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# Find how many of each coin is there

kevin and randy have a jar containing 57 coins all of which are either quarters or nickels the total value of the coins is \$8.85 how many of each type of coin is there please help

My answer is pending review again.

### 5 Answers by Expert Tutors

George T. | George T.--"It's All About Math!"George T.--"It's All About Math!"
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Herbert

Let n=the number of nickels and q=the number of quarters.

From the problem statement, we know two things:

The total number of coins is 57; This means that n+q=57
The total value of the coins is \$8.85; Since each nickel is worth .05 and each quarter is worth .25, this means that .05n+.25q=8.85

We now have two equations with two unknowns.

We can solve for n in the first equation, n=57-q
Plugging this in for n in the second equation, we get .05(57-q)+.25q=8.85
Multiplying through and simplifying:  .05*57-.05q+.25q=8.85
2.85   +    .20q    =8.85
.20q   =8.85-2.85
.20q   =6
q   = 6/.20
q   = 30 (This is the number of quarters)

Since n+q=57; n=57-q
n=57-30
n=27 (this is the number of nickels)

We can now double check the results by taking the number of quarters (30) times the value of each quarter (.25) and adding this to the the number of nickels (27) times the value of each nickel (.05):

30*.25+27*.05 = 7.50+1.35 = 8.85, which is the value stated in the problem.

Let me know if you have any questions.

George T.
Adel E. | Math for EveryoneMath for Everyone
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Let q be the number quarters,n the number of nickles.
q     +n   =  57  ...........................(1)
25q +5n =  885 .....................     .(2)
solve for q &n by either get rid of one of them.Lets omit n by multiply eq 1 by 5

5q+5n =  285  (1)'
25q+5n  =885  (2)
subtract (1)' from (2)
20 q = 600
q=30 ,substitute in any eq. n =27
30 quarter= \$7.5
27 nickles= \$1.35
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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Hi Herbert;
x=quantity of nickles
y=quantity of quarters

\$0.05x+\$0.25y=\$8.85

x+y=57

Let's solve for y using the easier equation...
x+y=57
y=57-x

Let's proceed with the first equation...
\$0.05x+\$0.25y=\$8.85
Let's cancel the \$
0.05x + 0.25y=8.85
Let's plug-in the new equation for y...
0.05x + 0.25(57-x)=8.85
0.05x+14.25-0.25x=8.85
14.25-0.20x=8.85
Let's subtract 14.25 from both sides...
-14.25+14.25-0.20x=8.85-14.25
-0.20x=-5.40
Let's divide both sides by -0.20...
-0.20x/-0.20=-5.40/-0.20
x=27

Let's now solve for y...
x+y=57
27+y=57
y=30

Let's check our work...
\$0.05x+\$0.25y=\$8.85
\$0.05(27) + \$0.25(30)=\$8.85
\$1.35+\$7.50=\$8.85
\$8.85=\$8.85

There are 27 nickles and 30 quarters in the jar.
4.9 4.9 (226 lesson ratings) (226)
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Hey Herbert -- try a 30-cent pack ... \$9 would have 30Q and 30N => drop 15 cents
by removing 3N ==> 30Q and 27N for 7.50 + 1.00 + 0.35 ... Best wishes, sir :)