Q: If you have 20 coins worth 1.35 and you only have nickels and dimes how many do you have of each coin?

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ms lynch has 21 coins in nickels and dimes. their total value is 1.65. how many of each coin does she have

20 dimes and 30 nickles.

If one-fifth of the coins are dimes two fifteeths are nickles and two-thirds are pennies how much of each coin do you have for a total of one dollars worth of coins.

There are 83 coins. If there are N nickels then there are (83 - N) dimes. Davida has nickels worth 5N and dimes worth 10(83 - N) but. 5N + 10(83 - N) = 695 5N + 830 - 10N = 695 5N = 830 -695 = 135 therefore N = 135/5 = 27 : D = 83 - N = 83 - 27 = 56 Davida has 27 nickels and 56 dimes.

7 nickels, 4 dimes, 3 quarters

7 nickels, 4 dimes, and 3 quarters.

40 Dimes and 8 Nickles

20 pennies = 20 08 nickels.. = 40 02 dimes... = 20 --- ................ --- 30 ................ 80

20 dimes = 2.00 20 nickels = 1.00 Total = 3.00

There were no bicentennial designs for cents, nickels, or dimes. If your coin is from 1976 it's worth exactly 10 cents.

Chen has 8 nickels, 2 dimes, and 6 quarters.

there r 40 nikels

Different coins are made of different things. Dimes, quarters, and nickels are made of nickel, and pennies are made of copper and nickel.

The combinations depend on how many are dimes and how many nickels.

U.S. or British? The only U.S. coins with the single date of 1976 are dimes, nickels and cents. Post new question

The coins in a cash register amount to $12.50. One coin combination that would produce this total is 40 quarters, 19 dimes, 2 nickels, and 50 pennies. Another combination is 20 quarters, 50 dimes, 45 nickels, and 25 pennies.

There are 11 combinations. Beginning with the largest coin, they are:1) 3 quarters, 1 dime, and 2 nickels2) 2 quarters, 4 dimes, 1 nickel3) 2 quarters, 3 dimes, 3 nickels4) 2 quarters, 2 dimes, 5 nickels5) 2 quarters, 1 dime, 7 nickels6) 1 quarter, 6 dimes, 2 nickels7) 1 quarter, 5 dimes, 4 nickels8) 1 quarter, 4 dimes, 6 nickels9) 1 quarter, 3 dimes, 8 nickels10) 1 quarter, 2 dimes, 10 nickels11) 1 quarter, 1 dime, 12 nickelsIf you do not need to use all three coins, there are 15 more combinations:12) 3 quarters, 2 dimes13) 3 quarters, 4 nickels14) 2 quarters, 9 nickels15) 1 quarter, 7 dimes16) 1 quarter, 14 nickels17) 9 dimes, 1 nickel18) 8 dimes, 3 nickels19) 7 dimes, 5 nickels20) 6 dimes, 7 nickels21) 5 dimes, 9 nickels22) 4 dimes, 11 nickels23) 3 dimes, 13 nickels24) 2 dimes, 15 nickels25) 1 dime, 17 nickels26) 19 nickels

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'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....

Ten (10) nickels and Three (3) quarters.

Depends. First, I'm assuming you mean a dime folder (one where you put in the dimes in holes and not just a generic book about dimes) ' Secondly, does it have coins in it? The folder itself is worth only about $3 if brand-new and is worth about 50 cents if used. But any value comes from the coins themselves. A coin within a Whitman album is worth the same as a coin not in a Whitman album.

No US nickels or half dimes dated 1822, please take another look at the coin