One antifreeze solution is 26% alcohol and another is 21% alcohol. How much of each mixture should be added to make 60 L of a solution that is 24% alcohol?

1 Answer
Jan 19, 2018

Answer:

Use 36 L of 26 % alcohol and 24 L of 21 % alcohol.

Explanation:

Two conditions must be satisfied:

(a) The volume of the two solutions must add up to the total volume and
(b) The amount of alcohol in the two solutions must equal the amount of alcohol in the combined solution.

This gives us two equations:

#bb((1))color(white)(m)V_26 + V_21 = V_24 = "60 L"#
#bb((2))color(white)(m)c_26V_26 + c_21V_21 = c_24V_24#

From Equation 1,

#bb((3))color(white)(m)V_21 = "60 L - "V_26#

From Equation 2,

#bb((4))color(white)(m)0.26V_26 + 0.21V_21 = "0.24 × 60 L = 14.4 L"#

Substitute Equation 3 into Equation 4

#0.26V_26 + 0.21("60 L - "V_26) = "14.4 L"#

#0.26V_26 + "12.6 L - 0.21"V_26 = "14.4 L"#

#0.05V_26 = "1.8 L"#

#V_26 = "1.8 L"/0.05 = "36 L"#

#V_21 = "60 L - 36 L = 24 L"#

∴ You would mix 36 L of 26 % alcohol and 24 L of 21 % alcohol and get 60 L
of 24 % alcohol.

Check:

#0.26 × 36 color(red)(cancel(color(black)("L"))) + 0.21 ×24 color(red)(cancel(color(black)("L"))) = 0.24 × 60 color(red)(cancel(color(black)("L")))#

#"9.36 + 5.04 = 14.4"#

#14.4 = 14.4#

It checks!