NEW SET of good PS(3)

The solution provided by you is well explained.. However for the text marked in red - is there a faster way for solving the eq for k...? Since time crunch is a big issue in GMAT exam? Please let me know your views...

Thanks,
JT

hi Bunuel,
could you please, clarify "evenly divisible" does not mean the quotient should be multiple of 2?
or evenly divisible is equivalent to completely divisible in this context.
thanks.

Bunuel.. Can you explain the solution of Problem #1

I'm curious as well. The language of this question really threw me off.

The language is from GMATPrep

10. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?[/b]
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84

It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...

From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.

We'll have 8 sets of collinear points of three:
3 horizontal {(1,1),(2,1),(3,1)} {(1,2)(2,2)(3,2)}...
3 vertical
2 diagonal {(1,1)(2,2)(3,3)}{(1,3)(2,2)(3,1)}

So the final answer would be; 9C3-8=84-8=76

Answer: B.

Hope it's clear.

Bunuel,

1- Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) -1 )-- Not so sure on this though ...

2- I see a similar Question in OG12 PS Q.229- The method explained here in the above example does not seems to fit too well there. basically in the question we have -4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question....

TIA ~ Yogesh

if anyone can help please to clarify the methos:

let's say that the Q was:
How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6?
how will it be solved:
will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance

if anyone can help please to clarify the methos:

let's say that the Q was:
How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6?
how will it be solved:
will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance

yangsta,
i liked your solution for 4. I didnt know we can use the definition of linear equation to solve such problems.

I used the guessing method.
we have two relationships...6--30 and 24---60.
This means when R is increased 4 times, S increases 2 times, so if R is increased 2 times S will increase 1 time.
Now, 30*3 ~ 100, so 3 times increase in S will have atleast a 6 times increase in R, i.e. R should be something greater than 36..closest is 48

I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind).
Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method.

How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6?

Ok, so we have a total of 36 co-ordinates (as shown below by the red and black dots). We need to make triangles so we need to select a triplet of co-ordinates out of these 36 which can be done in 36C3 ways. Out of these, we need to get rid of those triplets where the points are collinear. How many such triplets are there?
Look at the diagram:


The Black dots are the outermost points. Red dots are the inside points. Now each of these red dots is the center point for 4 sets of collinear points (as shown by the red arrows). Hence the 4*4 = 16 red dots will make 16*4 = 64 triplets of collinear points.
These 64 triplets account for all collinear triplets except those lying on the edges. Each of the 4 edges will account for 4 triplets of collinear points shown by the black arrows. Hence, there will be another 4*4 = 16 triplets of collinear points.
Total triplets of collinear points = 64 + 16 = 80
Therefore, total number of triangles you can make = 36C3 - 80


Similarly you can work with 1<=x<=5 and -9<=y<=3.
The number of red dots in this case = 11*3 = 33
So number of collinear triplets represented by red arrows will be = 33*4 = 132
Number of black arrows will be 3 + 11 + 3 + 11 = 28
Total triplets of collinear points = 132 + 28 = 160
Total triangles in this case = 65C3 - 160

It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw.
my way:
no: of collinear points=?
horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs.
hence horz and vert. lines give a total of 2*6*6C3.
next 2 diagonals give same no: of such possibs.
consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).

Ma'am,
It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw.
my way:
no: of collinear points=?
horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs.
hence horz and vert. lines give a total of 2*6*6C3.
next 2 diagonals give same no: of such possibs.
consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).

Can you please elaborate on the bolded part in details...

Shouldn't the answer to Question 2 be B?