He would have 2 quarters and 8 dimes. (2 x 0.25) + (8 x 0.10) = 0.50 + 0.80 = 1.30
Chen has 8 nickels, 2 dimes, and 6 quarters.
4 Qs 12 Dimes 20 Nickles
7 quarters 2 dimes 4 nickels
7 nickels, 4 dimes, 3 quarters
7 nickels, 4 dimes, and 3 quarters.
Chen has 8 nickels, 2 dimes, and 6 quarters.
11 ways. You need an even number of quarters and the rest dimes; so 0,2,4,6,8,10,12,14,16.18 or 20 quarters with the balance dimes in each case.
4 Qs 12 Dimes 20 Nickles
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
7 quarters 2 dimes 4 nickels
70 - 30 = 40, 40/2 - 20, so there are 20 dimes and 50 quarters
12 quarters 6 dimes
30 quarters equals $7.50 42 dimes equals $4.20 That adds up to 72 coins with a value of $11.70
Sally has 76 Quarters and 24 Dimes. I hope that I am not doing your homework for you =)
7 nickels, 4 dimes, 3 quarters
it took a few mins to figure it out but its 36 quarters and 39 dimes
Each quarter is 25, and he has X quarters. Each dime is 10, and he has 2X dimes. Each nickel is 5. He has 12 coins in total, and all of the ones that are not quarters or dimes are nickels. Since quarters and dimes have to come in sets of 3, that's 12-3X. 25X + 10(2X) + 5(12-3X) = 120 25X + 20X - 15X + 60 =120 30X + 60 = 120 30X = 60 X = 2 He has 2 quarters, 4 dimes, and 6 nickels. To double check let's add them up: (2 * 25) + (4* 10) + (6 * 5) = 120