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DS - Looked easy but I got them wrong.

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DS - Looked easy but I got them wrong.

by gmatrant » Fri Oct 19, 2007 9:45 am
1. One hour after Chris leaves her home, Charlie leaves his home to intercept her. If
Charlie travels twice as fast as Chris, how long will it take Charlie to intercept Chris?
(1) Chris lives 90 miles due south of Charlie
(2) Charlie’s rate is 30 mph

Ans: E

2. What is the greatest common divisor of positive integers a and b?
(1) a and b share exactly one common factor
(2) a and b are both prime numbers

Ans: A

3.Does positive integer a equal 5?
(1) a is a not factor of 6006
(2) 5 is the largest divisor of a

Ans : B

4.In the rectangle coordinate system, triangle ABC has a vertex at point (0, 56). If
point B is at the origin, then how many points on line AC have integer values for
both their x and y values?
(1) The third vertex of triangle ABC lies on the x-axis, and the triangle has an area of
196
(2) Point A has a positive x coordinate and a y coordinate of zero

Ans : A
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Source: — Data Sufficiency |
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by mschling52 » Fri Oct 19, 2007 11:02 am
1. I agree with E for this one. If we combine both statements, we know the rates of each person and their relative starting positions, but we don't know what direction Chris traveled in (so we don't know how far she was from Charlie to begin with).

2. All positive integers will share the common factor 1. Statement (1) tells us that a and b have only one common factor, so we know it must be 1, which makes 1 the greatest common factor of a and b. Statement (2) tells us that a and b are both prime, so the greatest common factor will have to be 1 or a and b could be the same prime number and the gcf would be a (=b). Therefore, (2) alone is not sufficient b/c the possiblity of a=b is not ruled out.

3. Clearly (1) alone does not give us enough info since there are plenty of numbers in addition to 5 that are not factors of 6006. However, (2) tells us that 5 is the largest divisor of a. Since all positive integers have the divisors 1 and themselves, the largest divisor of a positive integer will always be the number itself. So, for 5 to be the largest divisor of a, a must be equal to 5 and statement (2) alone is sufficient.

4. I would do this one by drawing it out on a graph. Put a point at (0,56) to start. Statement (1) tells us another vertex is at the origin, so add a point at (0,0). Then for the area of the triangle to equal 196 we can solve for the x-coordinate of the third point.

196 = (1/2)*56*x
x = 7

So, add a point at (7,0). If we then draw a line connecting A and C, we would be able to count the number of times it intersects the graph grid (where both x and y will be integers). We don't have to actually do this, just know that we could and that (1) is sufficient.

Statement (2) gives us enough info to place point C somewhere on the positive x-axis, but we don't know where and therefore, we can't know how many times the line AC will intersect the grid. For example if we put point C at (0,0.5) the line will never have a point with integer x and y coordinates. On the other hand, if point C is at (56,0), then AC will intersect with lots of points with integer coordinates (55,1) (54,2) ... (1,55)... therefore, (2) alone is not sufficient
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by gmatrant » Sat Oct 20, 2007 4:53 am
outstanding explanation.. thanks a ton
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