XAT 2020 Question 67
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XAT
Given, a+b+c=50 and a+b+c+x+y+z=190 => x+y+z=140.
Also, let E1=k => E2=2k and E3=3k
E1+E2+E3= 6k=190+50=240 => k=40.
Option A, the number of students choosing E1 is 40.
Option B, number of students choosing either E1 or E2 or both, but not E3 = total - E3 = 190-120 = 70.
Option C, number of students choosing both E1 and E2 => this can not be obtained.
Option D, number of students choosing E3 = 3x = 120.
Option E, number of students choosing exactly one elective = Out of 190, 50 are choosing two electives, hence 190-50 = 140 are
choosing exactly one elective.