yngwie

That may be true, but in this question, we dont need to calculate the individual probabilities of picking each hat. Picking 4 hats is one event. That's the probability we need. It doesnt matter that the probability of picking a second red hat is reduced if my first hat was red. All that matters is how many ways can I pick 3 red hats in a series of 4 picks. In other words, this is not required - The probability of RRRB is (4/8)*(3/7)*(2/6)*(4/5) = 2/35 This is required - Considering the scenario of picking 3 red hats, Probability of picking hats in the order RRRB = RRRB/{RRRB, RRBR, RBRR, BRRR} = 1/4 I stand by the answer C.

Let's look at it this way: As a counting method, the number of ways (outcomes, possibilties - I'm not talking about probability yet) I can pick exactly 3 red hats out of this pool is 4 (RRRB, RRBR, RBRR, BRRR). Each set of 4 picks is an individual outcome. I start picking. First hat I get is Red. I lay it aside - no replacement Second hat is also red. I lat it aside Third hat is blue and fourth is red. I picked 4 hats, got exactly 3 red hats. That's one favorable outcome. Probability of this particular outcome (picking RRBR in that order) occurring is - 1/4. Any other picking order has the same probability since I pick four hats without replacement and I'm done with the exercise. So in one set of four picks, the probability of getting 3 red hats is 1/4. In the same way, the probability of getting exactly 3 blue hats in a set of 4 picks is also 1/4. So the probability of picking exactly 3 red or exactly 3 blue hats is the sum of their individual probabilities. I think you are calculating the probability of each pick instead of counting 4 picks as one event. One pick does not tell us if the outcome will be favorable or not. For example, in calculating the probability of picking red hats, if I picked BBBB, then that's not a favorable outcome and must be discarded (not counted).

Agree. A looks like the best choice

Shouldnt it be "broke over ....something". Like "The two kids fought over the toy" E seems to be the best choice

I think C is the answer. I would have gone with E if the comma was after 'disturbed', instead of after 'and'.

Definitely B. Agree with Sidnim's explanation

You pick from the pool of hats 4 times. Each time the hat can either be red or blue. So for each pick there are two possibilities. For 4 picks, there are 2^4 = 16 possibilities. Now out of 4 picks, you can pick EXACTLY 3 red hats in the following ways - RRRB, RRBR, RBRR, BRRR. So the possible outcomes = 4 Probability of picking exactly 3 red hats = 4/16 = 1/4 Similarly picking EXACTLY 3 blue hats in 4 tries, we get BBBR, BBRB, BRBB, RBBB. Possible outcomes = 4, and probability = 4/16 = 1/4 So probability of picking exactly 3 red or exactly 3 blue hats in 4 picks = 1/4 + 1/4 = 1/2 Answer: C

You're right. Dont know why that didnt occur to me.... That's the whole basis of DS.

Where does the underlining end? Is the question like this - Soaring television costs accounted for more than half the spending in the presidential campaign of 1992, a greater proportion than it was in any previous election. "Soaring television costs" is plural, so eliminate any choice that refers to costs as "it" - A,E D implies the presidential campaign of 1992 is greater, so wrong. Between C and B, I would go with B. In C, "they have been" seems redundant. I would say the answer was B.

There is no negative marking. Why dont you just guess instead of skipping. I think your score will be low if you dont answer all the questions. Anyway, depends on what score you are looking for.

Stmt 1: Between 10 and 99, if X and Y have the same two digits, but in reverse order, their difference will always be a multiple of 9. You can check with (x,y) pairs like (12,21), (36,63), (27,72), (13,31) etc. Sufficient. Stmt 2: Again, you can construct (x,y) pairs that satisfy the condition - (20,13) - In this case, (x-y)/9 is not an integer (31,13) - In this case, (x-y)/9 is an integer So stmt 2 alone is insufficient. Answer: A

Stmt 1: 20 behind Pam and 20 in front of Ed. But we dont know how many are between Pam and Ed. Insufficient Stmt 2: 5 in between Pam and Ed. We dont know how many are behind or ahead of the two people. Insufficient Combine 1 and 2 together: The question doesnt tell us who is in front of whom. There can two cases where the number of people in the line will differ. So this is also insufficient. Case 1: Pam is behind Ed in the line. The line will look like this -------20------P--5---E------20------- Number of People = 20+Pam+5+Ed+20 = 47 Case 2: Ed is behind Pam in the line. Then the line will look like ------14------E--5--P-----14------- Number of people in line = 35 Answer: E

Perimeter = 360 ==> l + w = 180 -----------------(1) Stmt 2: As far as data sufficiency is concerned, it doesnt really matter how you interpret the line - "difference between length and width". If you take it as l-w = 60, then l = 60+w. Substituting this in Equation 1, you get 60 + w + w = 180 2w = 120 or w=60. Then length becomes 120. If you interpret it as w-l = 60, then w = 60+l Substituting in equation 1, l+60+l = 180 2l = 120 or l = 60. Then w = 120. Depending on your interpretation, length can be 120 or 60. But the main point is length can be calculated. That's all we need to say Stmt 2 is sufficient. The problem of whether the larger value is always the length or not does not affect the outcome of our result. For example, I can say Stmt 2 gives me l=60 and w=120 (even though that doesnt meet your definition of a rectangle, since w>l). Then I have answered the question "what is the length?", which is what the question asks. Answer : D

How did you get 0.2x = 0.3y? Let's say Club A has 5 members - A1, A2, A3, A4, A5 and Club B has 10 members - B1, B2, B3, B4, B5, B6, B7, B8, B9, B10 Stmt 1: 20% of club A (1 out of 5, let's say A1) joined club B also. Stmt 2: 30% of club B (3 out of 10, say B6, B7 and B8) joined club A So how can A1 = B1, B2, B3? In this example, the ratio of A/B = 5/10 = 1/2 We can also come up with another ratio if A had 12 members and B had 20 members, in which case A/B = 12/20 = 3/5 In my opinion, the answer should be E.

Here's how I do it.... Taking electrolyte's example: x^2-7x+6 The co-efficient of the first term is 1 (call this a) The co-efficient of the middle term is -7 (call this b) The last term (the constant) is 6 (call this c) Find two numbers that when : - multiplied equal the product of first and last terms - added equal the second term These two numbers will be among the factors of a*c So in this case, find two numbers that when - multiplied, equal 1*6 (product of first and last) = 6 - added, equal -7 Factors of a*c (1*6) are 2,3;1,6 -1 and -6 are the numbers that satisfy these two conditions since -1*-6 = 6 and -1 + (-6) = -7. So now we can write the equation as: x^2-1x-6x+6 Take out the common factor x-1 x(x-1)-6(x-1) = (x-1)(x-6) are the factors. ================================== Another example: 3x^2 - 8x + 4 = 0 a=3, b=-8, c=4 Find 2 numbers that when multiplied equal a*c (that is 4*3 = 12) and when added equal b (-8) Factors of a*c are 4,3;6,2;12,1 Clearly, -6 and -2 are the two numbers we need -6*-2 = a*c = 12 -6+(-2)= b = -8 Write the equation as: 3x^2 - 8x + 4 = 0 3x^2 -6x -2x + 4 = 0 3x(x-2) -2(x-2) = 0 (3x-2) (x-2) = 0

Dkumar, This was my first thought. Then I figured that I needed half the area, not half the radius. Of course, half the area is wrong because it's half the radius that determines if the point is closer to the circumference or not. 3/4 is the correct answer.

Not sure if this is the way to approach it, but here goes - Take a circle or radius 6. Area will be 36pi. Draw a smaller concentric circle inside the main circle with area equal to half of the big circle. i.e., area of smaller circle = 18pi. A random point is closer to the circumference of the main circle than its center if it lies outside the smaller circle. So probability of the point lying outside the smaller circle and inside the bigger circle = (Area of big circle - Area of small circle)/Area of big circle = (36pi - 18pi)/36pi = 1/2 Am I right?

Thanks, but not really. I've seen your answers. You're not bad yourself. This one just fell into place so nicely in my mind. Hope it happens on the real test.....

I agree with Serena and Rookie. Answer is C. If it is true that reckless drivers prefer red cars, banning red cars will not get rid of the reckless drivers. They'll still cause the accidents in a car of another color.

I'll go for E. For those choosing D, what's wrong with C? It's the same as D.

Bronco, the argument does not assume that flux of retirees impacts ALL local businesses. It assumes that the flux of retirees impacts only those businesses that cater to retirees. There is no mention in the argument of other businesses (or local business in general) suffering if the number of retirees decreased.

D is correct. C says there are more businesses in Florida that cater to tourists. How does this weaken the argument that if less retirees move into Florida, the businesses that cater to retirees will suffer?

That's correct. That's how I solved it.

1 A starting line up of a team consists of x men and y women. There are also 4 reserve players, 2 of whom are men. If one of the starting players is unable o play and needs to be replaced by one of the reserves, what is the probability that the number of women on the starting team will increase? 1) x+y = 12 2) x/y = 1/3 The reserve team consists of 4 people, 2 men and 2 women. The only way the number of women can increase on the main team is if the injured player is a man AND a woman is picked from the reserve team. i.e., P(Women increasing in main team) = P(Man leaving main team) * P(Woman picked from reserve team) Now, the number of ways a woman can be picked from the reserve team is 2 ways out of 4 i.e., 2/4 =1/2 We need to find out the probability of a man leaving the main team i.e. picking a man out of the main team = Number of men in the main team/Number of players in the main team Stmt 1: Gives us the total number of players in the main team, but we dont know how many are men. So insufficient Stmt 2: Gives us the ratio of men to women in the main team. This is enough to figure out the probability. You can take (men=1, women=3) or (men=3,women=9) or any ratio that equals 1/3. Then the probability of a man leaving the main team = 1/1+3 = 1/4, or 3/3+9 = 3/12 = 1/4. So, the probability that the number of women will increase is 1/2 * 1/4 = 1/8 So Statement B is sufficient. Of course, we wouldnt have to really solve to get the actual answer. Knowing that the statement B can give us the answer is enough.

7. If a, b and c are integers, and a+b-c > 0 , is c > 4 ? 1) a > b - c - 2 2) a Stmt 1: a > b - c -2 and from the question, a+b-c > 0. For a=7,b=10,c=3, both equations are satisfied and c For a=11,b=15,c=5, both equations are satisfied and c > 4. Insufficient. Stmt 2: a a+b From question, a+b-c > 0. From above line, a+b can at most be 4. So a+b-c > 0 is possible only if c is equal to or less than 3. Hence we can say for sure that c is not greater than 4. Sufficient. Answer B