# Can you figure this out? (word problem)

• Put a quark in it
Of course, in real life, you would probably have to get change back from the cashier so the actual amounts would be different.In summary, the original amount of money was $99.98 and it was spent on something, leaving half as many dollars as cents. After spending half of the money, the person had the same number of cents as they did dollars. The solution is unique and the person may have gotten change back from the cashier. Put a quark in it Greetings, On the board today in my calculus class, the teacher had a word problem that has stumped some other teacher, and wanted to see if we could figure it out (I realize this is the PREcalculus board, but I'm sure calculus isn't required to solve the problem). I don't remember it word for word, but I know the gist of it... "You go into a store with a certain amount of dollars and cents. You spend half of your money. After spending half of your money, you now have half as many dollars as you had in cents, and the same number of cents as you had in dollars. How much money did you have to start with?" Just in case that isn't clear, here's an example of what an answer COULD look like... original:$48.66
new: $33.48 See how the$33 in the "new" amount is half of the cents that was in the original, and the cents in the "new" amount is exactly the amount of dollars in the "original"? Of course, $48.66 isn't the right answer, because 33.48 isn't half of$48.66. I imagine I could come up with an answer after some "brute force", but I'd rather not guess and check. Good luck!

I hope the answer isn't $0.00 Oh nevermind I got it.$99.98

KoGs, how was the money split up? How many dollars and how many cents? Also, how was it spent? I'm kinda curious... I'm working this out right now as well.

Originally the person started with $99.98 since. After spending exactly half, he/she was left with$49.99. Note that 49 is exactly half of 98, and 99 = 99.

As for how this person spent it...I dunno. Dinner maybe? :)

Damnit, I was accidentally counting one cent as 100 dollars

I haven't glanced at your post (post #5), so I can solve it by myself first...

EDIT: Looks like I have the same thing as you...

For a second I didn't realize that she could get change from the cashier...lol

So, Put a quark in it, set up the total amount and then try to set half of that equal to the new amount...

Last edited:
moose said:
Damnit, I was accidentally counting one cent as 100 dollars

So, Put a quark in it, set up the total amount and then try to set half of that equal to the new amount...

I understand in theory how it should work, I just don't know how to go about doing it on paper.

Use algebra and define 2 variables. One for cents and one for dollars.
Now you have 2 equations with 2 unknowns. You also have a relationship between the first equation and the 2nd equation.

Well, from what I can see, you can make a system of 4 equations in 5 unknowns:
Let:
D: Number of dollars before spending
C: Number of cents before spending
T: Total amount before spending, measured in cents
D*: Number of dollars after spending
C*: Number of cents after spending.

Thus, I get the four equations:
100D+C=T
100D*+C*=T/2
D*=C/2
C*=D

From what I see, this gives an infinity of whole numbered solutions.

However, the reasonable, but by no means necessary, inequality C<100, along with the tacit assumption T>0 yields a unique solution.

Last edited:
Here's how I did it. Let C be the number of cents initially, D the number of dollars. Of course, the way the problem is worded, as arildno pointed out, I might have, say, $5 in bills and another$2.50 in "cents"! In order to get a single solution we have to assume "a certain amount of dollars and cents" means $D.C whether the "dollars" are in bills or coins! In order to be able to spend half, C must be even. Assume, first, that D is also even. Then D*, how many dollars I have now, is D/2 and C*, how many cents I have now, is C/2. "You now have half as many dollars as you had in cents": D*= C/2 or D/2= C/2 so D= C. "The same number of cents as you had in dollars": C*= D or C/2= D so C= 2D. The only numbers satisfying both are D= C= 0. Now suppose D is odd. Then D*= (D-1)/2 and C*= 50+ C/2. "You now have half as many dollars as you had in cents": D*= C/2 so (D-1)/2= C/2 so C= D-1 or D= C+ 1. "The same number of cents as you had in dollars": C*= D so 50+ C/2= D or 50+ C/2= C+1. Then C/2= 49 so C= 98 and D= 98+ 1= 99. You must have started with$99.98. After spending half of it, you have \$49.99 left. The number of dollars now, 49 is half the number of cents you initially had. The number of cents you have now, 99, is the same as the number of dollars you initially had.

## 1. What is the first step in solving a word problem?

The first step in solving a word problem is to read the problem carefully and identify the key information and variables.

## 2. How do you determine the correct operation to use in a word problem?

To determine the correct operation, you need to identify the relationship between the given information and what needs to be found. For example, if the problem involves adding groups of numbers, you would use addition.

## 3. What should you do if you get stuck while solving a word problem?

If you get stuck while solving a word problem, it is helpful to break it down into smaller, simpler steps. You can also try using a different strategy or asking for help from a teacher or classmate.

## 4. How do you know if your answer is correct in a word problem?

You can check your answer by plugging it back into the original problem to see if it satisfies all the given conditions. You can also use estimation to get an idea of whether your answer is reasonable.

## 5. Are there any common mistakes to avoid when solving word problems?

Some common mistakes to avoid when solving word problems include misreading or misinterpreting the problem, using the wrong operation, and making simple calculation errors. It is important to carefully read and understand the problem and double-check your work for accuracy.

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