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I think the answer should be 11. Why are 0 & 11 excluded?. Please explain.
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maheshsrini
I think the answer should be 11. Why are 0 & 11 excluded?. Please explain.

You are right. thanks.
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Ok Including 0 we will get an Int but how will 11 help?
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First develope the equation of line with coordinates given:-
Y = MX + C

When X =0, Y=30, putting the values in above equation of line gives the value of C
30 = C

When Y=0, X=30, putting the values in above equation of line gives the value of M
M= - 3/5

The equation of Line is Y= (-3/5)X +30, Now from this equation it is clear that if we want to have both X and Y to be integers, then all values of X has to be multiple of 5, so starting from X=0 to X=50 (Coordinate limits of line PQ), we note that there are 11 integer values of X for which 11 integer values of Y exists in line PQ
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GMATD11

In the figure above, how many of the points on line segment PQ have coordinates that are both integers?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 20

Attachment:
Untitled.png

We have -

\(3x + 5y = 150\)
\(5y = 150 - 3x ---> y = 30 - (3x/5)\)
As, \(x & y\) has to be integers x has to be multiple of y. Also, \(x & y\) has to be postie as it lies in the first quadrant.
\(x = {0,5,10,15,20,25,30,35,40,45,50}\)
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The alternate method would be usiing the slope to find the points on the line

Since we know that slope is \(\frac{-3}{5}\) we know that y intercept is 30 and x intercept is 50.

We have negative slope
If we increase x then x decreases. So if we start at point (0,30) then we can say that next point on this line would be 3 units down from 30 and 5 units right of 0 we will get our point at (5,27). Now other point on this line would be 3 units down from 27 and 5 units right to 5 which would be (10,24) Similarly we see that we get

(15, 21) , (20, 18). ( 25, 15), (30,12) (35,9) (40 ,6) (45,3) (50,0)

So in all we have 11 Points
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Using INTERCEPT Form of straight line : x/xo + y/yo = 1
(xo = x intercept & y0 = y intercept)

So here eq is : x/50 + y/30 = 1
=> x = (1-y/30) 50
=> x = 50-5y/3
Hence all values of y divisble by 3 will give an integer value
now y range is [0,30], so 11 values
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Shouldn’t our approach be to also find the range of Y and then select the answer which would be the lesser of the two? ( the question over here has range 11 for both X and Y but had the ranges been different we would have to calculate ranges for both X and Y - hence a little more time had to be invested in this quest.)

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Asked: In the figure above, how many of the points on line segment PQ have coordinates that are both integers?

Equation of the line PQ: -

(y - 0) = (30-0)/(0-50) (x-50)
y = -3/5 (x-50)
5y = - 3x + 150
3x + 5y = 150

(x,y) = {(0,30),(5,27),(10,24),(15,21),(20,18),(25,15),(30,12),(35,9),(40,6),(45,3),(50,0)}: 11 points

IMO D
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When you get the equation of the line and you have determined that for y to be an integer x must be a multiple of 5, instead of listing which may cost you some seconds just divide 50 by 5. That gives 10, then add x is 0 to give you 11 as your answer.

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Lesson learned : how to create line equation from two coordinates , 5y + 3x = 150
then, x = (150-5y)/3
in order to make x an integer , y must be a multiple of 3 and since y can be from 0 to 30 inclusive, so y can be 0,3,6,9,12,15,18,21,24,27,30 i.e 11 values

GMATD11

In the figure above, how many of the points on line segment PQ have coordinates that are both integers?

(A) 5
(B) 8
(C) 10
(D) 11
(E) 20

Attachment:
Untitled.png

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