Let us assume that he answered x question correctly. Marks scored by him in x question = 4x
Wrong answer would be 200 - x

According to question,
4x - (200 - x) = 200
or 4x - 200 + x = 200
or 5x = 200 + 200 = 400
or x = 80

Option method:
The best way to go through options. You will find that when you consider 150 questions you cannot get 75 marks anyhow.

Alternatively:
Since 75 is an integer, he must have to do (75 + 4x) problems, where x is a whole number. Now we put value of x (0, 1, 2, 3, 4, 5 .........) to get correct answer.
As he gets net marks in integer it means when he does 3n question wrong, he losses actually 4n marks.

$$\eqalign{ & {112 \times {5^4}} \cr & = 112 \times {\left( {\frac{{10}}{2}} \right)^4} \cr & = \frac{{112 \times {{10}^4}}}{{{2^4}}} \cr & = \frac{{1120000}}{{16}} \cr & = 70000 \cr} $$

LCM = 45 HCF
LCM + HCF = 1150
45 × HCF + HCF = 1150
46 × HCF = 1150
HCF = 25
Then,
LCM = 45 × 25 = 1125
Now, use formula,
1^st number × second Number = LCM × HCF
125 × second number = 1125 × 25
Second Number = 225

Since, $${3^4} = 81$$  gives remainder 1 on divided by 80
So, $$\frac{{{3^{4n}}}}{{80}}$$  gives remainder 1
Thus, $$\frac{{{3^{67!}}}}{{80}}$$  will also give remainder as 1
Since, 67! = 4n for a positive integer n

5^{(x + 3)} = 25^{(3x - 4)}
Or, 5^{(x + 3)} = 5^{2 × (3x - 4)}
Now, comparing powers,
(x + 3) = 2 × (3x - 4)
Or, x + 3 = 6x - 8
Or, 6x - x = 8 + 3
Or, 5x = 11
Hence, x = $$\frac{{11}}{5}$$

Going through options:
LCM of 204 and 170 = 1020.

$$\eqalign{ & \frac{{{{65}^{203}}}}{7} \cr & Or,\,\frac{{{{\left( {63 + 2} \right)}^{203}}}}{7} \cr & 63\,{\text{is}}\,{\text{divisible}}\,{\text{by}}\,7, \cr & {\text{so}}\,{\text{remainder}}\,{\text{will}}\,{\text{depend}}\,{\text{on}}\,{\text{}}\,{\text{powers}}\,{\text{of}}\,2 \cr & Or,\,\frac{{{2^{203}}}}{7} \cr & {\text{Its}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{2^3}}}{7} \cr & {\text{Now,}}\,{\text{Required}}\,{\text{Remainder}}\,{\text{wold}}\,{\text{be}}, \cr & \frac{8}{7} = 1\cr & {\text{Required}}\,{\text{remainder}} = 1 \cr} $$
Note: We have manipulated powers in form of (4x + n). It means 203 is taken as,
203 = 4x + n = 4 × 50 + 3.
We neglect power which is in multiple of 4.

1 - 3 + 5 - 7 + 9 - 11 .......... + 197 - 199
= (- 2) + (-2) + (-2) + .......... + (-2) (50 times)
= 50 × (-2) = -100

Let numbers be 17x and 17y where x and y are co-prime.
LCM = 17xy
Now, 17xy = 714
Or, xy = 42 = 6 × 7
?x = 6 and y = 7
Or, x = 7 and y = 6
1^st number = 17 × 6 = 102
2^nd number = 17 × 7 = 119
Sum = 102 + 119 = 221