Question

PLS SQDANCEFAN. PLEASE SHOW ALL WORK!!! A bag contains 27 coins, all of which are either quarters or dimes. If their total value is $5.55, how many quarters and how many dimes are there?

I WILL MARK BRAINLIEST!!!!!!

Answer 1

Answer:

There are 19 quarters and 8 dimes.

Step-by-step explanation:

We will let x denote the amount of quarters.

Since there are 27 coins in total, (27-x) will be the amount of dimes there are.

Since each quarter is worth $0.25, the total value of quarters is given by 0.25x.

And since each dime is worth $0.10, the total value of dimes is given by 0.1(27-x).

We know that the total value of all 27 coins is $5.55. So, we can write the following equation:

$0.25x+0.1(27-x)=5.55$

So, we will determine the value of x, the amount of quarters there are.  

First, let’s multiply everything by 100 to remove the decimals. So:

$100(0.25x+0.1(27-x))=100(5.55)$

Distribute:

$25x+10(27-x)=555$

Distribute:

$25x+270-10x=555$

Combine like terms:

$15x+270=555$

Subtract 270 from both sides:

$15x=285$

Divide both sides by 15:

$x=19$

So, the value of x is 19.

This means that there are 19 quarters.

So, the amount of dimes must be 27-19 or 8 dimes.

Answer 2

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Answer:

• 19 quarters
• 8 dimes

Step-by-step explanation:

When working mixture problems, it can be convenient to write a single equation using the variable to represent the highest-value contributor to the total value.

Here, we can let q represent the number of quarters. Then the number of dimes is 27-q.

The total value (in cents) is ...

  25q +10(27 -q) = 555

  15q +270 = 555 . . . . . . . collect terms

  15q = 285 . . . . . . . . . subtract 270

  q = 19 . . . . . . . . divide by 15

  27 -19 = 8 . . . . . the number of dimes

There are 19 quarters and 8 dimes.

_____

Check

19 quarters have a value of 19×$0.25 = $4.75

8 dimes have a value of 8×$0.10 = $0.80

Then the total value of the 27 coins is $4.75 +0.80 = $5.55

_____

Additional comment

This is equivalent to working the problem by substitution where you write two equations ...

• d + q = 27
• 10d +25q = 555

Then solve for d and substitute, leaving the variable q.

  d = 27 -q

  10(27-q) +25q = 555 . . . . the equation we wrote above

Of course, this pair of equations can be solved any convenient way, including elimination, matrix methods, and graphing.

You will find that if you eliminate the 'q' variable, you may end up dealing with negative coefficients in the equation. That is why I prefer to keep 'q' and eliminate 'd'.