Question

A 40% dye solution is to mixed with a 53% dye solution to get 260L of a 50% solution. How many liters of the 40% and 50% solutions will be needed

Answer 1

60 liters of 40 % dye solution is mixed with 200 liters of 53 % dye solution to get 260 liters of 50 % dye solution

Solution:

From given,

Final solution is 260 liter

Let x be the liters of 40 % dye solution

Then, (260 - x) is the liters of 53 % dye solution

Therefore, according to question,

x liters of 40 % dye solution is mixed with (260 - x) liters of 53 % dye solution to get 260 liters of 50 % dye solution

Thus we frame a equation as:

40 % of x + 53 % of (260 - x) = 50 % of 260

Solve for "x"

$\frac{40}{100} \times x + \frac{53}{100} \times (260-x) = \frac{50}{100} \times 260\\\\0.4x+ 0.53(260-x) = 0.5 \times 260\\\\0.4x + 137.8 - 0.53x = 130\\\\0.13x = 137.8 - 130\\\\0.13x = 7.8\\\\Divide\ both\ sides\ by\ 0.13\\\\x = 60$

Thus 60 liters of 40 % dye solution is used

Then, (260 - x) = 260 - 60 = 200

Thus 200 liters of 53 % dye solution is used