greinsanity

Hello! I would like to share a priceless tip that will help you solve even the wildest workers' productivity problems. Just remember that (Number of workers1 x Days1)/(Work done1) always equal (Number of workers2 x Days2)/(Work done2) So no matter which numbers you have here, you just plug in those into the formula substituting the unknown for x. When the work done equals to 1 barn, 1 wall, 1 house, or anything else 1, you put 1 in both denominators, hence, leaving (Number of workers1 x Days1) =(Number of workers2 x Days2). So for example, If 6 workers can build 4 cars in 2 days, then how many days would it take 8 workers to build 6 cars? (6x2)/4=(8xY)/6 3=(8xY)/6 Y=18/8 And, It takes three workers four days to build a wall. How long will it take two workers to build the same wall? (3x4)/1=(2xY)/1 Y=6

When you are given a table with values of x and their relative frequency, is there any fast way to see what the median is? Because for me, it takes too much time to write down each value n-number of times and then see which one stands in the middle. Thank you.

Hello! I was wondering about the format of the score result at the end of the test. Are you offered it or not? If yes, is it a specific score or a something similar to the previous format like showing only 750-800 range ?

In the official GRE notes, they have an example. on p. 37 where they ask about points of intersection of a parabola and a straight line. The only way to solve that problem is to get the quadratic equation right because the answers are in decimal points.

Thank you for this insightful paragraph. It surely helps. I must say, you have very strong scores!

I highly recommend that you do because very often I encounter quadratic equations that are not easy to solve by other methods. Quadratic formula is very quick and easy to use.

Please explain the logic behind the solution: When five (not necessarily distinct) integers are put in ascending order, the median is 16 and the average of the smallest and largest integers is 15. Furthermore, when the smallest and largest numbers are removed from the set, the average of the new smallest and largest integers is 14. What is the smallest value that the largest of the original five integers could have? Ans: 18.

Almost manually, I found that the answer is -12. I solved for the first 8 sequences and found a pattern that repeats itself every 6 numbers. Then I took 99/6=16.5 I summed up the first 6 numbers, multiplied them by 16 and added the first three numbers of the sequence. Got -12.

It looks like zero to me, NOT O. So, B is right then.

I have a different answer. We want to find an X such that: 2a-3.6=X or a=(X+3.6)/2 Note that 4.1 Hence, when I plug in the possible values for X, I find that for b), c), d), and e) my a lies within the range 4.1

The area of the square: x*x=x^2 the radius of the inscribed circle: x/2. Hence the area of the circle by using formula Pi*r^2=Pi*(x/2)^2=(Pi*x^2)/4=3.14/4 x^2=0.785 x^2. x^2-0.785x^2=0.215 x^2 0.215 or 21.5% is less than 25%

Please provide less obvious but still no-formula-use examples like this.

Can I ask you solve this problem for me step by step? I see a lot of problems like this in practice GRE tests. Is there a universal approach on how to tackle such questions where large amount of numbers is used?

4B+6G=10 Total: (4*3*6*5)/(2*1*2*1) (10*9*8*7)/(4*3*2*1) You have divisors in numerator and denominator because you show that the order of those disks doesn't matter to you. This all adds up to 3/7.

The solution is straight forward and does not require the use of the rectangle. So you are asked to find the area of the triangle with the height and the base given. If you know those two values, no matter which form the triangle is, the area is (height x base)/2. Hence, (5x7)/2=35/2=17.5