Acollection of nickels, dimes, and quarters totals $6.00. if there are 52 coins together and twice as many dimes as nickels,

4 answers
Question:

Acollection of nickels, dimes, and quarters totals $6.00. if there are 52 coins together and twice as many dimes as nickels, how many of each kind of coin are there?

Answers

For this case, the first thing we must do is define variables:
 x: number of nickels
 y: number of dimes
 z: number of quarters
 We now write the system of equations:
 [tex]0.05x + 0.10y + 0.25z = 6

x + y + z = 52

y = 2x[/tex]
 Solving the system of equations we have:
 [tex]x = 14

y = 28

z = 10[/tex]
 
 14 nickels
 28 dimes
 10 quarters

Option D

q = 10

d = 28

n = 14

Step-by-step explanation:

Given : A collection of nickels, dimes and quarters totals $6.00. If there are 52 coins altogether and twice as many dimes as nickels.

To find : How many of each kind of coin are there?

Solution :

Let n be the nickels

Let d be the dims

Let q be the quarters.

According to question,

There are twice as many dimes as nickels - [tex]d=2n[/tex]

Total number of coins = 52

So,  [tex]n+d+q=52[/tex]

Substitute [tex]d=2n[/tex]

[tex]n+2n+q=52[/tex]

 [tex]3n+q=52[/tex]

[tex]q=52-3n[/tex]

Coins have a total value of $6 which is equal to 600 cents.

We know,

one nickel is worth $0.05= 5 cent,

one dimes are worth $0.10=10 cent

one quarters are worth $0.25=25 cent.

So, [tex]25q+10d+5n=600[/tex]

[tex]5q+2d+n=120[/tex]

Now, put d=2n and q=52-3n

[tex]5(52-3n)+2(2n)+n=120[/tex]

[tex]260-15n+4n+n=120[/tex]

[tex]260-10n=120[/tex]

[tex]-10n=-140[/tex]

[tex]n=14[/tex]

Substitute n in q and d,

[tex]q=52-3n\\q=52-3(14)\\q=52-42=10[/tex]

[tex]d=2n\\d=2(14)=28[/tex]

There are 10 quarters, 28 dimes, and 14 nickels

Therefore,  Option D is correct.

D=2n:  there are twice as many dimes as nickels.

n+d+q=52:  a total of 52 coins, using d from above in this gives you:

n+2n+q=52  combine like terms on left side

3n+q=52  subtract 3n from both sides

q=52-3n

Then you are told that the coins have a total value of $6 which is equal to 600 cents.  So we can say:

25q+10d+5n=600  divide all terms by 5

5q+2d+n=120, and from earlier we saw q=52-3n and d=2n so you have:

5(52-3n)+2(2n)+n=120  expanding...

260-15n+4n+n=120  combining like terms

260-10n=120  subtract 260 from both sides

-10n=-140  divide both sides by -10

n=14, since q=52-3n and d=2n

q=(52-3(14))=52-42=10 

d=2(14)=28

So there are 10 quarters, 28 dimes, and 14 nickels or as your choices put it:

q=10, d=28, n=14  (the correct answer is d.)

The answer is d. q=10 d=28 n=14

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