# Linear Equation type of question (fuel mixture)

• Jun 11th 2006, 11:09 AM
usm_67
Linear Equation type of question (fuel mixture)
here is the question:

Premium gasoline sells for 78.9cents/Liters. Regular gas sells for 71.9 cents/ litres. To boost sales, a middle octane gasoline is formed by mixing premium and regular. If 1000 L of this middle octane gas is produced, and is sold at 73.9 cents/Litres, then how much of each type of gasoline can you assume was used in the mixture?

Answer : 600 L premium and 400 L regular.........i got the answer from the back of the buk.....but i m not able to solve this question and get this answer

help plz
• Jun 11th 2006, 11:27 AM
earboth
Quote:

Originally Posted by usm_67
here is the question:

Premium gasoline sells for 78.9cents/Liters. Regular gas sells for 71.9 cents/ litres. To boost sales, a middle octane gasoline is formed by mixing premium and regular. If 1000 L of this middle octane gas is produced, and is sold at 73.9 cents/Litres, then how much of each type of gasoline can you assume was used in the mixture?
Answer : 600 L premium and 400 L regular.........i got the answer from the back of the buk.....but i m not able to solve this question and get this answer
help plz

Hello,

I've difficulties to get the answers you mentioned above, because:
600 * 78.9 + 400 * 71.9 = 76100. That means 1 ltr of this mddle octane gas will cost 76.1 and not - as it is said in your problem - 73.9.

Let be x = amount of prem. gas.
Let be y = amount of reg. gas.

Then you get a system of linear equations:
x + y = 1000
78.9*x + 71.9*y=73.1*1000

Solve this system and you'll get:

x = 285.7 ltr
y = 714.3 ltr.

Greetings

EB
• Jun 11th 2006, 11:40 AM
usm_67
sir can u plzz solve by all steps so i can understand how u got dat anwer.
• Jun 11th 2006, 09:08 PM
earboth
Quote:

Originally Posted by usm_67
sir can u plzz solve by all steps so i can understand how u got dat anwer.

Hello,

I don't know which method you are used to. I'll sho you one which is here convenient to solve this system of equations:
(A and B are labels of an equation)

$A:\ x+y=1000$
$B:\ 78.9x+71.9y=73.9 \cdot 1000$

Solve A for y: y = 1000-x (****)
Plug in this term in B:

$B:\ 78.9x+71.9(1000-x)=73900$

$B:\ 78.9x+71900-71.9x=73900$. Now subtract 71900 from both sides:

$B:\ 7x=2000\ \Longrightarrow\ x=\frac{2000}{7}\approx 285.7$

Plug in this value into (****) and yo get $y=1000-\frac{2000}{7}\approx 714.3$

Greetings

EB