10 dimes 5 quarters
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?
Helen has twice as many dimes as nickels and five more quarters than nickels the value of her coins is 4.75 how many dimes does she have?
12 quarters 6 dimes
7 quarters 2 dimes 4 nickels
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?
The question is incomplete. Please post a new version with the rest of the problem.
Let d= number of dimesq= number of quartersd=2q.25q+.10d=4.50.25q+.10(2q)=4.50.25q+.20q=4.50.45q=4.50q=10d=2qd=2(10)d=20Therefore, there are 10 quarters totaling $2.50 and 20 dimes totaling $2.00.
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
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.
Twelve. Here is how: Using cents rather than decimal dollars: 5n + 10d + 25q = 640 Express the numbers of nickels and quarters in terms of the number of dimes: # of nickels = twice the # of dimes ==> n = 2d # of quarters = 5 less than the # of dimes ==> q = d-5 Substitute for n & q in original equation: 5(2d) + 10d + 25(d-5) = 640 45d - 125 = 640 45d = 765 d = 17 ==> 170 = $1.70 n = 34 ==> 170 = $1.70 q = 12 ==> 300 = $3.00 . . . . . . . . . Total = $6.40
There are fifteen (15) nickels.
There are 10 nickels, 20 dimes and 40 quarters in the cash register. The 10 nickels is 10 x 5 cents or 50 cents. The 20 dimes is 20 x 10 cents or 200 cents. The 40 quarters is 40 x 25 cents or 1000 cents. Converting and adding these, we get $0.50 + $2.00 + $10.00 = $12.50, which is the sum given in the question. Let's work through it. The number of nickels is N, the number of dimes is D and the number of quarters is Q. These are our variables in this problem. We don't know how many of them there are, and their numbers could vary. That's why we call them variables. We might also call them unknowns, too. A nickel is 5 cents, so the value of the nickels is the number of nickels, which is N, times the value of the nickel, which is 5 cents. That's 5N here. A dime is 10 cents, so the value of the dimes is the number of dimes, which is D, times the value of the dime, which is 10 cents. That's 10D here. A quarter is 25 cents, so the value of the quarters is the number of quarters, which is Q, times the value of the quarter, which is 25 cents. That's 25Q here. The sum of the values of the coins was given as $12.50, or 1250 cents, because we are working with coins, whose values are measured in cents. Further, we can write this expression as 5N + 10D + 25Q = 1250 on our way to the answer. Of the last two facts, the first was that there were twice as many dimes as nickels. We could write that as D = 2N because said another way, there are twice the number of dimes as nickels. We might also say that for every nickel, there are 2 dimes, so doubling the number of nickels will give us the number of dimes. The last fact is that there were twice as many quarters as dimes. We could write that as Q = 2D because said another way, thre are twice the number of quarters as dimes. We might also say that for every dime, there are 2 quarters, so doubling the number of dimes will give us the number of quarters. The last two bits of data we have allow us to solve the problem, because the do something special for us. Each bit of data expresses one variable in terms of another. That means we can make substitutions in our expressions for the sum of the values of the coins. Let's put up or original expression, and then do some substitutions. 5N + 10D + 25Q = 1250 This is the original expression. We know that D = 2N, so lets put the 2N in where we see D and expand things a bit. 5N + 10(2N) + 25Q = 1250 5N + 20N + 25Q = 1250 We changed the "look" of the expression, but we didn't change its value. Let's go on. We know that Q = 2D, so lets put that in. 5N + 20N + 25Q = 1250 5N + 20N + 25(2D) = 1250 5N + 20N + 50D = 1250 We're almost there. Remember that D = 2N, and we can substitute that in here. 5N + 20N + 50D = 1250 5N + 20N + 50(2N) = 1250 5N + 20N + 100N = 1250 Groovy! We have substituted variables and now have an expression with only one variable in it! Let's proceed. 5N + 20N + 100N = 1250 125N = 1250 We're close! N = 1250/125 = 10 N = 10 The number of nickels is 10, and because the nickel is 5 cents, the value of these coins is their number times their value, or 10 x 5 cents = 50 cents = $0.50 We were told the number of dimes was twice the number of nickels. This means that since there are 10 nickels, there will 2 x 10 or 20 dimes. And 20 x 10 cents = 200 cents = $2.00 We were also told the number of quarters was twice the number of dimes. This means that since there are 20 dimes, there will be 2 x 20 or 40 quarters. And 40 x 25 cents = 1,000 cents = $10.00 If we add the values of the coins, we should get the $12.50 that we were told was in the register. $0.50 + $2.00 + $10.00 = $12.50 We're in business. The value of each denomination of coins adds up to the given value of all the coins in the register. Piece of cake.
12 quarters and 6 dimes 2(.25x) + .10x = 3.60 50x + 10x = 360 60x = 360 x = 6
You have 22 dimes and 11 pennies.
You have 22 dimes and 11 pennies.
Let the number of nickels, dimes and quarters be n, d, q respectively. Then n +d + q = 30 5n + 10d + 25q = 550 But d = 2n, so: n + 2n + q = 30 => 3n + q = 30 5n + 10(2n) + 25q = 550 => 25n + 25q = 550 => n + q = 22 Which gives two simultaneous equations to solve, resulting in: n = 4, q = 18 So there are 18 quarters (plus 4 nickels and 8 dimes).
16 nickels, 8 dimes, 4 quarters. 25+25+25+25+10+10+10+10+10+10+10+10+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5= 260
8 quarters and 16 nickels
In order to eliminate the decimal point, let us work with cents. So, let $ 4.50 = 450 cents. If the number of nickels is x, then the value of x nickels is 5x cents. Since the number of dimes is twice the number of nickels, then we can write the number of dimes as 2x, and the value of 2x dimes as 10(2x) or 20x cents. So we have: 5x + 20x = 450 25x = 450 divide by 25 to both sides x = 18 the number of nickels 2x = 2 x 18 2x = 36 the number of dimes Thus, Cameron has 18 nickels and 36 dimes. Check: 18 nickels = 18 x 5 cents = 90 cents or $ 0.90 36 dimes = 36 x 10 cents = 360 cents or $ 3.60 0.90 + 3.60 = 4.50? 4.50 = 4.50 True
The X.X1 at the end of the number implies that there could be 1, 11, 21, 31, etc. pennies. Now, we could go through and test the instances. .01 means that there is 1 penny. Therefore, there are 2 dimes to make .31, which does not work. .11 means that there are 11 pennies. Therefore, 22 dimes are required, which make $2.31. This works, so there are 22 dimes.
14 Quarters = $3.50 28 nickels = $1.40 To get this answer you simple add 2 nickles to one quarter which = 35 cents divide 4.90 by 35 which equals 14 14 will be the number of quarters and double that, 28 will be the number of nickels
Yes. Up till 1964 all US dimes, quarters, halves, and dollars were made 90% silver and the amount in each was in the same ratio as the denominations. So a 90% silver half had exactly twice as much silver as a 90% quarter.
One kilogram is about 2.2 pounds. Therefore, a kilogram of dimes is worth more than twice than a pound of dimes.