GMAT Problem Solving Discussions

MBA
361
I have used the approach (Total number of ways - Not application ways). Total number of ways is the Universal set including when all 7 need to be played and when less than 7 are played.

There u are ..
wwww(xxx) such type of combinations are there in those 2^4 ... which shouldn't be
Infact you are missing out the cases in which only 5,6 or 4 matches are played
Hope I m clear
362
There u are ..
wwww(xxx) such type of combinations are there in those 2^4 ... which shouldn't be
Infact you are missing out the cases in which only 5,6 or 4 matches are played
Hope I m clear

There's a word which you missed - UNIVERSAL SET which means you dont look at only the cases which you want.
363
mealiennothuman Says
There's a word which you missed - UNIVERSAL SET which means you dont look at only the cases which you want.


Universal Set is not 64 dude .. that is what I am trying to say
Please think once again ... seven games need not be played in the first place(see the italic n underlined part below)

The first team to win four games wins the series and no subsequent games are played. If you have no special info about either of teams what is the probability that the world series will consist of fewer than 7 games?
364
Universal Set is not 64 dude .. that is what I am trying to say
Please think once again ... seven games need not be played in the first place(see the italic n underlined part below)

Ok. I am not saying that you are wrong. Just as you are not able to understand my point of view, I am not able to understand yours. Let the answer come. If it does turn out that I am wrong then good I will not make the mistake again. Not replying anymore till then.
365
Universal Set is not 64 dude .. that is what I am trying to say
Please think once again ... seven games need not be played in the first place(see the italic n underlined part below)




I think I understood the approach taken by man-h, lemmi try to explain.


Question asked: Probability that the tournament finishes before 7 games are completely played ( p(
This is equal to = 1 - p(7)
i.e. 1 minus probability that all 7 games played.

Now we need to find p(7) and so we don't need to bother about p(4), p(5), p(6).

To find the probability for all 7 games to be played in the tournament we must know
1) total possible ways 7 games could be played, without being bothered about the conditions in given in the tournament ( universal set).
2) Number of ways that satisfy the condition of the problem.


Guess I made it clear now ?
366
I think I understood the approach taken by man-h, lemmi try to explain.


Question asked: Probability that the tournament finishes before 7 games are completely played ( p(
This is equal to = 1 - p(7)
i.e. 1 minus probability that all 7 games played.

Now we need to find p(7) and so we don't need to bother about p(4), p(5), p(6).

To find the probability for all 7 games to be played in the tournament we must know
1) total possible ways 7 games could be played, without being bothered about the conditions in given in the tournament ( universal set).
2) Number of ways that satisfy the condition of the problem.


Guess I made it clear now ?

puys ... sorry misinterpreted ...
It is clear now ...
367

the answer is indeed D, that is 68.75. The approach used by man-h is the right one where you find out the probability of each team winning three matches out of first six and taking its complement.

368

here is one more, that should revise your set theory fundae

Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?

(A) 256
(B) 260
(C) 316
(D) 320
(E) It cannot be determined from the given information.

369
here is one more, that should revise your set theory fundae

Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?

(A) 256
(B) 260
(C) 316
(D) 320
(E) It cannot be determined from the given information.

Here are the values:
Bags containing only Raisins = 200
Bags containing only Peanuts = 20
Bags containing only Almonds = 100
Bags containing only Peanuts and Raisins = 5

Bags containing only one kind of item = 200 + 20 + 100 = 320.
371

good work, but if you had put up explanation it would have benefitted others as well.

the explanation goes as, there are seven possibilities that make up the 435 bags with possible values ranging from 0 to 435. R, P, A, R&P;, R&A;, P&A;, R&P;&A; are the possibilities. I hope you would be able to deduce what they mean since i am not illustrating it with a venn diagram. Let us assume P=x, then R= 10x and A=5x. Also R&P; = x/4. But it is given that the union of all, that is sum of all these seven possibilities is 435, while that of A, R&A;, P&A; and R&P;&A; is 210 hence subtracting gives R, P and R&P; together as 225. We know their values in terms of x and solving this linear equation gives x as 20. Now in terms of x the required quantity is 10x+x+5x, so 320. phew.............

372
Help !!!

Two identical trains A and B running in opposite directions at same speed take 2 mins to cross each other completely. The number of bogies of A are increased from 12 to 16. How much more time would they now require to cross each other?
a.40s b.50s c.60s d.20s e.30s

Ans given in the study material I have is a.40s

I disagree. I belive it ought to be d.20s

Would you agree?


total distance to be traveled is 2x and now it becomes 7x/3. Equating expressions of speed we get new time as 140 secs which is 20 secs more. I agree with your answer, unless a same increase in number of bogies of B follows incrase in A' , 20 secs must be the answer.
373
dumbJoe Says
................... Now in terms of x the required quantity is 10x+x+5x, so 320. phew.............

Now you know why I didnt give any explanations. 😃
374

i do :idea:

375

Hi there. Can you help me wit this one-

Does the integer K have a factor p such that 1

4!
2. 13!+2

376
Hi there. Can you help me wit this one-

Does the integer K have a factor p such that 1

4!
2. 13!+2


Please do not post the same question in two threads. Anyways, this questions is a DS question. Please edit your post and post a PS question.
377
50% of the aparments in a certain building have windows and hardwood floors. 25% of the apartments without windows have hardwood floors. If 40% of the apartments do not have
hardwood floors, what percent of the apartments with windows have hardwood floors?

a.10
b.16.66
c.40
d.50
e.83.33

please post the explanations so that others can benefit as well.
378
50% of the aparments in a certain building have windows and hardwood floors. 25% of the apartments without windows have hardwood floors. If 40% of the apartments do not have
hardwood floors, what percent of the apartments with windows have hardwood floors?

a.10
b.16.66
c.40
d.50
e.83.33
please post the explanations so that others can benefit as well.

Let number of apartments with,
Only Hardwood floors = x
Only windows = y
Both = z
None = w
Hence, x+y+z+w = 100. -- (I).
From the statement "50% of the aparments in a certain building have windows and hardwood floors." we get z = 50 -- (1)
From the statement "40% of the apartments do not have hardwood floors" we get,
100 - (x+z) = 40. Hence x = 10 -- (2)
The statement, "25% of the apartments without windows have hardwood floors." gives us,
25% of (w+x) = x. Solving we get, w = 30. -- (3)
From (I),(1),(2) and (3) we get, y = 10.
Hence we have,
Only Hardwood floors = 10
Only windows = 10
Both = 50
None = 30
Required answer = Both/(Both+Only windows) = 50/60 = 83.33%. Hence (E).
379

Hi all,
Pls help with this question.

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
6346

Regards

380
Hi all,
Pls help with this question.

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
6346

Regards



Are they 13 and 15 ?

If correct then will post my solution or else it will add to more confusion.
381
Are they 13 and 15 ?

If correct then will post my solution or else it will add to more confusion.


Bingo!
You are right.
Please post your explanation.

Thanks