a collection of 52 dimes and nickels is worth $4.50. How many nickels are there?

Question

please help!

Answer 1

10d + 5n = 450
d + n = 52 [*5]
5d + 5n = 260

subtract the eq.

5d + 0n = 190
d = 190/5
d = 38

52 - 38 = 14 nickels

Answer 2

First convert everything to pennies for simplicity. $4.50 = 450 pennies 1 dime = 10 pennies 1nickle = 5 pennies Now let d = the number of dimes and n = the number of nickles then we can say that d +n =52 at the same time that 10d + 5n = 450. Now in order to get rid of one of the unknowns we multiply both sides of the first equation by 5 and then subtract it from the second equation. 5 [d +n ] = 5 [52 ] =5d + 5 n = 260 . Now [10d + 5n =450 ] - [5d +5n = 260 ] or 5d + 0 = 190 and d = 190 / 5 = 38. Now d + n = 52 and [ substituting 38 for d ] 38 + n = 52 and n = 52 - 38 = 14. So 14 is the number of nickles there is.

Answer 3

let n be number of nickels and d be number of dimes then n + d = 52 because you have 52 coins.
so d = 52-n

and since nickels are worth 5cents each and dimes are 10cents each
5n + 10d = 450
replace d with 52-n
5n + 10(52-n) = 450
5n + 520 - 10n = 450
-5n = -70
n = 14 so 14 nickels
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Answer 4

14

Answer 5

let the number of dimes be x
the number of nickels-52-x.so
10x+5(52-x)=450c
5x=450-260
5x=190
x=38 dimesANS.
nickels=52-38=14.ANS.

Answer 6

(x+y=52) = (x = 52-y)
5x+10y=450

5(52-y) + 10y= 450
260 - 5y + 10y = 450
5y = 190
y= 38

38 dimes and 14 nickels

Answer 7

dimes = $0.10
nickels = $0.05

systems of equations:

d + n = 52
.1(d) + .05(n) = $4.50

d = 52-n

.1(52-n) + .05(n) = $4.50

5.2 -.1n + .05n = $4.50

put like terms together

-.05n + 5.2 = 4.5
put terms together

-.05n = -0.7
divide -.7 by -.05 to obtain n

n = 14

go back to original equation

d + n = 52
d + 14 = 52
solve
d = 38

*****NICKELS = 14********
*****DIMES = 38*********