How many correct predictions were made by D?__

Question

How many correct predictions were made by D?___

Options:

Answer: Option A
:

From the given condition "if any person had predicted that a single horse would win each of the eight games, he would not have gained or lost any amount". It means, every horse won 2 races.
From the given condition "B received the maximum amount of 16000". Let B predicted x races and y races incorrect.
3000 x-1000 y=16000 ...................(1) And x+y=8......................... (2)
From equation (1) and (2), we can determine x=6 and y=2
From the second condition "Had C made one more correct prediction and B made one more incorrect prediction.
The amounts gained by them at the end of the races would have interchanged". We can conclude C predicted 5
races correct and 3 races incorrect. And E neither gained nor lost any amount. So, E predicted 2 races correct and 6
races incorrect.
$\begin{matrix} P e r s o n s & W i n & L o s e & S c o r e B & 6 & 2 & 16, 000 C & 5 & 3 & 12, 000 E & 2 & 6 & 0 \end{matrix}$
Now, from the given table, we can easily conclude below table :-
$\begin{matrix} R_{1} & R_{2} & R_{3} & R_{4} & R_{5} & R_{6} & R_{7} & R_{8} C h e e t a h & B i n g o & A z u r e & B i n g o \end{matrix}$
Because B's 6 predictions are correct. Out of 8 predictions, 4 times B predicted "Dylan". And we know Dylan won only two times. It means B's predictions for Race $R_{2}, R_{4}, R_{5}$ and $R_{8}$ were correct.
$\begin{matrix} R_{1} & R_{2} & R_{3} & R_{4} & R_{5} & R_{6} & R_{7} & R_{8} A z u r e & C h e e t a h & D y l a n & B i n g o & A z u r e & C h e e t a h & D y l a n & B i n g o x & (B, C) & (A, B, C, D) & (A, B, D, E) & (A, B, C) & (A, C, D) & (B, C, D) & (B, E) \end{matrix}$
Now, we can conclude below table. Because, E's 2 predictions were correct and Cheetah won 2 times. So, cheetah
has to win Race $R_{5}$ . C's 5 predictions were incorrect. So, C cannot be winner of race $R_{1}$