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manikandanp
Manikandan P

6 Followers 15 Following 0 Pings 675 Posts 918 Karma Score

manikandanp's posts

manikandanp replied to [9 Jan 2013] CAT 2012 Results : Analysis, Discu...

my frnd got 77 percentile.. General category.. what are the college for which the deadline is not over yet and he can apply?
thanks in advance

11:18 AM, 11 Jan '13

[9 Jan 2013] CAT 2012 ...

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my frnd got 77 percentile.. General category.. what are the college for which the deadline is not over yet and he can apply?

thanks in advance

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manikandanp commented on a post

There are five fruits - Banana, Apple, Orange, Pear and Mango - which are to be served among six boys. If no fruit can be served to more than one boy, and the fruits are served randomly, find the probability that all the five fruits were served to exactly three boys.
PLEASE EXPLAIN THE APPROAC...

1) 5/8
2) 25/54
3) 55/108
4) 125/324

17 users have answered this question.

5 Comments

joker29

01:30 PM, 29 Oct '12

Permutations and Combi...

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There are five fruits - Banana, Apple, Orange, Pear and Mango - which are to be served among six boys. If no fruit can be served to more than one boy, and the fruits are served randomly, find the probability that all the five fruits were served to exactly three boys.

PLEASE EXPLAIN THE APPROACH..this is an AIMCAT QUESTION.

1) 5/8
2) 25/54
3) 55/108
4) 125/324 Submit Submitting Skip

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manikandanp commented on a post

what is the sum of coefficients in the expansion (3+2x)^99
plz give approach

1) 5^99
2) 3^99
3) 2^99
4) 6^99

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anirudh1

08:58 AM, 29 Oct '12

Number Systems, Progre...

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what is the sum of coefficients in the expansion (3+2x)^99

plz give approach

1) 5^99
2) 3^99
3) 2^99
4) 6^99 Submit Submitting Skip

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manikandanp replied to Official Quant Thread for CAT 2012

start - 7x 6x 5x
end- 6y 5y 4y
they gamble among themselves
18x = 15y
y = (6/5) x
first person - mod(7x - 6y) = 7x -36/5*x=1/5x=12->x=60
he bought 7x which is 420$

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04:06 PM, 27 Oct '12

Official Quant Thread ...

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@catanuj said:
Three men are gambling in Casino Royal. They start with sums of money in the ratio 7 : 6 : 5 and finish with sums of money in the ratio 6 : 5 : 4, in the same order as before. One of them won $ 12. How many dollars did he start with ? [The three men gambled amongst each other only]a)1080 b)420 c)210 d)None of these

start - 7x 6x 5x
end- 6y 5y 4y

they gamble among themselves
18x = 15y
y = (6/5) x

first person - mod(7x - 6y) = 7x -36/5*x=1/5x=12->x=60

he bought 7x which is 420$

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manikandanp replied to Official Quant Thread for CAT 2012

let d first digit be 4.. 4 _ _ _ _ other 4 digits .. 4 can occupy in any of d remaining places(4 ways). other 3 - 9*9*9 -> 729*4=2916
let d first be a digit other than 4.. X _ _ _ _ no of ways of putting 4 is 6 ways and total such nos is 8*9*9 is 3888
answer is 6804??

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03:45 PM, 27 Oct '12

Official Quant Thread ...

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@anudsweet said:
5 digit no. having exactly two 4's in them?

let d first digit be 4.. 4 _ _ _ _ other 4 digits .. 4 can occupy in any of d remaining places(4 ways). other 3 - 9*9*9 -> 729*4=2916

let d first be a digit other than 4.. X _ _ _ _ no of ways of putting 4 is 6 ways and total such nos is 8*9*9 is 3888

answer is 6804??

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manikandanp replied to Official Quant Thread for CAT 2012

while from I - 0.3
P(not B| II) = 0.8 , so the other half would be B from I - P(B∩ I) = 1-0.8 - 0.2

02:25 PM, 27 Oct '12

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@piyushrohella12 said:
There are two boxes – I and II. Each contains balls of two colours – White and Black. A ball is selected at random. It is known that P(W ∩ I) = 0.3 and P(B'|II) = 0.8, where W, B, I and II represent the events White ball selected, Black ball selected, ball selected from Box I and ball selected from Box II respectively. Find P(I|B).

while from I - 0.3
P(not B| II) = 0.8 , so the other half would be B from I - P(B∩ I) = 1-0.8 - 0.2

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manikandanp replied to Official Quant Thread for CAT 2012

we want 21st term right. which would be a+20d..
series is a,a+d,a+2d,..
cannot be answered. because we wouldnt know d speed before 1st stopover and after 9th stopover.. we are given oly wid speed btwn stopovers..

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02:13 PM, 27 Oct '12

Official Quant Thread ...

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@pinaknagaon said:
y 41 nd we want a1+21d1:a2+21d2

we want 21st term right. which would be a+20d..
series is a,a+d,a+2d,..

@furious_chap said:
@YouMadFellow Pls answer the following:DS:How long did Mr. X take to cover 5000 km journey with 10 stopovers?I. The ith stopover lasted i2 minutes.II. The average speed between any two stopovers was 66 kmph.

cannot be answered. because we wouldnt know d speed before 1st stopover and after 9th stopover.. we are given oly wid speed btwn stopovers..

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manikandanp replied to Official Quant Thread for CAT 2012

11x+7=8x+1=13x+11
only number satisfying in d options is 777

02:09 PM, 27 Oct '12

Official Quant Thread ...

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@kannabal said:
N = 7777………………7777, where the digit 7 repeats itself 429 times. What is the remainder left when N is divided by 1144?1. 913 2. 1129 3. 777 4. None of these Pls help

11x+7=8x+1=13x+11
only number satisfying in d options is 777

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manikandanp replied to Official Quant Thread for CAT 2012

have to substitute n as 41..
(a1+ [n-1]/2*d1)/(a2+ [n-1]/2*d2) = (7n-17)/(4n+16)we want a1+20d1/a2+20d2
to get dat n shld be 41.. subs that getting 3/2..

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02:03 PM, 27 Oct '12

Official Quant Thread ...

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@pinaknagaon said:
bro... its d sum ... n(2a+(n-1)d/2

have to substitute n as 41..

(a1+ [n-1]/2*d1)/(a2+ [n-1]/2*d2) = (7n-17)/(4n+16)we want a1+20d1/a2+20d2
to get dat n shld be 41.. subs that getting 3/2..

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manikandanp replied to Official Quant Thread for CAT 2012

(a1+ [n-1]*d1)/(a2+ [n-1]*d2) = (7n-17)/(4n+16)
we want
a1+20d1/a2+20d2 which can be got be substituting n - 21
so, ans is 13/10

01:53 PM, 27 Oct '12

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@pinaknagaon said:
The sum of n terms of two series in A.P. are in the ratio of (7n-17) : (4n+16). Find the ratio of their 21st terms.

(a1+ [n-1]*d1)/(a2+ [n-1]*d2) = (7n-17)/(4n+16)

we want
a1+20d1/a2+20d2 which can be got be substituting n - 21

so, ans is 13/10

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