Official Quant thread for CAT 2013

MBA CAT 2022
2401
@airirm said:
@Ibanez Bro, the answer is not 4.. You were almost right!
Why did you say the ans is not 4? It is 4.
2402
@MI8888 Terrific! Great bro..
2403
@Ibanez said:
Approach:Sk = first term/(1- common ratio) = (k-1)/k! div by 1- 1/k = 1/(k-1)!Now expand the series,1/1! + 1/1! + 1/2!+ 5/3! + 11/4!+.......this is approximately 2+0.5+0.8 + xx > 3the first part 100^2/100! = 100/99! = neglegible. our ans is > 3 so it will converge to 4.
Arre...for the first term when k =1, S = 0
because the first term is k-1/k!
So tumhara answer bhi 3 hi hona chhaiye...
2404
@airirm said:
@MI8888 Terrific! Great bro..
Mujhe kyu bola fir 4 is wrong??
2405
@Ibanez said:
Even I got 4. by direct approximation. I'm sure 4 is correct.
I changed my answer
woh S1 = 0 hoga...
2406
@Ibanez Bro refer to @MI8888 's post. The answer is 3.
2407
@crazy_mn2002 said:
Find the missing term in the series :(1). 64 128 92 220 148 404 ?(2). 5 ? 4 7.5 17 45
1)soln: 64,92,148,260 and other series goes like 128,220,404
actually 148=92+2(92-64) and
260=148+ 2(148-92)
so no. is 260
2)nt getting
2408
@MI8888 said:
Arre...for the first term when k =1, S = 0because the first term is k-1/k!So tumhara answer bhi 3 hi hona chhaiye...
Are S is not zero for k=0. 0!= 1 it is 1 only. 1/0! = 1 . mera answer 4 hi hai
2409
@Ibanez It's a 2010 IITJEE Question :)
2410
@Ibanez said:
But how's it not 4? I got 1+1+0.5+0.8+... this is >3 and only 4 is >3 in the options.
sk=1/(k-1)!

E((k^2-3k+1)/(k-1)!)
=>E((k-1)^2-k)/(k-1)!)
=>E((k-1)/(k-2)!-k/(k-1)!)

=>2/1!-1/0!+2/1!-3/2!+3/2!-4/3!+4/3!-5/4!+...-100/99!
=>3-100/99!

100^2/100!+3-100/99!
=>100/99!+3-100/99!
=>3

@airirm
2411
@crazy_mn2002 said:
Find the missing term in the series :(1). 64 128 92 220 148 404 ?(2). 5 ? 4 7.5 17 45
1)soln: 64,92,148,260 and other series goes like 128,220,404
actually 148=92+2(92-64) and
260=148+ 2(148-92)
so no. is 260
2)nt getting
2412
@Ibanez said:
Are S is not zero for k=0. 0!= 1 it is 1 only. 1/0! = 1 . mera answer 4 hi hai
Han but numerator mein (k-1) hai...
2413
@krum Perfect man. It's 3. Beautiful question..
2414
@MI8888 said:
Han but numerator mein (k-1) hai...
a= (k-1)/k!
r = 1/k

Sk = a/(1-r)= (k-1)/k! / 1-1/k = 1/(k-1)!

Isme for k=1, we get 1/0! = 1

BUT if the first term is (k-1)/k! and we put k=1, we get 0

Yeh kaise contradictoin I'm doing some really facepalm mistake
2415
@Ibanez said:
a= (k-1)/k!r = 1/kSk = a/(1-r)= (k-1)/k! / 1-1/k = 1/(k-1)!Isme for k=1, we get 1/0! = 1BUT if the first term is (k-1)/k! and we put k=1, we get 0Yeh kaise contradictoin I'm doing some really facepalm mistake
(k-1)/k! / (1-1/k) = k(k-1)/(k-1)k!
Now we can cut k-1 from numerator and denominator only when k is not equal to 1
for k=1 , k-1=0
2416
@Ibanez Bro, when we cancel out (k-1), it implies K is not equal to 1 .. So we need to be doubly careful..
2417
@MI8888 said:
(k-1)/k! / (1-1/k) = k(k-1)/(k-1)k!Now we can cut k-1 from numerator and denominator only when k is not equal to 1for k=1 , k-1=0
Oh yes how did I cancel it for k-1=0. of course I can't... Thanks.
itni bevakufi
2418
@krum said:
sk=1/(k-1)!E((k^2-3k+1)/(k-1)!)=>E((k-1)^2-k)/(k-1)!)=>E((k-1)/(k-2)!-k/(k-1)!)=>2/1!-1/0!+2/1!-3/2!+3/2!-4/3!+4/3!-5/4!+...-100/99!=>3-100/99!100^2/100!+3-100/99!=>100/99!+3-100/99!=>3@airirm
Beautiful soln bhai...
2419
@airirm said:
@Ibanez Bro, when we cancel out (k-1), it implies K is not equal to 1 .. So we need to be doubly careful..
so for k=1, it's 0. then u get 3 as OA by direct approximation. Won't have to expand and cancel all terms.
2420
@Ibanez said:
Oh yes how did I cancel it for k-1=0. of course I can't... Thanks.itni bevakufi
@pankaj1988 said:
Beautiful soln bhai...
bhai main to latka hua tha, ibanez bhai ne Sk ko simplify kia 1/(k-1)! tab hua