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Q: The coins in the store's cash register total 4.00 The cash register contains only nickels dimes and quarters There is an equal number of each coin How many dimes are in the cash register?

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The minimum number of quarters, pennies, and nickels needed to make up 123 cents is 4 quarters, 4 nickels, and 3 pennies.

There are five nickels (5¢) in a quarter (25¢) so the rule is: 1) Divide the number of nickels by 5 to get the number of quarters 2) The remainder, if any, is the number of nickels left over For example, if you have 17 nickels, 17/5 = 3 rem 2, so that means you have 3 quarters with 2 nickels remaining. To confirm, 17 nickels are worth a total of 85¢ (5 * 17); 3 quarters = 75¢ so 10 cents - i.e. 2 nickels - would be left over.

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.

19 Quarters & 11 Nickels Let X = number of quarters & Y = number of nickels So, X = Y +8 (8 more quarters than nickels) 0.25*X + 0.05*Y = 5.30 Substitute first equation for x into second equation: 0.25*(Y+8) + 0.05*Y = 5.30 0.25Y + 2 + 0.05Y = 5.30 0.30Y + 2 = 5.30 0.30Y = 3.3 Y = 11 nickels, so the number of quarters X = 19 We can check these values with the second equation: 0.25*19 + 0.05*11 4.75 + 0.55 = 5.30

25

Call the number of nickels n and the number of quarters q. From the problem statement, q = n + 78 and 25q + 5n = 3240. Substituting the first equation into the second yields 25(n + 78) + 5n = 3240, or 30n = 3240 - 1950 or n = 1290/30 = 43 nickels.

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

It depends on the size of the box and the number of nickels in it.

The easiest way is to convert everything to cents first. - Seven nickels = 35 cents - Two quarters = 50 cents - so 7 nickels to 2 quarters is the same as 35/50, or 0.7 as a decimal number.

3 quarters

20 - 10 of each

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.

Let X equal the number of quarters X * 25 is the value of the quarters ((X+8) * 5) is the value of the nickels25X + 5X + 40 = 610 so 30X + 40 = 610 .subtract 40 from both sides , divide both sides by 30X = 19There are 19 quarters and 27 nickels in the piggy bank

You would have to know the exact number of coins per denomination. Modern U.S. quarters weigh 5.67 grams. Dimes 2.27 grams. Nickels 5 grams.

well it depends on how you want it. If you want it as, like having 4 quarters and counting one way for four quarters then counting another as having the same 4 quarters but in different order (if you understood any of that) there is 293 possibilities. but if you want it the other way, We can use either 0, 1, 2, 3, or 4 quarters. If we use 0 quarters, we can use from 0 up to 10 dimes, and the rest, if any, in nickels. That accounts for 11 ways. If we use 1 quarter, we can use from 0 up to 7 dimes, and the rest in nickels. That accounts for 8 ways. If we use 2 quarters, we can use from 0 up to 5 dimes, and the rest, if any, in nickels. That accounts for 6 ways. If we use 3 quarters, we can use from 0 up to 2 dimes, and the rest in nickels. That accounts for 3 ways. If we use 4 quarters, that's the whole dollar, so that accounts for 1 way. So the total number of ways = 11+8+6+3+1 = 29 You weren't asked to list them, but here is the list of all 29 ways: 1. 0 quarters, 0 dimes, and 20 nickels. 2. 0 quarters, 1 dime, and 18 nickels. 3. 0 quarters, 2 dimes, and 16 nickels. 4. 0 quarters, 3 dimes, and 14 nickels. 5. 0 quarters, 4 dimes, and 12 nickels. 6. 0 quarters, 5 dimes, and 10 nickels. 7. 0 quarters, 6 dimes, and 8 nickels. 8. 0 quarters, 7 dimes, and 6 nickels. 9. 0 quarters, 8 dimes, and 4 nickels. 10. 0 quarters, 9 dimes, and 2 nickels. 11. 0 quarters, 10 dimes, and 0 nickels. 12. 1 quarter, 0 dimes, and 15 nickels. 13. 1 quarter, 1 dime, and 13 nickels. 14. 1 quarter, 2 dimes, and 11 nickels. 15. 1 quarter, 3 dimes, and 9 nickels. 16. 1 quarter, 4 dimes, and 7 nickels. 17. 1 quarter, 5 dimes, and 5 nickels. 18. 1 quarter, 6 dimes, and 3 nickels. 19. 1 quarter, 7 dimes, and 1 nickel. 20. 2 quarters, 0 dimes, and 10 nickels. 21. 2 quarters, 1 dime, and 8 nickels. 22. 2 quarters, 2 dimes, and 6 nickels. 23. 2 quarters, 3 dimes, and 4 nickels. 24. 2 quarters, 4 dimes, and 2 nickels. 25. 2 quarters, 5 dimes, and 0 nickels. 26. 3 quarters, 0 dimes, and 5 nickels. 27. 3 quarters, 1 dime, and 3 nickels. 28. 3 quarters, 2 dimes, and 1 nickel. 29. 4 quarters, 0 dimes, and 0 nickels. Hope this helped!

There are fifteen (15) nickels.

apples

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.

It depends on the number of nickels.

No. It's almost (17*5 cents + 10 cents = 95 cents) but not exactly. If you think about it for a few seconds, a dollar contains an even number of nickels (20), and a dime is also an even number of nickels (2) but 17 is an odd number. An odd number (17) plus an even number (2) is always an odd number so the total can't possibly also be an even number.

1/35 or 2.86%

This problem can be solved by solving the system of equation. Total worth of coins: $2.65 Total number of coins: 33 n= number of nickels q= number of quarters since we know that there are 33 coins total, we can set the equation like this: number of nickels + number of quarters = total number of coins => n+q=33 We also know that the worth of these coins is $2.65. each nickel is worth of $0.05 each quarter is worth of $0.25 therefore we can set the equation: 0.05 x number of nickels + 0.25 x number of quarters = total worth of coins. 0.05n+0.25q=2.65 However, for convienience, we should multiply the equation above by 100 to get rid of decimals. Thus it is 5n+25q=265 you will now have a following set of 2 equations: n+q=33 5n+25q=265 Use the SUBSTITUTION METHOD to solve either n or q for solving n: (replace q with n if you're willing to solve q instead) n+q=33 => n=33-q (since n is equal to 33-q, we can -q -q substitue n in the other equation.) 5(33-q)+25q=265 => 165-5q+25q=265 => 20q=100 => q=5 -165 -165 /20 /20 There are 5 quareters as a result.(or 28 nickels) since you know that q=5 you can substitute q in the first equation. n+(5)=33 => n=28 - 5 -5 therefore, there are 5 quarters and 28 nickels. ELIMINATION METHOD: n x -5 + q x -5 = 33 x -5 => -5n-5q=-165 5n+25q=265 + 5n+25q=265 ------------- 20q=100 => q=5 /20 /20 Or simply we can say: if we have x quarters, we have .25x value of them. So the value of nickels will be 2.65 - .25x. Since we have 33 coins, and x quarters, then the number of nickels will be 33 - x. So the value of all nickels would be also .05(33 - x). Thus, we have:2.65 - .25x = .05(33 - x)2.65 - .25x = 1.65 - .05x2.65 - 1.65 - .25x + .25x = 1.65 - 1.65 - .05x + .25x1 = .20x1/.20 = .20x/.20x = 5 the number of quarters 33 - x= 33 - 5= 28 the number of nickels. Thus, we have 5 quarters and 28 nickels.

The idea is to write two equations, one for the number of coins, one for the amount of money. Then solve the equations.Assuming "n" is the number of nickels, and "q" the number of quarters, the equations for the coins, of course, is quite simply: n + q = 64 And the equation for the money (I'll use cents; you can just as well use dollars if you prefer): 5n + 25q = 740 You can solve the first equation for "n", then replace that in the second equation.

More information is needed. Please post a new, separate question giving the total number of cents.

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

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