The difference between the number of male and two times the female students studying course E is x. Which of the following is true?

It is giving that 1800 students are studying five different courses -A, B, C, D and E.

The number of male students is $$57\frac{1}{7}\%$$ more than that of female students.

Let number of female students be 100x, thus number of male students will be 100x+(400/7)x = 1100/7x

Total students = 1800

100x+1100x/7 = 1800

1800x/7 = 1800

x = 7

Thus, number of female students = 700 and number of male students = 1100

It is given that Out of male total students, 24% are studying course B = 24% of 1100 = 264

the ratio of the number of female students to male students studying this course is 5:6.

Number of female students studying course B = (264/6)*5 =220

Total students studying course B = 264+220 = 484

It is given that the total number of students studying course C is 342, out of which female students are 104. Hence, number of male students = 342 - 104 = 238

It is given that 18% of females are studying course A = 18% of 700 = 126

Also, the ratio of male students to female students studying A is 4:3.

Hence, the number of male students = (126/3)*4 = 168

Total students studying A = 168+126 = 294

It is also given that One-fifth of total males are studying course D = 20% of 1100 = 220

Also, the number of female students studying D is 25% less than that of males studying D = 220*0.75 = 165

Total number of students studying D = 165+ 220 = 385

Therefore, we make a table from the above information-

The difference between the number of male and two times the female students studying course E is x

=> 210-2*85 = x

=> 210-170 = x

=> x = 40

Option C is correct answer.