SOLUTION: Kiran reaches into her pocket and finds 28 coins with a total value of $4.10. The coins are nickels, dimes or quarters only. There are twice as many quarters as dimes. How many of

Question 1170563: Kiran reaches into her pocket and finds 28 coins with a total value of $4.10. The coins are nickels, dimes or quarters only. There are twice as many quarters as dimes. How many of each type of coin does Kiran have?
Found 3 solutions by math_tutor2020, ikleyn, greenestamps:
Answer by math_tutor2020(3820)   (Show Source): You can put this solution on YOUR website!

n = number of nickels
d = number of dimes
q = number of quarters
n,d, and q are nonnegative whole numbers

1 nickel = 5 cents
n nickels = 5n cents

1 dime = 10 cents
d dimes = 10d cents

1 quarter = 25 cents
q quarters = 25q cents

(n nickels) + (d dimes) + (q quarters) = (5n cents) + (10d cents) + (25q cents)
(n nickels) + (d dimes) + (q quarters) = (5n+10d+25q) cents
5n+10d+25q = 410
Since $4.10 is equivalent to 410 cents

We can divide all terms of this equation by 5 to go from
5n+10d+25q = 410
to
n+2d+5q = 82

We are told "There are twice as many quarters as dimes", so,
q = 2d
Whatever the number of dimes there are, we double it to get the number of quarters.

Lastly, we have 28 coins total that are either nickels, dimes or quarters
This means n+d+q = 28. Solve for n to get n = 28-d-q.

----------------------------------------------------

The system of equations we can have is
n+2d+5q = 82
q = 2d
n = 28-d-q

Let's substitute equation (3) into equation (1)
n+2d+5q = 82
28-d-q+2d+5q = 82 ... replace n with 28-d-q
28+d+4q = 82
d+4q = 82-28
d+4q = 54

Then plug in q = 2d
d+4q = 54
d+4(2d) = 54
d+8d = 54
9d = 54
d = 54/9
d = 6 .... there are 6 dimes

Use this value of d to find q
q = 2d
q = 2(6)
q = 12 .... there are 12 quarters

We can then find the number of nickels
n = 28-d-q
n = 28-6-12
n = 22-12
n = 10 .... there are 10 nickels

So we get
n = 10
d = 6
q = 12

----------------------------------------------------

Let's check the answers
n = 10 nickels leads to 5n = 5*10 = 50 cents
d = 6 nickels leads to 10d = 10*6 = 60 cents
q = 12 quarters leads to 25q = 25*12 = 300 cents
We get a total of 50+60+300 = 410 cents = $4.10
That takes care of one condition

Another condition is that we must have 28 coins
n+d+q = 10+6+12 = 16+12 = 28
So that condition is satisfied as well.

Finally, the number of quarters must be double that of the dimes. This is true since q/d = 12/6 = 2. Or we can see that (d,q) = (6,12) satisfies the equation q = 2d by plugging in the values.
q = 2d
12 = 2*6
12 = 12

All three conditions are met, so we've confirmed the answers.

----------------------------------------------------
Answers:
10 nickels
6 dimes
12 quarters

Answer by ikleyn(52942)   (Show Source): You can put this solution on YOUR website!
.
Kiran reaches into her pocket and finds 28 coins with a total value of $4.10.
The coins are nickels, dimes or quarters only.
There are twice as many quarters as dimes. How many of each type of coin does Kiran have?
~~~~~~~~~~~~~~~~~~~~~~

Let x be the number of dimes. Then the number of quarters is 2x, according to the condition; and the number of nickels is (28-x-2x) = (28-3x): they are the rest of coins. Now you write the total money equation 10x + 25*(2x) + 5*(28-3x) = 410 cents. Simplify it step by step 10x + 50x + 140 - 15x = 410 10x + 50x - 15x = 410 - 140 45x = 270 x = 270/45 = 6. ANSWER. 6 dimes, 2*6 = 12 quarters and the rest coins, 28-6-12 = 10 are nickels. CHECK. Total money is 6*10 + 12*25 + 10*5 = 410 cents. ! Correct !

Solved.

----------------

The remarkable fact is that the problem is solved using only one equation in one unknown.

Using this approach has that advantages that the problem becomes accessible for 5th grade and 6th grade students
who are not familiar yet with systems of equations;

it really teaches young students to THINK working on setup, and it provides the solution as simple as it should be.

I am 129% convinced that the real purpose and real destination of this and other similar problems is to teach young students
to solve problems in the most simple way.

==================

To see many other similar solved problems (your TEMPLATES), look into the lesson
    - Advanced coin problems
in this site.

Answer by greenestamps(13216)   (Show Source): You can put this solution on YOUR website!

Certainly, if a formal algebraic solution is required, taking the time to analyze the problem to set up the problem using a single variable is much better than using several variables.

But if a formal algebraic solution is not required, an informal solution using logical analysis and simple arithmetic can get you to the answer faster; and it can give you some very good brain exercise (i.e., valuable problem solving experience).

(1) Because the number of quarters is twice the number of dimes, we can group the quarters and dimes into groups each containing two quarters and one dime.
(2) The value of each such group is 60 cents; that means the total value of the quarters and dimes is a multiple of 60 cents.
(3) The largest multiple of 60 cents that is less than the total of $4.10 (410 cents) is 6 times 60 cents, or 360 cents, or $3.60. That leaves 50 cents to be made using the nickels.

Check that answer. The 6 groups of quarters and dimes contain a total of 6*3=18 coins; the 50 cents in nickels contains 50/5 = 10 coins. 18+10 = 28; the number of coins is right, so the solution is correct.

ANSWER: 6 dimes and 12 quarters; and 10 nickels.

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