SOLUTION: Susan cashed a check at the bank, but the teller made a mistake. He gave her as many dollars as he should have given her cents and as many cents as he should have given her dollars

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Question 1164209: Susan cashed a check at the bank, but the teller made a mistake. He gave her as many dollars as he should have given her cents and as many cents as he should have given her dollars. Susan took the money without realizing the error and spent $3.50. She then had twice as much money as the actual amount of the check. What was the amount of the check?
Found 5 solutions by ankor@dixie-net.com, Edwin McCravy, greenestamps, MathTherapy, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Susan cashed a check at the bank, but the teller made a mistake.
He gave her as many dollars as he should have given her cents and as many cents as he should have given her dollars.
Susan took the money without realizing the error and spent $3.50.
She then had twice as much money as the actual amount of the check.
What was the amount of the check?
;
let c = correct no. of cents
let d = correct no. of dollars
then
(d+.01c) = amt of the check
and
(c+.01d) = amt paid by teller
:
c + .01d - 3.50 = 2(d+.01c)
c + .01d - 3.50 = 2d + .02c
c -.02c + .01d - 2d = 3.50.
.98c - 1.99d = 3.50
Set up as an equation that we can graph on our calc and check the table
y=c; x=d
y = %281.99x%2B3.50%29%2F.98
only one, two digit, integer solution
d = 14, c = 32
Actual check value: $14.32, teller paid $32.14
:
:
See if that checks out
$32.14
-3.50
-------
$28.64 which is twice $14.32

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The other tutor's answer is correct, but he didn't really solve it.  He used
trial and error.  Here is how to solve it completely, using no trial and
error.  This is a "Diophantine equation", i.e. one that is solved only in
integers.
D = dollars
C = cents
100D + C = amount check was for (in pennies)
100C + D = amount the teller gave Susan (in pennies)
100C + D - 350 = how much she had after spending $3.50 (in pennies)

The cents can only be a 1 or 2 digit number, and since they were switched
the dollars can also only be a 1 or 2 digit number. 

So 

100C+%2B+D+-+350+=+2%28100D+%2B+C%29

That simplifies to

98c+=+199d+%2B+350

Solve for c

c+=+%28199%2F98%29d+%2B+%28350%2F98%29}

c+=+%282%263%2F98%29d+%2B+%283%2656%2F98%29

c+=+%282%2B3%2F98%29d+%2B+%281%2B4%2F7%29

c+=+2d+%2Bexpr%283%2F98%29d+%2B+3+%2B+4%2F7%29

Isolate the fractions on the right:

c-2d-1=expr%283%2F98%29d+%2B+4%2F7

The left side is an integer, so the right must also be
that same integer.  Let that integer be A

c-2d-1=A   
                 
expr%283%2F98%29d+%2B+4%2F7=A
3d+%2B+56+=+98A 

Solve for d

3d+=+98A-56

d+=+expr%2898%2F3%29A+-+56%2F3

d+=+%2832%262%2F3%29A+-+%2818%262%2F3%29

d+=+%2832%2B2%2F3%29A+-+%2818%2B2%2F3%29

d+=+32A+%2B+expr%282%2F3%29A+-+18+-+2%2F3

Isolate fractions on the right

d-32A%2B18+=+expr%282%2F3%29A+-+2%2F3

The left side is an integer, so the right must also be
that same integer.  Let that integer be B.

d-32A%2B18+=+B

expr%282%2F3%29A+-+2%2F3=B
2A+-+2+=+3B

Solve for A

2A+=+3B+%2B+2

A+=+expr%283%2F2%29B+%2B+1

A+=+%281%261%2F2%29B+%2B+1

A+=+%281%2B1%2F2%29B+%2B+1

A+=+B+%2B+expr%281%2F2%29B+%2B+1

Isolate the fraction on the right:

A+-+B+-+1+=+expr%281%2F2%29B

The left side is an integer, so 

The left side is an integer, so the right must also be
that same integer.  Let that integer be E (since we've
used c and d).

A+-+B+-+1+=+E

expr%281%2F2%29B=E

B=2E

A+-+2E+-+1+=+E

A+=+3E%2B1

Substitute 3E+1 for A and 2E for B in

d-32A%2B18+=+B

d-32%283E%2B1%29%2B18+=+2E

d-96E-32%2B18+=+2E

d-96E-14=2E

d=98E%2B14

We can tell that E=0, otherwise d would be more than a 2 digit number.

So E=0 and d = 98(0)+14 = 14

Substitute 0 for E in

A+=+3E%2B1

A+=+3%280%29%2B1

A+=+1

Substitute 14 for d and A = 1 in

c-2d-1=A  

c-2%2814%29-1+=+1

c-28-1=1

c=28%2B1%2B1

c=32

Therefore d=14 dollars and c = 32 cents.

The check was for $14.32.

Checking:  $32.14 - $3.50 = $28.64 which is twice $14.32

Edwin

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let d and c be the numbers of dollars and cents on the original check.

Then the value of the check (in cents) is 100d+c.

The amount of money the bank teller gave her had the dollars and cents reversed; the amount of money Susan got was 100c+d.

After spending $3.50, the amount Susan had left was 100c+d-350.

That amount was twice the amount of the original check:

100c%2Bd-350+=+2%28100d%2Bc%29

This is a Diophantine equation -- a single equation in two unknowns. It can be solved for a distinct set of answers (in many cases, such as this, a single answer) because of the fact that the unknowns must have integer (often positive integer) values.

In this example, we know c and d are both positive integers less than 100.

Solve the equation for one variable in terms of the other.

100c%2Bd-350+=+200d%2B2c
98c+=+199d%2B350
c+=+%28199d%2B350%29%2F98
c+=+199d%2F98%2B350%2F98

Perform the divisions on the right to get whole numbers plus remainder fractions.


c+=+%282d%2B3%29%2B%283d%2F98%2B56%2F98%29+=+%282d%2B3%29%2B%283d%2F98%29%2B%284%2F7%29

Now c is an integer; and d is an integer, so 2d+3 is an integer. That means the expression

3d%2F98%2B4%2F7

must be an integer.

The "first" time it will be an integer is when 3d/98 is equal to 3/7:

3d%2F98+=+3%2F7
3d%2F98+=+42%2F98
3d+=+42
d+=+14

Then with d=14,

c+=+%282d%2B3%29%2B%283d%2B56%29%2F98+=+%282%2814%29%2B3%29%2B98%2F98+=+31%2B1+=+32

ANSWER: the amount of the check was $14.32

CHECK:
Original amount: $14.32
Amount teller gave: $32.14
Amount spent: $3.50
Amount left: $28.64
Twice the amount of the check: $28.64

The Diophantine equation has other solutions; however, none of the others satisfy the requirement that both c and d are positive integers less than 100.


Answer by MathTherapy(10553) About Me  (Show Source):
You can put this solution on YOUR website!
Susan cashed a check at the bank, but the teller made a mistake. He gave her as many dollars as he should have given her cents and as many cents as he should have given her dollars. Susan took the money without realizing the error and spent $3.50. She then had twice as much money as the actual amount of the check. What was the amount of the check?
Let amount of dollars and cents he should’ve given her be D, and C, respectively
Therefore, he should’ve given her: “100D + C” cents
Then, he gave her C dollars and D cents, or “100C + D” cents
She spent $3.50, or 350 cents
Therefore, we get: 100C + D - 350 = 2(100D + C)
100C + D - 350 = 200D + 2C
100C - 2C = 350 + 200D - D
98C = 350 + 199D
matrix%281%2C3%2C+C%2C+%22=%22%2C+%28350+%2B+199D%29%2F98%29
Converting this to a DIOPHANTINE EQUATION, we get:
Now, 2 * 98 = 196, and the integer that can be substituted for D in order to make 154 + 3D a DIVIDEND of 98 is 14. This gives us:
So, with D being 14, we get:
Therefore, D or number of dollars he should’ve given her was: 14, and number of cents he should’ve given her was: 32. Therefore, amount on check was: highlight_green%28%22%2414.32%22%29

Answer by ikleyn(52807) About Me  (Show Source):
You can put this solution on YOUR website!
.

For linear  Diophantine equations  (one equation with integer coefficients for two integer unknowns)  the trial an error method
finding a solution is absolutely and totally  LEGAL  (and standard)  method.


For such equations,  any version of trial and error search is  ALLOWABLE,  and nobody of mature mathematicians
do not really care by  WHICH  concretely searching method the solution is obtained.


You may use an  Excel table/spreadsheet for it and to find the solution in seconds  (as I usually do it . . . )

Nobody cares about it . . .


Or,  alternatively,  you can find it via  30  lines formula transformations.   It is also good,
but again,  NOBODY CARES about it . . .


What is  REALLY  IMPORTANT  for such equations,  it is the necessary and sufficient condition for  highlight%28existing%29  the solution.

This condition  REQUIRES  that the right side constant term must be a MULTIPLE of the Greatest Common Divisor of the coefficients
of the equation at  "x"  and  "y".

If this condition is satisfied,  then the solution  DOES EXIST,  and all the troubles just do stop at this point.