Type a is 5% ethanol & 10% tol. Type b is 15% ethanol & 8% tol. Type c is 10% ethanol & 4% tol. Type d is (e)% ethanol and (t)% tol. how much of each type to make 100 gallons of type D?
it's dealing with percentages & whole numbers
Since we are dealing with integers, let us assume that the quantities of A, B and C to make up the 100 galls of final solution are also integer values.
Let the quantities of solutions of type A, B and C be X, Y and Z galls respectively.
Type A: 5% Ethanol, 10% Tol – X galls
Type B: 15% Ethanol, 8% Tol – Y galls
Type C: 10% Ethanol, 4% Tol – Z galls
Type D: E% Ethanol, T% Tol – 100 galls
Then we have: X + Y + Z = 100 -------------------------- (1)
In the X galls of Type A solution, 5% of that is Ethanol, i.e. 5% * X galls
In the Y galls of Type B solution, 15% of that is Ethanol, i.e. 15% * Y galls
In the Z galls of Type C solution, 10% of that is Ethanol, i.e. 10% * Z galls
In the 100 galls of Type D solution, E% of that is Ethanol, i.e. E% * 100 galls
Then we have: 5%*X + 15%*Y + 10%*Z = E%*100
Or,
5X + 15Y + 10Z = 100E -------------------------- (2)
Similarly, for the Tol,
10X + 8Y + 4Z = 100T ---------------------------(3)
Multiplying (2) by 2,
10X + 30Y + 20Z = 200E -------------------------- (4)
10X + 8Y + 4Z = 100T ---------------------------(5)
Subtracting (5) from (4),
22Y + 16Z = 100(2E – T)
11Y + 8Z = 50(2E – T) ----------------------------(6)
The above equation is a far as we can go without further information about E and T.
It is assumed that E and T will be integer values.
As such, eqn (6) is a Diophantine equation, with integer coefficients and integer solution values.
To show the method solution for Eqn (6), let us assume some probable values for E and T.
Let E% = 10% (average value of 5%, 15% and 10%)
Let T% = 7% (average value (approx) of 10%, 8% and 4%)
Then 50(2E – T) = 50(20 – 7) = 50*13 = 650, giving,
11Y + 8Z = 650 ------------------------------(7)
To solve the above Diophantine eqn, we create a relationship between the coefficients, 11 and 8. Watch closely,
(1) 11 = 1*8 + 3 (Coeff 11 = multiple of coeff 8 plus remainder)
(2) 8 = 2*3 + 2 (Coeff 8 = multiple of coeff 3 plus remainder)
(3) 3 = 1*2 + 1 (Coeff 3 = multiple of coeff 2 plus remainder)
Now rearrange (1), (2) and (3).
Using (3), 1 = 1*3 – 1*2
Using (2), 1 = 1*3 – 1*(1*8 – 2*3)
1 = 3*3 – 1*8
Using (1), 1 = 3*(1*11 – 1*8) – 1*8
1 = 3*11 – 4*8
650 = 1950*11 – 2600*8
11(1950) + 8(-2600) = 650 ---------------------------------(8)
Comparing (8) with (7), a solution is,
Y_0 = 1950, Z_0 = -2600
The general solution is,
Y = Y_0 – 8k, k = +/- 0,1,2,...
Z = Z_0 + 11k, k = +/- 0,1,2,...
(If you substitute the X and Y solutions into (17), the k-values disappear)
Y = 1950 – 8k, k = +/- 0,1,2,...
Z = -2600 + 11k, k = +/- 0,1,2,...
A more convenient solution set is, (using k = 243)
Y = 6 – 8k, k = +/- 0,1,2,...
Z = 73 + 11k, k = +/- 0,1,2,...
Using eqn (1), X + Y + Z = 100,
X = 100 – (Y + Z)
X = 100 – (6 – 8k + 73 + 11k)
X = 100 – (79 +3k)
X = 21 – 3k
We now have
X = 21 – 3k
Y = 6 – 8k
Z = 73 + 11k
Substituting for X, Y and Z into (2), 5X + 15Y + 10Z = 100E
5(21 – 3k) + 15(6 – 8k) + 10(73 + 11k) = 100*10 (10 is the assumed value for E)
105 – 15k + 90 – 120k + 730 + 110k = 1000
925 – 25k = 1000
k = -3
Therefore,
X = 21 – 3k = 21 +9 = 30
Y = 6 – 8k = 6 + 24 = 30
Z = 73 + 11k = 73 – 33 = 40
The mix we require is: 30 galls of type A, 30 galls of type B and 40 galls of type C
Check
Let [X, Y, Z] = [30, 30, 40]
Then we have:
X galls of Type A = 5% * X galls = 5% * 30 = 1.5 galls of Ethanol.
Y galls of Type B = 15% * Y galls = 15% * 30 = 4.5 galls of Ethanol.
Z galls of Type C = 10% * Z galls = 10% * 40 = 4.0 galls of Ethanol.
1.5 + 4.5 + 4.0 = 10 galls Ethanol in 100 galls of final solution = 10% -- Correct
And,
X galls of Type A = 10% * X galls = 10% * 30 = 3.0 galls of Tol.
Y galls of Type B = 8% * Y galls = 8% * 30 = 2.4 galls of Tol.
Z galls of Type C = 4% * Z galls = 4% * 40 = 1.6 galls of Tol.
3.0 + 2.4 + 1.6 = 7 galls of Tol in 100 galls of final solution = 7% – Correct.