Search found 15 matches
rephrase the question: P(red marble) > P(w marble) ==> r/(r+w+b) > w/(r+w+b) ==> Is r > w? Let us begin with Stmt 2 as it looks easier: b-w>r ==> b>w+r. But this says nothing about relative number of red and white marbles. So insuff. Stmt1: r/(b+w) > w/(b+r) ==> (b+w)/r < (b+r)/w Now add 1 to both s...
- by gmatcoach
Wed Sep 02, 2009 10:43 am- Forum: Data Sufficiency
- Topic: Jars
- Replies: 1
- Views: 1419
I agree with gmat550's solution. I would solve it with a little more elaborately: first consider in how many ways 3 people can be arranged linearly: ABC BCA CAB BAC CBA ACB As seen above it is 3! = 6. So in general you can arrange n people linearly in n! ways. Now consider in the ways in which 3 peo...
- by gmatcoach
Tue Sep 01, 2009 8:43 am- Forum: Problem Solving
- Topic: permutation
- Replies: 12
- Views: 7631
- by gmatcoach
Tue Sep 01, 2009 8:15 am- Forum: Problem Solving
- Topic: Please further explain MGMAT's explanation
- Replies: 5
- Views: 1556
Problem 1: Answer = 320 Please refer to the attached venn diagram: R - # of bags with raisins P - # of bags with peanuts A - # of bags with almonds r - # of bags with raisins only p - # of bags with peanuts only a - # of bags with almonds only m - # of bags with raisins and peanuts only n - # of bag...
- by gmatcoach
Tue Sep 01, 2009 4:26 am- Forum: GMAT Math
- Topic: QUANT + DS Questions
- Replies: 2
- Views: 1629
5^13 - 5^12 = 5^12(5-1) because :
Just split the powers:
5^13 = (5^12)(5)
5^13 - 5^12 = (5^12)(5)-5^12(1) (factor the powers: so they have 5^12 as a common factor)
= 5^12(5-1)
- by gmatcoach
Tue Sep 01, 2009 3:39 am- Forum: Problem Solving
- Topic: Please further explain MGMAT's explanation
- Replies: 5
- Views: 1556
Ah Yes, 158 is correct. I used n to denote the number of terms in the sequence. n is indeed the the last term of the sequence.
- by gmatcoach
Tue Sep 01, 2009 3:26 am- Forum: Problem Solving
- Topic: Maths Question
- Replies: 5
- Views: 2041
If you mean consecutive even numbers between 1 and n, we can solve it the following way: let us express the terms in the sum as 2X_i as they are all even. X_i are consecutive #s. Sum(2X_i) for i=1 to n is 79*80. This implies: 2 * Sum(X_i) (for i=1 to n) = 79*80. Since X_i are consecutive, Sum(X_i)(f...
- by gmatcoach
Mon Aug 31, 2009 10:09 am- Forum: Problem Solving
- Topic: Maths Question
- Replies: 5
- Views: 2041
distance D-C = L = 5^13 - 5^12 = 5^12(5-1) = 5^12(4).
C - A = 2xL => A = C - 2L = 5^12 - 5^12(4)(2) = 5^12(1-8) = -7x5^12. So D is the answer
- by gmatcoach
Mon Aug 31, 2009 7:32 am- Forum: Problem Solving
- Topic: Please further explain MGMAT's explanation
- Replies: 5
- Views: 1556
Problem 1: 2+2+2^2+2^3+2^4+2^5+2^6 +2^7+ 2^8 = Rule: Sum of the series: 2^0+2^1+2^2....2^n = 2^(n+1)-1. The simplest way to prove this is to just evaluate each term and sum them. Problem 2: 5^21 x 4^11 = 2 x 10^N. What is N? Left hand side:5^21 x 4^11 = 5^21 x (2^2)^11 = 5^21 x 2^22 Right hand side:...
- by gmatcoach
Fri Aug 28, 2009 3:37 pm- Forum: Problem Solving
- Topic: Some GMAT Prep questions
- Replies: 2
- Views: 1438
The last term P(E and F) is there regardless of discrete or continuous probability. If events E and F are mutually exclusive, then P(E and F) is 0. Then P(E or F) = P(E) + P(F); For example, if there are 3 green balls, 4 white balls and 2 red balls, if your draw 1 ball what is the probability that i...
- by gmatcoach
Fri Aug 28, 2009 2:50 pm- Forum: GMAT Math
- Topic: Probability OR
- Replies: 3
- Views: 5462
uptowngirl92 Notice that stmt 2 says 1.5 more cups of flour. So correcting the bold part of the equation: f = s + s(1 1/2) = s + (3*s)/2 = (5*s)/2 ------ f/s = 5/2 ---- Suff.. to: f = s+1.5. If s = 5, f = 7.5, ratio is 7.5:5 = 3:2 If s = 1, f = 2.5, ratio is 2.5:1 = 5:1 So stmt is ambiguous, hence i...
- by gmatcoach
Mon Aug 24, 2009 10:13 am- Forum: Data Sufficiency
- Topic: GMAT Prep Question Flour to Sugar
- Replies: 4
- Views: 5646
shrek1089, Are you referring to the q. corresponding to a square/rectangle? If so, I have provided the solution in my previous post. In short, 5C3 takes into account 2 cases where three points lie on a straight line (along the diagonals) and cannot form a triangle. So correct solution is to subtract...
- by gmatcoach
Mon Aug 24, 2009 9:51 am- Forum: Problem Solving
- Topic: Interesting Geometry Question / Triangle
- Replies: 19
- Views: 3673
A rule on median of a combined set
The standard rule is: Always the median of a combined set will lie between the medians of the individual sets (endpoint inclusive). So for Q.2, median of combined set lies between 12, 18 (inclusive). Largest element of Set S could be as big as 18 or as small as 12. So we cannot say unambiguously whe...
- by gmatcoach
Sun Aug 23, 2009 8:40 am- Forum: Data Sufficiency
- Topic: interesting data sufficiency problems
- Replies: 5
- Views: 2174
Q.6 solution
My 2 cents. 6. Is the median of the numbers in set S less than the average of the numbers in the set? (1) The median is of the sum of all of the numbers in set S. (2) There are 9 numbers in set S. Answer: (C) Stmt 1: median = sum(#s). average = sum(#s)/L. (L = length of series) If L>1, stmt is suffi...
- by gmatcoach
Sun Aug 23, 2009 8:28 am- Forum: Data Sufficiency
- Topic: interesting data sufficiency problems
- Replies: 5
- Views: 2174
Extrending the problem
My 2 cents: I agree with the 6c3 explanation. It should be 5C3 when the center point is removed. If we extended this problem to a square ABCD or a rectangle ABCD and denote the center as E(point of intersection of diagonals), then lets ask the question: how many triangles can be formed with A,B,C,D,...
- by gmatcoach
Sun Aug 23, 2009 7:06 am- Forum: Problem Solving
- Topic: Interesting Geometry Question / Triangle
- Replies: 19
- Views: 3673