Can non-HDT members attend the meet????

srikvarmap Says

guys , is there any meet in the near future ?

If there is, I'd surely like to be a part of it

guys , is there any meet in the near future ?

AIMCAT 1207

Let p,q,r be the respective numbers which show up when three fair dice D1,D2,D3 are rolled. If y = (p-1)(q-4)(r-5), the probability that y<= 1 is

a. 109/216 b. 5/12
c. 95/216 d. 93/216

Plz koi ye samjha do .... nt able to get from the solution provided

P=1, Q = 1,2,3,5,6 and R=1,2,3,4,6
=> Prob = 1*5*5/6^3 = 25/216
Same way when Q=4.. 25/216 and when R=5.. 25/216
=> 3*25/216 = 75/216

Now, when P=1, Q=4, R=1,2,3,4,6 = 5/216. Same way for other two = 2*5/216 => 3*5/215 = 15/216

When p=1, Q=4 and R=5 = 1/216

When P=2, Q=(3,5); R=(4,6) = 2*2/216 = 4/216

=> total probability = (75+15+1+4)/216 = 95/216

AIMCAT 1207

Let p,q,r be the respective numbers which show up when three fair dice D1,D2,D3 are rolled. If y = (p-1)(q-4)(r-5), the probability that y<= 1 is

a. 109/216 b. 5/12
c. 95/216 d. 93/216

Plz koi ye samjha do .... nt able to get from the solution provided

Aa gaya samajh

AIMCAT 1207

Let p,q,r be the respective numbers which show up when three fair dice D1,D2,D3 are rolled. If y = (p-1)(q-4)(r-5), the probability that y<= 1 is

a. 109/216 b. 5/12
c. 95/216 d. 93/216

Plz koi ye samjha do .... nt able to get from the solution provided

Part-2

2. Improving the option marking ability a.k.a. how to test oneselfhttp://cdn.pagalguy.net/pagalguy/smilies/icon_mrgreen.gif]
Common complaint?
How do we address that? Well, most shall agree that a considerable element of gut feel comes into the picture here. We must know our abilities well and then take a call on attempting such questions.

Now let's see how to test oneself:

Funda: Say, you take up a set of 4 RCs and place a time limit of 30-40 minutes to solve them.
40 minutes over? Done with marking the options?

1. Drink some Glucon D-nariyal paani-vodka cocktail http://cdn.pagalguy.net/pagalguy/smilies/icon_mrgreen.gif] and get back after 10 minutes.
2. Don't touch the answer key yet! Now for the dirty part. Review each question and the answer you have marked.

Say you have marked 20Q out of the given 30.
Rate your option.
#Surely Correct
#Most probably correct
#Dicey
#WTF have I marked!

Now check the 10 Qs and mark an option for each one of them.
Rate all of them as #tukka
They have not been marked either because you could not get to them or because you chose to play safe.

Anyway, these 10Q are not for evaulation as you would not have marked them in the actual exam.

Now check your key and see how many you got correct.
Say you got 12 questions correct out of the 20 attempted. Now you know the wrong option for each of the other 8 questions. Attempt them again; now easier because you have three options for each Q instead of the earlier 4.
Say this time, out of the 8, you got 6 correct. Repeat again-this time you'll have two options to choose from for the remaining 2 Qs. I'm sure-we shall get all the 20 right by this time. If not-tough luck and look at it as a plain aberration http://cdn.pagalguy.net/pagalguy/smilies/icon_mrgreen.gif]


Now you know which and how many Qs of each rating were correct and wrong in which round. Analyse, how and why you missed out on the correct answer in the first go in case of the wrong ones. For the correct answers in dicey Qs- check why you were hesitant in marking it.
After a few rounds and having tested yourself in diverse areas- you shall have a sound knowledge of your abilities in each area: sports, philosophy, politics, blah, etc. and should be mindful of it in the examination.

This shall drastically improve your discerning abilities and you'll be very aware while attempting an RC. You won't rush through the passage; take your own sweet time and smartly breeze through it.


Quote:
E.g. I have a tremendous record of screwing up philosophy passages- I better be very careful about marking the options-rather I would not attempt the dicey Qs at all. On the other hand, I appear amazingly comfortable in sports based passages, I'll definitely take my chances in the dicey Qs
What about the 10Qs we had rated #Tukka?
Well, as the name suggests, it checks your blind guesswork; lays bare your gut feel and you can analyse this and use it to check how good you are for the guesswork you do when you take a risk and go all out attempting all 30Q in section-2 If you consistently get >6 out of 10 correct, i'll say take the chances!

CAT has always been about diligence and I find these two fundas diligent ways of improving and testing one's ability!
Hope this helped.

Regards
Anish Nambisan

here is a post on "how to improve on RCs" by anishnambisan:
"
We discussed quite a few things about VA in the last confy. It's basically the RC part that most of us face a solvable problem in. There are two strategies I know which we can follow to improve our RC scores. I have classified into two actually; was passed on by TIME teachers to me to which I added slight fundas of my own

I personally feel solving countless RCs isn't the best way to improve one's RC score as it is a way of testing rather than building the skill. RC area of a test checks our comprehension powers. Why not build the skill of comprehension first?


1. Increasing the comprehension power
Many among us keep returning to the 1st paragraph of a passage while we are yet to finish the first read! That's because our mind is untrained to comprehend and retain stuffs; add to it a time gun and we are

Funda: Take a newspaper's editorial-usually we get diverse subjects there- preferrably online. Time yourself and read it. One go. Don't move onto the next para if you have not understood the current one. Don't get back to the earlier para. Say, you took 6 minutes to do this.
Next day, give yourself a time-limit of 5 and half minutes to do the same. After you are done reading, write out a summary of the same. It might be pathetic but it is all right. All first summaries are like that. But here is the important part: Check what all points you could not cover which are given in the article. Also, check what all things you added to the summary which are not given in the passage. This happens sometimes as we assume things to be a given and questions in RC test this too. Now repeat this everyday! You'll see that after 1 week, the quality of the summary has improved. Now reduce your time limit by 30 seconds. You might find on the first day that the quality of summary has declined again! But be patient and repeat it. After another week, you'll be comfortable with a time limit of 5 minutes.

Repeat this cycle every week and you might see that you end up comprehending a passage in 3-4 minutes very well by Oct 15. Hence you reduce the time spent per passage drastically. "

Krazy bhai aap bhi hain yahan :D
pata nahin tha hamein.... welcome hai...!!

and ofourse, all your answers are perfect

Some exercise question for all of us:
1. What are the last two digits in 1372^482
PS : i got that question wrong in this AIMCAT silly max mistake.

1372^482/100 = 72^482/100 => 72^482/25*4
=> 72^482/4 = 0 remainder
=> 72^482/25 => E(25)=20
=> 72^2/25 => -3^2/25 = 9 remainder

4x=25y+9 => x=21,y=3 => remainder = 84

2. remainder when 1234^5678/91

1234^5678/91 => 1234 and 91 are coprimes. Hence E(91)= 72

=> 1234^62*(1234^72)^78/91 => 1234^62/91
1234^62/91 => 91=7*13
=> 1234^62/7 = 2^62/7 = 2^2/7 = 4 remainder
=> 1234^62/13 = -1^62/13 => 1 remainder
=> 7x+4 = 13y+1 => 7x+3=13y => x=7, 13=4 => 52+1 = 53 remainder

3. Remainder when 12^34 * 56^78/9

12^34*56^78/9 = 0 remainder as 12^34 is divisible by 9

Rahul3 Says

dude we can't separate them it's not addition so only one 9

Rahul bhai, u can very well separate them... and since the operator between them is Multiplication, multiply the results.

10*10/8 = 2*2 = 4

Rahul3 Says

(1234^12)^473*1234^2/13 = 1234^2/13 =(1235-1)^2=1235k+1/13 =13q+1

YES, I checked it, it is 1 only, My Bad!

i'll do the third one before the second cuz this looks easier :)
3)So, 12^34 *56^78 /9
This can be separated in to 12^34/9 and 56^78/9
Now, 12^34/9 = 3^34/9
3^even is always divisible by 9, thus 3^34 mod 9 = 0

I dont think we need to find the 56^78/9 part now,

So, 12^34 * 56^78 mod 9 = 0

Am I right here?

dude we can't separate them it's not addition so only one 9

2)(1234^567/91
We will have to use the CHinese Remainder Theorem here,
91=13*7

Now, (1234^567/7 and 1234^5678 / 13 need to be done separately

1234^5678 / 7= using Fermat's theorem.
1234^6 mod 7 = 1, hence
(1234^946)^6 * 1234^2 /7
(1234^946)^6 / 7 is 1, and 1234^2/7 is nothing but 1234 * 1234 /7, which by normal division itself gives 2*2/7, hence remainder 4. Thus,1234^5678 mod 7 is 7p + 4

And similarly, 1234^5678 mod 13 = 13q + 4

Now, 7p+4=13q+4, we get
p=13 and q=7
SO, the remainder is 91

AM I right, I have just learnt the CRT, and I am not sure if I have done correctly!

(1234^12)^473*1234^2/13 = 1234^2/13 =(1235-1)^2=1235k+1/13 =13q+1

I'm gtn 53

1234^5678/91

91=7*13

1234^5678/7 = (1232+2)^5678/7 = 2^5678/7 = (2^3)^1892*2^2/(2^3 -1) applying rem theorem x^1892*4/x-1 => rem is (-1)^1892*4 = 4

so 7a+4

1234^5678/13 = (1235-1)^5678/13 = 1235K+1/13 => rem is 1

so 13b+1

7a+4=13b+1

for a=7,b=4 eqn satisfies

rem is 7*7+4=13*4+1=53

can non-HDT members join in?
i dint knw hyd has an active DT team....! wud have joined long back :(. anyways looking forward to have a wonderful time with u all
cheers :D

Yes you can, I am not a HDT member myself!

can non-HDT members join in?
i dint knw hyd has an active DT team....! wud have joined long back :(. anyways looking forward to have a wonderful time with u all
cheers

Mohit bhai, perfect hai...
Bas ek question hai, how can remainder be 91 when divided by 91 :P
Actually if its leaving 4 with both the factors, it will leave 4 only when divided by 91.

remainder is 7p+4 = 91+4 = 95 = 4 when divide by 91

Others are right.

for the first one where you were stuck at 72^482/100, here it goes:
Rem(1372^482/100)
Rem(72^482/100)
Rem(72^2/100) - reducing the power using Eulers totient function.
Rem(5184/100) = 84

So, Last two digits = 84.
Last digit = 4
Second Last digit = 8

haha thanks Mishra Ji, yes it cant be 91, I just didnt keep my eyes open, but the procedure is right na?I mean I have just the CRT and I wasnt sure of it first.

Anyways, woh Euler's wala, even I thought of doing it the way you did, E(100) is 40 and hence 72^482 reduces to 72^2,lekin ek problem hai shayad, for Euler's to be applicable, both the numerator and denominator should be co-prime na, but 72 and 100 are not,COrrect?

P.S: I am assuming the Euler's Totient function is the normal Euler's we do, or is it any different?Kyu ki I do not know the original name of the Euler's theorem:)

2)(1234^567/91
We will have to use the CHinese Remainder Theorem here,
91=13*7

Now, (1234^567/7 and 1234^5678 / 13 need to be done separately

1234^5678 / 7= using Fermat's theorem.
1234^6 mod 7 = 1, hence
(1234^946)^6 * 1234^2 /7
(1234^946)^6 / 7 is 1, and 1234^2/7 is nothing but 1234 * 1234 /7, which by normal division itself gives 2*2/7, hence remainder 4. Thus,1234^5678 mod 7 is 7p + 4

And similarly, 1234^5678 mod 13 = 13q + 4

Now, 7p+4=13q+4, we get
p=13 and q=7
SO, the remainder is 91

AM I right, I have just learnt the CRT, and I am not sure if I have done correctly!

Mohit bhai, perfect hai...
Bas ek question hai, how can remainder be 91 when divided by 91 :P
Actually if its leaving 4 with both the factors, it will leave 4 only when divided by 91.

remainder is 7p+4 = 91+4 = 95 = 4 when divide by 91

Others are right.

for the first one where you were stuck at 72^482/100, here it goes:
Rem(1372^482/100)
Rem(72^482/100)
Rem(72^2/100) - reducing the power using Eulers totient function.
Rem(5184/100) = 84

So, Last two digits = 84.
Last digit = 4
Second Last digit = 8

2)(1234^567/91
We will have to use the CHinese Remainder Theorem here,
91=13*7

Now, (1234^567/7 and 1234^5678 / 13 need to be done separately

1234^5678 / 7= using Fermat's theorem.
1234^6 mod 7 = 1, hence
(1234^946)^6 * 1234^2 /7
(1234^946)^6 / 7 is 1, and 1234^2/7 is nothing but 1234 * 1234 /7, which by normal division itself gives 2*2/7, hence remainder 4. Thus,1234^5678 mod 7 is 7p + 4

And similarly, 1234^5678 mod 13 = 13q + 4

Now, 7p+4=13q+4, we get
p=13 and q=7
SO, the remainder is 91

AM I right, I have just learnt the CRT, and I am not sure if I have done correctly!

i'll do the third one before the second cuz this looks easier :)
3)So, 12^34 *56^78 /9
This can be separated in to 12^34/9 and 56^78/9
Now, 12^34/9 = 3^34/9
3^even is always divisible by 9, thus 3^34 mod 9 = 0

I dont think we need to find the 56^78/9 part now,

So, 12^34 * 56^78 mod 9 = 0

Am I right here?

Mohit bhai: guess you are right.. thanks for pointing it out :)

Sufal bhai, did you get the concept thoda sa bhi?

and here are the 3 links where you will find previous years and this years other coaching institute's mock cat papers..
includes SIM CATS, AIMCATS, Proc CATs..

Abhishek Jain on Scribd | Scribd
http://www.mediafire.com/rahuliit2003#in4jxj7ejj88u,1
4shared - TimeAimcat2010 - shared folder - free file sharing and storage

All the best hai!

and if someone is giong for some particular test, do let me know.. i might give it also and we can discuss.

Thanks and all the best

Rohit

PS : Sufal bhai, sorry hai for posting these late, there was no power once i came back from office at 12 last night.

Haan Mishra Ji, Got the concept. will solve the questions related to the concept which you posted in the thread today. And thanks a lot for those links.

Some exercise question for all of us:
1. What are the last two digits in 1372^482
2. remainder when 1234^5678/91
3. Remainder when 12^34 * 56^78/9

1)There are two ways of doing this, one is to find the remainder when 1372^482 / 100.Since, when divided by 100, we get the last two digits.So, I tried doing this and got stuck at 72^482/100 =??.Couldnt go any further, anybody has got it doing this way?

The second way is to find the digits separately.The last digit of 13726482 is nothing but the last digit of 2^482.looking at the cyclicity of 2, we get 4 as the last digit.

The second last digit can be found out using using the method Ravi Handa had posted a while ago on PG, this is the link to it:

Winning Silver: Figuring out the second last digit in Quantitative Aptitude | PaGaLGuY.com - Indias biggest website for MBA in India, International MBA, CAT, XAT, SNAP, MAT

So, accordingly, the second last digit of 1372^482 is:
(1372^10)^48 * 1372^2
= (72^10)^48 * 72^2 (since the first two digits become irrelevant here)
= 24^48 * (xx84) ( since xxx2^10 always ends with 24)
= xx76 * xx84 (since 24^even always ends with 76)
= 8
Thus, the second last digit is 8 and we had already got last digit as 4, so last two digits are 84.Infact, I guess we dont even need to the first step. that is the mod 10 one, since we can directly get the last two digits from (xx76 * xx84).

Correct me if I am wrong guys!

Mohit bhai: guess you are right.. thanks for pointing it out :)

Sufal bhai, did you get the concept thoda sa bhi?

and here are the 3 links where you will find previous years and this years other coaching institute's mock cat papers..
includes SIM CATS, AIMCATS, Proc CATs..

Abhishek Jain on Scribd | Scribd
http://www.mediafire.com/rahuliit2003#in4jxj7ejj88u,1
4shared - TimeAimcat2010 - shared folder - free file sharing and storage

All the best hai!

and if someone is giong for some particular test, do let me know.. i might give it also and we can discuss.

Thanks and all the best

Rohit

PS : Sufal bhai, sorry hai for posting these late, there was no power once i came back from office at 12 last night.

mohit07 Says

Ok, I saw the missile destroying the bridge solution given by TIME and I couldnt get much of it.could somebody help me out!

yeah, time's solution is a bit confusing. Let me tell you how I did.

See there are 4 options.
1.7
2. 8
3. 14
4. 15
And question is find the maximum number of shots in which probablity is > 99%. That means, probablity of not destroying is less than 1%

So, if you start with 7, lets find the probablity that the bridge does not get destroyed. So, since we need atleast 3 shots. So, there are three cases:
1.0/7 shots hit.
probability 7C0*(1/2)^7
2. 1/7 shots hit
7C1*(1/2)*((1/2)^6)
3. 2/7 shots hit
7C2*(1/2)^2*(1/2)^5

So total probability will be (7C0 + 7C1 +7C2)/(2^7). This is > 1%
Similarly for 8 will be (8C0 + 8C1 +8C2)/(2^ This is also > 1%
Similarly for 14 will be (14C0 + 14C1 +14C2)/(2^14). This will be < 1%.
So, 14 will be the answer.

Sufal bhai you need to get well versed with chinese remainder theorem and Eulers number concept and you can solve these questions easily.

something similar has been discussed at the starting of this thread. go thru that...
In very short i will try to explain you:
Euler number is given after u have spplit you number into prime factors.
like 164 = 41*2^2

now E(164) = 164(1-1/2)*(1-1/41) = 80



Now any number to the power of 80 when divided by 164 will give 1 as remainder.

so 2^164/164 = (2^80)^2 * 2^4/164 = 2^4/164 = 16

Mishra Ji,

Ek doubt , for the Euler's theorem to be applicable, the dividend and divisor should be co-prime na?I mean, suppose, a^n/p, so here 'a' and 'p' need to be co-prime right to apply Euler's theorem.SO I guess, we cant apply Euler's directly to 2^164/164.We will have to reduce it to 2^82/ 41 and then multiply the obtained remainder by 4.Am I right?

Ok, I saw the missile destroying the bridge solution given by TIME and I couldnt get much of it.could somebody help me out!

Ok, I saw the missile destroying the bridge solution given by TIME and I couldnt get much of it.could somebody help me out!

There are different ways to tackle remainder questions, depends on personal preferences (refer TG-Number Systems for different methods).

For this probability question
Prob. not hitting bridge in 1 shot = 1/2
Prob. not hitting bridge in 2 shots = 1/2*1/2
Prob. not hitting bridge in 7 shots = 1/(2^7) = 1/128

Now prob. of hitting bridge in 7 shots = 1-1/128 =127/128 > 0.99
note that we need to find x such that 1/(2^x) should be less than .01 (1/100)

Not able to remember what i marked for it, was 7 an option there??

Hey,

The actual answer,according to the TIME key is 14 and not 7.

Mumbo Jumbo Says

Hey everybody..... Aimcat 07 results are out

Yep!I just managed to clear the overall cutoff by 1 mark!

Hey everybody..... Aimcat 07 results are out

Hey Mohit.. Thanks for the link.. But this dosen't have an index/content. So could not find the solutions...... Does this ebook has the solutions/answers as well...
Guys is there any other material from where we can practice CR/FIJ

hey,

yes it has solutions.It doesnt have an index but the solutions are just after all the questions.

sufal Says

Thanks, Rohit Bhai. I have not given any mock apart from AIMCATS. Any other than that works fine for me. Btw can you please explain that remainder question again in detail? I didn't get it. I always loose out on such questions.

Sufal bhai you need to get well versed with chinese remainder theorem and Eulers number concept and you can solve these questions easily.

something similar has been discussed at the starting of this thread. go thru that...
In very short i will try to explain you:
Euler number is given after u have spplit you number into prime factors.
like 164 = 41*2^2

now E(164) = 164(1-1/2)*(1-1/41) = 80

Now any number to the power of 80 when divided by 164 will give 1 as remainder.

so 2^164/164 = (2^80)^2 * 2^4/164 = 2^4/164 = 16

This is one way of solving. This concept is extremely important as it reduces big powers to small values.

now to make it little complex lets find the remainder for 2^164^164/164

Rem(2^164^164/164) = Rem(2^4^164/164) - this is using E(164) = 80 as above.

Ab aar dhang se dekhein toh hum log power ka remainder nikal rahe hain when it is divided by E(164)= 80

toh lets find that, i.e. Rem(4^164/80) = Rem(4^162/5) = 1

this reduces ou problem from 2^164^164/164 to 2/164 = 2


Lemme know if somethin confuses you here...

there is another way of solving it, CRT... read that and solve these questions again.


Some exercise question for all of us:
1. What are the last two digits in 1372^482
2. remainder when 1234^5678/91
3. Remainder when 12^34 * 56^78/9

questions are self created... might be abrupt. Try kariye sab log and full approach post kariyega...

PS : i got that question wrong in this AIMCAT silly max mistake.

no link available for aimcat 1206 in my page ...... guys is there anyone who already attempted this aimcat ?

Hey Puys...

Please find attached a list of common English confusing words.
Also attached is a set of 80 questions for your practice (or practise :P) and to make usage of many of them clear.


Please go thru them today and we can discuss anything here itself!

Thanks Mishra Ji,

This is indeed a very important part of verbal and one of the best collection of confusing words for CAT is in one of the TIME verbal booklets

mohit07 Says

I found the CR from this helpful.Maybe some of you might have already have,its not the latest edition.

Hey Mohit.. Thanks for the link.. But this dosen't have an index/content. So could not find the solutions...... Does this ebook has the solutions/answers as well...
Guys is there any other material from where we can practice CR/FIJ

hey Sri bhai

Actually point of giving that question was this only.
actually, if u try to draw the parallelogram with diagonal 5 as other diagonal, u will see that the dimension values wont satisfy that.

All,
please find attached a list of phrasal verbs that are more commonly used. Thoda sa help hoga...

M planning to start my preparation for the final leg for CAT from today :$ hope it starts :(

Sufal bhai, will post the link for other MOCK CATS soon. Infact i was thinkiing we all can decide on one of the mock cat that most of us have not given and give that together and talk abt it in a chat session or here... whatsay??

Also, can anyone tell me what other exam dates and foems are out. I think i will lose out on something important :P

All the best bhai logon!

Thanks, Rohit Bhai. I have not given any mock apart from AIMCATS. Any other than that works fine for me. Btw can you please explain that remainder question again in detail? I didn't get it. I always loose out on such questions.

hey Sri bhai

Actually point of giving that question was this only.
actually, if u try to draw the parallelogram with diagonal 5 as other diagonal, u will see that the dimension values wont satisfy that.

All,
please find attached a list of phrasal verbs that are more commonly used. Thoda sa help hoga...

M planning to start my preparation for the final leg for CAT from today :$ hope it starts :(

Sufal bhai, will post the link for other MOCK CATS soon. Infact i was thinkiing we all can decide on one of the mock cat that most of us have not given and give that together and talk abt it in a chat session or here... whatsay??

Also, can anyone tell me what other exam dates and foems are out. I think i will lose out on something important :P

All the best bhai logon!

Thanks a lot Mishra ji :D
Have been searching for Phrasal verbs

Friends... try this:

What is the area of a parallelogram with an angle 45 degrees, height 4 cm and a diagonal 5 cm?

PS: little disappointed with the way the posts are coming on this thread. This thread is next to dead if i may say so. If nothing to discuss or post, we can just share a link, someone else's post, some material or even a joke... all of us can always savor a nice joke, cant we...?
The idea is to increase the regualrity of people on this thread and increase dependency on each other... more people come here, faster will the issues get resolved... all for biger better good. Sincere request, post anything, but make sure that u are known to the group.

Sorry if that feels like a jargon, ignore aur groan maar dena agar too much ho gaya ho toh :)

28 should be the answer when the value 5 is taken for the diagonal opposite to the angle 45 ..... but if this value is taken for the other diagonal then the answer may change and that will be a comparatively tough question ... i think this is the ambiguity in this question .. correct me if iam wrong ....... but when two different diagonals are possible , how can we assume that , this value 5 should be considered for a particular diagonal and not the other ?

Yes , it was there

mohit07 Says

Thanks , I think I got the drift of the second one and Mishra Ji just further explained the first.The third question actually was, The probability of a missile hitting a bridge is 1/2 and it takes three missiles to destroy a bridge.So, what is the minimum number of missiles so that the probability of the bridge getting destroyed exceeds 0.99?

There are different ways to tackle remainder questions, depends on personal preferences (refer TG-Number Systems for different methods).

For this probability question
Prob. not hitting bridge in 1 shot = 1/2
Prob. not hitting bridge in 2 shots = 1/2*1/2
Prob. not hitting bridge in 7 shots = 1/(2^7) = 1/128

Now prob. of hitting bridge in 7 shots = 1-1/128 =127/128 > 0.99
note that we need to find x such that 1/(2^x) should be less than .01 (1/100)

Not able to remember what i marked for it, was 7 an option there??

For 1st question, my approach is

2^12%164=-4
(-4)^13*2^8%164 = -64*256%164 = 256*100%164 = 16 as remainder

For 2nd qn

Find the first point of meet (d/1280-d = 24/39)
Based on this get the relationship between sucessive points of meeting.

Not able to remember 3rd qn

Thanks , I think I got the drift of the second one and Mishra Ji just further explained the first.The third question actually was, The probability of a missile hitting a bridge is 1/2 and it takes three missiles to destroy a bridge.So, what is the minimum number of missiles so that the probability of the bridge getting destroyed exceeds 0.99?

Wow, that is as simple as it can get,Thanks Mishra Ji!:D

mohit07 Says

Could you explain your approach for the remainder question, like how do you know when you look at it that we gotta take 2^12, cuz that is 4096, so you divide it by 164 and check the remainder , as in the pattern method?

Mohit... u know the fact that remainder of a/b = c, then remainder of ka/kb = kc.

eg: 5/4 = 1 rem
and 25/20 = 5*1 = 5

Now looking at the question, 2^164/164, we can cancel 4 from numerator and denominator. that leaves 2^162/41

Now E(41) = 41(1-1/41) = 40
therefore 2^40/41 = 1

or, Rem(2^164/164) = 4* Rem(2^162/41) = 4* Rem(2^2/41) = 4*4 = 16

btw, I found this link useful:

http://www.pagalguy.com/forum/quantitative-questions-and-answers/23536-concepts-total-fundas-4.html

Could you explain your approach for the remainder question, like how do you know when you look at it that we gotta take 2^12, cuz that is 4096, so you divide it by 164 and check the remainder , as in the pattern method?

Btw, now that I guess the 1207 window has closed, I guess I can safely ask this here, did anyone get the answer for these questions from the paper:
1)Remainder of 2^164 / 164 ?On a general note, are there any fixed methods to solve REM sums, I mean I tried the Fermat theorem, the Pattern method and the Remainder theorem for this one,it didnt work!
2)The speed, distance and time question where two people are running and their speed changes after every meet?
3)The probability of destroying the bridge question?

P.S: I know the TIME key should be out soon, but I just wanted to see if people had multiple ways of answering them and probably try to find the best out of them!

For 1st question, my approach is

2^12%164=-4
(-4)^13*2^8%164 = -64*256%164 = 256*100%164 = 16 as remainder

For 2nd qn

Find the first point of meet (d/1280-d = 24/39)
Based on this get the relationship between sucessive points of meeting.

Not able to remember 3rd qn

Btw, now that I guess the 1207 window has closed, I guess I can safely ask this here, did anyone get the answer for these questions from the paper:
1)Remainder of 2^164 / 164 ?On a general note, are there any fixed methods to solve REM sums, I mean I tried the Fermat theorem, the Pattern method and the Remainder theorem for this one,it didnt work!
2)The speed, distance and time question where two people are running and their speed changes after every meet?
3)The probability of destroying the bridge question?

P.S: I know the TIME key should be out soon, but I just wanted to see if people had multiple ways of answering them and probably try to find the best out of them!

This is the 12th edition.A faculty from TIME had said that the CR from a GMAT material should be enough, and It is working in my case, my accuracy in CR has improved. But could anyone help me in FIJ type questions, I suck at them!

I found the CR from this helpful.Maybe some of you might have already have,its not the latest edition.

birbal4cat Says

Mishra ji, 28 hai kya iska answer??

Mujhe bhi 28 aaya!

Friends... try this:

What is the area of a parallelogram with an angle 45 degrees, height 4 cm and a diagonal 5 cm?

Mishra ji, 28 hai kya iska answer??

Testfunda has provided free access to their rank booster program for 3 days.

Visit the link for some practise
Extra Practice

Friends... try this:

What is the area of a parallelogram with an angle 45 degrees, height 4 cm and a diagonal 5 cm?

PS: little disappointed with the way the posts are coming on this thread. This thread is next to dead if i may say so. If nothing to discuss or post, we can just share a link, someone else's post, some material or even a joke... all of us can always savor a nice joke, cant we...?
The idea is to increase the regualrity of people on this thread and increase dependency on each other... more people come here, faster will the issues get resolved... all for biger better good. Sincere request, post anything, but make sure that u are known to the group.

Sorry if that feels like a jargon, ignore aur groan maar dena agar too much ho gaya ho toh

Guys... this is an exhaustive list of AIMCAT words till last year. Stumbled on these in PG only somewhere. Meanings are alongside :)

What i would suggest is to go thru these on a regular basis and learn specially the confusing ones - eccentric/accentric types... these generally repeat themselves.


And Sufal bhai, have never appeared for any other Test series other than AIMCATs... but for AIMCAT, i m sure you wont get the missed papers. Max what we can do ourselves is to share these exam papers. You can give them at home. Ofcourse percentlie factor wont be there, but u can post ur score in some forum and ask for expected scores.

telling this coz i feel too many cooks spoil the broth. Having another test series to take care of with just 50 days left might spoil your future strategies.

Look at it this way, you have 6 AIMCATS stil left... another test series will have atleast 10 Mocks... so u will have to give 16 tests in 45 days (keeping last 5 days as reserve for going thru everything u wanna revise)- 1 every 3 days - i.e. 3 days for giving test... analysing it... working on your weak areas... and coming up with strategy for the next one.

Just make sure that if you are enrolling for a new one, you are not over doing it for urself.... decide a study plan (rough) and then see how many tesets u can fit in that plan. And in case the need for test is to improve accuracy, speed and get a feel of exam environment (rather than knowing your personal standing) then you can go to sites like chalobolo.com for free tests.

having said that, dont wanna contaminate your thoughts with mine . BTW, Testfuda tests are exact replica of CATs.

Yup ... You can give the previous mocks also, they will be available till CAT is conducted Waise how is the preparation going? MY QA+DI is going on a big time log .. suggest something dude.
And one more clarification. How do we add the signature in our posts :P

Preparation is going okay. Not getting enough time to prepare, thanks to office n stuff. I am just giving mocks and then analyzing them these days.

The DI concepts are generally simplest of all, but the most time taking ones. If you are having issues with understanding the question, I would just say go through the basics once again and with the help of previous mocks, do atleast one question of each. That will make sure that your fundamentals are good. Other than that, try to learn speed maths concepts( I have started from today from TIME material, its pretty good). With Speed maths, we would need to practice it regularly( read: daily) to get the best out of it.

Best way of learning QA, I believe is analyzing the mocks very deeply. Try to solve a question from different techniques, as many as possible. If you get answer wrong through one of them, try to think that why the logic is wrong. Also, one more technique which I do while analyzing is, sometimes, I dont take paper and pen while solving them. Try eliminating options and use shortcuts where-ever possible. With a good amount of practice while analyzing, you will realize that you are automatically using these tricks in exams. Also, before attempting a question, read the question very carefully. Many a times, there is a fine print in the question, that we miss while reading.

As for signatures, I think, right now it is only enabled for few special creatures like pg madcapz, moderator etc and its not for Aam Junta like us.

Thanks man for the advice. Just a small clarification
Wao, So I can also give proc mock-1 to 9 also, whose test dates are over? Or can I give only those whose test-window is open? If it is, then it would be pretty kool.

Yup ... You can give the previous mocks also, they will be available till CAT is conducted Waise how is the preparation going? My QA+DI is going on a big time low .. suggest something dude. Help Needed team mate
And one more clarification. How do we add the signature in our posts

TheMask03 Says

Dude I have joined CL and its pretty good. You can even give the mocks which you have missed, as for not able to go to the centre problem even i give all the mocks at my home. The proctored mocks remain open for 3 more days i.e. monday,tuesday and wednesday so you can very well give it on those days. Go for CL Up Kaptaan :D

Thanks man for the advice. Just a small clarification
Wao, So I can also give proc mock-1 to 9 also, whose test dates are over? Or can I give only those whose test-window is open? If it is, then it would be pretty kool.

Puys, need your advice. Is there anyone here who is giving mocks apart from TIME ones like CL or Testfunda?

I want to join 1 more test series. either CL or Testfunda, just the thing is I wont have much time to go to the test center. So, its preferable if I give it at home. Is it possible at CL? Also, which one of the two is closer to the actual mock? Also, if possible, I would like to give old mocks, which I would have missed, if I join now. So, which do you think of the two is better for me?

Dude I have joined CL and its pretty good. You can even give the mocks which you have missed, as for not able to go to the centre problem even i give all the mocks at my home. The proctored mocks remain open for 3 more days i.e. monday,tuesday and wednesday so you can very well give it on those days. Go for CL Up Kaptaan