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?

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- Jan 20th 2009, 04:47 PMtmac11522Coin 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 AMGrandadCoin problem
Hello tmac11522Number of pennies = $\displaystyle p$. Each penny is worth one cent. So the value of the pennies = $\displaystyle 1\times p = p$ cents

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

Number of dimes = four more than the number of nickels = $\displaystyle (2p + 4)$. 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 $\displaystyle Ap + B$, where $\displaystyle A$ and $\displaystyle B$ are two whole numbers. That will be your final answer, if you don't have any more information.

Can you complete it now?

Grandad