To answer a question like this,you need to think a little bit. The question is really "What combination of numbers of coins will add up to 70, and has a difference of 30 between the two numbers?" For example, lets say old Tanisha had 10 dimes and 60 quarters - 10 plus 60 indeed adds up to 70, for the number of coins we're looking for, but the difference between 10 and 60 is 50, not quite the 30 we are looking for. But we're on the right track. How about 20 and 50? Hmm... adds up to 70... and has a difference of 30. So ol' Tanisha has 50 quarters and 20 dimes. Not a bad stash.
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.
Mula has 16 dimes and 11 quarters.
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
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
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
16 % of the coins are dimes. 4 of a total of 25.
The five coins are 3 quarters and 2 dimes.
there are 23 quarters 37 dimes to do this use simultaneous equations to make this simpler i will assign variables to amount of quarters and dimes x-quarters y-dimes you know you have 60 coins, so amount of quarters plus dimes is 60 x+y=60 you also know the amount is equal to $9.45 since quarters are $0.25 and dimes are $0.10 another equation can be made 0.25x+0.10y=9.45 this equation shows what you do when you count your change now solve the first equation for a variable using algebra x=60-y you can then plug this new value into your second equation 0.25(60-y)+0.1y=9.45 you can now solve for y, which is the number of dimes y=37 now plug this number back into x=60-y to get the amount of quarters x=60-(37) x=23 23 quarters 37 dimes