Work in cents.
10d + 25q = 355 and d + q = 25 so d = 25 - q
Substitute this last: 10(25 - q) + 25q = 355 ie 250 - 10q + 25q = 355
ie 15q = 105 so q = 7, making d = 18
Check: 7 x 25c + 18 x 10c = $1.75 + $1.80 = $3.55
And there you have it...
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.
Peggy had three times as many quarters as nickels. She had $1.60 in all. How many nickels and how many quarters did she have?
You need to define variables for the different types of coins, write the corresponding equations, then solve them. One equation for each fact. Here are the equations:5N + 10D + 25Q = 1250 D = 2N Q = 2D
4 quarters and one nickel
Nickles - 10 $0.50dimes - 20 $2.00quarters - 40 $10.000.50+2.00+10.00 = $12.50 containing 40 quarters---Here's how to solve this with Algebra :Let N be the number of nickels, so that2N is the number of dimes, and2(2N) or 4N is the number of quarters.A nickel is 5 cents, a dime is 10 cents, and a quarter is 25 cents,and the total in the cash register is $12.50Multiplying by their cents values, we have5(N) plus 10(2N) plus 25 (4N) = 12505N + 20N + 100N = 1250125 N = 1250N= 10So the number of nickels is 10, dimes 20, and quarters 40.
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.
The coins in the store's cash register total $12.50. The cash register contains only nickels, dimes, and quarters. There are twice as many dimes as nickels. There are also twice as many quarters as dimes. How many quarters are in the cash register?
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.
Eighteen
15 quarters, 3 dimes
16 % of the coins are dimes. 4 of a total of 25.
8 quarters, 5 dimes
There are 39 combinations of dimes and quarters that will total 19.75 from 1 quarter and 195 dimes to 77 quarters and 5 dimes.
111 quarters, zero dimes, zero nickels 110 quarters, two dimes, one nickel 109 quarters, four dimes, two nickels
If you have 37 coins and 18 are quaters ($4.50) and 19 are dimes ($1.90) then the amount is $6.40.
Not possible, the maximum value of 1 dime and 56 quarters is $14.10. Increasing the number of dimes reduces the total value by 15c per dime