He has 165 cents.
$1.65
Pat has 3.80 in nickels and dimes. She has 51 coins in all. Pat has 26 nickels of the 51 coins.
If n is the number of nickels and d the number of dimes, then the equations are:n + d = 160 (total number of coins) 5n + 10d = 1050 (total value). And I have thought through to the answer.
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.
A standard US roll of nickels contains 40 coins.
One bank box contains $100 worth of nickels, which is 2,000 coins.
3 dimes, 2 nickels, and 3 pennies
Pat has 3.80 in nickels and dimes. She has 51 coins in all. Pat has 26 nickels of the 51 coins.
You could have fourteen dimes and one nickel or twenty-seven nickels and one dime or any of the combinations in between. This problem needs to know the number of coins.
20 - 10 of each
47 Quarters 83 Nickels
26 nickels
Since the smallest of these currency values (the nickel) is equal to 5 cents, the number of five cent coins that go into a dollar is equal to 20. It is impossible to have a combination of 35 nickels and dimes whose sum is exactly equal to a dollar.
Ten it each group.
12 x 100/80 ie 15%
You cannot. With US coins, you are limited to pennies, nickels, and dimes. 5 pennies = 5 cents 4 nickels = 20 cents Total of 9 coins and 25 cents. It is impossible to use any greater number of coins and impossible to use 10 coins in today's US currency.
Coins are weighed in grams so to start, you need to know that a US ounce is 28.35 grams. US nickels weigh 5.00 gm so a pound would be 28.35 / 5 = 5.7, or 6 coins rounded to the nearest whole number. Current Canadian nickels weigh 3.95 gm; 28.35 / 3.95 = 7.2, or 7 coins rounded to the nearest whole number.
If n is the number of nickels and d the number of dimes, then the equations are:n + d = 160 (total number of coins) 5n + 10d = 1050 (total value). And I have thought through to the answer.