CAT 2012 Kerala study group

MBA CAT 2024
641
assume time taken to read a passage to be 1x.

then time taken to solve the 27 qns =
27*x/12= 2.25x (he can solve 12 qns in 'x' time)

total time spent on reading passages = 4x

Therefore total time spent in reading the 4 passages AND solving all the qns= 4x+2.25x= 6.25x


10% of total time = .625x

i.e,
we have to cut time spent by 0.625x ,

This would make the time spent on reading the 4passages =
4x-0.625x = 3.375x.

Earlier, he read 1 passage in 1x time.
Now, he reads 4x/3.375 = 1.1851 passages in 1x time i.e
vaibhav increases speed of reading by 0.1851 passage/time ie,
18.5%
642

@[533596:Kevin88] i just drew the thing and counted. 1st got 15 and then when i checked over here, understood that i was wrong

@[319787:vinmohcetvmklm] Thanx for the solution.
643
@seetharam7 said: My goodness, this thread has become really active. Anyways, this is how i solved kevin's CAT ques.
LCM of 5,8,6 =120
time to read passage =120min
time for 1 ques = 10min
Total time spent = 120*4 + 27*10 =750
Reducing 10%, we get time = 675
Now time for ans questions is same (ie 270) and so
675=4x+270
x=101.25
% change = (101.25-120) / 120 *100 =15.625%
Hows the answer 18.5 ??
Dude, the question is to find the percentage increase in speed of reading.
So till the last step ur method is perfect.
U found out the percentage decrease in time instead.
So actually it has to be (1/101.25)-(1/120)/(1/120) = 18.5
if he read 1 rc in 120 mins initially, ie 1 min he read (1/120)th of the RC
later he read 1/101.25th of the RC in 1 min
Hope it is clear.
644

@[514195:seetharam7]

He..he @[397955:rohan_bhasker] Initiative to improve activesness is officially a success!!

Hope u're clear abt how its 18.5%.. @[518068:raku1989] & @[526706:pradyothcjohn] have given explanations for the same.
Regarding the no of squares.. the ans is 15
All we need now are more concepts, queries & a myriad of questions from the rest of the team...
Even if u guys find gud Q's..post it here for the benefit of the entire team !!!

645

8) There are 3 boxes A, B, and C in which 9 balls comprising 3 identical red balls, 3 identical blue balls and 3 identical green balls are to be placed such that there are exactly 3 balls in each box and no box contains all three balls of the same colour. In how many ways can this be done?


1)31 2)35 3)25 4)49

646
@Kevin88 said: 8) There are 3 boxes A, B, and C in which 9 balls comprising 3 identical red balls, 3 identical blue balls and 3 identical green balls are to be placed such that there are exactly 3 balls in each box and no box contains all three balls of the same colour. In how many ways can this be done?
1)31 2)35 3)25 4)49
a) 31???

647
@Kevin88 said: 8) There are 3 boxes A, B, and C in which 9 balls comprising 3 identical red balls, 3 identical blue balls and 3 identical green balls are to be placed such that there are exactly 3 balls in each box and no box contains all three balls of the same colour. In how many ways can this be done?
1)31 2)35 3)25 4)49
31 ways???
648

@[518068:raku1989] Ya.. its 31.. any other way than listing ??

You are on a roll brother !! Replying correctly to almost everything i ask !!!
649
@raku1989 said:
31 ways???
3 different posibilities
1) all three balls are of different colour in all three boxes - 1 way
2) 2 balls are of same colour in all three boxes - 24 ways
3) 1 box with all three different and 2 boxes with 2 balls same colour - 6 ways

Total 31 ways??
650

@[518068:raku1989] Cud u elaborate on how u got the case of 2 balls of the same colour in all 3 boxes = 24 ways ??


651
@Kevin88 said: @raku1989 Cud u elaborate on how u got the case of 2 balls of the same colour in all 3 boxes = 24 ways ??
Consider box A
U can choose the 2 balls of same colour in 3 ways. And the third ball in 2 ways.
So a total of 6 ways.
Now u consider box B
where u can choose 2 balls of same colour in 2 ways. and again the third ball in another 2 ways.
So a total of 4 ways.
Third box will get filled with the remaining balls.
Total ways = 6*4 = 24 ways
652

Nice one!! @[518068:raku1989]

Honestly speaking, i was kind a lost after reading the question !! He..he
653

What is the remainder when 33^34^35 is divided by 7.

options
5
4
6
2

654

How many integer values of x and y will satisfy he eqn.
4x + 7y = 3
where mod x

284
285
286
Cannot be determined
655
What is the remainder when 33^34^35 is divided by 7.

raku1989 Is the sol 2 ?

656
@Kevin88 said: @raku1989 Is the sol 2 ?
yup :)
657

@[533596:Kevin88]

Quote the question while replying the answer
So that later people wont get confused. :)
658

33^34^35 /7 is similar to (5^34^35)/7

5^N /7 will give remainders of 5,4,6,2,3,1 & repeats again..

Therefore are next objective will be to find the remainder of (34^35) / 6
(6 bcoz we observe a pattern of 6 from the earlier step)

Now (34^35) / 6 is the same as 4^35 /6... will always give 4 as a remainder.

Now the final sol : 33^34^35 /7 is similar to 5^(6k+4) /7
So Remainder is 2 .
659
@raku1989 said: How many integer values of x and y will satisfy he eqn. 4x + 7y = 3 where mod x
284
285
286
Cannot be determined
Is the ans 286
660
@Kevin88 said:
Is the ans 286
nop...its 285!!
check again. 143+142 :)