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y = 2x+20, where y is an integer and 0 < x < 19.
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26 Feb 2025, 02:07
26 Feb 2025, 02:07
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45% (01:04) correct
55% (00:54) wrong based on 20 sessions
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\( y = 2x+20\), where y is an integer and \(0 < x < 19\).
Quantity A |
Quantity B |
The number of different possible values of y |
18 |
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Re: y = 2x+20, where y is an integer and 0 < x < 19.
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26 Feb 2025, 09:58
26 Feb 2025, 09:58
2
It's a bit tricky because if you assume that x must be an integer, then there are 18 possible values of y.
However, since y = 2x + 20, for every "half" integer value of x (e.g., when x = 0.5, 1.5), we get an integer value for y.
E.g., if x = 0.5, y = 2(0.5)+20 = 21
Since 21 is an integer, it satisfies the given constraints.
Therefore, the minimum possible value of x is 0.5 and the max is 18.5, and there are 18*2 = 36 total possible values of y.
However, since y = 2x + 20, for every "half" integer value of x (e.g., when x = 0.5, 1.5), we get an integer value for y.
E.g., if x = 0.5, y = 2(0.5)+20 = 21
Since 21 is an integer, it satisfies the given constraints.
Therefore, the minimum possible value of x is 0.5 and the max is 18.5, and there are 18*2 = 36 total possible values of y.
gmatclubot
Re: y = 2x+20, where y is an integer and 0 < x < 19. [#permalink]
26 Feb 2025, 09:58
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