Official Quant thread for CAT 2013

adwaitjw · January 31, 2013, 11:35am

@joyjitpal said:
what is the formula plz mentioni am confused

e.g. to find no of 3 digits in base 10 u find [100,1000) i.e. [10^2, 10^3)

Similarly here
3 digit no.s in base 7 are [49, 343)
And 4 digits or above in base 6 is [216, infinity)

So required range is [49, 215] =167

joyjitpal · January 31, 2013, 11:38am

@adwaitjw said:
e.g. to find no of 3 digits in base 10 u find [100,1000) i.e. [10^2, 10^3)Similarly here3 digit no.s in base 7 are [49, 343)And 4 digits ar above in base 6 is [216, infinity)So required range is [49, 215] =167

thanx

bs0409 · January 31, 2013, 11:39am

@adwaitjw said:

167 ?

@vijay_chandola said:

three digit numbers in base 7 = [7^3, 7^4) = [343, 2401)

4 digit numbers in base 6 = [6^4, 6^5) = [1296, 7984)

==> required numbers = 343, 344......1295 = 953.

OA-167

How many natural numbers less than 100 when squared and then divide by 24 leave a remainder of 1?

vijay_chandola · January 31, 2013, 11:41am

@adwaitjw said:

Wont those be 4 digit n 5 digit no.s resp which u are mentioning?

you are right sir :thumbsup:

That would be for 4 digit and 5 digit respectively. :)

@joyjitpal said:

what is the formula
plz mention

i am confused

compare this situation with 10 base.

range of three digits numbers = [10^2, 10^3)

similar is the case with base 6 and 7 :)

adwaitjw · January 31, 2013, 11:42am

@vijay_chandola said:
you are right sir That would be for 4 digit and 5 digit respectively. compare this situation with 10 base.range of three digits numbers = [10^2, 10^3)similar is the case with base 6 and 7

P.S. no sir

Sorry for spam

vijay_chandola · January 31, 2013, 11:43am

@bs0409 said:

OA-167

How many natural numbers less than 100 when squared and then divide by 24 leave a remainder of 1?

1 and all prime numbers less than 100. It's a fact. :)

EDIT: all prime numbers except 2 and 3 :)

joyjitpal · January 31, 2013, 11:45am

A school having 135 students provides facilities for playing four games €“ Cricket, Football, Tennis and Badminton. There are a few students in the school who do not play any of the four games. It is known that for every student in the school who plays at least N games, there are two students who play at least (N €“ 1) games, for N = 2, 3 and 4. If the number of students who play all the four games is equal to the number of students who play none, then how many students in the school play exactly two of the four games?
(a) 20 (b) 30 (c) 60 (d) 90

mailtoankit · January 31, 2013, 11:46am

@bs0409 said:
OA-167How many natural numbers less than 100 when squared and then divide by 24 leave a remainder of 1?

all prime no. except 2 and 3...so 23??..

leonidas_26147187 · January 31, 2013, 12:06pm

@bs0409 this is the right answer. please explain.

bs0409 · January 31, 2013, 12:12pm

@mailtoankit said:

all prime no. except 2 and 3...so 23??..

ANS is 33........

mailtoankit · January 31, 2013, 12:15pm

@bs0409 said:
ANS is 33........

kaise sir?..explain kardo..

codecatcher · January 31, 2013, 12:18pm

@joyjitpal 30 ???

bs0409 · January 31, 2013, 12:22pm

@joyjitpal said:

A school having 135 students provides facilities for playing four games €“ Cricket, Football, Tennis and Badminton. There are a few students in the school who do not play any of the four games. It is known that for every student in the school who plays at least N games, there are two students who play at least (N €“ 1) games, for N = 2, 3 and 4. If the number of students who play all the four games is equal to the number of students who play none, then how many students in the school play exactly two of the four games?(a) 20 (b) 30 (c) 60 (d) 90

Let students who play all 4 games=students who play none=x

Then students who play at least 3 games=2x

Then students who play 3 games=x

Similarly,

Then students who play 2 games=2x

Then students who play 1 game=4x

So, 9x=135

x=15

exactly 2 games=30

joyjitpal · January 31, 2013, 12:24pm

@codecatcher said:
@joyjitpal 30 ???

@bs0409 said:
Let students who play all 4 games=students who play none=xThen students who play at least 3 games=2xThen students who play 3 games=xSimilarly,Then students who play 2 games=2xThen students who play 1 game=4xSo, 9x=135x=15exactly 2 games=30

bs0409 · January 31, 2013, 12:27pm

@mailtoankit said:

kaise sir?..explain kardo..

Numbers of the form 24k+1,24k+5,24k+7,24k+11,24k+13,24k+17,24k+19,24k+23

when squared give 1 as remainder on dividing by 24

tootahuadil · January 31, 2013, 1:02pm

n! has x number of zeroes at the end and (n+1)! has x+3 zeroes at the end. Find the number of possible values of n if n is a three digit number.

logrhythm · January 31, 2013, 1:17pm

@TootaHuaDil said:
n! has x number of zeroes at the end and (n+1)! has x+3 zeroes at the end. Find the number of possible values of n if n is a three digit number.

0's increase with powers of 5...so 5^3 ke multiples pe aesa hoga..

124 has 28 zeros while 125 has 31

125*2 = 250

125*3 = 375

125*4 = 500

125*6 = 750

125*7 = 875

so, total 6 such 3 digit numbers...

PS - 122*5 = 5^4 hence 4 more zeroes wld be there hence did not consider it...

swapnil4ever2u · January 31, 2013, 1:20pm

@ravi.theja its 214 .. check it ..

swapnil4ever2u · January 31, 2013, 1:21pm

@The_Loser plz check.. its OA is 214..

tootahuadil · January 31, 2013, 1:29pm

@Logrhythm said:
0's increase with powers of 5...so 5^3 ke multiples pe aesa hoga..124 has 28 zeros while 125 has 31125*2 = 250125*3 = 375125*4 = 500125*6 = 750125*7 = 875so, total 6 such 3 digit numbers...PS - 122*5 = 5^4 hence 4 more zeroes wld be there hence did not consider it...

125*1 ...125*8 and 125*9 kyu nhi?? samjha do