# Coins in a jar word problem

• Jul 23rd 2010, 10:46 AM
rex
Coins in a jar word problem
I just joined today, thank you for making this forum. I will be coming here often, again thank you all so much.

Here is the problem.
The problem starts with coins in a jar and I must come up with the number of coins of each denomination. The number of quarters is unknown, number of dimes unknown, number of nickels unknown, and number of pennies unknown.

Here on the known parts of the equation.
- There are a total of 471 coins worth $43.13. - Two Susan B. Anthony dollars, a bunch of quarters, a whole lot of dimes, a good number of nickels and a fair quantity of pennies. - The number of nickels is equal to the number of quarters plus the number of dollars.....{so I figure this, n = q + 2} The number of pennies is only one more than the number of dimes. .. {so I figure this, p + 1 = d} A. Generate and solve the linear equations for this problem. B. How many coins, of each type (Susan B Anthony dollars, quarters, dimes, nickels, and pennies) are in the jar? Thank you all in advance. I will make a second post right after this one showing what I have so far. Please tell me what I do and do not have correct. • Jul 23rd 2010, 10:51 AM rex pennies, p = d - 1 dimes, d = p + 1 nickels, n = q + 2 quarters, q = n - 2 and the number of dollar coins is known, there are 2 susan b anthony coins. So I think from this I make my main equation which is this, p + d + n + q + 2 = 471 Once I have the number of coins for each unknown I will have their worth, the total worth of all the coins in the jar is$43.13.
so now I believe I use this equation,
p + d + n + q + 2.00 = $43.13 I know I should just now put all my values listed above into the larger equation but am running into trouble. Please some guidance everybody, thanks again in advance. • Jul 23rd 2010, 11:43 AM Soroban Hello, Rex! Welcome aboard! Quote: There are a total of 471 coins worth$43.13 in the jar: two Susan B. Anthony dollars,
some quarters, some dimes, some nickels and some pennies.

The number of nickels is equal to the number of quarters plus the number of dollars.
The number of pennies is one more than the number of dimes.

A. Generate and solve the linear equations for this problem.

B. How many coins of each type are in the jar?

The jar contains:

. . $\displaystyle \begin{array}{cc}2 & \text{dollars} \\ Q & \text{quarters} \\ D & \text{dimes} \\ Q+2 & \text{nickels} \\ D+1 & \text{pennies} \end{array}$

We have this chart:

. . $\displaystyle \begin{array}{c|cc|ccc|}& \text{coins} & && \text{value} \\ \cline{2-3} \cline{4-5} & 2 & \text{dollars} && 200 \\ & Q & \text{quarters} && 25Q \\ & D & \text{dimes} && 10D \\ & Q+2 & \text{nickels} && 5(Q+2) \\ & D+1 & \text{pennies} && D+1 \\ \cline {2-3} \cline{4-5} \text{Total:} & 2Q+2D+5 & & & 30Q + 11D + 211 \end{array}$

We have equations: .$\displaystyle \begin{array}{ccc}2Q + 2D + 5 &=& 471 \\ 30Q + 11D + 211 &=& 4313 \end{array}$

Solve the system: .$\displaystyle \begin{array}{ccc} Q + D &=& 233 \\ 3Q + 11D &=& 4102 \end{array}$

. . and get: .$\displaystyle \begin{Bmatrix} 2\text{ dollars} \\ 81 \text{ quarters} \\ 152\text{ dimes} \\ 83\text{ nickels} \\ 153\text{ pennies} \end{Bmatrix}$

• Jul 23rd 2010, 12:07 PM
rex
Thank you so much Soroban.
I was complicating it more and using four variables, you showed me that I really only had to worry about two. Thanks again.

Your chart was very helpful to me. How do I go from the equation, which I now know how to build, to finding each coin type?
I go from; 2Q + 2D + 5 = 471
to 1; Q = -D -233
to 2; 2(-D -233) + 2D = 471 and this gives me no answer cause I'm left with nothing.

Where am I messing up? I still don't really understand how to go from equation to coins.
• Jul 23rd 2010, 08:01 PM
Wilmer
Quote:

Originally Posted by rex
I go from; 2Q + 2D + 5 = 471
to 1; Q = -D -233

Rex, are you learning on your own, or do you attend math classes?
That should be Q = 233 - D.

Have a peek here:
SOLVING EQUATIONS
• Jul 23rd 2010, 08:20 PM
rex
I have not been in a math class for some long time.
But I do see my obvious error now. Thanks Wilmer.

Still, I don't quite understand how I go from Q = 233 - D to all the numbers of coins.
I see that the equations work to get it. But my seeing the whole process is still off.
• Jul 25th 2010, 07:46 PM
rex
2Q+2D+5=471
Q=-D+233
D=-Q+233

30Q+11D+211=4313
Q = (-11D + 4102) / 30
D = (-2 (15Q - 2051)) / 11

This is what I come up with still.
How do you go from the equations and systems you have, to finding the number of coins and the values?
I am pretty sure I am doing algebra correctly here. Given the equations that is all I can come up with. So do I split the 233 in two values? One is +2 than the other? I am still fully confused on the problem and would really appreciate more explanation on this problem. Thank you.
• Jul 25th 2010, 09:17 PM
Wilmer
Ok Rex, time to get you out of your misery!

Forget about the 2 $1 coins: we have 469 coins totalling$41.13

Always a good idea to number equations; easier to refer to; we have:
Q + D + N + P = 469 [1]
N = Q + 2 [2] : The number of nickels is equal to the number of quarters plus 2
P = D + 1 [3] : The number of pennies is only one more than the number of dimes

Next step is substitute [2] and [3] in [1]:
Q + D + Q+2 + D+1 = 469
2Q + 2D = 466
Q + D = 233
D = 233 - Q [4]

Next step is getting P in terms of Q; substitute [4] in [3]:
P = 233-Q + 1
P = 234 - Q [5]

Now that we have D, N and P in terms of Q, we do the "value" work:
.25Q + .10D + .05N + .01P = 41.13 ; multiply by 100 to get rid of decimals:
25Q + 10D + 5N + 1P = 4113 [6]

Now, substitute [4], [2] and [5] in [6]:
25Q + 10(233 - Q) + 5(Q + 2) + 1(234 - Q) = 4113
25Q + 2330 - 10Q + 5Q + 10 + 234 - Q = 4113
19Q = 1539
Q = 1539/19 = 81 : so we have 81 quarters.

[2]: N = 81 + 2 = 83
[4]: D = 233 - 81 = 152
[5]: P = 234 - 81 = 153

Hope that was clear...
• Jul 25th 2010, 09:38 PM
rex
omg, thank you so much Wilmer.

That i get a whole lot easier.
I'm still going to do it on paper and force the learning into my brain.

THANK YOU!