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