How to solve XAT 2015 Question 33?

This question can be solved by applying the core concept explained in the solution above, followed by step-by-step calculation and elimination of options.

Which concept is tested in this question?

This question tests fundamental concepts commonly asked in XAT exams.

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XAT 2015 Question 33

This is a detailed step-by-step solution for XAT 2015 Question 33. Understand the concept, common mistakes, and expert tips to solve similar questions in XAT exam.

XAT

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.

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