 5:
Jay= 5(2J- 25)
Also with the information given I know that the Total payment from
the boxes=
12J.
I will let E= the sum of money paid by everyone else who purchased a
box
14J- 150 is the amount the 3 boys paid for their boxes combined. I
got that number by simplifying the equation I used for Jay and
adding it to Leroy and Joseph's equation:
Jay= 5(2J- 25)
Leroy= 3J- 25
Joseph= odd amount (I will use J to represent the amount Joseph paid)
5(2J- 25) = 10J- 125
10J- 125+ J+ 3J-25 = 14J- 150
It was given in the problem that the difference between what the biys
paid and what everyone else paid was $4. I wrote this as the
equation:
14J- 150- E= 4
which can be simplified to: 14J- E= 154
This equation represents the difference between what the boys and
everyone else purchasing the boxes paid.
The sum of what the boys paid added to the sum of what everyone else
paid
equals the total amount paid for all of the boxes.
It is given that the total amount paid was 12 J, so I set these two
equations
http://mathforum.org/pows/print/office/submissions.htm?publication=1926&status=&pagesize=50
Page 25 of 26
The Math Forum @ Drexel University
11/24/10 12:44 PM
equal to each other
14J- 150+ E= 12J
Then I subtract 12 J from both sides and get 2J- 150+ E= 0
Next I add 150 to both sides to get my final equation:
2J+ E= 150
Next, I add these two equations together to cancel the E variable and
solve for J.
14J- E= 154
+2J +E= 150
which equals 16J= 304
Therefore J= 19
So Joseph paid 19 dollars for his box.
Since I know the equation i need to find the total amount paid for
all the
boxes is 12J, i can now solve it.
12 * 19= 228
The amount Leroy paid is 3J:
3 * 19= 32
And finally Jays box price can be found by using the equation:
5(2J- 25)
5(2* 19)- 25= 65
Therefore Jay paid 65 dollars for his box.
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