Search found 19 matches
This question is simpler than many of the explanations make it seem. Use the following formula twice. X + Y - XY = 100 (where X is one group, Y is a second group, and XY is people who are in both groups) In reality, X + Y - XY could be less than 100, but since we want to find the smallest XY, we nee...
- by jcnasia
Mon Jul 09, 2012 11:59 pm- Forum: Problem Solving
- Topic: least value venn diagram
- Replies: 94
- Views: 54324
If k is an integer, and 35^2-1/k is an integer, then k could be each of the following, EXCEPT (A) 8(B) 9(C) 12(D) 16(E) 17 This question is written poorly. As written, k must equal 1 which isn't an answer. We'll assume the poster meant to write (35^2 - 1)/k To solve this problem, we want to find po...
- by jcnasia
Mon Jul 09, 2012 12:42 am- Forum: Problem Solving
- Topic: If k is an integer
- Replies: 56
- Views: 28257
Here's a fun proof to show that if x + 2|y| = 0 and y + 2|x| = 0 then |x| + |y| = 0. x + 2|y| = 0 and y + 2|x| = 0 => x + y + 2|x| + 2|y| = 0 //add the two equations to each other => -1/2(x + y) = |x| + |y| //rearrange the equation -1/2(x + y) <= |-1/2(x + y)| //since absolute value of a number is n...
- by jcnasia
Sun Jul 08, 2012 11:44 pm- Forum: Data Sufficiency
- Topic: absolute value
- Replies: 37
- Views: 21576
1)for all a>1 this is true...no issues here 2)for 0<a<1 let a=0.5 0.5^1>0.5^2 here the above relation is not true as 1>2 (false) 3)for -1<a<0 again let a=-0.5 -0.5^1<-0.5^2 hence 1<2...the relation holds valid here 4)for a<-1 a=-2 -2^1>-2^2 but here 1>2 (which is false)hence the above relation not ...
- by jcnasia
Mon Jul 02, 2012 3:49 am- Forum: Data Sufficiency
- Topic: Exponents theory 2
- Replies: 3
- Views: 960
I don't get it. Why do you assume that all the girls will pick the marbles first before the boys do or vice versa? If the girls and boys picked a marble each in turns, the result would be different. Could anyone please explain? Thanks. -BM- It doesn't matter what order they pick the marbles in. The...
- by jcnasia
Mon Jun 25, 2012 6:58 am- Forum: Problem Solving
- Topic: Probability - 10 marbles
- Replies: 143
- Views: 55336
I'm going to first assign variables. Bn = Bobby now Jn = Johnny now Bb = Bobby before Jb = Johnny before Then, from the prompt, I can write the following equations. Jn = Bb Jb = 1/2 * Bn Statement 1: I can write the following equations. Bn = 4*(Bn - Jn) since (Bn - Jn) is the age difference between ...
- by jcnasia
Sun Jun 24, 2012 3:48 am- Forum: Data Sufficiency
- Topic: Bobby & Johnny
- Replies: 3
- Views: 1353
The question wants you to find the equation that fits the line exactly. Answer (b) doesn't fit it exactly, because I can find an x value that doesn't fall on the highlighted number line. For example, x = 4. |4| <= 5 but 4 is not between -5 and +3. Answer (e) is correct. |x+1| <= 4 --> x+1 <= 4 and x...
- by jcnasia
Sat Jun 23, 2012 5:06 pm- Forum: Problem Solving
- Topic: need some explanation - Inequalities
- Replies: 1
- Views: 804
I used algebra to solve this problem. If m + n = p, then mn = (m+n) + 1 mn - m = n + 1 m(n - 1) = n + 1 m = (n + 1) / (n - 1) At this point, I realized that n = 2 or n = 3. Otherwise, m will not be a positive integer. So, (m,n) = (3,2) or (2,3) Statement 1: Both m and n are prime numbers. (m,n) = (2...
- by jcnasia
Sat Jun 23, 2012 4:22 am- Forum: Data Sufficiency
- Topic: DS - Prime
- Replies: 4
- Views: 1376
If a, b, c, d, and e are consecutive integers and a+b+c+d+e=560, then the next five integers are...
a+5, b+5, c+5, d+5, and e+5, and they add up to...
a+b+c+d+e+(5*5) = 560 + 25 = 585
- by jcnasia
Sat Jun 23, 2012 3:57 am- Forum: GMAT Math
- Topic: Question on A.P. formula
- Replies: 6
- Views: 7054
You can also solve this problem by taking the log of both sides.
log(5^21 * 4^11) = log(2 * 10^n)
21 * log 5 + 11 * log 4 = log 2 + n * log 10
21 * log 5 + 11 * 2 * log 2 - log 2 = n
21 * log 5 + 21 * log 2 = n
21 * log(5*2) = n
21 = n
- by jcnasia
Fri Jun 22, 2012 7:06 am- Forum: GMAT Math
- Topic: Equations with variables in the exponents
- Replies: 4
- Views: 1658
The denominator is # of golfers.dhonu121 wrote:If we had to find the chance that golfer also skies, what is the formula that we would have used ?
As in, what quantity was supposed to come in the denominator ?
The numerator would have number of golfers who ski.
- by jcnasia
Fri Jun 22, 2012 6:56 am- Forum: Data Sufficiency
- Topic: Golfers
- Replies: 5
- Views: 1644
- by jcnasia
Fri Jun 22, 2012 6:33 am- Forum: Sentence Correction
- Topic: Despite the recent election of a woman
- Replies: 8
- Views: 3344
Here's how I solved the problem... Statement 1: Case 1: 15 Golfers: Andrew, Ben, Carla, Della, Ella, F, G...O 5 Tennis players: Andrew, Ben, Carla, Della, Ella 14 Surfers: Andrew, Ben, Carla, and F, G, ... 0, and Pamela Skiers: Andrew, Ben, Carla, F, G, H, I, J 8 of the golfers also ski so the chanc...
- by jcnasia
Fri Jun 22, 2012 5:09 am- Forum: Data Sufficiency
- Topic: Golfers
- Replies: 5
- Views: 1644
Si = stocks initial Sf = stocks final Bi = bonds initial Bf = bonds final The question is: Is Sf > Si? Statement 1: Sf / (Sf + Bf) > Si / (Si + Bi) Let Sf = 1, Bf = 0, Si = 1, and Bi = 1,000,000, then the above statement is true, but Sf = Si. Let Sf = 2, Bf = 0, Si = 1, and Bi = 1,000,000, then the ...
- by jcnasia
Fri Jun 22, 2012 12:02 am- Forum: Data Sufficiency
- Topic: DS Ratio GMATPrep CAT
- Replies: 2
- Views: 965