String of challenging DS questions

[crak.gmat]

Seeking tips on solving these problems.. take a look at the attached.

OA

1. A
2. C
3. E Actually someone answered this already, so you may ignore. (The 2 angles could be 45-90 OR 60-120)
4. E
5. E


Thanks for your help in advance!!

[crak.gmat]

Sorry here is the attachment

[gmat_nov_2008]

Hi...New to the post. My answers are as follows:

1. (1)
(1) gives R/(R + Remaining) > W (W + Remaining). Therefore probability of R is greater.
(2) gives B = R + W. Is not sufficient.

2. (3)
(1) gives us that n is not an EVEN integer.
(2) gives us n is not a multiple of 3.
Combining the two n can be 1,5,7,11...and substituting always gives a remainder of 0.

3. (3) My thoughts:
since the sum of internal angles of a quad is 360.

Therefore 180 + 2x + x = 360 gives x = 60, and other angle = 120.

4. (3)
(1) gives us R0 = 3MO (Ofcourse the absolute value)
(2) gives us M - R = 12 (ABsolute value)
Also M = M0 and R = R0 (Distance from 0 on the number line)

Combining we can solve the above equations.
5. (2)
I thought an intersection should be at tree points. But I may be wrong.

Thanks crak.gmat

[crak.gmat]

Thanks for your responses.

The one on #1 and 2 are right.

#3, 4 & 5 are wrong.

Official Answers listed above. Still looking for explanations for 4 & 5

[ashishjha100]

1. is " r/(b+w+r) > w/(b+w+r)" ? or is " r > w"?
st1: r(b+w) > w(b+w)
rb+r^2-wb-w^2>0
b(r-w)+r^2-w^2>0
b(r-w)+(r-w)(r+w)>0
(r-w)(b+r+w)>0
==> r>w
So A.

2. Agree with reasoning by crack.gmat

3.Just draw and see .......different possible angles can be obtained from 1 & 2 statements ......(90,90,45,135) (90,90,120,60)

4. st1: we cannot say whether r is +ve or -ve as we need specific value of r. or |r| = 3|m|

st2. many sets of (m,r)...(11,13), (10,14) etc......ie average of m&r is 12

combining: m+r = 24 and r=3m solving we get r=18

5. E ...i cant explain.....just draw in a piece of paper.

[mim3]

[quote="ashishjha100"][b]1[/b]. is " r/(b+w+r) > w/(b+w+r)" ? or[b] is " r > w"?[/b]
st1: r(b+w) > w(b+w)
rb+r^2-wb-w^2>0
b(r-w)+r^2-w^2>0
b(r-w)+(r-w)(r+w)>0
(r-w)(b+r+w)>0
==> r>w
So [b]A[/b].

2. Agree with reasoning by crack.gmat

3.Just draw and see .......different possible angles can be obtained from 1 & 2 statements ......(90,90,45,135) (90,90,120,60)

4. st1: we cannot say whether r is +ve or -ve as we need specific value of r. or |r| = 3|m|

st2. many sets of (m,r)...(11,13), (10,14) etc......ie average of m&r is 12

combining: m+r = 24 and r=3m solving we get r=18

5. E ...i cant explain.....just draw in a piece of paper.[/quote]





4. I agree that R can be 18, but it can also be 36, thus E is the answer. If we look at a number line we can see both scenarios that satisfy both 1 & 2:

m r
--------------------(0)-------(6)------(12)------(18)

m r
-------(-12)-------(0)------(12)----------------(36)

Hope that makes sense.

5. This one is tricky, but think of an intersection as any time that the triangle touches the circle but doesn't pass through...(kind of like an intersection on a road) with this in mind, you can see how a triangle can "intersect" the circle 1,2,3,4,5 or 6 times.

[rosh26]

for problem 1, i dont see where this step follows from:

b(r-w)+(r-w)(r+w)>0
(r-w)(b+r+w)>0

Please explain.

[rosh26]

Can someone explain 4 furthur?

It looks like everyone is choosing C when the answer is E