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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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∙ 14y ago
There are 10 of each coin in the cash register. 10 quarters (2.50) plus 10 dimes (1.00) plus 10 nickels (.50) = 4.00 The formula to solve this problem is: (X x .25) + (X x .10) + (X x .05) = 4.00 X (.25+.10+.05) = 4.00 X = 4.00/.4 = 10
Woah that is really hard math
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∙ 14y ago
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How do I write the rule to find the number of quarters for any number of nickels?
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.
What is the ratio of 7 nickels to 2 quarters?
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.
A piggy bank has quarters and nickels in it There are eight more nickels than quarters if the total value of the coins is 6.10 how many of each coin are there?
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
A jar contains 18.50 in dimes and quarters There are 110 coins in the jar How many quarters are there?
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
The coins in a cash register total 12.50 and there are only nickels dimes and quarters. There are twice as many dimes as nickels and there are twice as many quarters as dimes. How many of each coin?
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.
A cash register contains 12.50 and there are only nickels dimes and quarters There are twice as many dimes as nickels and twice as many quarters as dimes How many quarters are in the cash register?
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.
What is the minimum number of quarters pennies and nickels needed to make up 123 cents?
The minimum number of quarters, pennies, and nickels needed to make up 123 cents is 4 quarters, 4 nickels, and 3 pennies.
How do I write the rule to find the number of quarters for any number of nickels?
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.
If there are twice as many nickels as quarters and the coins are worth 4 dollars and 90 cents then how many quarters are there?
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
Provided you have quarters and nickels together equaling 5.30 dollars in which there are 8 more quarters than nickels how many nickels and quarters are there?
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
What number contains 5 nickels?
25
What is the ratio of 7 nickels to 2 quarters?
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.
How many nickels would you have if you have 32.40 of only quarters and nickels and you have 78 more quarters?
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.
How much does a box of nickels cost?
It depends on the size of the box and the number of nickels in it.
Tejendra has a bag containing 35 nickels and quarters. The total value of these coins is less than 2.50. What is the maximum number of quarters that meets these conditions?
3 quarters
If you have the same number of nickels and quarters The total value is 3.30 What is the total number of coins?
20 - 10 of each
what possible ways can change make 1 dollar?
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!
