problem on Solutions

IMO B. I am not very sure if my approach is correct. What is the OA?

what's your solution?

Solve it using diagram::

Sugar Water
10 90
x
16 84


Total soln : 100 lt
sugar 10lt, Water 90 lt
1/4 will have sugar as 10/4 and water as 90/4
replaced by x/4 and 100-x/4 water
so sugar added by 6 and water reduced by 6

so 1/4 of soln contains 6ltr of sugar and so the whole solb will contain : 6*4 i.e. 24% sugar

IMO A (My earlier post was done with approximation, considering the limited time in real time GMAT exam, but I think I got it wrong). I am going to substantiate my answer this time. Hopefully I am right.

Consider both the solutions have 100 liters (because we are dealing with percentages). If 1/4 of the solution is replaced by another solution x, then 7.5 sugar & 77.5 lit of water will remain. If x/4 of another solution is brought into replace, the equation will be:

7.5+x/4 = 16%(100) i.e. comparing sugar quantity
==> x/4 = 8.5
==> x = 34

So, 34% of the second solution was sugar.

let original solution be 100 lires, sugar: 10 ltrs

1/4 of this solution (25 ltrs) replaced by another solution=>total volume remains same, only the volume of sugar changes

1/4*100*10/100=2.5 litres of sugar removed
let x litres of sugar be added

10-2.5+x/100=16/100
=>x=16-7.5=8.5 ltrs

8.5 lts in 25 ltr of new solution=>8.5/25*100=34%

hence, A

IMO A.

Sugar is 10 initially. After mixing, it is 16. Considering 4 units and applying law of average, the solution should contain 10 + (6 * 4) sugar.
10,10,10,10 => average 10
(10+6*4),10,10,10 => average 16