@[495113:joethaliath] Think of it like this.Fresh grape contains more water than raisins.So you can mix raisins with water to produce fresh grapes.So we are mixing a 20% material with a 100% material ( pure water) to get an 84% material.Hence using alligations raisins and water must be mixed in the ratio 1:4.Since we have 80 Kg fresh grapes, 1/5th of it would be raisins.So ans is 16 Kg
@[397955:rohan_bhasker]
@[526706:pradyothcjohn]
yea 😃 got it. thanks :D
@joethaliath said: Fresh grapes contain 84% water while raisins contain 20% water.How many kg of raisin can be made from 80kg of fresh grapes>a)16 kgb)18 kgc)20 kgd)22 kg
sorry to barge in..but could stop myself from solving this question...
As fresh grapes contain 84% water.. this means that 80 kg of fresh grapes would contain 67.2 kg water and 12.8 kg dry mass.
So this 12.8 kg of dry mass would constitute for 80% of raisin(as raisins contain 20 %water) Therefore (12.8/80)*100 would be the reqd answer ...i.e 16 kg
@[573636:Debasis070] thanks 😃
my cat date ..... Oct 30th , 3:15 pm , Chennai ....
@[149515:dead_alive] are you working in chennai?
Hey All, My CAT date 25th oct- 3:15 pm, Ekm.. This time i don't think i am going to get any good percentile. My consistency has been very poor. Percentile dips n rises on a big margin
.. 60- 90.. never crossed 90 yet. Need some serious preparation. If not being able to crack this time, which i am rather very sure i wouldn't, den i'm Gearing up to be part of Kerala DT 2013..
By then i'll have an year of Job experience too. Currently Graduate trainee in LNt, kolkata.
I have a query.Even though I would be giving CAT only in 2013 , some of my friends are giving it this year even though they are not eligible to join ( 4th year of integrated 5 year course i.e pre-final year) .
They say they won't be responding to any mails/calls from IIMs(provided they manage to get a good percentile) and thus it won't create any issue next year, when we really will become eligible to join.
By doing so they will be able to get an idea of where they stand of now.
Is it risky ? Should I also give it a try ? Or should I just keep solving mock tests and build basics up by 2013? Or should I take XAT(that also would give me approximate idea of where I stand) ? Please give your opinions
@spdspd said: I have a query.Even though I would be giving CAT only in 2013 , some of my friends are giving it this year even though they are not eligible to join ( 4th year of integrated 5 year course i.e pre-final year) .They say they won't be responding to any mails/calls from IIMs(provided they manage to get a good percentile) and thus it won't create any issue next year, when we really will become eligible to join.By doing so they will be able to get an idea of where they stand of now.Is it risky ? Should I also give it a try ? Or should I just keep solving mock tests and build basics up by 2013? Or should I take XAT(that also would give me approximate idea of where I stand) ? Please give your opinions
I don't think it'll be a problem,even i know people who've done that..
@[533596:Kevin88] @[495113:joethaliath] I changed my date. 30th morning.
Nice initiative to discuss questions here.
Gave AIMCAT 1308 today
QA/DI 16A 10C 6W == 24
VA/LR 26A 18C 8W == 46
OA 42A 28C 14W== 70
Feels miserable. Ever since this season started, i have been clearing Section II cut offs every time and missing Section I cut off !! So, the only thing i have been doing is QA/DI. Still my marks end up between 20-26 for 5 consecutive AIMCATs !!
As for %iles
AIMCAT 1313 86.41
AIMCAT 1312 80.3
AIMCAT 1311 84.92
AIMCAT 1310 86.25
AIMCAT 1309 Dint take
AIMCAT 1308 Looks like the same thing is gonna happen 
Team.. A few queries from my end...
1) If the first two terms of the HP are 2/5 and 12/13 respectively, which of the following terms is the largest term?
a) 4th Term b) 5th Term c) 6th Term d) 7th term e) 8th Term
2)Suppose that a and b are two integers, not necessarily
distinct, chosen from the set S = 1, 2, 3, · · · , 200. What is the
probability that the unit digit of 3a + 7b is 6
a)5/16 b) 3/16 c) 3/17 d) 7/17 e) 3/13
Any idea on how we can PM the entire group rather than adding guys one by one ?
@ all the active Mallu friends here!!
Great going guys!!!....i have been visiting PG from 2010(even though i am not an active member in any threads). I may say this is the first time i see a group exclusive for mallus being so lively and brisk inside PG. Keep this momentum up till the D-day.
I am planning to analyze all Mocks(CL and Time) in the month of september. Also ill try my level best to give all MCs and CRTs uploaded in the time homepage in the same month.
In the month of October, i will be concentrating much to tighten the Basics of all the portions covered under CAT. As many of you guys might be knowing, in CAT they try to measure your depth of knowledge from each areas. They wont give much importance in testing your strength in mugging up formulas or your Voccab build.
Also my sincere advise to the fresh takers here , dont get demotivated by the difficulty of quant questions in AIMCATS. I swear, the actual CAT quant difficulty will be far less. So practice more and more questions from whichever available resource( Arun Sharma/ Arihant/ Time materials/......)and get your base strong and rigid. Because i have seen lot many smiling faces after attempting sec 1 in actual CAT. (They would have never attempted such an easy Sec 1 during the entire year while giving AIMCATS)
And for all the repeat takers like me, this should be your last attempt. You have lot many other things to do in life rather than giving mocks, analyzing them and posting your score in PG every year. Life of all of us is meant for doing better things than this. So get rid of the formality of cracking CAT this year itself.
So in the coming 50-60 odd days, the effort you are planning to put in, will definitely reflect on your final score. So utilize each and every second available for you and prepare to the level best....ATB!! :)
@[533596:Kevin88] My take is 1) Option e 8th term
@Kevin88 said: Team.. A few queries from my end...1) If the first two terms of the HP are 2/5 and 12/13 respectively, which of the following terms is the largest term? a) 4th Term b) 5th Term c) 6th Term d) 7th term e) 8th Term2)Suppose that a and b are two integers, not necessarily distinct, chosen from the set S = 1, 2, 3, · · · , 200. What is the probability that the unit digit of 3a + 7b is 6a)5/16 b) 3/16 c) 3/17 d) 7/17 e) 3/13
1) Find the correspond terms in AP. here it is 5/2 and 13/12.
Evidently common difference is negative. that is -17/12. so after 2nd term, all the terms in AP will be negative. So the biggest term of this HP will be the second term.
But out of the options available, the biggest term will be 4th term( lowest negative value)
@[533596:Kevin88] For question 2) I have done this problem on the prep thread.I dont know the exact solution,but made a guesswork to get to the right answer.Now probability = No of favorable outcomes/ Total no of possible outcomes.Here a and b can be chosen in 200 x 200 ways. ie 40000 ordered pairs.This comes as the denominator.Considering the options, the only factor of 40000 in the denominator of the any option is 16.Now three of the options have 3 in the numerator.so i decided to go with 3/16 which is the right answer.I think perhaps pugna,rakesh or pradyoth might be able to give you the actual solution
@Kevin88 said: Team.. A few queries from my end...1) If the first two terms of the HP are 2/5 and 12/13 respectively, which of the following terms is the largest term? a) 4th Term b) 5th Term c) 6th Term d) 7th term e) 8th Term2)Suppose that a and b are two integers, not necessarily distinct, chosen from the set S = 1, 2, 3, · · · , 200. What is the probability that the unit digit of 3a + 7b is 6a)5/16 b) 3/16 c) 3/17 d) 7/17 e) 3/13
2)option 2) 3/16???
@raku1989 said:2)option 2) 3/16???
@[397955:rohan_bhasker]
@[533596:Kevin88]
U can apply the concept of MOD here....u dont need to consider whole 40000 possibilities.
Since u are finding out the posibility of 3a+7b=6,
Consider the powers of 3 and 7 mod 10
3^1 mod 10 = 3
3^2 mod 10 = 9
3^3 mod 10= 7
3^ 4 mod 10 =1
the same pattern repeats
Same is the case with 7.(7, 9, 3, 1)
So a total of 16 outcomes are there where the sum is 6 only in 3 cases
(3,3) (9,7) and (7,9)
So total probability = 3/16
@[518068:raku1989] @[397955:rohan_bhasker]
Solutions: 1) a - 4th Term
2) b - 3/16
For 1 : Even i got it as the 8th term but sadly it wasn't right.
Pls post the steps for Q2.
Rakesh , in the AP format the 4th term would be the largest (-21/12) but while in an HP shouldn't the 8th term be the largest (-12/89) > (-12/21) ?