Official Quant thread for CAT 2013

@veertamizhan said:
@sails how did I do? what is the wrong with my approach? how did you solve this?
The error is just that you have said "n = 0 is possible" but the question clearly states Natural Numbers, so 1 se chalu hona hai!

regards
scrabbler

@veertamizhan said:
@sails damn. I take into consideration the fact that these are natural number which begin with 1. I am a god damn idiot. thanks anna (i am not tamil)
Dunt get disheartened...no1 here is a pro, just keep practising..:)
@veertamizhan said:
"If a student weighing 70 kgs joins a group of n students, the average of the group increases by 1 kgs. If the new student weighed 55 kgs, the average of the group would have declined by 2 kgs. Find n"3, 4, 5, 6, 7
@techsurge said:
(n+1)(x+1) =nx+70n+x+1=70(n+1)(x-2)=nx+55x-2n-2=553n =12n=4EDITED : Calculation mistake. thanks @Logrhythmany shortcut via allegation anyone ???
No alligation needed....no massive equations either.

Just consider the difference in the 2 cases. For (n+1) students, a decrease in 15 kg total caused a decrease of 3 in average. So (n+1) = 5 and n = 4. Makes sense?

regards
scrabbler

@scrabbler so I was right, except for the fact that this was not needed. Thank guys. You guys are way to patient and kind.
@veertamizhan said:
@sails damn. I take into consideration the fact that these are natural number which begin with 1. I am a god damn idiot. thanks anna (i am not tamil)

Lol, just don't repeat the mistake! Better go wrong here than in the exam right?

regards
scrabbler

@bs0409 said:
Find a number consisting of 9 digits in which each of the digits from 1 to 9 appears only once. This number should satisfy the following requirements:a. The number should be divisible by 9.b. If the most right digit is removed, the remaining number should be divisible by 8.c. If then again the most right digit is removed, the remaining number should be divisible by 7.d. etc. until the last remaining number of one digit which should be divisible by 1.
147653289

by hit and trail
@negiSannu said:
Find the total number of sets of factors of 264 which are co-prime to each other.post approach?
264*(1/2)*(2/3)*(10/11)

80
@gnehagarg said:
147653289by hit and trail
the 6th digit is not divisble by 6..after 5 the next number should be even..
@bs0409 said:
Solve this:264=a*b ,gcd(a,b)=1How many solutions?
(2,33)
(4,33)
(8,33)
(1,66)
(1,132)
(1,252)
(3,88)
(11,24)

8 solutions
@pavimai said:
N! is completely divisible by 13^52.what is sum of the digits of the smallest such number N
13*52=676

check with 670,660,650,640
637-640

637 is smallest
@bs0409 said:
Ki baba....!!! Kemon cholche tomar??So, u applied to any univ or waiting for CAT'13?I am assuming your company joining letter will come soon enough.Find the total number of values of n for which n^2 + 24n + 21 is a perfect square.Where n is natural number.
(n+12)^2-4=k^2-13^2
(n+14)*(n+10)=(k+13)*(k-13)

n=odd k=even
n=even k=odd

no value satisfies it


A 10 digit number N has among its digit one 1, two 2's , three 3's and four 4's. Can N be a perfect square ?

MY PROFILE IS SO HUMANITIES BASED..TOO AWAY FROM QUANT...
higher secondary...humanities...graduation in philosophy...and i have studied in a malayalam state board school...this year cat...OA 62, Isec 33 and second section 90. next year i want to bell the cat....and it is possible only if i improve my QA mark..please suggest me how to improve quantitative ability..i don't know the basics of function, coordinate geometry etc. i am not in a situation to attend coaching class...please direct me to the way.. i will walk on it...
If this thread is not for this pls suggest me the apt thread....

@ravi.theja said:
A 10 digit number N has among its digit one 1, two 2's , three 3's and four 4's. Can N be a perfect square ?
can be yaar not sure...sum of digit is 30..so may be a five digit multiple of 3... but digit sum is 3..so cannot be

total 4 digit no's tat can be formed that have the digit 5 used exactly once in them?



total 4 digit no's tat can be formed using digit 5 exactly once?

total 4 digit no's tat entails 5 evrytym exactly once?

total 4 digit no's tat entails 5 evrytym exactly once?
@ravi.theja said:
A 10 digit number N has among its digit one 1, two 2's , three 3's and four 4's. Can N be a perfect square ?
It can never be becoz digital sum is 3 whereas for perfect square digital sum shld be 1,4,7,9
@The_Loser said:
total 4 digit no's tat entails 5 evrytym exactly once?
For 4 digit no's starting with 5 possibilities are 1*9^3
For other no's it is 3*(8*9^2)
Add them up
Two pipes A and B would fill a cistern in 37.5 minutes and 36 minutes respectively. Both pipes being opened, find when the 2nd pipe must be turned off, so that the cistern may be filled in half an hour?