Official Quant thread for CAT 2013

@audiq7 said:
similar Q, how many 7 digit no.s can be formed having digit 3 three times and digit 0 4 times?
frst digit should be 3 always...jaise hi fix kia this question will be like our previuos question..so
3xxyyyy
=>
!6/!2!4=15 (where x=3 and y=0)
@amresh_maverick said:
If P =K^3 / N^2 by what % will it change if K and N is increased by 20% each ?
20%
@amresh_maverick said:
If P =K^3 / N^2 by what % will it change if K and N is increased by 20% each ?
OA inc by 20%
@audiq7 said:
in an exam, max marks for 3 papers are 50 each. max marks for 4th paper is 100. find the no.of ways with which a student can score 60% marks in aggregate?
if he scores maximum marks or zero marks than 5 ways.
other wise cannot be determined. since he may score any marks in any subject.
correct me?
@maddy2807 said:
if he scores maximum marks or zero marks than 5 ways.other wise cannot be determined. since he may score any marks in any subject.correct me?
bhai kuch aise nhi hoga isme ? :
a+b+c+d = 120
Total solutions 123C3

Now we have to subtract the cases where each of a,b or c can have the values from 51 to 120, isme mein atak raha hu :banghead:
@amresh_maverick aap bhi dekho jara

If all the 5 letter words are formed from the letters of MATHEMATICS , in how many of them will the letter E occur ?


6945
630
960
6990
7012

@maddy2807
@rachit_28 yar OA is 110551
@rachit_28 said:
bhai kuch aise nhi hoga isme ? :a+b+c+d = 120Total solutions 123C3Now we have to subtract the cases where each of a,b or c can have the values from 51 to 120, isme mein atak raha hu @amresh_maverick aap bhi dekho jara
bhai 60% is 150 marks
@rachit_28 said:
bhai kuch aise nhi hoga isme ? :a+b+c+d = 120Total solutions 123C3Now we have to subtract the cases where each of a,b or c can have the values from 51 to 120, isme mein atak raha hu @amresh_maverick aap bhi dekho jara
it shud be a+b+c+d=150
and remove the case when a,b,c wil be 51 to 150 and cases when d wil be 101 to 150
which is very tedious.
@audiq7 said:
in an exam, max marks for 3 papers are 50 each. max marks for 4th paper is 100. find the no.of ways with which a student can score 60% marks in aggregate?
p1,p2,p3

p4

p1 + p2 + p3 + p4 = (0.6)*(250) = 150

Normal ways = 153C3

But we need to remove cases when p1,p2,p3>50 or/and p4 >100

Case 1: p4 > 100
p4 = 101 + a (where a>=0)

a + p1 + p2 + p3 = 49 => Number of ways = 52C3

Case 2: p1 or p2 or p3 > 50
p1 = 51 + a

a + p2 + p3 + p4 = 99 => Number of ways = 102C3

But, there may be cases when both p1,p2 are more than 50
=> a + b + p3 + p4 = 48 => Number of ways = 51C3

Now, we need to use the concept of set theory

All - (When p4 >100) - (When p1 /p2/p3 > 50) + ( When any two of p1/p2/p3 are more than 50)

=> 153C3 - 52C3 - 3*(102C3) + 3C2*(51C3) I think..

PS: Answer comes out to be 110551..
@amresh_maverick said:
bhai 60% is 150 marks
oops haan ..
@amresh_maverick said:
If all the 5 letter words are formed from the letters of MATHEMATICS , in how many of them will the letter E occur ?694563096069907012
kaise ho ga yaar..p&c; ne to le rakhi hai..
@amresh_maverick is that 6990
@maddy2807 said:
it shud be a+b+c+d=150and remove the case when a,b,c wil be 51 to 150 and cases when d wil be 101 to 150which is very tedious.
isliye to maine ise haath nahi lagaya

frm a set of 3 capital and 5 small consonants and 4 small vowels, how many wrds can be formed each starting with a capital consonant and containing 3 small consonants and two small vowels?

@audiq7 IS that 21600
@amresh_maverick Awesome Strategy
@audiq7 said:
frm a set of 3 capital and 5 small consonants and 4 small vowels, how many wrds can be formed each starting with a capital consonant and containing 3 small consonants and two small vowels?
21600??
@Nihilist1002 kaise?

select the capital = 3C1

select the conso. = 5C3
selcet the vow. = 4C2

now these cons. and vow. can be arranged in 5! ways

so total = 3C1*5C3*4C2*5!