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Explanation:

Let QJ type electives, only Grade-oriented electives and only Quantitative-oriented electives be zy and x respectively.

Note: y > x

∴ Number of QG type electives = z + 2

∴ Number of JG type electives = z + 4

The given data can be represented in the form of the following Venn diagram,

Thus, the following equation is obtained,

3z + 2x + y = 13

(x, z ≥ 1 and y ≥ 2 )

For z = 1, 2x + y = 10 and possible values of (x, y): (1, 8), (2, 6) and (3, 4).

For z = 2, 2x + y = 7 and possible values of (x, y): (1, 5) and (2, 3).

For z = 3, 2x + y = 4 and possible values of (x, y): (1, 2).

Now, G = 2z + y + 7

If z = 1, G = 17, 15, 13

If z = 2, G = 16, 14

If z = 3, G = 15

Only 5 unique values of G are possible i.e. 13, 14, 15, 16 and 17.

Hence, option (b).

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