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If we have A gallons of type a we have A/20 gallons of ethanol and A/10 of tol.

If we have  B gallons of type b we have 3B/20 of ethanol and 2B/25 of tol.

If we have C gallons of type c we have C/10 of ethanol and C/25 of tol.

A+B+C=100, so C=100-A-B; A/20+3B/20+(100-A-B)/10=e for ethanol; A+3B+200-2A-2B=20e;

-A+B=20e-200 (1).

A/10+2B/25+(100-A-B)/25=t for tol; 5A+4B+200-2A-2B=50t;

3A+2B=50t-200 (2).

3(1)+(2): 5B=60e+50t-800; B=12e+10t-160 and A=-8e+10t+40.

C=100+8e-10t-40-12e-10t+160=220-4e-20t.

 

by Top Rated User (1.1m points)
So much easier!

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.

 

by Level 11 User (81.5k points)

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