D = dimes
Q = quarters
D + Q = 70
Q = D + 30
Substitute
D + D + 30 = 70
2D = 70 - 30 = 40
D = 20
Hence Q = 20 + 30 = 50
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.
82 dimes
This question cannot be answered.Assume there is1 nickel. There must be 3 more than that in quarters - that makes4 quarters and there are13 dimes---- that adds up to18
The 8 coins are: 3 quarters, 2 dimes, 1 nickel and 2 pennies.
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
70 - 30 = 40, 40/2 - 20, so there are 20 dimes and 50 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.
Let ( q ) be the number of quarters and ( d ) be the number of dimes. We have two equations based on the problem: ( q + d = 102 ) (the total number of coins) ( 25q + 10d = 1710 ) (the total value in cents) Solving these equations, we find that there are 6 quarters and 96 dimes.
where x=number of quarters, and 110-x=number of dimes (because the number of dimes is 110 minus the number of quarters): .25(x) + .10(110-x) = 18.50 .25x + 11 - .10x = 18.50 .25x - .10x = 7.50 .15x = 7.50 x = 50 50 quarters, 60 dimes
Mula has 16 dimes and 11 quarters.
dimes : all coins = 6 : 12+6+18 = 6 : 36 = 1×6 : 6×6 = 1 : 6 Dimes are ⅙ of all coins.
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
To make $2.50 from quarters (worth 25 cents) and dimes (worth 10 cents), we can set up a system of equations. Let q represent the number of quarters and d represent the number of dimes. The equations would be 25q + 10d = 250 (representing the total value in cents) and q + d = 25 (representing the total number of coins). Solving these equations simultaneously, we find that there are 6 ways to make $2.50 using quarters and dimes.
The man has 13 dimes and 7 quarters, which equate to $3.05 $1.30 + $1.75 If he had 13 quarters and 7 dimes he would have $3.95 $3.25 + 70c The method used to work this out was dividing 90c by the difference between the value of a dime and a quarter- i.e. 15c . 90 / 15 = 6, so of 20 coins, 6 more were dimes than quarters. Subtract 6 from 20, then halve the result = 7 the lower number is 7, the higher number is 7+6, = 13. Please also note that 'Quarters' is spelled with 2 'r's
82 dimes