A 20 kilograms mixture of dry fruits consisting of x percent of almond

Total mixture = 20 kgs.
Amount of almonds in it = (20*x)/100

Weight of mix A taken = a
Weight of mix B taken = b
a+b = 20. |eq1

% of almonds in a = (a*y)/100
% of almonds in b = (b*20)/100

So, (a*y)/100 + (b*20)/100 = (20*x)/100
a*y + b*20 = 20*x

We need to find what is the weight of almonds in kilograms in mixture A
i.e a*y = ?
a*y = 20*x-20*b =?

(1) The amount of almonds in mixture B is three-fifths of the total weight of mixture A
b/5 = 3/5*a
b = 3a
Putting this into eq 1
we will get 4a = 20
a = 5
b = 15
We don't know the value of x
Not sufficient.

(2) If the weight of the final mixture is reduced by 25 percent, the almonds in the final mixture would reduce by 1.5 kilograms
20*(.75) = 15
So amount of almonds in it = 15x
We know 20x - 15x = 1.5
x = 0.3
But we don't know the value of B.
Not sufficient.

Combining helps will give us both the values of x and b so a*y can be found out.
Sufficient.

C is the answer.