gmatfundoo

VINDICATE: REWARDING The above appeared on the GRE this month. Don't have the options. Can someone come up with a bridge. I came up with: You VINDICATE (verb: To justify or prove the worth of, especially in light of later developments) someone by REWARDING the person The above meaning of VINDICATE is from vindicate - definition of vindicate by the Free Online Dictionary, Thesaurus and Encyclopedia. However, I was hoping someone could do better, and throw more light on this. thanks!

It is pretty evident if you take simple examples such as If a+b = 10, then maximum possible value of ab = 25 (5*5). As the values of a and b come closer, the product ab increases and reaches a maximum when both are equal. This can be proved algebraically as well: If a + b = 10, lets try and find the maximum value of ab b = 10 -a ab = a(10-a) = 10a - a^2 = 25 - (a-5)^2 Now 25 - (a-5)^2 would be maximum if (a-5)^2 is minimum that is when (a-5)^2 = 0 (Since minimum value of any square number is 0) Thus for ab to be maximum a = 5 and b = 5 Further, the value of ab (which is 25 - (a-5)^2) keeps on decreasing as the difference between a and 5 increases, or in other words when the difference between a and b increases. Hope this helps P.S. The same logic applies to the rule that " a rectangle of given perimeter has maximum area when it is a square"

Answer B If the sum of two positive numbers is known, then the closer the numbers, the higher their product Here, the sum of the each of the two pairs (100,210+90,021) and (100,021+90,210) is the same. But the pair in Column B is closer than that in Column A (or in other words the difference of the numbers in Column B is less). So Column B must be greater.

The total number of ways in which 5 people can be arranged around a table = (5-1)! = 24 So the answer must be less than 24 @ndakota I think anticlockwise and clockwise are treated as same only in case of a necklace, garland or any thing that can be flipped around. In case of a circular table clockwise and anticlockwise arrangements must be taken as different. Also, I would be interested to know whether you got the answer 8 by some different method. If so, I would really appreciate if you could share it with us

This is a circular permutation question: IMO answer should be 8 Total no. of ways in which 5 people can be arranged around a table = 4! = 24 Let the people A1,A2, B1, B2, C (The A’s and B’s are the sibling pairs) Its easier to do if we count the number of arrangements when at least one of the sibling pairs is together, and then subtract the same from the total arrangements Lets first take the case when A1,A2 are always together: For two things to be together we treat both as a single unit Let A =No. of arrangements when are A1,A2 together = 3! *2= 12 (we multiply by 2 since the two A's among themselves can be arranged in 2! ways) Let B = No. of arrangements when B1, B2 are together = 12 Let C = No. of arrangements when both pairs are together = 8 (these have been counted in both A and B) Therefore, no. of arrangements when at least one of the sibling pairs is together = A + B – C = 16 No. of arrangements when the siblings are separated = 24 – 16 = 8 Someone please correct me if I'm wrong What is the OA?

@CB One Slight error, Let's say you take (1,8) as a possible Maximum-minimum combination so as to get a range of 7, the other 4 numbers will be from the range 2 to 7, instead of all the remaining 8 numbers. Therefore a total of 6C2 combinations instead of 8C2, for each of the 3 cases This is because we have already fixed the maximum and minimum values. In the method you have applied, you may have tried to subtract the repetitions, but you have counted sets like (1,2,3,9), (1,2,3,10), (2,8,9,10) and so on. For such sets the range is more than 7. Hope this helps.

4. A In triangle ACD AD + CD >AC (sum of any 2 sides of a triangle must always be greater than the third side) therefore, AC Now, In isosceles triangle ABC angle ABC must be less than the other two angles Hence in triangle ABC angle ABC must be less than 60 and the other two angles (BAC and BCA) must be greater than 60 (since sum of the 3 angles =180) Answer A

3. Difficult to explain without drawing. Even then I'll give it a shot Lets first mark the points. Let center of the left circle be A and center of the right circle be B Also Let the upper and lower points of intersection be C and D respectively Now perimeter of the swimming pool = Length of major arc (CD) of the left circle+ Major arc (CD) of the right circle = 2 Length of major arc (CD) Form a triangle ABC: the triangle is equilateral with each side being equal to the radius. Therefore, angle CAB = 60 degrees angle CAD = twice angle CAB = 120 degrees larger (reflex) angle CAD = 360-120 = 240 major arc (CD) = 2 * pi * 3 * 240/360 = 4pi Perimeter of the pool = 8pi

1. No. of values from 337 to 350 = 14 No. of values from 1 to 350 = 350 Required probability = 14/350 2. This is an actual GRE question. If I am not mistaken the question is find (x,y) in terms of a, b and K y = a-k (evident from the graph For x: Slope of the line = -a/b Also slope (in terms of x and y) = (y-a)/x Both represent slope of the same line Therefore, -a/b =(y-a)/x x = bk/a Ans: x = bk/a; y = a-k will post the other explanations shortly

I think for each of the three cases, it should be 6C2. If we select (1,8) the other two numbers can be selected from 2 to 7 in 6C2 ways For (2,9), the other two numbers can be selected from 3 to 9 to 6C2 ways.. and so on @computer-bot could you please provide the OA

@accenture Could you please provide the OA if you have it. This question was slightly different from regular GMAT questions

Option B says "Nearby communities have not changed their sources of drinking water" Cant figure out how this weakens the spokesperson's argument. Could you please post the official explanation, if there is one.

IMO C Question: Is z-y = y-x => Is x+z = 2y This is a yes/no question 1 alone: (x + y + z)/3 (x + y +z) Insufficient 2 alone: Median of {x, y, z} = y Median of the set {x, y, z, 4}, should either be (x+y)/2 OR (y+4)/2 Since Median of the set {x, y, z, 4} Either (y+4)/2 OR (x+y)/2 Combining (1) and (2) Median of the set {x, y, z, 4} must be (y+4)/2 Therefore, we obtain the following two conditions: y> 4 (from 2) (x + y +z) Minimum value of y = 5 And so x+z Thus we obtain (X+z) is always less than 2y Ans: No Sufficient © However, I may still be wrong, considering the fact that I am solving DS questions at the end of an extremely hectic day What is the OA

IMO, C The way I have worked it out, on combining 1 and 2, the 24th tile should definitely be green. Will post the explanation if you confirm the OA, because the explanation is going to be a long one. This is a very time consuming DS. Could you also provide us the source of this question. Thanks in advance

Range = maximum – minimum. The following Max-min combinations would give a range of 7: (1, 8), (2, 9), (3, 10) Now for each of these combinations, the other two numbers can be selected, from among the 6 numbers in between, in 6C2 ways. Therefore, total number of selections with range 7 = 3 * 6C2 = 45 Total number of selections of 4 numbers = 10C4 = 210 Required Probability = 45/210 = 3/14 Kindly do provide the OA, in case I may have missed something.