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Solve by using a matrix.

A collection of nickels, dimes and quarters totals $6.00. If there are 52 coins altogether and twice as many dimes as nickels, how many of each kind of coin are there?
a.
q = 5
d = 60
n = 30
c.
q = 15
d = 28
n = 6
b.
q = 35
d = 10
n = 5
d.
q = 10
d = 28
n = 14

Respuesta :

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

Answer:

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.