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.
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?
The coins in a cash register amount to $12.50. One coin combination that would produce this total is 40 quarters, 19 dimes, 2 nickels, and 50 pennies. Another combination is 20 quarters, 50 dimes, 45 nickels, and 25 pennies.
7 quarters 2 dimes 4 nickels
In the US, we use pennies, nickels, dimes and quarters.
The question suggests that there are 24 coins. 13 of them are pennies, 14 are nickels, and 16 are dimes and the rest are quarters. To answer this question, One would add the number of pennies, nickels, and dimes and subtract the sum of those coins from 24. The difference of the two numbers would be the amount of quarters. However, 13+14+16=43. 24-43= -19 There can't be -19 quarters.
7 nickels, 4 dimes, 3 quarters
4 quarters. 100 pennies. 10 dimes. 20 nickels
you have 3 quarters 31 dimes and 65 pennies
4 Qs 12 Dimes 20 Nickles
16 % of the coins are dimes. 4 of a total of 25.
ten dimes. or two quarters, six nickels and two dimes.
10 Nickels 2 Quarters = $1.00 4 Dimes 7 Nickels 1 Quarter = $1.00 8 Dimes 4 Nickels = $1.00
7 nickels, 4 dimes, and 3 quarters.
12 quarters,7 dimes, and 3 nickels
Dimes and quarters, yes. Nickels, no. 1964 was the last year for silver dimes and quarters, and nickels only contained silver during WWII.
Chen has 8 nickels, 2 dimes, and 6 quarters.
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
There are 2 solutions (if you include the non-use of quarters): 1 Quarter, 2 Dimes, 2 Nickels, 45 Pennies No Quarters, 2 Dimes, 8 Nickels, 40 Pennies
2 quarters, 4 dimes, and 2 nickels.
All coins such as quarters, nickels, dimes, and half dollars
2 quarters 5 dimes 10 nickels 50 pennies
111 quarters, zero dimes, zero nickels 110 quarters, two dimes, one nickel 109 quarters, four dimes, two nickels