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14 Apr 2026, 19:00
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B
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Difficulty:
55%
(hard)
Question Stats:
65% (02:32) correct
35%
(02:47)
wrong
based on 92
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History
Date
Time
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Not Attempted Yet
If 3x + y = 41 and |2y| > 15, how many pairs of values {x, y} exist where x and y are both positive integers ?
A. 10
B. 11
C. 12
D. 33
E. 38
A. 10
B. 11
C. 12
D. 33
E. 38
BongBideshini
Joined: 25 May 2021
Last visit: 20 Apr 2026
Posts: 121
Given Kudos: 74
Location: India
Concentration: Entrepreneurship, Marketing
Schools: HBS '27 Stanford '27 Wharton '27 Sloan '27
GPA: 3.8
WE:Engineering (Government)
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14 Apr 2026, 22:19
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From
3x + y = 41 → y = 41 − 3x
|2y| > 15
Because y > 0
|2y| = 2y
So:
2y > 15
y > 7.5
y ≥ 8
Substitute y:
41 − 3x ≥ 8
3x ≤ 33
x ≤ 11
So,
x ≥ 1 and x ≤ 11
So x = 1 to 11
total 11 values
Each gives a valid y
Answer: B
3x + y = 41 → y = 41 − 3x
|2y| > 15
Because y > 0
|2y| = 2y
So:
2y > 15
y > 7.5
y ≥ 8
Substitute y:
41 − 3x ≥ 8
3x ≤ 33
x ≤ 11
So,
x ≥ 1 and x ≤ 11
So x = 1 to 11
total 11 values
Each gives a valid y
Answer: B
ExpertsGlobal5
14 Apr 2026, 23:29
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Bookmarks
From 3x + y = 41, we get y = 41 - 3x.
Since x and y must be positive integers, y ≥ 1 means 41 - 3x ≥ 1, so x ≤ 13. Also x ≥ 1. So x can range from 1 to 13, giving 13 possible pairs before applying the constraint.
Now applying |2y| > 15, which means |y| > 7.5, so since y is a positive integer, y ≥ 8.
y = 41 - 3x ≥ 8 means 3x ≤ 33, so x ≤ 11.
So x can range from 1 to 11, giving 11 pairs.
Answe B
Since x and y must be positive integers, y ≥ 1 means 41 - 3x ≥ 1, so x ≤ 13. Also x ≥ 1. So x can range from 1 to 13, giving 13 possible pairs before applying the constraint.
Now applying |2y| > 15, which means |y| > 7.5, so since y is a positive integer, y ≥ 8.
y = 41 - 3x ≥ 8 means 3x ≤ 33, so x ≤ 11.
So x can range from 1 to 11, giving 11 pairs.
Answe B
