I cant set up the equation for one of my students I am helping, I dont have a key I know the actually amounts just not how to set up problem. Its definitely a Monday.

Problem reads
26 bills total the total of the bills is $120 sum of the bills is 3 times as many ones as fives, 1 less ten than fives and the rest twenties.

Partially done, gotta go.

# of 1s = X
# of 5s = Y
10s = Z
20s = T (didn't see 20's in there at first...Mondays)


X + Y + Z +T = 26
X +5Y + 10z + 20T = 120

From the word problem:
X = 3Y
Z = Y - 1

It's written poorly, probably just as easy to iterate vs solving for Y & T.

Sorry problem I just got them to forward me the exact problem. Problem reads

Cashier has 26 bills consisting of three times as many ones as fives, one less ten than fives, and the rest twenties. If the total value is $120, how many of each each does she have.

mine works.
substitute for Z & X in the 2nd equation
You get 18y + 20t = 130
Only thing that satisfies is T=2, Y = 5
then
X = 15
Z = 4

Not sure about setting up an equation, but just logically thinking about it:

In order to have the possibility of a round number of dollars made up of 1's that is divisible by 3, it has to be 15 1's. 30 would obviously be too many.

The rest is trivial.

quote:

Originally posted by bobtheelf:
Not sure about setting up an equation, but just logically thinking about it:

In order to have the possibility of a round number of dollars made up of 1's that is divisible by 3, it has to be 15 1's. 30 would obviously be too many.

The rest is trivial.

Yeah, like I said - easier to iterate than do the algebra.
2 unknown variables is confusing to keep up with. I could see the answer before I could figure out how to get there & I still skip steps. I saw the 18y & needed something that made a 0 on the end.... 5 it is.

You have four unknowns so you need four equations which snidera gave you. Solve for one of the unknowns in terms of the others and keep eliminating in turn until you get one equation in one unknown.

Is there a locomotive involved?

equation 1 with substitution gives 18y + 20t = 130, which can be reduced to 9y + 10t = 65;
equation 2 with substitution gives 5y + t = 27;

multiplying #2 by 10 gives 50y + 10t = 270;
subtracting 9y + 10t = 65 gives
41y = 205, from which y = 5 and the rest is easy.

flashguy

^^^^^ that is one ugly cat!

flashguy

quote:

Originally posted by Sig2340:
Is there a locomotive involved?

There is always a locomotive. Have to know if it was moving toward the other locomotive or away from it.

quote:

Originally posted by Sig2340:
Is there a locomotive involved?

What about a flag pole and time of day?

Hmmm. The total is $120 and there are three times as many ones as fives. The number of ones must be multiple of five otherwise the total will not be divisible by five, let alone ten.

Just for giggles, let’s think it through with equations first (always hated algebra).

15 ones = $15
5 fives = $25
4 tens = $40

So far 24 bills and $80.

2 twenties = $40 for a total of $120. I guess you could call that iterative, but it popped out on the first iteration with a little logic applied.

Now let’s try the annoying stuff:

26 total bills, $120 total value. 3 times as many ones as fives, one less ten than fives. So let’s try one equation for the number of bills:

26 = 3F + F + (F - 1) + ? Dang question mark, no good idea for how to represent the number of twenties.

$120 = (3F * $1) + (F * $5) + ((F - 1) * $10) + (? * $20) Dang question mark again!

Hey wait, let’s solve the first equation for ?, then plug it into the second.

26 - (3F + F + (F - 1)) = ?
? = 26 - 5F + 1
? = 27 - 5F

Okay, let’s plug that into the second equation:

$120 = (3F * $1) + (F * $5) + ((F - 1) * $10) + ((27 - 5F) * $20)

Let’s lose the dollar signs and simplify a little:

120 = 3F + 5F + 10(F - 1) + 20(27 - 5F)

120 = 8F + 10F - 10 + 540 - 100F

130 = 540 + 18F - 100F

82F = 410

F = 410 / 82

F = 5

Dang, I hate algebra.

quote:

Originally posted by Cycler:
You have four unknowns so you need four equations which snidera gave you. Solve for one of the unknowns in terms of the others and keep eliminating in turn until you get one equation in one unknown.

Nope, only two unknowns: the number of fives and the number of twenties. Write up two equations, one for the the number of bills and the other for the value of the bills. Solve one of them for the number of twenties expressed as related to the number of fives. Then plug that expression into the other equation and solve for the number of fives.

slosig's method works, but I like mine better. Different strokes for different folks, I guess.

flashguy

quote:

Originally posted by slosig:

Nope, only two unknowns: the number of fives and the number of twenties. Write up two equations, one for the the number of bills and the other for the value of the bills. Solve one of them for the number of twenties expressed as related to the number of fives. Then plug that expression into the other equation and solve for the number of fives.

This right here.

quote:

Originally posted by snidera:
Partially done, gotta go.

# of 1s = X
# of 5s = Y
10s = Z
20s = T (didn't see 20's in there at first...Mondays)


X + Y + Z +T = 26
X +5Y + 10z + 20T = 120

From the word problem:
X = 3Y
Z = Y - 1

It's written poorly, probably just as easy to iterate vs solving for Y & T.

This. Do the substitutions for x and z in both of the top two equations.

The first equation becomes:

3y + y + y - 1 + t = 26 which solving for t becomes:

t = 27 - 5y

Do the substitutions now for x, z, and t into the second equation and you get one unknown:

3y + 5y + 10(y - 1) + 20(27 - 5y) =120

which solving for y gives y = 5.

15 ones, 5 fives, 4 tens, and 2 twenties.