The algebraic formula for this problem has the form
25x + 10(35-x) = 515
where x is the number of quarters (times 25 cents) and (35-x) is the number of dimes (times 10 cents)
With the result that 25x + 350 - 10x = 515, yielding 15x = 165, and x = 11
So James has 11 quarters and 24 dimes, (2.75 and 2.40) equalling $ 5.15 total.
8 of them.
If all coins were dimes he would have $1.30. Every quarter that replaces a dime increases the total by 15c. The total has to be increased by $1.20 which is 15c x 8. He has 8 quarters and 5 dimes.
11 dimes.
7 nickels, 4 dimes, and 3 quarters.
If 5 dimes to every 8 quarters that is 5 out of every 13 coins dimes and 8 of 13 coins quarters 5/13 x 520 = 200 dimes
8 of them.
8 quarters, 5 dimes
If all coins were dimes he would have $1.30. Every quarter that replaces a dime increases the total by 15c. The total has to be increased by $1.20 which is 15c x 8. He has 8 quarters and 5 dimes.
The 8 coins are: 3 quarters, 2 dimes, 1 nickel and 2 pennies.
11 dimes.
7 nickels, 4 dimes, and 3 quarters.
If 5 dimes to every 8 quarters that is 5 out of every 13 coins dimes and 8 of 13 coins quarters 5/13 x 520 = 200 dimes
If Keoki has 14 quarters and 8 dimes (for a total of 22 coins), she has $3.50 and $0.80 or $4.30 in coins. If Keoki has 15 quarters and 7 dimes (for a total of 22 coins), she has $3.75 and $0.70 or $4.45 in coins. If Keoki has 22 coins that are all dimes and quarters and their value in total is $4.35 as asked, there isn't a combination of coins that will permit her to have both 22 coins and $4.35 worth of coins.
Mula has 16 dimes and 11 quarters.
There are three possible combinations: 1 quarter and 13 dimes 3 quarters and 8 dimes 5 quarters and 3 dimes
30 quarters equals $7.50 42 dimes equals $4.20 That adds up to 72 coins with a value of $11.70
82 dimes