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

###### Question:

## 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