mmesarina

In your picture above , you are assuming that the Y point is on the radial line that crosses through P and X. The statement only says that Y is on the same xy plane. So could be for example in this position : ------4.5-----P-----4.5------X--------------9------------------------Y So could you explain why this case is excluded? thanks

Hi, can someone please show me how this is solved? A certain circular area has its center at point P and has a radius 4, and points X and Y lie in the same plane as the circular area. Does point Y lie outside the circular area? 1) The distance between point P and point X is 4.5 2) the distance between point X and Y is 9 official answer my answer: stmt1 : X is outside the circle because its distance to the center of the circle is > 4. But we do not know anything about Y so => insufficient. stmt2 : we do not know where x or y is with respect to circle so insufficient Both, insufficient because Y could be one the left or right side of the point x at different angles. Sometimes Y would be inside the circle , sometimes not. So my answer is E, but this iincorrect, so what the hell am i doing wrong?

Hi, can someone tell me which is right: instead of : is followed by a subject rather than : is followed by a verb or clause? or is it the opposite, can someone give examples?

I do not agree with the oa, can someone explain? If the integer n is greater than 1 , is n equal to 2? 1) n has exactly two positive factors 2) the difference of any two distinct positive factors of n is odd the oa is My answer: stmt1) n = a * b , a and b can be any numbers, so insufficient. The oa says that if n has exactly 2 positive factors then it is prime, but this is not true, this is true only if the factors are prime numbers. A factor of a number could be a multiple of some prime, so this is confusing me. I thought that the GMAT people differentiated between "factors" and "prime" factors. stm2) n = a * b *c * d * etc. and the difference of any two is odd, that is |a-b| = odd |a-c| = odd |a-d| = odd |b-c| = odd etc, etc from this , I deduce that n has to have only two factors, because otherwise then we can't just pick any two factors and have a difference that is odd. So n = a * b. Now if the difference of a and b is odd, then either of one is odd. Say a is odd, and b even. Then we can have a = 2 and b = 1, or a = 4, and b = 3, etc, etc. so this is insufficient, because n can be any even number. Combining, both I still don't have enough info. Only if they were to tell me that the factors of n are prime, could i say that the answer is B, because the only even prime is 2, but they do not tell me this, so I chose E. Any other explanations? -malena

this is the best explanation from my point of view thanks

Yes, you are right. Sorry, for the mistake. It looks like when I was reviewing my mistakes, since the DS section was section 4, I took column 4 as being the solution column for the DS section, when in fact the colum was number 3, and colum 4 was for section 4. It looks like i made this mistake when correcting some of the problems I got wrong. I guess I was too tired . -Malena

Hi Bob, it seems like you know hwo to solve this problem. Could you provide an explanation when you have time, thank you. As you can see we are all puzzled. -Malena Thank you Bob, I just looked again, and yes the asnswer key is B. The problem was section 4, problem number 11. When I was looking at the answer key, i must have inadvertedly seen the answer key for section 5, problem 11 which is right next to this problem asnwer key, and that one is E. Sorry guys for the mistake, all my fault. Well, that is good for me, looks like I got that right -:) -Malena

The question appeared in testcode 52, from ETS, which is one of the real paper test versions the gmat.com sells. The OA is , which as you guys , I do not agree with . I found already about three OA's in the several paper tests that I found and others to be doubtfully correct. -M

I put the OA behind the SPOILER quotes , it is

thansk guys

thank you!

Can someone please tell me why stmt2 is not sufficient stmt1: (t-r)/r = 0.15 and (t-r) = 3 so r = 3/0.15, sufficient stmt2 : the area of the concrete section is pi * r^2 = 1.29 pi , so r^2 = 1.29 , sufficient so D is my answer but the oA is Sorry for forgetting to attach the question, here it is.

Could someone please tell me what am I doing wrong? thanks My answer: Line equation: y = mx + 30 m = (x2-x1)/(y2-y1) m = (50-0)/(0-30) m = -5/3 so line equation: y = -5/3x + 30 So x has to be a multipel of 3 in order for y to be an integer too. And x is between 0 and 50, 0 Trying numbers for x and stoppin when y goes below 0, since these are the only integer points on the line. x = 3 y = 25 x = 6 y = 20 x = 9 y = 15 x = 12 y = 10 x = 15 y = 5 x = 18 y = 0 therefore, there are only 6 points, but this answer is not one of the answers given, so what did I do wrong? thanks

A wildlife preserve is being planned for 3000 rhinoceroses. The preserve is to contain a total of 10,000 acres of watering area, plus 100 acres of grazing area for each rhinoceros. If the number of rhinocerose is expected to increase by 10 percent, how many thousand acres should the preserve have in order to provide for the increased population? A) 340 B) 330 C) 320 D) 310 E) 300 oa is

Can someone please give a detail explanation? If Carmen had 12 more tapes , she woudlhave twice as many tapes as Rafael. Does Carment have fewer tapes than Rafael? 1) Rafael has more than 5 tapes 2) Carmen has fewer than 12 tapes

Hi, I have no clue why the answer is not B, can someone please explain? Is (3x+8)/(x+2) an integer? 1) x is an integer 2) x = 0 Stmt1: Not the best way to check this stmt1 but I tried replacing x with numbers 0 to 16: with x = 0 (3x+8)/(x+2) = 4 , integer with x = 1 through 16,i got all fractions, non integer so insuficient stmt2: x = 0, so the divison is an integer, sufficient. So my answer is B. But the OA is PS: by the way 0 is considered an integer right? thanks

hi , could someone please explain why my thinking is wrong of the four numbers representend on the numberline above,is closest to zero? 1) q = -s 2) -t stmt1 q = -s , since q mirrors s on the negative side , then the 0 origin must be between q ands, and so r is closest to 0, sufficient stmt2) -t

Hi, I am not sure how to interpret this information, can someone please help A rectangualr tabletop consists of a piece of laminated wood bordered by a thin metal strip along its four edges. The surface area of the tabletop is x square feet, and the total length of the strip before it was attached was x feet. If the tabletop is 3 feet wide, what is its approximate length, in feet? a) 12 b) 10 c) 9 d) 8 e) 6

thanks for noting that. Could you provide your solution when you have time? -malena

My answer They are asking whether we can write K = 2^r Stmt1: K = 2^6 * M , where M can be any integer so possible values: K = 2^6 K = 2^6 * 3 K = 2^6 * 5.. etc so K can be exactly written as K= 2^r some times, but some other times not, because we do not know what M is, so Insufficient Stmt2: K is an even number, since it is not add, so K can be: K = 2 = 2 ^1 K = 4 = 2^2 K = 6 = 2 ^1*3 so K again can be written as K=2^r or K=2^r * M , so insufficient. My answer is E For the second question: they are asking whether the median is greater than the average. Median= number below that partitions set in half. half numbers below and half numbers above. Average = (1 + 3 + 8 + 12 + x)/5 = (24 + x)/5 Stmt1: x > 6 so we have the following possibilities for the position of x a) 1 3 x 8 12 x = 7 b) 1 3 8 x 12 x = 9, 10 or 11 c) 1 3 8 12 x x = 13, 14 ... For case a) : median = x, x = 7 avg = (24 + 7)/5 = 31/5 = 6 1/5 so median > avg For case b) : median = 8 x = 9 , avg = 33/5 = 6 3/5 => median > avg median = 8, x = 10, avg = 34/5 = 6 4/5 => median > avg median = 8, x = 11, avg = 35/5 = 7 => median > avg For case c): median = 8, x = 1000, avg = 1024/5 = 204 4/5 => median Since median > avg fro case a) and b) but median Stmt2: same info as Stmt1 so insufficient. Combined insufficient. so E It was a good exercise for me to do this problem. At first I didn't see immediately why the median could be below or above average, but once I wrote it down, I realized that it is so because the average can be either very small or huge depending on what X adds to the sum of the numbers. But the median will always be either 7 or 8

I still don't get why you say b = 3 * c where c is an integer. b = 3a/5, and you just said that a/5 need not be an integer from stmt1, so I am confused as to your conclusion. However , from 5b = 3a, i can see that for this equation to hold, since 5 and 3 are prime, then 'b' has to be a multiple of 3 and 'a' has to be a multiple of 5, then that would make n/18 = a/5 = k5/5 = k, where k is an integer , and n/18 = b/3 = 3P/3 = P , where is an integer, concluding that n/18 is an integer, and so the answer is C.

Hi, thanks for your answer. I followed everything you said until you said b = 3c, where did you get this from? Combining both you only know a/5=b/3, from this you can only conclude that a = 5b/3 or b = 3a/5, so why are you saying that 3 * c where c is an integer. You would be concluding that a/5 is an integer, but you have not proved this. Could you explain ? thank you -Malena

Can someone tell me why I am wrong? Is n/18 integer? 1) 5n/18 is an integer 2) 3n/18 is an integer From stem: they are asking whether n is a multiple of 18 or is n = 18K where K is an integer number. Stm1) 5n/18 is an integer 5n/18 = K n = 18K/5 = 2 * 3^2 * K, where k is any integer. So here n would be a multiple of 18 only if K is a multiple of 5, since k can ben anything then this is insufficient. Stm2) 3n/18 n = 18K/3 = 2 * 3 * K , so n is a multiple of 6, n could be 6 or 18 or 24, so sometimes it could be a multiple of 18 and sometimes not, so insufficient Combined: n is a multiple of 18K/5 and 6k, so n must be divisible by both of these. n should contain then the LCM of both, so 18/5, thus n is multiple of 18K/5. Again, since we do not know what K is then n could be or could be not multiple of 18. when K is a multiple of 5, n is a multiple of 18, but when k is not a multiple of 5 then it is not. so my answer is E, but this seems to be wrong. Can someone explain?

Thanks, but my question was regarding why in one case there are multiple solutions and in the other just one. I know that there is only one answer for the pencils problem, this is a problem that I posed and was solved before. But the main thing is to understand why in one case, there are multiple solutions and in the other just one. Then we can solve similar problems in the future the same way. -Malena

Hi GMAT-HELP. I found this problem similar to the following pencil problem, where at the end one equation with two unknowns turned out to be solvable. Could you explain why in this cupcake problem the equation c*0.3 + d*0.4 = 6 cannot be solved but in the cupcake problem below the equation 23 *x + 21 * y = 130 could be solved? The link for this thread is http://www.www.urch.com/forums/gmat-data-sufficiency/55399-pencils-powerprep.html. Is the reason that in the equatin above we have decimals whereas in the cupcake case we have only integers? Your input is appreciated. thank you -Malena Martha bough several pencils. If each pencil was either 23-cent pencil or 21-cent pencil , how many 23-cent pencils did Martha buy? 1) Martha bought a total of 6 pencils 2) The total valu of the pencils Martha bought was 130 cents SPOILER: B My answer: From stem: 23 * x + 21 * y = Total cost in cents They are askin for x Stmt1) x + y = 6 but we do not know total cost, so can't solve for x and y with the stem equation and this equation Stmt2) 23 *x + 21 * y = 130 Insufficient, we need another equation involving x and y Both, sufficient, since we have two equations and two unknowns. Can solve for x, so C