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**Neha****Member**- Registered: 2006-10-11
- Posts: 173

there were 35 nickels , dimes , and silver dollars in the pile.

their value was $12.25 . how many coins of each kind were there if there were twice as many silver dollars as nickels ?

N + D + S = 35

0.05N + 0.1D + S = 12.25

subtract -0.05N both sides

0.1D + S = 12.2

subtract 0.1D from both sides

S = 12.1

0.05N + 0.1D + (12.1) = 12.25

subtract 12.1 from both sides

0.05N + 0.1D = 0.15

subtract 0.1D from both sides

0.05N = 0.05

N = 1

0.05(1) + 0.1D + (12.1) = 12.25

0.05 + 0.1D + 12.1 = 12.25

subtract 12.1 from both sides

0.05 + 0.1D = 0.15

subtract 0.05 from both sides

0.1D = 0.1

D = 1

S = 12.1

N = 1

D = 1

0.05N + 0.1D + S = 12.25

0.05(1) + 0.1(1) + (12.1) = 12.25

0.05 + 0.1 + 12.1 = 12.25

12.25 = 12.25

CORRECT

but it is not 35cents????????

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**pi man****Member**- Registered: 2006-07-06
- Posts: 251

N + D + S = 35

0.05N + 0.1D + S = 12.25

subtract -0.05N both sides

0.1D + S = 12.2

subtract 0.1D from both sides

S = 12.1

You were doing fine until you started subtracting. When you subtract 0.05N from both sides, it does "cancel out" the existing 0.05N on the left side of the equation. But on the right side you would have 12.25 - 0.05N. You simply subtracted .05 rather than subtracting .05N.

You need to use the statement that there are twice as many silver dollars as nickles.

S = 2N

Going back and doing some substitution:

N + D + 2N = 35

3N + D = 35 (A)

0.05N + 0.1D + S = 12.25

0.05N + 0.1D + 2N = 12.25

2.05N + 0.1D = 12.25 (B)

Let's mutliple both sides of the equation I denoted as (B) by 10

20.5N + D = 122.5 (B1)

Now subtract A from B1

17.5N = 87.5

N = 5

So you have 5 nickles ($.25) and 10 silver dollars ($10). Since you have 35 coins total, you must have 20 dimes ($2). And that works out: $10 + $2 + .25 = $12.25

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That's because you've just guessed the number of nickels and dimes there are and then worked out how many silver dollars there must be to make the total value work. You don't even have a whole number of silver dollars.

You need to find values of N, D and S that will work for:

N+D+S = 35

0.05N+0.1D+S = 12.25

S=2N

First, I'm going to multiply the middle equation by 20, to get rid of all those decimals because they're annoying me.

N+2D+20S = 245

And now we have 3 simultaneous equations with 3 variables.

We are already given S in terms of something else, so we can use the 3rd equation into the other 2 to make:

3N+D = 35

2D+41N = 245

Multiplying the first equation by 2 gives 6N+2D = 70

Subtracting the two equations gets you 35N = 175

Therefore, N = 5. (There are 5 nickels)

You are told that there are twice as many silver dollars as nickels, so that means there are 10 silver dollars.

And that's 15 coins accounted for and you wanted 35, which means there must be 20 dimes.

Check: 10 dollars + 20 dimes + 5 nickels = $10 + $2 + $0.25 = $12.25.

Success!

*Bah, pi man got there first. Well, at least it means that we've confirmed each other's answers.*

Why did the vector cross the road?

It wanted to be normal.

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**pi man****Member**- Registered: 2006-07-06
- Posts: 251

Bah, pi man got there first. Well, at least it means that we've confirmed each other's answers.

Mathsyperson - That's happened to me more than once also. And I still have less than 100 posts. Maybe the next time Mathisfun is doing someweb redesign work, there could be another option added next to "post reply" that's something like "Will post reply". And then the thread could be flagged somehow with something like "Pi Man is working on a reply". The flag would be cleared when I do post my answer or after some time limit (1-2 hours?). Being flagged wouldn't prevent anyone else from replying, just informing them that someone(s) else is already working on it.

Or I guess for solutions which we might be spending a few minutes working out and typing up, we could just post a reply which states we're working on it? Too simple!

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Some other forums have a list of users who are viewing whatever topic you're looking at, with emphasis on people who are posting.

eg.

right or wrong

Neha

there were 35 nickels , dimes , and silver dollars in the pile.

their value was $12.25 . how many coins of each kind were there if there were twice as many silver dollars as nickels ?Viewing this topic: mathsyperson,

pi man

So then I'd see that you were replying and so wouldn't reply myself. I'm not sure how easy it would be to put that into this forum though, as it's PunBB. It we could, it would be extremely useful.

Why did the vector cross the road?

It wanted to be normal.

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**Neha****Member**- Registered: 2006-10-11
- Posts: 173

hard to do a bit hard to do .. but thanks

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