Official Quant thread for CAT 2013

MBA CAT 2024
13483
@TONYMBA said:
if we consider that question posted by him answer is coming 3....yeah the question should be other way around..
Then the answer will be the whole expression itself If I'm not wrong.

13*7! > (14*7! + 14*13!)
Hence, Remainder would have been 13*7! itself. :splat:
@Logrhythm said:
thats what....it should be the other way arnd. 7! it should be in that case........
Yeah. Have done that before so knew. :)
13484
@vbhvgupta said:
find remainder 2(8!) - 21(6!) / 14(7!) + 14(13!)
Nr=13*7!
Dr=14*7!+14*13!=14*7!(1+8*9*10*11*12*13)

Cancelling 7!, we get remainer=13

So ans=13*7!
13485
@vbhvgupta said:
From Arun sharma...
You should have written it correctly.

As far as I remember, the Question reads -
What is the remainder when 2(8!) - 21(6!) divides 14(7!) + 14(13!) .

Anyway, Bring More. :)
13486
@bs0409 said:
Nr=13*7!Dr=14*7!+14*13!=14*7!(1+8*9*10*11*12*13)Cancelling 7!, we get remainer=13So ans=13*7!
Ans is 7!
13487
@vbhvgupta said:
Ans is 7!
Not possible.......What is your explanation...??
13488
@Estallar12 said:
You should have written it correctly.As far as I remember, the Question reads -What is the remainder when 2(8!) - 21(6!) divides 14(7!) + 14(13!) .Anyway, Bring More.
Are yar Q bhi yad hai...

Have edited the Q
13489
@bs0409 said:
Not possible.......What is your explanation...??
sawal ulta likh diya tha bhai ne.....other wise answer was coming 3 by calculation
13490
@Estallar12 said:
You should have written it correctly.As far as I remember, the Question reads -What is the remainder when 2(8!) - 21(6!) divides 14(7!) + 14(13!) .Anyway, Bring More.
Kitna aaya CAT mein???
Mera to vaat laag gaya..!!!!!!!!

If a,b,c,d are positive real numbers then find a possible value of a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)
(a)11/5 (b)17/3 (c)2 (d)11/9

13491
@bs0409 said:
Not possible.......What is your explanation...??
Yar I posted the Q with some mistake...check it now...
13492
@bs0409 said:

If a,b,c,d are positive real numbers then find a possible value of a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)(a)11/5 (b)17/3 (c)2 (d)11/9
Max. Value the expression can have will be when a = b = c = d.

=> Expression = 4/3
Only Option 4 is less than 4/3.

Hence, 11/9 :roll:

@bs0409 said:
Kitna aaya CAT mein???Mera to vaat laag gaya..!!!!!!!!
Bhai, Pucho mat. Bahut bura Kataa hai mera. :embarrassed:
Aap CAT nahin de rahe the na ? Bhar diya tha kya end main ? :O
Kitna bana toh bhi ? :splat:
13493
@bs0409 said:
Kitna aaya CAT mein???Mera to vaat laag gaya..!!!!!!!!If a,b,c,d are positive real numbers then find a possible value of a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)(a)11/5 (b)17/3 (c)2 (d)11/9
a=1
b=2
c=3
d=4
1/7 + 2/6 + 3/9 + 4/8 = 1/7 + 1/3 + 1/3 + 1/2 = 1/7 + 2/3 + 1/2 = 6+24+21/42 = 51/42 ~ 11/9

so option d by any chance??
13494
@bs0409 said:
Not possible.......What is your explanation...??
Arrey woh Num and Denom ulte hain actual question main. :P
@TONYMBA said:
sawal ulta likh diya tha bhai ne.....other wise answer was coming 3 by calculation
Otherwise, 7! aayega Bro. :)
13495
@Estallar12 said:
Arrey woh Num and Denom ulte hain actual question main. Otherwise, 7! aayega Bro.
maine 7! common liya tha both in numerator & denominator..hence they cancel out with numerator 3 and denominator in 34xxx....so remainder came 3....sayad mera approach galat hoga...
13496
@bs0409 said:

If a,b,c,d are positive real numbers then find a possible value of a/(a+b+d)+ b/(b+c+a) + c/(b+c+d) + d/(a+c+d)(a)11/5 (b)17/3 (c)2 (d)11/9

max value of expression(at a=b=c=d) is 4/3
so by options 11/9

13497
@bs0409

Shuld be D (11/9)

=> a/(a+b+d) + b/(b+c+a) +c/(b+c+d) +d/(a+c+d) > a/(a+b+c+d) + b/(a+b+c+d) + c/(a+b+c+d) +d/(a+b+c+d) = 1

=>Again, a/(a+b+d) + b/(b+c+a) +c/(b+c+d) +d/(a+c+d)

=>Thus, the range of the Eqn is (1,2)

Only, Option D falls..
13498
@TONYMBA said:
maine 7! common liya tha both in numerator & denominator..hence they cancel out with numerator 3 and denominator in 34xxx....so remainder came 3....sayad mera approach galat hoga...
bhai jabh den > num then remainder has to be the num

like what is the remainder when 3 is divide by 5....it is 3 itself
13499

@Estallar12

Yes,same method I used.

@Logrhythm

I think the method of putting a=b=c=d is better as it gives the maximum value and you can easily eliminate other options from it.

You luckily got the answer,but would have consumed time had some random value appeared after calculation.
13500
@TONYMBA said:
maine 7! common liya tha both in numerator & denominator..hence they cancel out with numerator 3 and denominator in 34xxx....so remainder came 3....sayad mera approach galat hoga...
If you took it common and cancelled it, then multiply the final Remainder again with it.
We cannot cancel like this.

For Eg :
3^2*2^30 / 27
Cancel 3^2 with 27. => 2^30/ 3 = 1
So, Overall Remainder = 1*3^2 = 9.
13501
@Logrhythm said:
bhai jabh den > num then remainder has to be the numlike what is the remainder when 3 is divide by 5....it is 3 itself
yes i agree...but while cancelling out the common in numerator and denominator i was getting 3/34xxx....anyways question galat post hua tha na...have been edited now...so lets move to next question...
13502