How much pure alcohol should be added to 400ml of strength 15% to make it's strength 32%?
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Answered by
3
Let x = the required milliliters of alcohol.
Then we have
400[15%] + x[100%] = [400+x][32%]
400×[0.15] + x[1] = [400+x][0.32]
Simplify and solve for x
60+x = 128 + 0.32x
60 + 0.68x = 128
0.68x = 68
⇒ x = 100 ml of pure alcohol required.
Then we have
400[15%] + x[100%] = [400+x][32%]
400×[0.15] + x[1] = [400+x][0.32]
Simplify and solve for x
60+x = 128 + 0.32x
60 + 0.68x = 128
0.68x = 68
⇒ x = 100 ml of pure alcohol required.
Answered by
0
400/32=25/2
25/2×17/15=85/6
85/6×400=17/120
120-17=103
the 32 becomes 3 because the 2 divides 120 in 4 parts.
So it is even we cut it and subtract 3 from 103
103-3=100
100 ml required
25/2×17/15=85/6
85/6×400=17/120
120-17=103
the 32 becomes 3 because the 2 divides 120 in 4 parts.
So it is even we cut it and subtract 3 from 103
103-3=100
100 ml required
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