Official Quant thread for CAT 2013

MBA CAT 2022
2381
@airirm said:
I chanced upon this question now. Floored! Options: 1, 2, 3, 4
sorry it will be 4. not 3.
2382
@airirm said:
I chanced upon this question now. Floored! Options: 1, 2, 3, 4
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 > 3

the first part 100^2/100! = 100/99! = neglegible. our ans is > 3 so it will converge to 4.
2383

Find the unit digit in LCM of 7^3001-1 and 7^3001+ 1

a) 1 b) 2 c) 4 d) 8 e) none of these
2384
@pankaj1988 said:
Find the unit digit in LCM of 7^3001-1 and 7^3001+ 1a) 1 b) 2 c) 4 d) 8 e) none of these
@pankaj1988
c.4
2385
@pankaj1988 said:
Find the unit digit in LCM of 7^3001-1 and 7^3001+ 1a) 1 b) 2 c) 4 d) 8 e) none of these
(7^6002-1)/2
last 2 digits of 7^6002 => 49*(49^2)^3000 =>49*01=49
units - (9-1)/2=4

4
2386
@Ibanez Bro, the answer is not 4.. You were almost right!
2387
@gautam22 Bro, tell me your approach.. The answer is not 1 though.
2388
@airirm said:
@gautam22 Bro, tell me your approach.. The answer is not 1 though.
3
2389

Find the missing term in the series :


(1). 64 128 92 220 148 404 ?
(2). 5 ? 4 7.5 17 45

2390
@pankaj1988 said:
Find the unit digit in LCM of 7^3001-1 and 7^3001+ 1a) 1 b) 2 c) 4 d) 8 e) none of these
4
2391
@airirm said:
@Ibanez Bro, the answer is not 4.. You were almost right!
If it's not 4 then it's 3.
2392
@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)260

2393
@airirm said:
@Ibanez Bro, the answer is not 4.. You were almost right!
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.
2394
@pankaj1988 said:
Find the unit digit in LCM of 7^3001-1 and 7^3001+ 1a) 1 b) 2 c) 4 d) 8 e) none of these
7^6002 - 1

=> 9-1 => 8 ?
2395
@krum Bro explain your approach.. Your answer is correct.
2396
@airirm said:
I chanced upon this question now. Floored! Options: 1, 2, 3, 4
Put S = a/1-r (For infinite GP)
S = (k^2 - 3k + 1)/(k-1)!
= [(k-1)^2 - k]/(k-1)!
= (k-1)/(k-2)! - k/(k-1)!

now for k=1,2 , k^2 - 3k + 1 is -ve , So the mod will come into effect

S3 + S4 + --- +S100 = (2/1! - 3/2!) + (3/2! - 4/3!) + --- + (99/98! - 100/99!)
= 2 - 100/99! = 2 - 100^2/100!

S1 = 0
S2 = 1

100^2/100! + 0 + 1 + 2 - 100^2/100! = 3
2397
@krum can you please share the approach
2398
@airirm said:
@krum Bro explain your approach.. Your answer is correct.
The answer to that question has to be 4 only.
2399
@pankaj1988 said:
Find the unit digit in LCM of 7^3001-1 and 7^3001+ 1a) 1 b) 2 c) 4 d) 8 e) none of these


2nd try -

HCF x LCM = (7^3001 - 1)(7^3001 + 1)

LCM = (7^6002 - 1) / 2 => 4


2400
@MI8888 said:
Put S = a/1-r (For infinite GP)S = (k^2 - 3k + 1)/(k-1)! = [(k-1)^2 - k]/(k-1)! = (k-1)/(k-2)! - k/(k-1)!now for k=1,2 , k^2 - 3k + 1 is -ve , So the mod will come into effectS3 + S4 + --- +S100 = (2/1! - 3/2!) + (3/2! - 4/3!) + --- + (99/98! - 100/99!) = 2 - 100/99! = 2 - 100^2/100!S1 = 1S2 = 1100^2/100! + 1 + 1 + 2 - 100^2/100! = 4
Even I got 4. by direct approximation. I'm sure 4 is correct.