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# Straight Line

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Manager
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Straight Line [#permalink]  12 Dec 2007, 09:03
Can you please provide a way to solve the question in the attachement
Attachments

13.doc [23.5 KiB]

Manager
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I think it is D. First try to define a function of this graph which is y= slope*x + y-intercept
slope = (y2-y1)/(x2-x1) = (0- 30)/(50-0) = - 3/5
y=-3/5x+30
thus all multiples of 5 up to 50 inclusive and 0 are the values of x which would make y an integer.
Director
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d'oh! Agree with D. I was trying to force an answer using the same technique but dividing 30 by 5...instead of 50 by 5. not sure why.
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Re: Straight Line [#permalink]  13 Dec 2007, 23:15
smily_buddy wrote:
Can you please provide a way to solve the question in the attachement

equation of the line is y=-3x/5 +30.

Not sure how to do this any other way then to try some values and elim some choices.

for X: 0 , 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 work. Y=30, 27, 24, 21, 18, 15, 12, 9, 6, 3, 0 . Notice it jumps 3.

-3x/5+150/5 (150-3x)/5, if it makes things easier.

Counting the values I get 11.
D.
Intern
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My answer is C. I think that one of the ways to solve it is to find all similar triangles that are smaller that present one in coordinates. I have found 10.
CEO
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Expert's post
agree with D

y=-3/5*x+30

x=5k ==> N=(50-0)/5+1=11
CIO
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Agree with D.

I like to think of slope as "over 1, up (or down) slope"

So the slope here is -3/5, so that means "over 1, down 3/5"

What I mean by that is that as you move positively along the x axis by 1, y will always change by the slope.

Since we want integers, you can sync up the ratio so that it now reads: "over 5, down 3"

So if we start at (0,30), we go over 5, we go down 3, we get (5,27). Then (10,24), etc...

Now you just need to ask how many 5's there are between 0 and 50. As stated above, that's ((50 - 0)/5) + 1, or 11.
Manager
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ian7777 wrote:
Agree with D.

I like to think of slope as "over 1, up (or down) slope"

So the slope here is -3/5, so that means "over 1, down 3/5"

What I mean by that is that as you move positively along the x axis by 1, y will always change by the slope.

Since we want integers, you can sync up the ratio so that it now reads: "over 5, down 3"

So if we start at (0,30), we go over 5, we go down 3, we get (5,27). Then (10,24), etc...

Now you just need to ask how many 5's there are between 0 and 50. As stated above, that's ((50 - 0)/5) + 1, or 11.

Cool Guys.. Very good Explanations... The OA is D
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