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How many ordered pairs of positive integers (x,y) satisfy th

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How many ordered pairs of positive integers (x,y) satisfy th [#permalink] New post 18 Nov 2018, 10:50
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Question Stats:

71% (00:26) correct 28% (00:40) wrong based on 7 sessions
How many ordered pairs of positive integers (x,y) satisfy the inequality \(2x+3y < 10\)?

A. One

B. Two

C. Three

D. Four

E. Five
[Reveal] Spoiler: OA
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Re: How many ordered pairs of positive integers (x,y) satisfy th [#permalink] New post 25 Nov 2018, 11:24
Explanation:

Here (x, y) can take integer values from 1 as they are positive integers.

So lets put the values one at a time, such that the resulting value is less than 10

1. If x = 1 and y =1, then 2*1 + 3*1 = 5 < 10

2. If x = 2 and y = 1, then 2*2 + 3*1 = 7 < 10

3. If x = 3 and y =1, then 2*3 + 3*1 = 9 < 10

4. If x = 4 and y =1, then 2*4 + 3*1 = 11 > 10 This is not possible

5. If x = 1 and y = 2, then 2*1 + 3*2 = 7 < 10

6. If x = 1 and y = 3, then 2*1 + 3*3 = 11 > 10. This is not possible

So in all, we have 4 pairs: (1, 1), (1, 2), (2, 1), (3, 1) i.e. Option D
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Re: How many ordered pairs of positive integers (x,y) satisfy th   [#permalink] 25 Nov 2018, 11:24
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