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Bunuel
How many pairs positive integers x and y satisfy 2x + y < 40, if y is a multiple of x?

A. 90
B. 91
C. 92
D. 93
E. 94

Are You Up For the Challenge: 700 Level Questions
It's much harder and lengthy than expected questioning GMAT so I can only call it a fine practice to do just to clear concepts

Answer: Option C

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Bunuel
How many pairs positive integers x and y satisfy 2x + y < 40, if y is a multiple of x?

A. 90
B. 91
C. 92
D. 93
E. 94
 
Since x and y are positive integers and y is a multiple of x, then y ≥ x.

If x = 1, we have:

2(1) + y < 40

y < 38

So y can be any integer from 1 to 37, inclusive. Therefore, there are 37 pairs of positive integers x and y (or solutions) when x = 1.

If x = 2, we have:

2(2) + y < 40

y < 36

So y can be any even integer from 2 to 34, inclusive. Therefore, there are 17 solutions when x = 2.

If x = 3, we have:

2(3) + y < 40

y < 34

So y can be any multiple of 3 from 3 to 33, inclusive. Therefore, there are 11 solutions when x = 3.

If x = 4, we have:

2(4) + y < 40

y < 32

So y can be any multiples of 4 from 4 to 28, inclusive. Therefore, there are 7 solutions when x = 4.

If x = 5, we have:

2(5) + y < 40

y < 30

So y can be any multiples of 5 from 5 to 25, inclusive. Therefore, there are 5 solutions when x = 5.

If x = 6, we have:

2(6) + y < 40

y < 28

So y can be any multiples of 6 from 6 to 24, inclusive. Therefore, there are 4 solutions when x = 6.

If x = 7, we have:

2(7) + y < 40

y < 26

So y can be any multiples of 7 from 7 to 21, inclusive. Therefore, there are 3 solutions when x = 7.

If x = 8, we have:

2(8) + y < 40

y < 24

So y can be any multiples of 8 from 8 to 16, inclusive. Therefore, there are 2 solutions when x = 8.

If x = 9, we have:

2(9) + y < 40

y < 22

So y can be any multiples of 9 from 9 to 18, inclusive. Therefore, there are 2 solutions when x = 9.

If x = 10, we have:

2(10) + y < 40

y < 20

So y can only be 10. Therefore, there is only 1 solution when x = 10.

Similarly, when x = 11, 12 or 13, there is 1 solution for each of these x-values. If x ≥ 14, then there are no solutions. (For example, if x = 14, 2(14) + y < 40 → y < 12. However, recall that y ≥ x since y is a multiple of x.)

Therefore, the total number of solutions is:

37 + 17 + 11 + 7 + 5 + 4 + 3 + 2 x 2 + 1 x 4 = 54 + 18 + 12 + 4 + 4 = 72 + 20 = 92

Answer: C­
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