CAT4MBA

Let the amount of first solution taken be x ltrs and second solution be y ltrs.now
total amount of beer when both solutions are mixed:-3/5x + 7/10y.
total amount of water when both solutions are mixed :-2/5x + 3/10y.

now final ratio we need is 2:1.

(3/5x+7/10y)/(2/5x + 3/10y) =2/1

solving for x/y we get x/y=1/2.

Thanks Rashmi.. Thats the right answer. If possible can you solve the below problem, it is also from Mixtures and Alligation...

Ordinary salt contains 32% impurities, while processed salt has 18% impurities. How many kilograms of ordinary salt must be processed to yield 204 kilograms of processed salt?

a) 272 b) 245 c) 264 d) 246

Thanks Rashmi..  Thats the correct answer..  I tried in the same way before dont know why I didnt get the answer. Can you try the below problem which is from the same chapter Mixtures and Alligation.

Ordinary salt contains 32% impurities, while processed salt has 18% impurities. How many kilograms of ordinary salt must be processed to yield 204 kilograms of processed salt ?

a)  272      b)  245      c) 264      d) 246

hi.. here it goes

let there is 100 kg ordinary salt, so it contains 32kg impurity & 68 kg pure salt
when this is purified, it contains 18% impurity, now the amount of pure salt remains the same i.e 68kg, and let amount of impurity be x kg, then
x/(68+x)*100 = 18%

so x=1224/82 kg of impurities and 68+1224/82 kg of purified salt is wat we get from 100 kgs of impure salt.

100 kg impure---->68+1224/82 pure salt
n kg impure----> 204 kgs pure salt

when solved, we get 246kgs as the ans...

any other easier method???:-)