Official Quant thread for CAT 2013

MBA CAT 2022
681
@realslimshady

13!(2+14)=16*13!=16(13!-1)+16
so remainder is 16...
682
@surajsrivastav i did manage to solve it, and i was about to post the answer!
thanks a lot though. :D
683
@krum you're getting it wrong, N is a digit of the number.
so if N=7 then N5 will mean 75 and not 7*5.. maybe i posted it in a confusing manner.. if this helps and you're able to solve this then do let me knw :D
i'm sorry but i dont have any similar questions
684
@realslimshady said:
@krum you're getting it wrong, N is a digit of the number. so if N=7 then N5 will mean 75 and not 7*5.. maybe i posted it in a confusing manner.. if this helps and you're able to solve this then do let me knw i'm sorry but i dont have any similar questions
question kya hai. :P
685
@realslimshady said:
@krum you're getting it wrong, N is a digit of the number. so if N=7 then N5 will mean 75 and not 7*5.. maybe i posted it in a confusing manner.. if this helps and you're able to solve this then do let me knw i'm sorry but i dont have any similar questions
2.if (N5)^2 + (N^3) + (1N) is divisible by 7 and N is non zero single digit, then what can be it's value
a.2
b.3
c.5
d.7
e.9
@milestogo3 , ^ question

(10N+5)^2+N^3+10+N
=>100N^2+25+100N+N^3+10+N
=>N^3+100N^2+101N+35

given that
(N^3+100N^2+101N+35)mod7=0
=>(N^3+2N^2+3N)mod7=0
=>N(N^2+2N+3)mod7=0

=>N=7k or N^2+2N+3=7K

hence d.7
686
@krum
A = {179, 180, 181, €Ĺő..,360}. B is a subset of A such that sum of no two elements of B is divisible by 9. The number of elements in B cannot exceed
easy one for you, but like the way yo're always busy solving here. 😁 this might keep you busy for another 2-3 mins. :/

a)81
b)82
c)85
d)102
e)101
687
@milestogo3 said:
@krumA = {179, 180, 181, β‚¬Εš..,360}. B is a subset of A such that sum of no two elements of B is divisible by 9. The number of elements in B cannot exceedeasy one for you, but like the way yo're always busy solving here. this might keep you busy for another 2-3 mins. :/a)81b)82c)85d)102e)101
9k-21; 9k+1-20; 9k+2-20; 9k+3-20; 9k+4-20; 9k+5-20; 9k+6-20; 9k+7-20; 9k+8-21;

we can choose only one from 9k
pairs not possible - (9k+1)&(9k+8); (9k+2)&(9k+7)....(9k+4)&(9k+5)

for max. elements take 21 from 9k+8 and 20*3 from 9k+5,9k+6,9k+7

so total = 1+21+60=82

ps- sorry, naruto dekh rha tha,, beech beech mien idhar aa jata hun :p
688
@krum said:
9k-21; 9k+1-20; 9k+2-20; 9k+3-20; 9k+4-20; 9k+5-20; 9k+6-20; 9k+7-20; 9k+8-21;we can choose only one from 9kpairs not possible - (9k+1)&(9k+8); (9k+2)&(9k+7)....(9k+4)&(9k+5)for max. elements take 21 from 9k+8 and 20*3 from 9k+5,9k+6,9k+7so total = 1+21+60=82ps- sorry, naruto dekh rha tha,, beech beech mien idhar aa jata hun
then,sorry to disturb, i thought you might be hungry for koshens. 😁 Enjoy.

689
@krum Awesome bro. thanks :D
690
@milestogo3 said:
@krumA = {179, 180, 181, β‚¬Εš..,360}. B is a subset of A such that sum of no two elements of B is divisible by 9. The number of elements in B cannot exceedeasy one for you, but like the way yo're always busy solving here. this might keep you busy for another 2-3 mins. :/a)81b)82c)85d)102e)101
179 to 360 182 numbers... we shall not have 9k number (but one),and can have either (9k+4 or 9k+5),(9k+1 or 9k+8),(9k+2 or 9k+7) (9k+3 or 9k +6).

from 179 to 360 there are :
21 9k numbers,
21 9k+8 numbes,
20 9k +1,.. upto 9k+7 numbers. so. (total 182),

take 1 9k number-->1
take 21 9k+8 numbers --> 21
rest 20,20,20 respectively. 82 numbers possible maximum.

691
Set A = { a,b,c,d,e}
Set B = { p,,q,r,s,t}
Find the No of bijective functions such that f(p) /= a and f(t) /= s ?
/= represents not equal to
692
@rkshtsurana said:
Set A = { a,b,c,d,e}Set B = { p,,q,r,s,t}Find the No of bijective functions such that f(p) /= a and f(t) /= s ?/= represents not equal to
78 ???

5!- 2* 4! + 3! ??
693
@Brooklyn said:
78 ???5!- 2* 4! + 3! ??
Bhai thoda aur explain karo. Ye Bijective-Injective sahi se nahi pada :splat:

@rkshtsurana : Achhe khase stud lgte ho, kahan Virat Kohli ki pic lagai thi DP mai πŸ˜› :D
694
@vijay_chandola : bijective means one one onto:ie each value from a maps to unique value in b and no of elements in set A must b =

so its like placing 5 people in 5 chair, with constraint a cant be placed in chair p so on
695
Find the sum of the following series:ďΕ₯ΕΌ

2/1!+3/2!+6/3!+11/4!+18/5!...........

a) 3(e-1) b) 3*e-1 c) 3(e+1) d) 3*e+1
696
@vijay_chandola a?
697
Harry and Jon plan to spend the afternoon at the fair. After paying the entrance price of Rs.5.00 each they entered the fair ground. Jon looked around and saw that the Dragon ride was Rs.3.50 and the Loppo was Rs.2.75. In addition, there were 3 activities he wanted to do which cost Rs.1.50 each. Jon guessed that snacks and drinks would cost Rs.3.50. Jon could see that he did not have enough money. Jon then borrowed Rs.6.00 from Harry. They noted that after Harry gave Jon Rs.6, Harry still had Rs.12.00 more than Jon. How much more money did Harry have than Jon had before Harry gave Jon Rs.6.00?
Rs.12
Rs.18
Rs.24
Rs.30
698
@vijay_chandola said:
Find the sum of the following series:ďΕ₯ΕΌ2/1!+3/2!+6/3!+11/4!+18/5!...........a) 3(e-1) b) 3*e-1 c) 3(e+1) d) 3*e+1
approx 2 + 1.5 + 1 + 0.5..... = ~5

option a

detailed approach me expansion use kre
e^x = 1 + x + x^2/2! + x^3 /3! +....
@vijay_chandola - teri fb pe pics dekhi. tu stud he re
699
@Shray14 24
700
@Shray14 said:
Harry and Jon plan to spend the afternoon at the fair. After paying the entrance price of Rs.5.00 each they entered the fair ground. Jon looked around and saw that the Dragon ride was Rs.3.50 and the Loppo was Rs.2.75. In addition, there were 3 activities he wanted to do which cost Rs.1.50 each. Jon guessed that snacks and drinks would cost Rs.3.50. Jon could see that he did not have enough money. Jon then borrowed Rs.6.00 from Harry. They noted that after Harry gave Jon Rs.6, Harry still had Rs.12.00 more than Jon. How much more money did Harry have than Jon had before Harry gave Jon Rs.6.00? Rs.12Rs.18Rs.24Rs.30
Rs.24

5 + 3*1.5 + 3.5 - 6 = 7 (amount initially with Jon)
3*1.5 + 3.5 = 8 (amount currently with Jon)
8 + 12 = 20 (amount currently with Harry)
5 + 20 + 6 = 31 (amount initially with Harry)

Difference = 31 - 7 = Rs. 24