The constituency in which B got lower number of votes compared to A and C is

Party	S1	S2	S3	S4	S5	Total
A						49
B						35
C						16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

How many votes does a winner need in one constituency?

• Suppose the winner got only 7. The other two parties would share the remaining 13, so at least one of them would also have 7 or more, and there would be no clear winner
• So the winner of any constituency needs at least 8 votes

C wins both S2 and S3, so C has at least 8 in each. That is already 16, which is exactly C's total, so there is nothing left over for the other three seats.

Party	S1	S2	S3	S4	S5	Total
A						49
B						35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

With C out of the picture in S1, S4 and S5, those three seats are straight A vs B fights over all 20 votes, so the winner there needs at least 11.

• S1 goes to A, so B gets at most 9 there
• S4 goes to B, so B has at least 11
• S5 goes to B, so B has at least 11, but B's votes are strictly increasing, so B in S5 must beat B in S4

"Number of votes obtained by B in S1, S2, S3, S4 and S5 are distinct natural numbers in increasing order."

That pushes B in S5 to at least 12.

Party	S1	S2	S3	S4	S5	Total
A						49
B	≤ 9			≥ 11	≥ 12	35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

Now S2 and S3, where C takes 8 of the 20 votes

• A and B share the remaining 12
• Neither of them can reach 8, otherwise C would not be the winner
• Two numbers adding to 12 with both below 8 must each be 5, 6 or 7

So B's votes in S2 and S3 both come from {5, 6, 7}, and being increasing, they add to at least 5 + 6 = 11.

Party	S1	S2	S3	S4	S5	Total
A						49
B	≤ 9	5-7	5-7	≥ 11	≥ 12	35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

Squeezing B's total of 35 from both sides

S4 and S5 already take at least 11 + 12 = 23, which leaves the first three seats at most 12:

B(S1) + B(S2) + B(S3) ≤ 35 − 23 = 12

Going the other way, B(S2) + B(S3) is at least 11 and B(S1) is a natural number, so it is at least 1. Those three add to at least 12.

Both bounds land on 12, so every value is forced:

• B(S1) = 1
• B(S2) + B(S3) = 11, and from {5, 6, 7} in increasing order that is 5 and 6
• B(S4) + B(S5) = 35 − 12 = 23, and with 11 and 12 as minimums that is exactly 11 and 12

Party	S1	S2	S3	S4	S5	Total
A						49
B	1	5	6	11	12	35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

Each column holds 20 votes, so A's row is simply whatever is left after B and C:

• S1: 20 − 1 − 0 = 19
• S2: 20 − 5 − 8 = 7
• S3: 20 − 6 − 8 = 6
• S4: 20 − 11 − 0 = 9
• S5: 20 − 12 − 0 = 8

Party	S1	S2	S3	S4	S5	Total
A	19	7	6	9	8	49
B	1	5	6	11	12	35
C	0	8	8	0	0	16
Total	20	20	20	20	20	100
Winner	A	C	C	B	B

A's votes come to 19 + 7 + 6 + 9 + 8 = 49, exactly as given.

Checking the winners once more: A takes S1 with 19, C takes S2 and S3 with 8 each, and B takes S4 and S5 with 11 and 12. Every condition is satisfied, so this grid is the unique arrangement and all questions can be read off it.

C has no votes at all in S1, S4 and S5, and in S3 B only ties A at 6. In S2, B's 5 trails both A's 7 and C's 8, so the answer is S2.