Question 201379: A coin bank contains 25 coins in nickles, dimes and quarters. There are four times as many dimes as quarters. The value of the coins is $2.05. How many dimes are in the bank?
Found 5 solutions by ptaylor, Theo, ikleyn, timofer, greenestamps:
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=number of quarters
Then 4x =number of dimes
And y=number of nickels
Now we are told that:
x+4x+y=25 or
5x+y=25--------------------eq1
also (we'll deal in pennies)
25x+10*4x+5y=205
65x+5y=205-------------------------eq2
multiply eq1 by 5 and then subtract it from eq2
(65x+5y=205)-(25x+5y=125) and we get
40x=80
x=2---------------number of quarters
4x=4*2=8 number of dimes
There must be 25-10=15 nickels
CK
2*25+8*10+15*5=205
50+80+75=205
205=205
We can also work this problem using only one unknown:
Let x=number of quarters
Then 4x =number of dimes
And 25-5x=number of nickels
So we have:
25x+10*4x+5(25-5x)=205 get rid of parens
25x+40x+125-25x=205 subtract 125 from each side
25x+40x+125-125-25x=205-125 collect like terms
40x=80---same as before
Hope this helps---ptaylor
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! I believe your answer is going to be
15 nickels (75 cents)
8 dimes (80 cents)
2 quarters (50 cents)
that adds up to be $2.05
here's how I did it.
let n = number of nickels
let d = number of dimes
let q = number of quarters
n + d + q = 25 (number of coins in total)
number of dimes is 4 times the number of quarters (n = 4d)
total value of coins is $2.05
$.05 * n + $.10 * d + $.25 * q = $2.05
take away the dollar sign and multiply both sides of the equation by 100 and you get
5n + 10d + 25q = 205
since d = 4q you can substitute to get
5n + 40q + 25q = 205
since n + d + q = 25 you can substitute 4q for d again to get
n + 5q = 25
solve for n to get
n = 25 - 5q
multiply both sides of the equation by 5 to get
5n = 125 - 25q
substitute in the equation 5n + 40q + 25q = 205 to get
125 - 25q + 40q + 25q = 205
simplify the left side of the equation to get
125 + 40q = 205
solve for q to get
40q = 80
q = 2
since d = 4q, then d = 8
since n + d + q = 25, then n = 15
your answer is
n = 15
d = 8
q = 2
as shown above.
Answer by ikleyn(52903)  (Show Source):
You can put this solution on YOUR website! .
A coin bank contains 25 coins in nickels, dimes and quarters.
There are four times as many dimes as quarters. The value of the coins is $2.05.
How many dimes are in the bank?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
At the first glance, this problem is for three unknowns.
But it is so only for the first glance.
Actually, this problem and many other similar problems admit elegant solution in one single unknown,
and it is the reason why the problem is designed in this way and proposed to you to teach you to true Math.
Let 'x' be the number of quarters.
Then the number of dimes is 4x, according to the problem,
and the number of nickels is 25 - x - 4x = 25-5x.
Write an equation for the total value
5*(25-5x) + 10*(4x) + 25x = 205 cents.
Simplify and find x
125 - 25x + 40x + 25x = 205,
40x = 205 - 125,
40x = 80,
x = 80/40 = 2.
Thus we found that there are 2 quarters.
Hence, the number of dimes is 4*2 = 8.
ANSWER. There are 8 dimes in the bank.
Solved elegantly, as promised.
It is a good style as the problem should be solved,
and a good style as the solution should be presented.
----------------------------------------------------
A few words about why I consider this approach correct and attractive.
First, it reduces the problem to one unknown and one equation, making the solution much simpler
than dealing with a system of three equations.
Second, it allows students to solve such problems at an earlier age, when they are not yet familiar
with systems of three equations. Therefore, it allows to accelerate the students' mathematical
education/development, which is a great benefit.
Third, this approach trains the students minds and makes their minds more flexible
(comparing with the stupid setup using 3 unknowns),
which is the important goal of mathematical education in middle and high schools.
Fourth, it is beautiful, simple and powerful at the same time, and this teaches students
to see the beauty and power in simplicity, which makes life more harmonious.
Answer by timofer(107)  (Show Source):
Answer by greenestamps(13215)  (Show Source):
You can put this solution on YOUR website!
The original solution from tutor @ikleyn was not correct, as it produced a total value of the coins that is greater than $2.05. But I see that she saw the error in her setup and has corrected her response.
Note that she recommends setting up the problem using a single variable; I concur. That requires a bit more effort than setting the problem up using two or even three variables; but it reduces significantly the amount of effort required to solve the problem.
Meanwhile, I will present an informal solution that can be obtained using mental arithmetic and logical reasoning. While of course the student should understand the solution using formal algebra, it should be noted that solving a problem like this informally can provide excellent brain exercise.
There are four times as many dimes as quarters. So consider groups of 5 coins consisting of one quarter and four dimes, with a total value of 65 cents. Then solve the problem by trying different numbers of those groups of 5 coins.
If there were one such group, with a total of 5 coins, the remaining value to be made with the nickels would be $2.05 - $0.65 = $1.40; but that requires 140/5 = 28 nickels, making a total of 5+28 = 33 coins, so that doesn't work.
If there are two such groups, with a total of 10 coins, the remaining value to be made with the nickels is $2.05 - $1.30 = $0.75. That requires 75/5 = 15 coins, which makes the correct total of 10+15 = 25 coins.
So there are two groups each containing one quarter and four dimes, making 8 the total number of dimes.
ANSWER: 8 dimes
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