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by balaze90 » Sat Apr 19, 2014 4:04 am
first of all, we have to deduce the equation of the line joining the two points.
y = mx + c

m = -1/3 => x + 3y = 60

Then, since the integer values of y which are lying on the line are 13,14,15,16 and 17.

Substitute each value of y in the equation to find if x is an integer. In all cases x is an integer. So there are 5 points on the line having x and y as integers.
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by GMATGuruNY » Sat Apr 19, 2014 5:44 am
[email protected] wrote:How many points (x, y) lie on the line segment between (22, 12 2/3) and (7, 17 2/3) such that x and y are both integers?


[spoiler]Ans:5[/spoiler]
The equation of a line is as follows:
y = mx + b, where m is the slope, and b is the y-intercept.

Step 1: Determine the slope
m = (y₂ - y�)/(x₂ - x�) = (17 2/3 - 12 2/3)/(7 - 22) = 5/-15 = -1/3.
Thus far, the equation of the line is as follows:
y = (-1/3)x + b.

Step 2: To determine the y-intercept, plug in one of the two given points
Substituting (7, 17 2/3) into y = (-1/3)x + b, we get:
17 2/3 = (-1/3)(7) + b
53/3 = -7/3 + b
60/3 = b
b = 20.

Thus, the final equation of the line is as follows:
y = (-1/3)x + 20

For y to be an integer, x must be a multiple of 3.
List the multiples of 3 between the two given x-values (7 and 22):
9, 12, 15, 18, 21.
Total options = 5.
Last edited by GMATGuruNY on Sat Apr 19, 2014 6:52 am, edited 1 time in total.
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by balaze90 » Sat Apr 19, 2014 6:49 am
y = (-1/3)x + 60 .... here 20 should be present instead of 60 (b = 20)
GMATGuruNY wrote:
[email protected] wrote:How many points (x, y) lie on the line segment between (22, 12 2/3) and (7, 17 2/3) such that x and y are both integers?


[spoiler]Ans:5[/spoiler]
The equation of a line is as follows:
y = mx + b, where m is the slope, and b is the y-intercept.

Step 1: Determine the slope
m = (y₂ - y�)/(x₂ - x�) = (17 2/3 - 12 2/3)/(7 - 22) = 5/-15 = -1/3.
Thus far, the equation of the line is as follows:
y = (-1/3)x + b.

Step 2: To determine the y-intercept, plug in one of the two given points
Substituting (7, 17 2/3) into y = (-1/3)x + b, we get:
17 2/3 = (-1/3)(7) + b
53/3 = -7/3 + b
60/3 = b
b = 20.

Thus, the final equation of the line is as follows:
y = (-1/3)x + 60

For y to be an integer, x must be a multiple of 3.
List the multiples of 3 between the two given x-values (7 and 22):
9, 12, 15, 18, 21.
Total options = 5.
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by GMATGuruNY » Sat Apr 19, 2014 6:54 am
balaze90 wrote:y = (-1/3)x + 60 .... here 20 should be present instead of 60 (b = 20)
Thanks for pointing out the typo, which I've corrected.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
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