# Coin problem

• Jan 20th 2009, 04:47 PM
tmac11522
Coin problem
The value (in cents) of the change in a purse that contains twice as many nickels as pennies, four more dimes than nickels, and as many quarters as dimes and nickels combined; p = number of pennies?
• Jan 21st 2009, 01:20 AM
Coin problem
Hello tmac11522
Quote:

Originally Posted by tmac11522
The value (in cents) of the change in a purse that contains twice as many nickels as pennies,

Quote:

Originally Posted by tmac11522
four more dimes than nickels, and as many quarters as dimes and nickels combined; p = number of pennies?

Number of pennies = $p$. Each penny is worth one cent. So the value of the pennies = $1\times p = p$ cents

Number of nickels = twice as many as number of pennies = $2p$. Each nickel is worth 5 cents. So the value of the nickels = $5\times 2p = 10p$ cents

Number of dimes = four more than the number of nickels = $(2p + 4)$. Each dime is worth 10 cents. So the value of the dimes = ??? cents.

Number of quarters = number of dimes plus number of nickels = ??. Each quarter is worth 25 cents, so the value of the quarters is ??? cents.

Add together all four value totals; remove the brackets and simplify the result. You'll then arrive at an expression like $Ap + B$, where $A$ and $B$ are two whole numbers. That will be your final answer, if you don't have any more information.

Can you complete it now?