Natalie has some nickles, Dirk has some dimes, and Quincy has some quarters. Dirk has five more dimes than Quincy has quarters. If Natalie gives Dirk a nickel, Dirk gives Quincy a dime, and Quincy gives Natalie a quarter, they will all have the same amount of money. How many coins did each have originally?
Quincy --- x quarters
Dirk ------ x+5 dimes
Natalie ---- y nickles
after giveaway:
Quincy ---- 1 dime + x-1 quarters
Dirk ---- 1 nickel + x+4 dimes
Natalie --- (y-1) nickels + 1 quarter
10 + 25(x-1) = 5 + 10(x+4) = 25 + 5(y-1)
10 + 25(x-1) = 5 + 10(x+4)
10 + 25x - 25 = 5 + 10x + 40
15x = 60
x = 4
5 + 10(x+4) = 25 + 5(y-1)
5 + 10x + 40 = 25 + 5y - 5
45 + 10(4) = 20 + 5y
65 = 5y
y = 13
Quincy had 4 quarters
Dirks had 9 dimes
Natalie had 13 nickels
check:
after giveaway
Quincy has 1 dime + 3 quarters ---> 85 cents
Dirk has 1 nickel + 8 dimes ---- > 85 cents
Natalie has 12 nickels + 1 quarter ---> 85 cents
YEAH!!!!
Let's represent the number of nickels Natalie has as N, the number of dimes Dirk has as D, and the number of quarters Quincy has as Q.
Dirk has five more dimes than Quincy has quarters, so we can represent that as: D = Q + 5.
If Natalie gives Dirk a nickel, Dirk's total amount will increase by 5 cents, and Quincy's total amount will decrease by 10 cents. This means we can represent the new amounts as follows:
N - 1, D + 0.05, Q - 0.10.
If Dirk gives Quincy a dime, Dirk's amount will decrease by 10 cents, and Quincy's amount will increase by 10 cents. At this point, we have:
N - 1, D + 0.05 - 0.10, Q - 0.10 + 0.10.
Simplifying further:
N - 1, D - 0.05, Q.
Finally, if Quincy gives Natalie a quarter, Quincy's amount will decrease by 25 cents, and Natalie's amount will increase by 25 cents. So we have:
N - 1 + 0.25, D - 0.05, Q - 0.25.
Since they all end up with the same amount of money, we can set up the equation:
N - 1 + 0.25 = D - 0.05 = Q - 0.25.
Comparing the first two parts of the equation:
N - 1 + 0.25 = D - 0.05,
N + 0.25 = D + 0.05,
N - D = 0.05 - 0.25,
N - D = -0.20.
Comparing the last two parts of the equation:
D - 0.05 = Q - 0.25,
D - Q = -0.25 + 0.05,
D - Q = -0.20.
Combining the two equations:
N - D = D - Q,
N + Q = 2D.
Since we know N - D = -0.20, we can substitute that into the second equation:
-0.20 + Q = 2D,
Q = 2D + 0.20.
Now, let's consider the possible values for D and Q. Since they represent the number of coins, they must be positive integers. Let's start testing values for D and see if we can find a solution that satisfies all the conditions.
Let's assume D = 1:
Q = 2(1) + 0.20 = 2.20,
N - 1 = -0.20,
N = 0.80.
However, we couldn't find a solution with integer values for N, D, and Q. Let's try another possible value for D.
Let's assume D = 2:
Q = 2(2) + 0.20 = 4.20,
N - 2 = -0.20,
N = 1.80.
Again, we couldn't find a solution with integer values for N, D, and Q. Let's continue testing values for D.
Let's assume D = 3:
Q = 2(3) + 0.20 = 6.20,
N - 3 = -0.20,
N = 2.80.
Still no integer solution. Let's try D = 4:
Q = 2(4) + 0.20 = 8.20,
N - 4 = -0.20,
N = 3.80.
Again, no integer solution. Finally, let's try D = 5:
Q = 2(5) + 0.20 = 10.20,
N - 5 = -0.20,
N = 4.80.
Finally, we have found a solution with integer values for N, D, and Q. Therefore, Natalie originally had 4 nickels, Dirk originally had 5 dimes, and Quincy originally had 10 quarters.
To solve this problem, we can start by assigning variables to the unknown quantities: let's say Natalie has "n" nickels, Dirk has "d" dimes, and Quincy has "q" quarters.
From the given information, we can derive a system of equations:
1) Dirk has five more dimes than Quincy has quarters:
d = q + 5
2) Natalie gives Dirk a nickel, Dirk gives Quincy a dime, and Quincy gives Natalie a quarter, so they all have the same amount of money:
(n - 1) + (d + 1) + (q - 1) = n + d + q
Let's simplify the equation:
n - 1 + d + 1 + q - 1 = n + d + q
n - 1 + 1 - 1 = n
n - 1 = n
We've reached a contradiction, which means there is no solution that satisfies the given conditions.
So, there are no coins for Natalie, Dirk, or Quincy initially that would make the situation possible.