Official Quant thread for CAT 2013

MBA CAT 2024
13423
@Estallar12 said:
Woh method chodho.Short waala explanation hi dekh lo.
bata do be...
samajh mein to aaye...
waise bhi bahut kuch chor ke gaya tha iss baar...aabki kuch nahin chorna hai...:banghead:
13424
Can any one can please make me understand why?
inequality of |3x+6|> -12 lies in between -infinitve>x>infinitive.
13425

7^99 / 2400 remainder???

13426
@rhinorocks said:
bata do be...samajh mein to aaye...waise bhi bahut kuch chor ke gaya tha iss baar...aabki kuch nahin chorna hai...
Okay. Wait. :)
13427
@Swapno said:
Can any one can please make me understand why?inequality of |3x+6|> -12 lies in between -infinitve
coz the value within mod will always be positive or zero...so its always greater than -12...so any value of x can be taken with the range of 0 to infinity
13428
@Swapno said:
Can any one can please make me understand why?inequality of |3x+6|> -12 lies in between -infinitve>x>infinitive.
Wrong signs used. :P

Anyway,
|3x + 6| > - 12.
As mod will always be greater than zero, hence, - Infinity
13429
@Estallar12 said:
Twas specifically mentioned in the question in Brackets (Include 1 and 3000).
did not notice.....thanks

@allan89 - srry, 1336 hoga yar. 😃
13430
@vbhvgupta said:
7^99 / 2400 remainder???
343..
7^4=2401..
so 7^96*7^3/2400 will give R(7^3/2400)=343..
waise CRT se bhi bana sakte hain...
13431
@Swapno said:
Can any one can please make me understand why?inequality of |3x+6|> -12 lies in between -infinitve
mod function | | always give a positive value, so for any value of x it will hold true. Basic definition of | |
13432
@vbhvgupta @TONYMBA

i cldnt figure out a better way

-35*29*37*(-37)*(-29)*13*15 = -(35*29^2*37^2*13*15) = -(xxx25) = 75

@Estallar12 the last 2 digits would be 75, not 25. pl check
13433
@vbhvgupta said:
7^99 / 2400 remainder???
7^4 = 2401.
=> 7^4 mod 2400 = 1
=> 7^96 mod 2400 = 1
=> 7^96* 7^3 mod 2400 = 343.
13434
@rhinorocks said:
the bold part i got sir..how u reduced it to 45*21*71*35..plz explain..
Oh LOL.

75 hoga Answer. Now don't ask my approach. Edit hi karta hun. :embarrassed:
13435
@vbhvgupta said:
7^99 / 2400 remainder???
343 :)
13436
@Estallar12 said:
Keep reducing N.45*21*71*3545*85= 25.To keep it short, the given N has two numbers ending with 5. So, 25 should be the last two digits.
Editing the Bold Part.

65*29*37*63*71*87*85
Multiplying each of the last two terms in short and keeping the last two digits -
25*23*27*63
75*21 = 75.
13437
@vbhvgupta said:
7^99 / 2400 remainder???
Is it 343??
2400 = 96*25
7^99%96
e(96) is 32
99%32 = 3
=> 7^3%96 = 55

7^99%25
e(25) is 20
99%20 = 19
7^19%25 = 18

=> remainder would be 96x+55 = 25y+18 or 343
13438
@Logrhythm said:
@vbhvgupta@TONYMBAi cldnt figure out a better way-35*29*37*(-37)*(-29)*13*15 = -(35*29^2*37^2*13*15) = -(xxx25) = 75@Estallar12 the last 2 digits would be 75, not 25. pl check
-(35*29^2*37^2*13*15) iss walle xpression se
-(xxx25) yahan tak kaise laye..??calculate kiya hai,,??
13439
@TONYMBA yes.I understand that it would take values form 0 to infinitive.I know basic definations of mod function.But In time materials that answer is given that the inequality lies between :- -infinitive
Is the approach is like this:-
|3x+6|.>0 or 3x+6 = 0
or,3x+6>0 or,3x=-6
or,x>-2 or,x=-2
so the value of x is going to be x>=-2.
So the value of x >=-2.
how the inequality is between like the given answer.
@Estallar12 :- Bro actually I am facing a bit problem with inequalities..
13440
@vbhvgupta said:
7^99 / 2400 remainder???
7^99//2400=7^3*7^96/2400=(343)(2401)^24/2400=343*1/2400=343?
13441

-infinitive > x > infinitive....@TONYMBA

13442
@rhinorocks said:
-(35*29^2*37^2*13*15) iss walle xpression se -(xxx25) yahan tak kaise laye..??calculate kiya hai,,??
hanji bhai, lattice method ya fir criss cross method for multiplication seekh lo youtube pe link mill jayega aapko.....usko use karke aese expressions ke digits find karna easy ho jata hai