Question

Solve by using a matrix.

Answer 1

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

Answer 2

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 - $d=2n$

Total number of coins = 52

So,  $n+d+q=52$

Substitute $d=2n$

$n+2n+q=52$

 $3n+q=52$

$q=52-3n$

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, $25q+10d+5n=600$

$5q+2d+n=120$

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

$5(52-3n)+2(2n)+n=120$

$260-15n+4n+n=120$

$260-10n=120$

$-10n=-140$

$n=14$

Substitute n in q and d,

$q=52-3n\\q=52-3(14)\\q=52-42=10$

$d=2n\\d=2(14)=28$

There are 10 quarters, 28 dimes, and 14 nickels

Therefore,  Option D is correct.