Answer: D
40% HCL is mixed with 60% HCL in the ratio 4:1 to get 44% hcl.44% and 32% are mixed in the ratio 2:1
Thus the average, we have (x-32)/(44-x) =2/1 => x=40%.
i.e 50 L of water has been added to a certain quantity of 40% hcl to get 20% hcl. The quantity has to be 50 L. Before the 25L was removed the quantity was 75 L. This, we got by mixing 25L of 32% hcl with 50% L of 44% hcl.
The 44% hcl is got by mixing 40% hcl and 60% hcl in the ratio 4:1. i.e 40L of 40% and 10 L of 60% hcl.
The data and calculations are tabulated below:-

Strength (%)	40	44	60	Strength (%)	40	44	60
Ratio of quantities	4	5	1	Quantity	40	50	10
Strength (%)	32	40	44	Strength (%)	32	40	44
Ratio of quantities	1	3	2	Quantity	25	75	50
Strength (%)	0	20	40	Strength (%)	0	20	40
Ratio of quantities	1		1	Quantity	50		50

Answer: A
Initial Quantity of lime and water in the jugs.
Jug P:- Lime = 25ml and water =75ml
Jug Q:- Lime = 20ml and water =80ml
Jug R:- Lime = 40ml and water =60ml.
After 40 ml is transferred from Jug P to Jug R, the quantity of lime and water in the beaker is as follows:-
Jug P:- Lime = 25-10 = 15 ml and water = 75-30 = 40ml.
Jug Q:- Lime = 20ml and water = 80ml.
Jug R:- Lime = 40+10 = 50ml and water = 60+ 30 = 90ml
Now, lime: water in Jug R = 5:9
After 20 ml is transferred from Jug R to Jug Q
Jug P:- lime = 15 ml and water = 40 ml
Jug Q:-lime = 20+10 = 30ml and water = 80+18 = 98ml
Jug R:-lime = 50-10 = 40 ml and water = 90-18 = 72 ml
Required ratio = 15:30:40 => 3:6:8

Answer: D
As, both the alloys A and B contains 60% of metal P, Alloys C will also contains 60% of metal P. Metal Q forms 30%.
P---------->Alloy-A(60%)----------Alloy-B(60%)-------------Alloy-C(60%)
Q---------->Alloy-A(__)------------Alloy-B(40%)-------------Alloy-C(30%)
R---------->Alloy-A(30%)----------Alloy-B(___)--------------Alloy-C(___)
S---------->Alloy-A(10%)----------Alloy-B(___)--------------Alloy-C(___)

Let the weights of Alloy A and alloy B be x and y. Let the percentage of R and S in alloy A is 30 % and 10% respectively. Weight of Q in alloy B must be equal to that is alloy C as alloy A does not contain Q.
40y/100 = 30(x+y)/100 => y= 3x.
Ratio of quantities of R in alloy A and Q in alloy B
=> 30x/100 : 40y/100 => 1:4