Search found 97 matches
Well, I went ahead and hit submit because I thought it might point me to the missing field, but the application ended up going through. I hope that meant everything was actually completed.
- by SoCan
Tue Jun 14, 2011 9:10 pm- Forum: The Application Process
- Topic: INSEAD Application - only 76 of 77 required fields?
- Replies: 2
- Views: 1464
INSEAD Application - only 76 of 77 required fields?
I want to submit my Jan 2012 application right now, but the application keeps telling me I've only filled 76 of the 77 required fields. I've quadruple checked the signature and it's fine - when I clear it I only have 75. I've used the "highlight field" function on Acrobat and gone through ...
- by SoCan
Tue Jun 14, 2011 8:24 pm- Forum: The Application Process
- Topic: INSEAD Application - only 76 of 77 required fields?
- Replies: 2
- Views: 1464
Don't worry. The relatively high scores in both sections will stand you in good stead. Congrats on the score. Really, you have to be proud of it. Just curious, what were your GMAT prep total scores? I don't remember. I was much more focused on the subscores than the overall score, so I didn't recor...
- by SoCan
Sun Jun 05, 2011 8:56 am- Forum: I just Beat The GMAT!
- Topic: Unbalanced 740
- Replies: 10
- Views: 3823
Unbalanced 740
This post is more about venting than anything else. I just took the GMAT this morning, and I'm "that person" that's not happy with a 740. Why? Because my 46Q is very disappointing (received a 46 on the verbal as well). My goal was a 50Q, and I received between 48-50 on the last few MGMAT C...
- by SoCan
Fri Jun 03, 2011 1:45 pm- Forum: I just Beat The GMAT!
- Topic: Unbalanced 740
- Replies: 10
- Views: 3823
Hi, Concentrations of the two solutions are 0.85 and 0.2 and that of mixture is 0.4 So, using principle of allegations, liquid1/liquid2 = (0.4-0.2)/(0.85-0.4)=4/9 So, liquid2 is 9/13 by volume. So, the portion of liquid1 replaced = 9/13 Hence, C I'm getting the hang of this way, I like it. Algebrai...
- by SoCan
Thu Jun 02, 2011 8:38 am- Forum: Problem Solving
- Topic: Mixture Problem**
- Replies: 3
- Views: 1100
Hi, Fraction of gold in bar1 = 2/5 Fraction of gold in bar2 = 3/10 Fraction of gold in alloy = 5/16 Using the principle of allegations, bar1/bar2 = (5/16-3/10)/(2/5-5/16) = 1/7 Total weight of alloy is 8 kg. So, weight of bar1 = 1kg Hence, A I've never used alligation before in this way, but it see...
- by SoCan
Thu Jun 02, 2011 7:46 am- Forum: Problem Solving
- Topic: A Good Mixture Problem
- Replies: 7
- Views: 1231
First gold bar is 4/10 gold, the second bar is 3/10 gold. We want to find the weights that give us an 8 pound bar with 5/16 gold.
(4/10)x + (3/10)(8-x) = (5/16)*8
x/10 + 24/10 = 5/2
x/10 = 1/10
x = 1
- by SoCan
Thu Jun 02, 2011 7:44 am- Forum: Problem Solving
- Topic: A Good Mixture Problem
- Replies: 7
- Views: 1231
A certain tree was first planned it was 4 feet tall then height incrsd by constt rate for next 6yrs at the end of 6th yr tree was 1/5 taller than it was at d end of 4 yr,by how many inches did the hite of tree inc each yr? 1.3/10 2.2/5 3.1/2 4.2/3 5.6/5 [spoiler]OA = 2/3[/spoiler] [spoiler] plz tel...
- by SoCan
Thu Jun 02, 2011 5:38 am- Forum: Problem Solving
- Topic: algebra equation rather dan backsolve
- Replies: 5
- Views: 1139
Ah yes, I was able to tell what you had written in the prior response, but for some reason was counting up to 17 the first few times. I must of missed one. Got it though, thanks. No problem - when you said 17, I counted again to make sure I hadn't left anything out, and I even counted 17 the first ...
- by SoCan
Wed Jun 01, 2011 8:17 pm- Forum: GMAT Math
- Topic: Factor of 2^k with n!
- Replies: 5
- Views: 1736
If m is a positive integer, does 24 divide m? (1) 54 divides m^2. (2) 32 divides m^2. See what you come up with! This has been answered a couple of times, but I'll restate things a bit. If 24 divides m, then m must contain at least the same prime factors as 24. Prime factors of 24 are 2^3, 3, so m ...
- by SoCan
Wed Jun 01, 2011 4:50 pm- Forum: Data Sufficiency
- Topic: DS OG 128
- Replies: 13
- Views: 2434
area of original square=25 hence side=5 let side of small square=a, then that of larger = 5-a thus (5-a)^2/a^2 = 9 taking root on each side, (5-a)/a=3 =>5-a=3a => a=5/4 Now that I've read it properly. Another way to look at it: Since larger square area is 9 times that of the smaller square, it's le...
- by SoCan
Wed Jun 01, 2011 3:55 pm- Forum: Problem Solving
- Topic: Square
- Replies: 12
- Views: 4071
Awesome. Yes, I meant 2^k, sorry. Now I understand, but I'm just a little confused about the number 18. I'm assuming that the multiples added up plus 0 is what makes it 18 rather than 17, right? And, thanks for the tip on where to post. I will surely be posting some new questions there soon. Thanks...
- by SoCan
Wed Jun 01, 2011 3:47 pm- Forum: GMAT Math
- Topic: Factor of 2^k with n!
- Replies: 5
- Views: 1736
Yikes - didn't read the question very closelyMBA.Aspirant wrote:The ratio is between the two "shaded" squares not the original square to the small one
area of big shaded square to that of small shaded one is 9:1
- by SoCan
Wed Jun 01, 2011 3:39 pm- Forum: Problem Solving
- Topic: Square
- Replies: 12
- Views: 4071
If area of larger square is 25 and is 9 times larger than the smaller square, then the area of the smaller square is 25/9. To get the length of the side, take the square root = 5/3. Or, because the area is the square of the sides, if the area is 9 times larger, you know the length of the side is 3 t...
- by SoCan
Wed Jun 01, 2011 3:07 pm- Forum: Problem Solving
- Topic: Square
- Replies: 12
- Views: 4071
The title says 2^k, but the question says 2k, which threw me off for a bit because I didn't notice the title. You want to know how many multiples of 2 there are in 20! You can ignore the odd numbers, obviously. Going from 2 to 20, you get 2 2^2 2 2^3 2 2^2 2 2^4 2 2^2 Add the multiples of 2 together...
- by SoCan
Wed Jun 01, 2011 2:35 pm- Forum: GMAT Math
- Topic: Factor of 2^k with n!
- Replies: 5
- Views: 1736