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Get access to the detailed solutions to the previous year's questions asked in XAT exam.
From the given information we draw the below Venn diagram:
GIven, the total number of electives are 20 and b>a.
also, x>2
Let the total number of G type electives be 'g'.
g=b+2x+3
so, 2a+b+3x+1=20 => 2a+b+3x=19.
Case 1: if x=3 => 2a+b=10
so, (a,b) can be (1,8),(2,6),(3,4) and g will be 17,15,13 respectively
Case 2: if x=4 => 2a+b=7
so, (a,b) can be (1,3),(2,3) and g will be 16,14 respectively
Case 2: if x=5 => 2a+b=4
so, (a,b) can be (1,2) and g will be 13.
.'. The values that 'g' can take are 13,14,15,16,17
So, 5 is the correct answer.