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Anybody here??? If yes, please help me! Thank you

Anybody can help to find out what is next after: 1) Approach (V) + which prepositions (on, to, for....) 2) Upheld (V) + which prepositions (on, to, for....) if each word have more than one combination of preposition word, please let me know including their meanings. Thank you very much.

Hi Vinay, I think I understand but want to make sure as follow: According to this question by Funcrazy, Suppose a bag contains 4 blue marbles and 6 red marbles. What is the probability that you can take out two marbles of the same colour? Here is her/his solution: P(B and B) = 4/10 X 3/9 = 12/90 P(R and R) = 6/10 X 5/9 = 30/90 P(P(B and B) U P(R and R)) = 12/90 + 30/90 = 42/90 = 7/15 Refer to Soo_raj's above solution method, P(B and B) = 4/10 X 3/9 = 12/90 should be P(B and B) = 4/10 X 3/9 X 2!/2! (b/c each time to pick 2 balls, it is a permutation - without any replacement.) And P(R and R) = 6/10 X 5/9 should be P(R and R) = 6/10 X 5/9 X 2!/2! . However, the answers of this question should be the the same b/c 2!/2! = 1. (no impact to insert 2!/2! Am I correct? Thanks.

1. Japan received huge sums of capital from the United States after the Second World War, using it to help build a modern industrial system. (A) Japan received huge sums of capital from the United States after the Second World War, using it to help build (B) Japan received huge sums of capital from the United States after the Second World War and used it to help in building ©Japan used the huge sums of capital it received from the United States after the Second World War to help build (D) Japan’s huge sums of capital received from the United States after the Second World War were used to help it in building (E) Receiving huge sums of capital from the United States after the Second World War, Japan used it to help build Correct answer is C. Is help in building a correct usage? How about to help build? help + V1 (without "to") ? 2. For many people, household labor remains demanding even if able to afford household appliances their grandparents would find a miracle. A) same B) despite being able to afford household appliances their grandparents would find a miracle C) even if they can afford household appliances their grandparents would have found miraculous The correct answer is C. But I think it is wrong b/c I've never seen any condition clause like "if S + present tense, S + would have + V3" As I know, it should be "if s + present, S + future tense. 3) Because natural gas is composed mostly of methane, a simple hydrocarbon, vehicles powered by natural gas emit less of certain pollutants than the burning of gasoline or diesel fuel. (A) less of certain pollutants than the burning of gasoline or diesel fuel (B) fewer of certain pollutants than burning gasoline or diesel fuel do © less of certain pollutants than gasoline or diesel fuel (D) fewer of certain pollutants than does burning gasoline or diesel fuel (E) less of certain pollutants than those burning gasoline or diesel fuel The correct answer is E; but are not pollutants countable???

Please refer to your previous explanation as follow: First, we can eliminate all the answer choices that use with communal eating and social facilities included or any other structure that follows this pattern: preposition + noun + participle This pattern is almost always wrong on the GMAT But, please read the below question: In December of 1987 an automobile manufacturer pleaded no contest to criminal charges of odometer tampering and agreed to pay more than $16 million in civil damages for car that were test-driven with their odometers disconnected. A) same as above B) cars that it had test-driven with their disconnected odometers C) its cars having been test-driven with disconnected odometers D) having test-driven cars with their odometers disconnected Based on your explanation, "D" must be wrong (preposition + noun + participle, with + their odometers + participle). Then, I choose "B" instead. But according to the solution, D is the correct answer. Anyone can help me to explain why D is correct (despite with that pattern). Thank you again.

Good job everyone

Hey Guys! Thank you so much. I think this method (I understand that this is a conventional method) is much more easier. Total Outcomes = 15C5 = 15!/5!10! = 21X13X11 Possible Outcomes = 5C2 X 10C3 = 10 X 120 = 1200 (not Possible Outcomes = 5C2 + 10C3 = 10 + 120 = 130) P of obtaining 2R and 3B = 1200/(21X13X11) = 1200/3003 = 400/1001 For sooraj_j's solving process, I still don't understand that he mentioned that the balls can be in different order so that's why we have to multiply 5/15 *4/14 * 10/13 * 9/12 * 8/11 by 5! /( 3! * 2!) RRGGG is not equal to GGGRR?? In my viewpoint, they are. I think I am wrong. Anyone can correct my thought? Thank you,

Oh! 0/1 = 0 but 1/0 = undefined. Am I correct?

Vinay, the more I look at coins question, the more I am confused. Now, I am posting a similar question from your recommended website. I tried to compare between your problem solving process and this method. But I don't understand how to apply the same logic as below: A coin is tossed five times. (This is the same as a sample size of five). What is the probability of obtaining exactly two heads in the five tosses? It is known, by prior knowledge, that the probability of a single success (probability of a head in one toss of a coin) is fifty percent. The question is looking for two successes or two heads in five tosses of a coin. A success is the outcome that is desired to occur. For this example: The number of trials = n = 5 The probability of a single event = p = 1/2 The number of successes that the question is seeking (x = 2). To arrive at the answer to the question the values are entered in the binomial formula. P(2 heads) = (5C2) *(1/2)^2 * (1/2)^3 =31.25% I understand that (1/2)^2 (2 heads) * (1/2)^3 (3 tails) but why we have to multiply it with (5C2) Thank you for your explanation

but if A4 = 0, so 1/0 =0 Am I correct?

An urn contains fifteen balls, five red and ten green. What is the probability of obtaining exactly two red and three green balls in drawing five balls without replacement? According to this question (..the probability of obtaining exactly two red and three green balls in drawing five balls without replacement?), I think we have to use MULTIPLICATION THEOREM to solve this problem. From your recommended website's solution, I think you already saw it. However, I want to solve it on my own way as follow: The chance to get 2 red: 5/15 * 4/14 = 2/21 The chance to get 3 green: 10/15 * 9/14 * 8/13 = 24/91 The question doesn't ask whether either 2 red or 3 green happens, but it ask both of them happen simultaneously. So, we have to use multiply (2/21) with (24/91) to get the answer? ..we can't combine them (if the question ask either of them..we can) However, once I finished calculating them. The answer is different from the answer on http://www.cqeweb.com/Chapters-HTML/Chap3_html/chapter3.htm Thank you for your help. :)

Hi Vinay, Why you mention that this first question's answer is infinity? Can you clarify that? Thanks!

WHAT IF THE QUESTION IS CHANGED TO BE: If A(n) = 1 + 1 , and A5 = 1, then A4 = ? .........A(n+1) so, A(4) = 1/1 + 1 A(4) = 2 I think the answer of modified question is A(4) = 2

Hi NY If A(4) = 1 / (1-1) then thats infinity not 0 right? ----------------------------------------- Infinity is different from 0...coz infinity means that there are more than one answer. But the answer of this question is only 0 Do you think so?

I have another shortcut: If An+1 = 1 + 1 , and A5 = 1, then A4 = ? ..............An so, 1 = 1 + 1 ......... A(4) 1-1 = 1 .......A(4) so, A(4) is 0