Let the amount of the first brand be x, and let the amount of the second brand be y.
0.09x + 0.14y = 240 * 0.13 .................(1)
x + y = 240 ..............(2)
y = 240 - x .......................(3)
Plugging the value for y from equation (3) into equation (1), we get:
$0.09x+0.14(240-x)=240\times0.13$ ...............(4)
Equation (4) simplifies to:
-0.05x = -2.4
giving the value for the required amount of 9% vinegar as 48 ml and the required amount of 14% vinegar as 240 - 48 = 192 ml.

7 0

2 years ago

how many gallons of a 50 antifreeze solution must be mixed with 90 gallons of 20% antifreeze to get a mixture that is 40% antifr

Tamiku [17]

Try this:
1) note that weight of pure antifreeze before mixing and after mixing is the same. So, if 'x' is weight of pure antifreeze in 50% solution, it is possible to make up equation before mixing: 0.5x+0.2*90.
2) there are 0.2*90=18 gal. of pure antifreeze in the 20% solution. If 'x' gal. is the weight of pure antifreeze in 50% sol. and 18 gal. is the weight of pure antifreeze in 20% sol., it is possible to make up an equation after mixing: 0.4(x+18).
3) using the both parts: 0.5x+0.2*90=0.4(x+18) ⇒ x=54 gal. of <u>pure</u> weight.
4) to find 50% solution of 54 gal. pure weight just 54:0.5=108 gal.
Answer: 108 gal.

3 0

2 years ago