IMO B .
Let out of 100 x be no of students playing Rugby or Football or both . From question stem ratio of (Either Rugby or Football or both/None playing either) = 1/3 . So Number of students out of 100 playing neither of any sport is 3x ==> we have sone simple eqn , x + 3x = 100 , gives x = 25 .
Now x = nof only Rugby + nof Football only + nof football and rugby
so number of Football = 25 - 10 = 15
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prindaroy
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yes the question is a little vague. So, essentially, for out of every 7, three play football as well? Are these the only people playing football? so if there are 70 rugby players then 30 of them play football as well and 40 play only rugby? And so, if 10% of the people play only rugby, then the people playing football should be less right?
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Talkativetree
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I love these hard problems. Oddly enough I think they're fun. Is there anywhere else I could find similar problems?
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Brent@GMATPrepNow
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You might try Magoosh.Talkativetree wrote:I love these hard problems. Oddly enough I think they're fun. Is there anywhere else I could find similar problems?
Of the 500 questions there, I'd say that about 100-150 of them are in the 650+ range. (some of them have already been posted in the other forums)
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lunarpower
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hey all.
regarding the "right triangle ABC" problem
(found here: https://www.beatthegmat.com/problem-solv ... 30325.html)
the solutions above would be nice in a geometry course; a lot of them are very pretty.
but - let's assume that this is a gmat problem. (NOTE: this would not conceivably be a gmat problem - it's too obscure/difficult to show up on the test.
if this were a gmat problem:
* it would be multiple choice, with five NUMBERS as answers - say, 30, 35, 40, 45, 50.
therefore:
* that NUMBER would be the same, regardless of the particular shape of the right triangle ABC.
therefore:
* if you can answer the problem for ANY SINGLE EXAMPLE of ABC, then you've got THE answer.
so:
we can just let ABC be a 45-45-90 triangle.
if that's the case, then BO will be a horizontal line, and it will be clear (from symmetry) that BO cuts right angle ABC into two 45-degree angles.
that's all you need.
again, BECAUSE THE PROBLEM WOULD BE MULTIPLE-CHOICE, there's no need to produce such elaborate general solutions; finding just one solution is sufficient.
regarding the "right triangle ABC" problem
(found here: https://www.beatthegmat.com/problem-solv ... 30325.html)
the solutions above would be nice in a geometry course; a lot of them are very pretty.
but - let's assume that this is a gmat problem. (NOTE: this would not conceivably be a gmat problem - it's too obscure/difficult to show up on the test.
if this were a gmat problem:
* it would be multiple choice, with five NUMBERS as answers - say, 30, 35, 40, 45, 50.
therefore:
* that NUMBER would be the same, regardless of the particular shape of the right triangle ABC.
therefore:
* if you can answer the problem for ANY SINGLE EXAMPLE of ABC, then you've got THE answer.
so:
we can just let ABC be a 45-45-90 triangle.
if that's the case, then BO will be a horizontal line, and it will be clear (from symmetry) that BO cuts right angle ABC into two 45-degree angles.
that's all you need.
again, BECAUSE THE PROBLEM WOULD BE MULTIPLE-CHOICE, there's no need to produce such elaborate general solutions; finding just one solution is sufficient.
Ron has been teaching various standardized tests for 20 years.
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Ian Stewart
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Great point, Ron - if you understand this, you can find simple solutions to some of the harder GMAT problems -- Q10, for example, in the diagnostic test at the beginning of OG12 (the one with the pentagram diagram). The solution given in the book introduces 10 unknowns, which is a bit crazy. But since the answers are all numbers (we don't have 'cannot be determined' in the answer choices), that guarantees that the answer is the same no matter how skew or symmetric the diagram; a GMAT question can only have one correct answer, after all. So you can freely assume the pentagon is regular, and thus has interior angles of 108, making all of the triangles isosceles with angles 72, 72 and 36, from which the answer comes immediately; 5*36 = 180.lunarpower wrote:hey all.
regarding the "right triangle ABC" problem
(found here: https://www.beatthegmat.com/problem-solv ... 30325.html)
the solutions above would be nice in a geometry course; a lot of them are very pretty.
but - let's assume that this is a gmat problem. (NOTE: this would not conceivably be a gmat problem - it's too obscure/difficult to show up on the test.
if this were a gmat problem:
* it would be multiple choice, with five NUMBERS as answers - say, 30, 35, 40, 45, 50.
therefore:
* that NUMBER would be the same, regardless of the particular shape of the right triangle ABC.
therefore:
* if you can answer the problem for ANY SINGLE EXAMPLE of ABC, then you've got THE answer.
so:
we can just let ABC be a 45-45-90 triangle.
if that's the case, then BO will be a horizontal line, and it will be clear (from symmetry) that BO cuts right angle ABC into two 45-degree angles.
that's all you need.
again, BECAUSE THE PROBLEM WOULD BE MULTIPLE-CHOICE, there's no need to produce such elaborate general solutions; finding just one solution is sufficient.
Note, however, that you could not use this approach on a question like Q179 in the Problem Solving section of OG12; there one of the answers is 'cannot be determined', which means that it is at least possible that the answer may be different depending on what numbers you start with (and indeed, in that question at least, it is different for different starting numbers).
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
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lunarpower
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totally true. but you can rest assured that there will be very, very few such problems; after all, there is already an entire half of the quant section (i.e., data sufficiency) dedicated to that exact principle.Note, however, that you could not use this approach on a question like Q179 in the Problem Solving section of OG12; there one of the answers is 'cannot be determined', which means that it is at least possible that the answer may be different depending on what numbers you start with
also - remember that they will pretty much ALWAYS write problems such that the "gee, i'm going to look at it and think this is the answer" answer is WRONG.
so, if you have a question like the one above - most people would look at it and say, "gee, i think that will probably have multiple values." so, if they take the trouble to include "cannot be determined" into a multiple-choice problem, then something is probably afoot; if the naive guess is "that quantity probably varies", then the quantity most likely doesn't vary. that's pretty much how they write these problems.
Ron has been teaching various standardized tests for 20 years.
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hansoo
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Hey all! I wanted to mention that hackers hacked Magoosh.com 4 months ago. Yet, we've heightened our security tremendously since then and have had no issues with hackers for 4 months. Rest assured that we are legitimate at Magoosh and are serious about keeping your information secure!
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gmatusa2010
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I got the first part, but the part regarding the factors of 2^3 and 7^1, I got
this:
2
22
222
2227
227
27
7
That's only 7 choices. What am I missing?
this:
2
22
222
2227
227
27
7
That's only 7 choices. What am I missing?
Musiq wrote:Word for word...thats exactly what I would have typed.Uri wrote:840=2^3 * 3 * 5 * 7sureshbala wrote:A factor of 840 is chosen at random. What is the probability that it is divisible by 15?
A. 1/4
B. 7/32
C. 1/2
D. 1/56
E. None of these
Total number of factors of 840 = (3+1) * (1+1) * (1+1) * (1+1)= 32
Now, all factors that are multiples of 15, will have 3*5 in them. So, if we find the total number of factors that we can have with 2^3 and 7, then that will give us the total number of factors divisible by 15.
Total number of factors with 2^3 and 7= (3+1)*(1+1) = 8
So, the probability of choosing any random factor of 840 that will be divisible by 15 is 8/32 = 1/4
Ans. (A)
Great minds think alike ... Uri?
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top_business_2011
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Very good question! It mixes up various concepts ranging from similarity of triangles to properties of different triangles- concepts that are frequently tested in the GMAT.sureshbala wrote:Folks, finally here is the solution....Thought of being more clear on this problem, so uploaded the image....Have a look at this....
While the problem is solved efficiently by sureshbala,I've found the answer to the problem without using any sin, cosine, or whatnots. This way, I hope, the question can be perceived more relevant to the GMAT.
Here is how I did that:
As I don't know how to put figures in here, I want you to draw a figure for yourself and follow me along.
On the original figure, draw a perpendicular line form C to line AP, and let's call the intersection point N. Similarly, draw a perpendicular line from A to line CP, and let's call the intersection point M.
Since triangle CNP is 30-60-90 right triangle, and line CP, which is the hypotenuse, has a length of 2, line CN = Square root 3 and line NP = 1.
Now look at triangle AMB; it is a 45-45-90 right triangle. Thus, Length MB= Length AM. Let's represent length of MP by X; Therefore MB=AM =1+X.
Now look closer. Triangle AMP and triangle CNP are similar triangles. As such, their corresponding sides must be proportional.
[Length of NP/Length of MP] = [Length CN/ Length AM]
1/X = Square root (3)/ (1+X)
X = (Square root(3) + 1)/ 2
Now, we can get length of CM = 2-X = (3- Square root(3))/2
and AM = 1+X = (Square root(3) + 3)/2
Therefore, we can find length of AC by the pythagorean theorem
(AC)^2 = (AM)^2 + (CM)^2
= Square root(6)
Now look closer at the figure. Triangle ANC is right triangle; so we can get the length of AN by the pythagorean theorem.
So, length of AN = Square root(3)
See that? Triangle ANC has two equal legs; so it must be a 45-45-90 triangle.
Therefore, angle NCA =angle NAC = 45 degrees. Finally, as we are asked to get angle measure of BCA, we've to add 30 to 45 = 75 degrees.
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Scott@TargetTestPrep
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Solution:ven4gmat wrote: ↑Thu Feb 12, 2009 1:55 amI am getting the answer for this as 70 days. So E. What is the OA?sureshbala wrote:Keeping in mind the feedback(I agree to the point that these questions are a bit tough but at the same time much within the basic concepts) that I got, here is the next one...
Edusoft, a software firm has engaged certain number of men to finish a project in the stipulated time. But after working for 14 days, 1/4 of them have left and to complete the remaining project the remaining members have taken as many days as the initial number of men would have taken to complete the entire project. In how many days was the project completed?
A. 48 B. 60 C. 56 D. 40 E. None of these
We can let the number of men originally working on the project be 4 and the total number of days to complete the project be x. Furthermore, if we consider the project as the number 1, then the daily rate of each of these 4 men is 1 / (4x).
Now, after working for 14 days, 1 - 14/x of the project would be left. Since only 3 men are left to finish the remaining portion of the project and it takes them x days, the daily rate of each of these 3 men is therefore (1 - 14/x) / (3x).
Since the daily rate of a man is always the same, we can create the equation:
(1 - 14/x) / (3x) = 1/(4x)
4x(1 - 14/x) = 3x
1 - 14/x = 3/4
1/4 = 14/x
x = 56
Now, not only is x the number of days it would take all 4 men to complete the project if no one leaves, but it is also the number of days it would take 3 men to complete the job after the initial 14 days. Therefore, the project was actually completed in 14 + 56 = 70 days.
Answer: E
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