Question 1152784: Roger has 27 standard U.S. coins worth a total of $3.65. No coin is worth more than 30¢ or less
than 3¢. How many dimes could Roger have? (NOTE: There are several possible answers, but
you only need to give TWO correct answers.)
Found 3 solutions by josgarithmetic, jim_thompson5910, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! Twenty-seven coins any of nickel, dime, quarter; total value $3.65
Choose either n as whole constant or q as whole constant. Look for possible values for d and the other.
.
.
.
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
List out the various coins
{penny, nickel, dime, quarter, half dollar, silver dollar}
We are told that no coin is less than 3 cents, so we can cross off the penny
{penny, nickel, dime, quarter, half dollar, silver dollar}
leaving us with this set
{nickel, dime, quarter, half dollar, silver dollar}
------------------------------------------------------
and we are also told that no coin is worth more than 30 cents, so we can cross off the half dollar and silver dollar coins
{nickel, dime, quarter, half dollar, silver dollar}
leaving us with
{nickel, dime, quarter}
------------------------------------------------------
Therefore, the coins we're dealing with are: {nickel, dime, quarter}
n = number of nickels
d = number of dimes
q = number of quarters
n+d+q = 27 because we have 27 coins
Solve for q to get q = 27-n-d
5n = value, in cents, of all the nickels
10d = value, in cents, of all the dimes
25q = value, in cents, of all the quarters
5n+10d+25q = total value of all the coins
total value = 365 cents = 3.65 dollars
5n+10d+25q = 365
------------------------------------------------------
5n+10d+25q = 365
5n+10d+25(27-n-d) = 365 ... plug in q = 27-n-d
5n+10d+25(27)+25(-n)+25(-d) = 365 ... distribute
5n+10d+675-25n-25d = 365
-20n-15d+675 = 365
-20n-15d+675-675 = 365-675 ... subtract 675 from both sides
-20n-15d = -310
-5(4n+3d) = -310
4n+3d = 62 .... divide both sides by -5
Now solve for n
4n+3d = 62
4n = -3d+62
n = (-3d+62)/4
n = (-3d+2+60)/4
n = (-3d+2)/4+60/4
n = (-3d+2)/4+15
Through trial and error, you'll find that d = 2 makes the expression (-3d+2)/4 result in an integer
n = (-3d+2)/4+15
n = (-3*2+2)/4+15
n = (-6+2)/4+15
n = -4/4+15
n = -1+15
n = 14
So we have d = 2 dimes pair up with n = 14 nickels.
If d = 2 and n = 14, then,
q = 27-n-d
q = 27-2-14
q = 11
--------
Now increase d = 2 to d = 6. This is an increase of +4, which is from the denominator of 4 in (-3d+2)/4
We have,
n = (-3d+2)/4+15
n = (-3*6+2)/4+15
n = (-18+2)/4+15
n = -16/4+15
n = -4+15
n = 11
Now use n = 11 and d = 6 to get,
q = 27-n-d
q = 27-11-6
q = 10
------------------------------------------------------
Two solutions are
(n,d,q) = (14,2,11)
and
(n,d,q) = (11,6,10)
For the solution (n,d,q) = (14,2,11), we can say
5n = 5*14 = 70 cents is from the nickels only
10d = 10*2 = 20 cents is from the dimes only
25q = 25*11 = 275 cents is from the quarters only
5n+10d+25q = 70+20+275 = 365 cents = $3.65 is the total combined value
This confirms the solution (n,d,q) = (14,2,11)
For the solution (n,d,q) = (11,6,10), we can say
5n = 5*11 = 55 cents is from the nickels only
10d = 10*6 = 60 cents is from the dimes only
25q = 25*10 = 250 cents is from the quarters only
5n+10d+25q = 55+60+250 = 365 cents = $3.65 is the total combined value
This confirms the solution (n,d,q) = (11,6,10)
There are 5 total ways to have nickels, dimes, and quarters combine to $3.65 such that we have 27 coins total
| Case | Nickels | Dimes | Quarters |
| A | 2 | 18 | 7 |
| B | 5 | 14 | 8 |
| C | 8 | 10 | 9 |
| D | 11 | 6 | 10 |
| E | 14 | 2 | 11 |
Each row adds to 27.
The solutions we found earlier are in cases D and E of this table.
So Roger could have 6 dimes (case D) or he could have 2 dimes (case E).
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! Roger has 27 standard U.S. coins worth a total of $3.65. No coin is worth more than 30¢ or less
than 3¢. How many dimes could Roger have? (NOTE: There are several possible answers, but
you only need to give TWO correct answers.)
The least number of VARIABLES you have, the easier these types of problems will be to correctly answer.
As these are standard US coins and none is > 30c (.30) and < 3c (.03), the coins MUST consist of quarters, dimes, and nickels.
By letting the number of quarters and dimes be Q and D, respectively, the NUMBER of nickels = 27 - Q - D, and the VALUE of the nickels would be: 3.65 - .25Q - .1D
We then get: 3.65 - .25Q - .1D = .05(27 - Q - D)
3.65 - .25Q - .1D = 1.35 - .05Q - .05D
3.65 - 1.35 = - .05Q + .25Q - .05D + .1D
2.3 = .2Q + .05D
46 = 4Q + D ------ Multiplying the above by 20 to get rid of decimals
46 - 4Q = D
From the above equation, Q CANNOT be 12, so Q MUST be ≤ 11, or ≥ 0 ====> 0 ≤ Q ≤ 11.
By letting Q, or number of quarters be 11, we get the  and nickels as  , and 14, respectively.
By letting Q, or number of quarters be 10, we get the and nickels as  , and 11, respectively.
Therefore, it can be deduced that as the number of quarters reduces by 1, the number of nickels reduces by 3, while the number of dimes increases by 4.
From above, you already have the 2 possible numbers of dimes that he could have, but altogether, there are a total of 5 different scenarios and if you go
by the statement above that I made about the alteration in the number of coins, you can easily find the other 3, if you so wish!
BE AWARE that you CANNOT have - 1 (< 0) nickels, which will produce 28 coins (CONTRADICTORY to the given info.)
|
|
|
