CAT 2012 Kerala study group

MBA CAT 2024
681
@Kevin88 said:@seetharam7 Gud suggestion, however if we try 2 chapters a day.. it wud result in confining ourselves to only a few Q's per chapter.. how abt we discuss 2 chapters per week; tht wud mean abt 1 chapter for say abt 2-3 days depending on the feedback from the team.
IAf any one has a particular weak section , we cud discuss more Q's based on the response & activeness of the team .
If each member of the team can post abt 5 different Q's relating to the respective chapter at the very least , we cud have a discussion of nearly 30-50 Q's
.
However this need not be compulsory, team members can post difficult Q's & any Q's they have a doubt in or which they encounter during self prep which in turn cud strengthen the core concepts of the entire team as a whole too..
@raku1989 @rohan_bhasker @joethaliath @pradyothcjohn @pugna @dead_alive @vinmohcetvmklm @brandnewrap @KatMann @sriraman89 : Any Suggestions ??

I am in for this plan!!!

I would suggest to club up 2 -3 areas in quant and give a span of 7-8 days for each Section

Quant topics can be reduced to 6 or 7 different sections

Check out the Prep thred where they have compilled it nicely.

Eg: 1) Time Speed and Distance, Work, Alligations & Mixtures

2) Mensuration, Geometry & Co-ordinate Geometry

3) Set Theory, Logarithms, Ratio Proportion & Variation

4) Permutations and Combinations, Probability

5)Algebra, Quadratic Equations, Inequalities, Functions

6) Averages, Percentages, Profit Loss and Interest

Same way we can combine together VA and LR sections too.

But considering the time we are left with i feel we have to do a section from all the three areas in one week time so that we can cover all the topics by OCT 1st week.

What say guys??

So 24th-31st Aug....what all sections can we cover?

682
The probablity of a bomb hitting bridge is 0.5 and two direct hits are required to destroy it. Find the least number of boms required so that probablity of the bridge being destroyed is 0.9
  • a) 11
  • b) 7
  • c) 9
  • d) 8
683

Lets go as per the list of topics mentioned in your post...

So 24-31 Aug :Time Speed and Distance, Work, Alligations & Mixtures
What say ??
Team, don't stop posting Q's you have doubts in too !!!
684
@pradyothcjohn said:
The probablity of a bomb hitting bridge is 0.5 and two direct hits are required to destroy it. Find the least number of boms required so that probablity of the bridge being destroyed is 0.9
  • a) 11
  • b) 7
  • c) 9
  • d) 8










b)7 ??
685

7 was not in the options actually,i added it after i saw an explanation 😁 So how do we get 7?

686

@[526706:pradyothcjohn] @[518068:raku1989] : Even i got 7


P = 1-(7c0 +7c1)/2^7 =0.9375
687
@Kevin88 said:@pradyothcjohn @raku1989 : Even i got 7
P = 1-(7c0 +7c1)/2^7 =0.9375
Same method here....check the probability of all 4 options.
I got 7 ,s probability around .93
688
@pradyothcjohn said:7 was not in the options actually,i added it after i saw an explanation So how do we get 7?
What is the correct ans bdw?
689

7c0+7c1, how do we get that? i would have just done 1- (1/2)^7 .. and got a negative 😁

690

@[526706:pradyothcjohn]

Probability of Bridge being destroyed = 1- Probability of No Hits - Probability of 1 Hit
Now P(Hit)=P(No Hit)=1/2
So when hell rains down on the bridge (lets say we send in 7 bombs)
P(7) = 1-(7c0 +7c1)/2^7 =0.937

P(6) = 1-(6C0 +6C1)/2^6 = 0.8906
691

7 should be the correct answer..

692
@pradyothcjohn said:7c0+7c1, how do we get that? i would have just done 1- (1/2)^7 .. and got a negative

My approach was

We need to find out the probability of atleast two hits

That would be equal to (1-(prob of 1 hit + prob of 0 hit))

Now if 7 bombs are fired,

U can get a total of 2^7 possible results (either hit or miss)

out of that there are 7 ways to hit one time

and 1 way so that none of the bombs hit the target

693

got it,thanks @[533596:Kevin88] and @[518068:raku1989] 😁

694
@raku1989 said:

My approach was

We need to find out the probability of atleast two hits

That would be equal to (1-(prob of 1 hit + prob of 0 hit))

Now if 7 bombs are fired,

U can get a total of 2^7 possible results (either hit or miss)

out of that there are 5 ways to hit one time

and 1 way so that none of the bombs hit the target

So the required prob will be

(1-(((7+1)/2^7))

=1-(8/2^7)

its approx 0.95

695

@[518068:raku1989] When 7 bombs are dropped,ther wud be 7 possibilities when a single hit is made (not 5)


0000001 (1-Hit & 0-Miss)
0000010
0000100
.
.
1000000
696
@Kevin88 said: @raku1989 When 7 bombs are dropped,ther wud be 7 possibilities when a single hit is made (not 5)
0000001 (1-Hit & 0-Miss)
0000010
0000100
.
.
1000000
ya I solved the same taking 7 possibilities and typed here 5. My mistake. Corrected now.
697

Sorry for violating our agenda....but need help on a qn from numbers...

How many of the first 2100 natural numbers are either prime to 6,15 or 35 ?

700

400

520

640

Answer is 700

Pls provide an explained soln for this

698

@[533596:Kevin88]@[526706:pradyothcjohn]@[397955:rohan_bhasker]@[544751:pugna]

699
@raku1989 said:

Sorry for violating our agenda....but need help on a qn from numbers...

How many of the first 2100 natural numbers are either prime to 6,15 or 35 ?

700

400

520

640

Answer is 700

Pls provide an explained soln for this

@[582772:brandnewrap] any idea?
700

@[518068:raku1989]

Coprime to 6 = 2100(1/2)(2/3) = 700
Coprime to 15 = 2100(2/3)(4/5) = 1120
Coprime to 35 = 2100(4/5)(6/7) = 1440
Coprime to 30 = 560
Coprime to 105 = 960
Coprime to 210 = 480

Numbers either prime to 6 or 15 or 35
= 700 + 1120 + 1440 - 2((560 - 480) + (960 - 480)) - 3*480 = 700

See attached figure : Find 1+3+7 only