# Blending remedies of specific toughness

While reviewing one capacity publication, I found this inquiry:

$35\%$ alcohol blended with $60\%$ alcohol to get a $50\%$ alcohol. In what proportion were they combined?

I invested 20 mins on this inquiry, yet I could not also identify what it suggests! Could any person clarify this to me?

Mixture1 : 60%

Mixture2 : 35%

Let us claim you take combination making use of a pot of 100 devices.

additionally you take Mixture1 'a' times and also Mixture2 'b' times making use of the pot.

$\rightarrow$. $\frac{(a 60 + b 35)}{(a 100 + b 100)} = \frac{50}{100}$

addressing this you get

$\frac{a}{b} = \frac{3}{2}$

- > 300 devices of Mixture1 and also 200 devices of Mixture2 will certainly offer you 500 devices of combination with alcohol focus of 50% due to the fact that 3 x 60+2 x 35 = 250.

Call the proportion you intend to locate r= (quantity of 35% alcohol) : (quantity of 60% alcohol). Intend that there are r litres of the 35% alcohol (so there are 0.35 L of alcohol and also 0.65 L of water), 1 litre of the 60% alcohol (so the proportion is r), and also 1+ r litres of the 50% alcohol. From there, you can construct a formula concerning the alcohol (or concerning the water) and also address for r.

Given `a`

litres of `35%`

alcohol remedy, and also `b`

litres of `60%`

alcohol remedy, what would certainly be the alcohol proportion after we'll blend this 2 remedies?

The `a`

litres are in fact `a*35/100`

alcohol, and also `a*(1-(35/100))`

water. Furthermore the `b`

litres has `b*60/100`

alcohol and also `b*(1-(60/100))`

water.

Blending them with each other would certainly offer us `a*35/100+b*60/100`

alcohol and also `a*(1-(35/100))+b*(1-(60/100))`

water. The inquiry is for which `a,b`

we have the proportion

```
((a*35+b*60)/100)/((a*65+b*40)/100) = 50% = 1/2
```

or

```
(a*35+b*60)/(a*65+b*40) = 1/2
```

I think you can take it from below ...