You can re-frame the question as:

= 3a+4b,Xwhere a>0, b>0 & x<20. Find all uniquefor which 3a+4b<20?X

Now, by choosing pairs for (a,b):

- (1,1) = 3(1)+4(1) = 7 < 20
- (1,2) = 3(1)+4(2) = 11 < 20
- (1,3) = 3(1)+4(3) = 15 < 20
**(1,4) = 3(1)+4(4) =***19*< 20- (2,1) = 3(2)+4(1) = 10 < 20
- (2,2) = 3(2)+4(2) = 14 < 20
- (2,3) = 3(2)+4(3) = 18 < 20
- (3,1) = 3(3)+4(1) = 13 < 20
- (3,2) = 3(3)+4(2) = 17 < 20
- (4,1) = 3(4)+4(1) = 16 < 20
**(5,1) = 3(5)+4(1) =***19*< 20

As you can see there are 10 unique values possible for * X*. So

**the answer is (D)**But is there a faster way to do this?

Under time constraints, mentally finding if the 3a+4b < 20 is all I would check and tend to not remember the values:

- It’s possible to choose (E) as one can assume they have yet to consider more possibilities
*(or)* - It’s possible to choose (C,D) as one can assume they have yet to find any common recurring values.

How do you suggest attack such problems?