Big Book Test 4 Section 3 Q29

GRE Quant Problem Solving
1

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You can re-frame the question as:

X = 3a+4b, where a>0, b>0 & x<20. Find all unique X for which 3a+4b<20?

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

  1. (1,1) = 3(1)+4(1) = 7 < 20
  2. (1,2) = 3(1)+4(2) = 11 < 20
  3. (1,3) = 3(1)+4(3) = 15 < 20
  4. (1,4) = 3(1)+4(4) = 19 < 20
  5. (2,1) = 3(2)+4(1) = 10 < 20
  6. (2,2) = 3(2)+4(2) = 14 < 20
  7. (2,3) = 3(2)+4(3) = 18 < 20
  8. (3,1) = 3(3)+4(1) = 13 < 20
  9. (3,2) = 3(3)+4(2) = 17 < 20
  10. (4,1) = 3(4)+4(1) = 16 < 20
  11. (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?

2

I think POE is the best strategy here.
List down few possible sums,
3+4 (7), 3+8 (11), 3+12 (15), 3+16 (19)
6+4 (10), 6+8 (14), 6+12 (18),
9+4 (13).

Note that I can stop listing here, since I know that there at least two numbers that cannot be made as a sum of positive multiple of 3 and 4 (1, 2) which will make the correct answer less than 19. Hence, option E is wrong. Also, the above list has more than 7 numbers so by POE, you can choose D.

3

Thanks for taking the time. Can you please elaborate a bit more on this?

4

The question is asking how many positive integers less than 20 and we know that there are 19 positive integers in this range (1, …, 19)

And we can’t get the sum of 1, 2, 3, 4, 5, 6 by adding any positive multiple of 3 and 4. Hence, 19 is wrong.