@towerSpider : please paste the question instead of link.

A chemist has one solution that is 30% pure salt and another is 60% pure salt. how many ounces of each solution must he use to produce 60 ounces of a solution that is 50 % salt.

Sol1 - s1
Sol2 - s2
Total = 60
s2 = 60 - s1
0.3 * s1 + 0.6 * s2 = 0.5 * 60
0.3 * s1 + 0.6 *(60 - s1) = 0.5 * 60
36 - 0.3 s1 = 30
0.3s1 = 6
S1 = 20
S2 = 60 - 20 = 40

Sol1 = 20 Oz
sol2 = 40 Oz

target2k9

this question can be solved even faster using "alligation rule"

I guess u would be aware of that already ?

towerSpider wrote:www.algebra.com/algebra/homework/word/m ... ers.1.html

btw nice link towerspider

thanks

Target2009 wrote:@towerSpider : please paste the question instead of link.

A chemist has one solution that is 30% pure salt and another is 60% pure salt. how many ounces of each solution must he use to produce 60 ounces of a solution that is 50 % salt.

Sol1 - s1
Sol2 - s2
Total = 60
s2 = 60 - s1
0.3 * s1 + 0.6 * s2 = 0.5 * 60
0.3 * s1 + 0.6 *(60 - s1) = 0.5 * 60
36 - 0.3 s1 = 30
0.3s1 = 6
S1 = 20
S2 = 60 - 20 = 40

Sol1 = 20 Oz
sol2 = 40 Oz

I am solving this problem with the use of alligation alternate method to avoid lengthy arithmetic calculations...

We want to find the amount of two ingredients to mix together to form a mixture with a given amount of pure salt solution.

There are only two ingredients: 30% pure salt and 60% pure salt - there is only one possible way to form a pair. The difference of 30% from the desired 50%, 20%, is assigned to the 60% pure salt, and the difference of 60% from the desired 50%, 10%, is assigned alternately to the 30% pure salt. The total amount, 60 ounces, is then divided by the sum 20 + 10 (%%) = 60/30 to yield 2, and the amounts of the two ingredients are 2*20=40 ounces of 60% pure salt and 2*10=20 ounces of 30% pure salt.

The key to approaching any problem with the alligation alternate method would be usage of hi-low method i.e. switching between high and low volume capacities.