sdasar

I agree with Anish. Good logic.

Yes. I'm approximating 9^2 as 10^2.

Yes. I mean E (both together are not sufficient)

Ans: D both are not enough Std deviation measures the distribution of population on either side of the mean. None of them are enough to prove the hypothesis.

1. C i) x^2 = Y^2 , not enough x/y can be -1 or 1 ii) xy> 0 , not enough x and y are of same sign but we don't know x/y combining both we know x/y = 1 2. B ; 9^2 * 10^15, the closest is 10^17

The Probability is 1/2. For 50% of the circle, A is > B. This is true for any circle that has origin at (0,0). See picture below. [ATTACH]5338[/ATTACH] Added Later: ------------ I stand corrected. I didn't notice the condition a and b are >0. As Manuu posted that this is only possible in 1st quadrant. However with in the circle and with in the 1st quadrant, only half the time the x coordinate is greater than y coordinate, and vice versa. So the probability is 1/8.

AB=BC and area ABD=ACD are not enough to conclude ABCD is parallelogram. See the two possible quadrilaterals in the diagram below that satisfy both conditions. First one is not a parallelogram but the second one is. [ATTACH]5289[/ATTACH]

Stmt1: x = y * ( 3y + 7) ; since y is an ineteger greater than 1 , x = a * y Suff. Stmt2: x * (x-1) = ay; x = y * a/ (x-1) ; x = 5 ,y =10 ; 20 is multiple of 10. But 5 is not multiple of 10. Not Suff. Ans: A

Shud, The way I see this is two triangles ABD and ACD with AD as base. Since both areas are same, the points B and C should be at the same height from base AD. Therefore BC is parallel to AD.

Stmt1: 30% of the class speak French. 30% students = 15. Total students = 50. Suff. Stmt2: All students speak English and no student speak more than two languages. The total students got to be at least 50. There could be students who speak only English. Therefore, we don't know the total students. Not Suff. Ans: A.

Ans: E Stmt1: AB = AC , two adjacent sides are equal. It doesn’t add much to the characteristic of parallelogram. Stmt2: Triagnle ABD = triangle ACD. All it proves is AD is parallel to BC. Combining both doesn’t as much. All it tells is that the quadrilateral is a trapezium.

Shud, How did you assume the diagonal is bisecting the angle? You are assuming it is rhombus. I don't want to venture into trigonometry, but that is the only way we can explain this problem. Let us say one diagonal makes x degrees with base and the other diagonal makes y degrees with base. As you eluded, the ratio is sin x / sin y. Even though I believe there is a relation between x and y, sin x / sin y varies with x. I strongly believe the answer is wrong.

Good one Manish! How did you figure that strange looking symbol is square root?:)

h(n) is divisible by all number from 2 through 50 ( and also by any exponent of 2 up to power 50) with remainder of zero. By adding 1 to h(n), it yields a remainder of 1 when divided by any number from 2 through 50. So none of the numbers (including primes) less than 50 can divide h(n) + 1. If at all it is divisible, that number has to be greater than 50. Like in your example 2*3*4 can only be divisible by prime greater than 4. Similarly 2*3**4*...50 can only divisible by prime greater than 50.

Stmt1: x^2(x-1) 0, (x-1) Stmt2: x^3(1-x) Combining both (x-1) 0. Therefore, x3 Ans: C

gschmilinsky, I don't see any other way. The value of the sqaure root has to be either (x-3) or -(x-3) based on whichever is positive. To get -(x-3) or 3-x , 3-x has to be positive. So the question translates to 'Is 3-x positive ? The rest of the steps and logic are as per CHIX's explanation above.

The issue I've with D and E is the modifier problem. I don't think the phrase 'an amount that is...' is modifying $169 billion. Interesting to see what the OA is.

Stmt1: 2x + 2 is cube of positive integer. Basically it results x = 4 * a^3 – 1, where a is an integer. X = 3, 31 , 107 Not Suff. Stmt2: The average of any x consecutive integers is integer. All this tells is x is odd. Not great help. Not Suff. Even combining, it is insufficient. I think the ans is E.

Actually you can simplify the problem further as 1 freshman, 1 soph, 1 junior, 1 Senior and we need to randomly select two students. That would be 4c2=6 ways and each combination would be one each from different class. So the probability of one such combination (1 freshman and 1 sophomore) would be 1/6.

I too agree with Ans: E The problem becomes unsolvable when one of the variables becomes negative, then the varaible that it is not associated to (based on the two euqations) can tilt the results in positive way for the two statements but act negatively on mt+sp. What I mean is, if ‘t’ is negative , then the variable 'm' can be high enough to make both statements true but can have negative influence on mt+sp. Similarly if ‘p’ is negative, then the magnitude of ‘s’ can dictate the final outcome. To gmatlove suggestion that if (p^2+s^2)/2 * mt + (m^2+t^2)/2 * ps > 0 then mt + ps > 0 is not true for exactly the reason I mentioned above. Let us see some examples: Let us say t = -5 , m = 10 , p=s=1 mp+ts = 10 – 5 > 0 sm+pt = 10 – 5 > 0 (p^2+s^2)/2 * mt + (m^2+t^2)/2 * ps = 1 * -50 + 125/2 * 2 = 75 > 0 But mt +sp = -50 + 1 = -49

ashraf, I don’t know why we need to consider other options when we know b = a-1. Just for completeness, let us look at other options. A. a is even. Not True. a = 7 , b =6 , x = (a*b - 1) = 41 or a = 6 , b =5 x = 29 B. x + b is divisible by a. Not true. It is x-b that is divisible by a. C. x – 1 is divisible by a. Same as above. D. b = a – 1. True E. a + 2 = b + 1. or b = a + 1. Not True.

x divided by a gives remainder b. Therefore a > b. x divided by b gives remainder a-2. Therefore b > a-2 a > b and a-b

I agree with you. The reason I thought that stmt2 is not sufficient is that it makes groups selected same as groups applied. But after re-reading this I agree that there is nothing in the stem that prevents it. Also the way the statements are constructed in stmt1 and stmt2, they are more of conjectures and doesn't really state the turth. It only helps us find number of people applied, but not necessarily number of people selected. Seems like bad construction.

Let a1, a2 ,,,an are the set. The range R is (an - a1) Stmt2: New mean is less than Range ( a1 + a2 + ...an + R ) / (n +1) a1 + a2 +...+an + (an-a1) a1+ a2 +.. + an (a1+a2+...an) /n If the original mean is ax (where a1 ax. if R 0. Not Suff Stmt1: a1 and an are > 0. R That is not enough to say whether R is greater than a1 or not. So we are not sure whether R lies between a1 and an. Combinig both, R ax > a1. Suff. Ans: C

So the terms are x , 5x/2 , 25x/4. The value of 3rd tem divided by the first term is 25/4 provided x is not 0. stmt1: x 0 ; sufficient. Stmt2: x is an integer divisible by 2. x can be zero too. Not suff. Ans: A