Ankost

(1/2)^3 is the probability that he won't be able to buy iced tea. 1-(1/2)^3=7/8 is the probability that he will be able to buy iced tea at one of 3 stores.

It is better to do as chanojmarian Combining x^2+2xy+y^2>6 xy=2 x^2+4+y^2>6 x^2+y^2>2

Xn+1 = 2Xn – 1 X20= 2X19– 1 2X19 – 1-X19=X19 -1 It is better to solve the problem as kluheve did.

When one of th statements says that x, y are integers that means there is a possibility that x and y are not integers. I missed that.

You are right guys

IMO B (x-y)(x+y)=23 23 is the prime number that is divisible only by 23 and 1 a. x-y=1 and x+y=23, y=11, x=12 b. -(x-y)=1 and -(x+y)=23, y=-11, x=-12 1. Insuff. Statement does not say whether x and y are positive or negative 2. x>0, x=12 Suff.

IMO C 1. 6 students could buy 6 books each, 1 student could by 4 books. The range 1 to 4. The median is 2.5. 5 students could buy 1 book each, 1 student 2 books and the 7th student 3 books. the range 1 to 3. The median 1.5 . Insuff. 2.Some students did not buy books, others bought 2 books each. the range 0 to 2. The median is 1. Some students bought 1 book each, some students bought 2 books each. the range 1-2. the median 1.5. Insuff. Combining, The range is 1 to 2. The median 1.5 Suff.

Consider x=y=sqrt2 1. (sqrt2+sqrt2)^2>6 true 2. sqrt2*sqrt2=2 true ((sqrt2)^2+(sqrt2)2 >6 wrong. C is insuff.

You are right.

2^20-2^19=2^19*2-2^19=2^19(2-1)= 2^19 Exponents are the product of the base (in this case base=2) n times (n in this case=20). 2*2*2*..... So 2^20 we can express as 2^19*2.

12 students - A movie. 12(A) 9 students - B. 9(B) 6 students -C. 6 © 2 students watched 3 movies. Total students 20. Students watched 2 movies - x. The formula is n(A U B U C) = n(A) + n(B) + n© - n(A intersect B) - n(B intersect C) - n(A intersect C)- 2*n(A intersect B intersect C) Where n(A intersect B) - n(B intersect C) - n(A intersect C) is x or number of students watched 2 movies. 2*2(A intersect B intersect C)= 4, this means 2 students watched 3 movies. We need to multiply the number of these students by 2, because they are included in 3 groups, we need to exlude them from 2 groups. 12(A)+9(B)+6©-4-x=20 x=3 Answer 3 students watched 2 movies.

Is ½x½ 1. The statement says that x has two values x=3 and x=1/3. In 1st case x is integer, in another one x is a fraction. Insuff. 2. The statement says that x is not equal to 3, but can be any other number. Insuff. Combining, x is not equal to 3, x is a fraction 1/3. Suff.

It is better to scetch the graph. The general rule is: x coordinate of the vertex -b/2a, y coordinate (4ac-b^2)/4a. Another form of quadratic equation is y=a(x-k)^2+h where k is value by which the graph is shifted to the right or to the left from y-axis, h is y coordinate of the vertex, a is the slope of the graph. In this case the graph is shifted to the right by 1 unit, the equation takes the form (x-1)^2. The slope is -1 because the graph is downwards. To find h, we can plug y-intercept point (0,3) in the above equation. 3=-1(0-1)^2+h, h=4 Actually, i did not do any calculations like this because it is time consuming. I just simply sketched the graph, found x-coordinate of the vertex, took a look how the graph is increasing from the point (-3,0) to the point (0,3). It is easy to find that from x=-3 to x=0, y increases by one point from y=0 to y=3. In this case the quadratic equation is as follows: y=-x+2x+3 or -(x-1)^2+3 Another link: Quadratic Functions(General Form)

E If denominator has other prime numbers at the base than 2 or 5, then deciaml is not terminated. A. 189=3^3*7 eliminated because prime numbers are 3 and 7 B. 196=2^2*7^2 eliminated because denominator has 7 C. 225=3^2*5^2 eliminated, 3 is in the denominator D. 144=2^4*3^2 eliminated, 3 is in the denominator E. 128=2^7 OK. No other prime numbers except of 2

Ranjeet, how you figure out that i am Sir? Sorry, i have to disappoint you, this is wrong answer.:) parabola has a quadratic equation like ax^2-bx+c=0. That calls quadratic because x has power of 2. Here is a link Quadratic Functions Unfortunately, i can't answer your question regarding whether quadratic graphs might be tested in GMAT exam. Probably, Bob 800 would answer it. Richita, by increasing x for one point on the graph, y also increases by one point. The graph crosses y axis at x=0 and y=3. So, adding 1 to x, we need to add 1 to y.