mixture (water and milk)

Algebra Level 1

There is a 40 litre solution of milk and water in which the concentration of milk is 72%. How much water must be added to this solution to make it a solution in which the concentration of milk is 60% ?

The answer is 8.

3 solutions

Unstable Chickoy
Jun 6, 2014

let $x$ be the volume of water to be added.

$40(1 - 0.72) + x = 0.6 (40 + x)$

$x = \boxed{8}$

Venture Hi
Apr 14, 2015

You can think of this as having 3 containers: 1. 40 liter solution of 28% water and 72% milk. 2. 100% water in a container. 3. The final solution where 40% is water and 60% is milk.

1 + 2 = 3

Rewriting this, we have

40(0.28) water + x (100%) water = (40+X) 40% water

Solve for x.

=> 11.2+x=16+0.4x

=> 0.6x= 4.8

=> $\boxed{x=8}$ liters of water

Ahmed Abdelbasit
Jun 5, 2014

the volume of milk = $\frac{7240}{100}$ = $28.8$ liters

*then .. the volume of water is $40-28.8 = 11.2$ liters

*the percent of water after addition is $40$ % ..

*let we add $x$ liters of water

*so : $40$ % $=$ $\frac{11.2 + x}{40 + x}$ $=$ $\frac{40}{100}$

*so , $1600 + 40x = 1120 + 100x$

*then : $60x = 480$

*and finally : $x = 8 liters$

mixture (water and milk)

The answer is 8.

3 solutions

*the volume of milk = 72 ∗ 40 100 \frac{72*40}{100} 1 0 0 7 2 ∗ 4 0 ​ = 28.8 28.8 2 8 . 8 liters

*then .. the volume of water is 40 − 28.8 = 11.2 40-28.8 = 11.2 4 0 − 2 8 . 8 = 1 1 . 2 liters

*the percent of water after addition is 40 40 4 0 % ..

*let we add x x x liters of water

*so : 40 40 4 0 % = = = 11.2 + x 40 + x \frac{11.2 + x}{40 + x} 4 0 + x 1 1 . 2 + x ​ = = = 40 100 \frac{40}{100} 1 0 0 4 0 ​

*so , 1600 + 40 x = 1120 + 100 x 1600 + 40x = 1120 + 100x 1 6 0 0 + 4 0 x = 1 1 2 0 + 1 0 0 x

*then : 60 x = 480 60x = 480 6 0 x = 4 8 0

*and finally : x = 8 l i t e r s x = 8 liters x = 8 l i t e r s

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