SOLUTION: A boy has 180 coins altogether . There are nickels dimes and quarters. Number of dimes is half the number of nickels and quarters. Value of coins is 16. How many of each kind of co

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Question 1119320: A boy has 180 coins altogether . There are nickels dimes and quarters. Number of dimes is half the number of nickels and quarters. Value of coins is 16. How many of each kind of coin does he have
Found 4 solutions by josgarithmetic, greenestamps, ikleyn, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!





Just this part of that system allows to eliminate two variables at once to find d.






This also means

.
.
.


and from



Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The statement "number of dimes is half the number of nickels and quarters" is ambiguous. It could mean the numbers of nickels and quarters are each 2x and the number of dimes is x; or it could mean the number of nickels and quarters together is 2x and the number of dimes is x.

A little analysis shows that the first interpretation leads to an inconsistent set of constraints; so the second interpretation must be correct.

In that case, the total number of coins is 3x; since the number of dimes is x, we know 1/3 of the coins are dimes. 1/3 of 180 is 60; we know the number of dimes is 60.

After accounting for 60 dimes worth a total of $6, we have 120 coins remaining, all nickels or quarters, with a total value of $10.

A traditional algebraic approach from here would be to let n be the number of nickels; then an equation that ways the total value of the nickels and quarters is $10 would be



That equation is solved relatively easily.

However, I like to solve the problem from here in a less formal manner, using some mental arithmetic, like this....

(1) If all 120 remaining coins were nickels, the value would be $6; that is $4 short of the actual remaining value of $10.

(2) Exchanging a nickel for a quarter keeps the total number of coins the same but increases the total value by 20 cents.

(3) To make up the $4 "shortfall" in the value of the nickels and quarters, the number of times I need to replace a nickel with a quarter is 400/20 = 20.

(4) Therefore the number of quarters is 20 and the number of nickels is 120-20 = 100.

Typing all those words of explanation makes that seem like a very long process. But if you are good with mental arithmetic, it takes far less time than writing and solving a formal algebraic equation.
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
A boy has 180 coins altogether. There are nickels dimes and quarters. Number of dimes is half the number of nickels and quarters.
Value of coins is 16 . How many of each kind of coin does he have
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Let x be the number of dimes.

Then the number of nickels and dimes (altogether) is 2x, according to the condition.


The total number of coins is 180 = x + 2x = 3x,  which implies that the number of dimes is  x =  = 60.


Thus you conclude that total number of nickels and quarters is 180 - 60 = 120, and their value is  16-60*0.1 = 10 dollars.


Now, let Q be the unknown number of quarters.

Then the number of nickels is (120-Q), and the value equation for nickels and quarters is


    5*(120-Q) + 25Q = 1000    (cents,  or $10).


Simplify and solve for Q:

    600 - 5Q + 25Q = 1000,

    20Q = 1000 - 600 = 400  ====>  Q =  = 20.


Answer.  20 quarters, (120-20) = 100 nickels and 60 dimes.

Solved.


Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

A boy has 180 coins altogether . There are nickels dimes and quarters. Number of dimes is half the number of nickels and quarters. Value of coins is 16. How many of each kind of coin does he have
Let number of dimes be D

Then number of nickels and quarters = 2D

We then get: D + 2D = 180

3D = 180



Therefore, number of nickels and quarters = 2(60), or 120

With N & Q being the number of nickels & quarters, we get: N + Q = 120__N = 120 - Q -- eq (i)



Since the number of dimes = 60, then value of the dimes = 60(.1) = $6. This means that the value of the nickels and quarters is $10 (16 – 6)

This gives us: .05N + .25Q = 10 ------- eq (ii)

.05(120 – Q) + .25Q = 10 -------- Substituting 120 – Q for N in eq (ii)

6 - .05Q + .25Q = 10

.2Q = 4





You can do your own check! 

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