Friday, November 26, 2010

CP #20

What is the number of integers from 1 to 1000 (inclusive) that are divisible by neither 11 nor by 35?
(C) 2008 GMAT Club - [t]m07#14[/t]
  • 884
  • 890
  • 892
  • 910
  • 945
To count the number of integers from 1 to [m]N[/m] (inclusive) that are divisible by [m]x[/m] , find the value of [m]\frac{N}{x}[/m] . Use only the integer part of the resulting number. Based on the formula, the number of integers divisible by 11 is [m]\frac{1000}{11} = 90.9 = 90[/m] (even though the result could be rounded off to 91, use 90). In the same way, the number of the integers divisible by 35 is [m]\frac{1000}{35} = 28.57 = 28[/m] .
Subtract the number of integers that are divisible by both 11 and 35, so that they are not counted twice.
Therefore, [m]\frac{1000}{11} + \frac{1000}{35} - \frac{1000}{11*35} = 90 + 28 - 2 = 116[/m] .
[m]1000-116 = 884[/m] .
The correct answer is A.
My answer was right. Wanted to keep this here to understand why they've used 1000/ 11*35. What I did was calculate 33*11 and 33*11*2 and saw that both are under 1000 and then counted 2. I should know this other method too.

CP #19

If Ben were to lose the championship, Mike would be the winner with a probability of [m]\frac{1}{4}[/m] , and Rob - [m]\frac{1}{3}[/m] . If the probability of Ben winning is [m]\frac{1}{7}[/m] , what is the probability that either Mike or Rob will win the championship (there can be only one winner)?
(C) 2008 GMAT Club - [t]m07#12[/t]
  • [m]\frac{1}{12}[/m]
  • [m]\frac{1}{7}[/m]
  • [m]\frac{1}{2}[/m]
  • [m]\frac{7}{12}[/m]
  • [m]\frac{6}{7}[/m]
The probability of Mike or Rob winning, conditional on Ben losing, is [m]\frac{1}{4} + \frac{1}{3}[/m] or [m]\frac{7}{12}[/m] . The unconditional probability (to take into consideration the probability of Ben winning) is then [m](1-\frac{1}{7}) * \frac{7}{12} = \frac{6}{7} * \frac{7}{12} = \frac{1}{2}[/m] .
The correct answer is C.
I have calculated it very differently and gotten 5/14 for an answer.


A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain?
(C) 2008 GMAT Club - [t]m07#6[/t]
  • 4
  • 5
  • 6
  • 7
  • 8
  • The amount of work in man-hours is: [m]110 * 10 = 1100[/m] mh. After 60 days with 5 days of rain (55 net days) the crew of 10 had completed 550 mh of work, and had 550 mh more to go. Of the remaining 40 days, [m]x[/m] were raining. The last 550 mh were completed by 16 men: [m](40 - x) * 16 = 550[/m] , and [m]x = 5.625[/m] . Round the number of rainy days down to 5 as with 6 rainy days the crew would have worked less than 550 mh and not have completed the job.

Tuesday, November 23, 2010

CR update

So after going through all these CR notes on the net, I tried 3 questions from OG12 (not too many left to choose from...yey!) - one of each type (S, W & A). Got the assumption question wrong, the others were right. Hmmm...going to practice slowly (i.e. identify question type, conclusion, premise(s) and imagine any linking assumptions) from my BTG account.

Wish me luck!

Edit: Tried a few BTG questions, and some other questions online...Hmmm...still not feeling satisfied, although I do have a methodical way of answering the questions now. *sigh*
I should go through other question types, and then do the remaining CR questions from OG12, right?

Monday, November 22, 2010

My erratic work schedule is producing deleterious effects on my GMAT prep.


Adjective: Causing harm or damage

Improving first, then moving forward.

Did not finish OG12. Reason is that I was going through my error log and realised I was getting certain types of questions consistently wrong in CR and SC - thought I should work on them. Posted it on to see if anyone could give me some tips on improving my accuracy.

For Quant, I've mainly been making silly mistakes. Also, I haven't gone through my FCs in really long so have forgotten some stuff like Compound Interest formula and how to calculate SD. Need to do that.
In addition, I've been getting Triangles questions wrong - so plan on going through this series on Triangles tricks.

Have spent most of today working on getting my basics right in CR. Will post an update soon.

GMAT love coming your way!

PS. Will take a month off from work before the GMAT because the way my work works, THIS is not going to work out! :) I've spoken to my boss and he said it'll not be an issue. So I am taking GMAT end of Jan, hopefully by 27th Jan 2011. I want to get my 800 already. :)

Friday, November 12, 2010

Slow & steady

Have been doing OG12 questions - 50 in one sitting. Has been going on slowly, but surely. Plan is to finish OG12 by next Thursday (6 more days). Have about half of OG left. Then I'll either work on my areas of weakness (as determined by error log analysis), or work through OG Quant & Verbal supplements (will have to buy them).

Work has been slow, and idle for the last 2 weeks. I wasn't expecting it at all, so have not made full use of it :(
But, things are looking up now in terms of studying so very happy :)

I love my life when I study, work, dance and do my chores consistently!!! :)

"Do or do not. There is no try."

Thursday, November 4, 2010


Question (from MGMAT Practice Test #1)
For positive integer k, is the expression (k + 2)(k2 + 4k + 3) divisible by 4?

(1) k is divisible by 8.

k + 1

 is an odd integer.

The quadratic expression k2 + 4k + 3 can be factored to yield (k + 1)(k + 3). Thus, the expression in the question stem can be restated as (k + 1)(k + 2)(k + 3), or the product of three consecutive integers. This product will be divisible by 4 if one of two conditions are met:

If k is odd, both k + 1 and k + 3 must be even, and the product (k + 1)(k + 2)(k + 3) would be divisible by 2 twice. Therefore, if k is odd, our product must be divisible by 4.

If k is even, both k + 1 and k + 3 must be odd, and the product (k + 1)(k + 2)(k + 3) would be divisible by 4 only if k + 2, the only even integer among the three, were itself divisible by 4.

The question might therefore be rephrased “Is k odd, OR is k + 2 divisible by 4?” Note that a ‘yes’ to either of the conditions would suffice, but to answer 'no' to the question would require a ‘no’ to both conditions.

(1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive ‘no’ to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4.

(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.

The correct answer is A.

I think the answer is D. In Statement (2), for (k+1)/3 to be an odd INTEGER, k+1 has to be divisible by 3. Also, k must be even. Therefore, k could be 2,8,... And if k is an even integer, (k + 1)(k + 2)(k + 3) will not be divisible by 4 for any of these values of k. Therefore, Statement (2) also gives us the answer. Please tell me where I'm going wrong.

CP #16 - have not understood why not my way?

Question (from MGMAT Practice Test #1)
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

We are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.

Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.

Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 – 5/21 = 16/21.

The correct answer is E. 

Wednesday, November 3, 2010

A Side Note to my Last Post

Some other reasons why you might want to buy BTG questions:
1. The cost: $99
The cost till 15th November: $49!
2. BTG is giving away 3 iPads! All Free and Premium users of Practice Questions who sign up by 15th November will be eligible to win.
So even if you are not sure of buying the product, atleast sign up for the free questions and you might just win an iPad!
3. 30-Day Money Back Guarantee. Seriously, it doesn't get any better than this. :)

Happy Gmatting!

Tuesday, November 2, 2010

Yes You Can! With the *new* BTG questions.

Want to practice only SC questions today? Have time to do only 10? Yes you can!
Have you mastered the fundamentals of Probability and want to target today's practice session only on 700+ level questions on the topic? Yes you can!
After a review of your error log, do you realize that you really need to work on those Weaken type Verbal questions? Across CR and RC? Yes you can!
Do you always get Integer Properties questions wrong only on the DS section? Have all day to work on it and want an (almost) unlimited supply of questions to be thrown at you till you crack it? Yes you can!
Tired of going through consecutive questions in the Official Guide which are at the same difficulty level? Want a practice session where the questions thrown at you are in an adaptive format? Yes you can!
Have you run out of CAT tests and want to create one of your own? Yes you can!

BTG has managed to create an (almost) perfect product. If you've run out of OG questions, and even CAT tests, this product's for you. [This is an improved product from BTG which is to be released today, November 2nd. It can be found here].

Totally customizable
Every practice quiz you take is totally customizable to your preferences. You can choose which section you want to target (Math or Verbal), which type of question to target (CR, RC, SC, PS, DS) and even the topic of the questions (Idioms, Weaken, Arithmetic...)!
You know what else I love? You can choose the number of questions you want to answer!
Have time only for 10 questions today? Your boss suddenly interrupts you during the 5th question with urgent work? No problem! You can choose to end the practice session there and see your results.
And this part, my fellow readers, is what I call an absolutely brilliant feature - the difficulty level. You have a choice of below 400 questions only, all the way upto 700+ level questions. And, that's not even the most brilliant part. Have you heard of..wait for it...adaptive questions on every quiz?! This one made me super happy!
A timer on the right hand side tells you how much time you've spent on the question during the quiz.

Create a New Test

You can choose to answer 37 Math questions, in an adaptive format. Once you've ended your session, head back to the "Practice" tab and create a session for 41 Verbal questions (adaptive mode). A brand new test has been created! :) You can also randomly pick any topics for the 2 essays from the OG, and you're all set to give a full 4 hour exam at home!

A breakdown of questions on each section
Math -
I found the difficulty level of questions pretty reasonable, not extremely easy and not so difficult that you'd never be able to crack it in 2 minutes (under 3 minutes for the really difficult ones).
Verbal -
Pretty much the same as for Math...

The Video Explanations
A friend of mine wanted to practice some SC questions and wanted me to help him out with understanding where he was going wrong. So I called him over to practice from my Reviewer's account of BTG questions. Once he was done with the questions, I realized I didn't have to say a word to explain his mistakes to him! The video explanations did all the talking! They even taught us a few things we didn't know, such as idioms that I wouldn't have caught if I was trying to explain it to him myself.
The Math explanations are clear, and I didn't find myself requiring any further explanation once the video was over.

The review section is pretty much the same as you'd find in any other similar product. But there are two parts that stand out for me:
First is the colour of the amount of time you've taken to answer the question. The colour is different according to whether you've outpaced others who answered the question, or were slower than them.
The second is the flagging the questions feature. In case you want to review the question again (irrespective of whether you got it right or wrong), you can flag it. What's the point of flagging it, you ask? *sly grin*

Prepare yourself for...

I was feeling pretty content and happy after having customized my practice session down to the topic of questions I wanted to answer. But before I clicked on the "Start the Session" button, I saw blue text that said "Show more options". More options?
Here they are:
- include questions that I have already answered (after going through 700 questions, are you really going to remember them all?)
- use only flagged questions (yup, include the questions you got right but should still do again)
- only use questions I got wrong (for obvious reasons)
And yes, you can select all 3 at the same time also.

Definitely worth it if you've run out of questions to practice from, or want to customize your practice sessions exactly according to your needs (especially if you want to do this!). I had hoped such a product would come out for OG12 questions (and it's Verbal & Math supplements) so that I could practice questions from those books in an adaptive customizable style. I hope this great effort from BTG gets a similar product out for Official questions soon! :) Till then, enjoy the power and remember, with BTG questions, YES YOU CAN!

PS. BTG is giving me a couple of subscriptions to this new product to give away. While I'm giving some to my friends who're also giving the GMAT soon, I might have 2 accounts left over to give away. So I'll do that on a first come first serve basis. HOWEVER, I will also snoop around and try ensuring that you don't already have an account that's been given away (I have my sources! :p). Once I'm satisfied that you do need one, I'll send you access to it! :)