Q: If you have 3.00 in dimes and nickels. The number of dimes you have is the same number of nickels. How many of each type of coin do you have?

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You would have 7 nickels and 13 dimes. (7 x $0.10) + (13 x $0.05) = $1.35

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infinite amounts of money... unless you have a limited amount of each coin

There's no single answer; it depends on how many of each coin you have. For example you could have any of the following:400 quarters1000 dimes2000 nickels500 dimes and 1000 nickels200 quarters, 400 dimes, and 200 nickelsand so on....

there r 40 nikels

Since each dime is equivalent to two nickels, you need 80 nickels.

There's no specific answer, except to say that the number of nickels must be even (otherwise the total would end in 5 rather than 0). You could have combinations like50 nickels and no dimes40 nickels and 10 dimes25 dimes and no nickels22 dimes and 6 nickelsand so on. Each one, and many other possibilities, all add up to $2.50

It's not possible to give a specific answer without knowing how the numbers of each coin are related. Otherwise there are hundreds of possible combinations; for example278 quarters, 0 dimes, 0 nickels277 quarters, 0 dimes, 5 nickels277 quarters, 1 dime, 3 nickels277 quarters, 2 dimes, 1 nickel276 quarters, 0 dimes, 10 nickelsand so on

You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.

There would be 15 of each coin. 15(.25) +15(.10)+ 15(.05)= 3.75+1.50+.75= 6.00

The volume of a dime is roughly half the volume of a nickel, so any specific number of dimes would take up about half as much space as the same number of nickels*. In other words, a half a trunk of dimes would have at least the same number of that coin as a full trunk of nickels. However each dime is worth twice as much so the half-trunk of dimes would be worth more. * Because dimes are smaller they'd be able to pack somewhat more tightly than nickels so you could probably get slightly more dimes into the half-trunk. However that doesn't affect the answer; it simply means the half-trunk would be worth even more.

The number of dimes can be anything between zero and 10 = 11 choices. For each choice, there is only one way to fill in the difference with nickels. So there are 11 ways. (This doesn't consider the number of different sequences in which the individual nickels and dimes can be selected, only the composition of the final bag of coins.)

A dime is 10 cents and a nickel is 5 cents, so each dime is the same as 2 nickels. That means 49 dimes are equivalent to 49*2 = 98 nickels.

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