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04 Mar 2017, 02:47
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D
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Difficulty:
75%
(hard)
Question Stats:
56% (02:29) correct
44%
(02:26)
wrong
based on 230
sessions
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Line M has x intercept as -40 and y intercept as 70. If the line M cuts x-axis at point P and y-axis at point Q,then how many points on line segment PQ have both x coordinate and y coordinate as integers?
a)40
b)70
c)10
d)11
e)30
a)40
b)70
c)10
d)11
e)30
04 Mar 2017, 04:00
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Bhalla
The equation of a straight line whose x and y intercepts are a and b, respectively, is: \(\frac{x}{a}+\frac{y}{b}=1\) (check for more here: https://gmatclub.com/forum/math-coordin ... 87652.html)
Thus, the equation of line M is -x/40 + y/70 = 1.
y = 7x/4 + 70. For y to be an integer 7x/4 must be an integer, thus x must be a multiple of 4, from 0 to -40 inclusive. There are 11 multiples of 4 in this range: 0, -4, -8, -12, -16, -20, -24, -28, -32, -36, and -40.
Answer: D.
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General Discussion
Updated on: 05 Mar 2017, 07:34
Originally posted by Diwakar003 on 04 Mar 2017, 09:07.
Last edited by Diwakar003 on 05 Mar 2017, 07:34, edited 1 time in total.
Last edited by Diwakar003 on 05 Mar 2017, 07:34, edited 1 time in total.
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Bhalla
We know two points through which the line PQ passes through: (-40,0) and (0,70).
Equation of a line passing through two points can be found using the formula (y-y1)/(y2-y1) = (x-x1)/(x2-x1).
Equation of the line PQ is 4y-7x=280.
y=7(40+x)/4.
For y to be an integer (40+x)/4 should be an integer. So 40+x should be divisible by 4. Here x can vary from -40 to 0. All multiples of 4 between 0 and -40 will satisfy the condition. So there are 11 such numbers, including 0.
Hence D.
Cheers!
