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Parallelogram ABCD is to be constructed in the xy-plane such [#permalink]

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05 Jul 2013, 04:08

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Parallelogram ABCD is to be constructed in the xy-plane such that the sides are parallel to x and y axis. The x and y coordinates of A, B, C and D are to be integers that lies between the lines |x-1| = 5 and |y+5| = 6. How many different parallelograms with these properties could be constructed? (A) 10 (B) 110 (C) 1980 (D) 4950 (E) 9900

|x-1| = 5 \(-4{\leq}x{\leq}6\) or 9 integral values or 9 lines parallel to y axis

|y+5| = 6 \(-11{\leq}y{\leq}1\) or 11 integral values or 11 lines parallel to x axis

Any 4 lines will give a parallelogram (2 lines parallel to x and 2 parallel to y axis) = \(^{9}C_2*{}^{11}C_2\) = 1980

1/4*n(n+1)*m(m+1), this formula is also correct but the here n is number of columns (not the number of points), so \(-4{\leq}x{\leq}6\) means n = 8 columns) and m = 10.

Re: Parallelogram ABCD is to be constructed in the xy-plane such [#permalink]

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05 Jul 2013, 09:08

I tried many a ways but couldn't get the correct answer. but here is my approach to the question:

|x -1| = 5 ==> -4 <= x <= 6; |y+5| = 6 ==> 1 <= y <= 11;

since the parallelogram should lie between the values formed by x and y, we cannot consider the boundary values hence there can be 9 distinct values of x and 9 distinct values of y;

selecting 2 distinct values of x and 2 distinct values of y = \(9C1.8C1.9C1.8C1 = 72 * 72 = 5184\)

So I assume 5184 is the answer. But I feel I am missing something here...

Please find attached picture as to why we need only 2 disctinct values of x and y instead of 4; (It is mentioned in the question that two sides are parallel to x and y)

Re: Parallelogram ABCD is to be constructed in the xy-plane such [#permalink]

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05 Jul 2013, 11:04

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GMATSPARTAN wrote:

I tried many a ways but couldn't get the correct answer. but here is my approach to the question:

|x -1| = 5 ==> -4 <= x <= 6; |y+5| = 6 ==> 1 <= y <= 11;

since the parallelogram should lie between the values formed by x and y, we cannot consider the boundary values hence there can be 9 distinct values of x and 9 distinct values of y;

selecting 2 distinct values of x and 2 distinct values of y = \(9C1.8C1.9C1.8C1 = 72 * 72 = 5184\)

So I assume 5184 is the answer. But I feel I am missing something here...

Please find attached picture as to why we need only 2 disctinct values of x and y instead of 4; (It is mentioned in the question that two sides are parallel to x and y)

Your approach is partially correct but you need to consider the end points as well so there are going to be 10 distinct values for x and 11 distinct values for y. So total possible number of combinations would be = 10 x 9 x 11 x 10 = 9900

But please do note that choosing x1 = 0, x2 = 4 is the same as choosing x1 = 4 and x2 = 0 so you need to divide the total number by 2

Hence answer is 4950.. Hope this helps..
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Re: Parallelogram ABCD is to be constructed in the xy-plane such [#permalink]

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05 Jul 2013, 15:10

Transcendentalist wrote:

GMATSPARTAN wrote:

I tried many a ways but couldn't get the correct answer. but here is my approach to the question:

|x -1| = 5 ==> -4 <= x <= 6; |y+5| = 6 ==> 1 <= y <= 11;

since the parallelogram should lie between the values formed by x and y, we cannot consider the boundary values hence there can be 9 distinct values of x and 9 distinct values of y;

selecting 2 distinct values of x and 2 distinct values of y = \(9C1.8C1.9C1.8C1 = 72 * 72 = 5184\)

So I assume 5184 is the answer. But I feel I am missing something here...

Please find attached picture as to why we need only 2 disctinct values of x and y instead of 4; (It is mentioned in the question that two sides are parallel to x and y)

Your approach is partially correct but you need to consider the end points as well so there are going to be 10 distinct values for x and 11 distinct values for y. So total possible number of combinations would be = 10 x 9 x 11 x 10 = 9900

But please do note that choosing x1 = 0, x2 = 4 is the same as choosing x1 = 4 and x2 = 0 so you need to divide the total number by 2

Hence answer is 4950.. Hope this helps..

If I consider end points as well, then it should be \(11C1.10C1.11C1.10C1 /2\)
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Re: Parallelogram ABCD is to be constructed in the xy-plane such [#permalink]

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09 Jul 2013, 05:28

The boundaries for |x-1| = 5 are: X=6 and X=-4. This have 11 co-odinates with X ordinate as integer hence 10 rows of single unit lenght (L) including the end points. The boundaries for |Y+5| = 6 are: Y=1 and Y=-11. This have 13 co-odinates with Y ordinate as integer hence 12 columns of single unit breadth (B) including the end points. As the Parallelogram are parallel to X and Y axis, its a rectangle.

No. of rectangle for L=1: 10*12 number with B=1, 10*11 with B=2, 10*10 with B=3, ........, 10*1 with B=12 Total = 10*12+10*11+10*10+....+10*1 = 10*[12+11+10+….+1] = 10*12*(12+1)/2 = 10*78

No. of rectangle for L=2: 9*12 with B=1, 9*11 with B=2, 9*10 with B=3, ........, 9*1 with B=12 Total = 9*12+9*11+9*10+....+9*1 = 9*[12+11+10+….+1] = 9*12*(12+1)/2 = 9*78

Similarly for L=3, 4, … 10 the value will be 8*78, 7*78,….., 1*78

Total no. of rectangles possible: =10*78+9*78+……+1*78 =78*[10+9+…..+1] =78*10*11/2 =78*55 =4290

If end points are not included then the number of parallelograms can be similarly calculated as 1980.

Re: Parallelogram ABCD is to be constructed in the xy-plane such [#permalink]

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23 Jul 2013, 13:47

The x and y coordinates of A, B, C and D are to be integers that lies between the lines |x-1| = 5 and |y+5| = 6.

" BETWEEN THE LINES ", onward I must read in between the lines much carefully
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Re: Parallelogram ABCD is to be constructed in the xy-plane such [#permalink]

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22 Aug 2013, 07:13

The stem states that the 4 sides are parallel to both x and y axis, so it's a quadrilateral or a square. |x-1| = 5 --> -4 <= x <= 6 (11 integers) |y+5| = 6 --> -11 <= y <= 1 (13 integers) But the vertices of the quadrilateral "lies between the lines" and are integers, so we cannot count those that stands or lies on one of the lines or borders, therefore we need to subtract 2 possibilities from each 'direction'. So have to rephrase the statement: -4 < x < 6 (9 integers) -11 < y < 1 (11 integers). As we concluded this is a quadrilateral, we need to choose 2 integer points from each range (2 points from the possible x integers and 2 from the y ones) and multiply them: ^{9}C_2*{}^{11}C_2 = 1980. (9¦2)*(11¦2)