@[571257:Estallar12] doesn't seem closed?

@[1:pagalguy] @[8964:deepu] Bhai...This Closed Quant Thread is also working now :|

@[272851:koolkanji]

what is the remainder when 73+75+78+57+107 is divided by 34. I need the solution?

catgirl2012 Says

In a series , the nth term tn=t(n+1)*n^2/(n2-1) , for all n>=2.If t5=6,find t25

is it 624/125?????

A,B,C,D each had some money.D doubled the amounts with the others.C then doubled the amounts with the others.B then doubled the amount with the others.A then doubled the amount with the others.At this stage,each of them had Rs 80.Find the initial amount with C.

A) 45
B) 65
C) 95
D) 85

My take 85

Start with the end and half the amount with everyone

Second last stage
A=200 B, C and D all have 40 each
Third Last stage
A=100 B=180 and C and D = 20 each
Fourth last stage
A=50 B=90 C= 170 and D=10
So in the first stage C=170/2= 85

A,B,C,D each had some money.D doubled the amounts with the others.C then doubled the amounts with the others.B then doubled the amount with the others.A then doubled the amount with the others.At this stage,each of them had Rs 80.Find the initial amount with C.

A) 45
B) 65
C) 95
D) 85

sir ji!
1000 pages to hone dete

catgirl2012 Says

Noooooooooooooooooo...10 more pages , till its 1000...

I understand your emotions puys ;)

The problem is the PG server. It becomes really a burden on a server when any thread nears 1000 pages. It is better to remove some burden off the server. I hope you understand

Puys!!! Please continue here:

http://www.pagalguy.com/forum/cat-and-related-discussion/83384-official-quant-thread-cat-2012-a.html#post3577256


Mods I request you to close the thread...:)

Noooooooooooooooooo...10 more pages , till its 1000...

Puys!!! Please continue here:

http://www.pagalguy.com/forum/cat-and-related-discussion/83384-official-quant-thread-cat-2012-a.html#post3577256


Mods I request you to close the thread...:)

sir ji!
1000 pages to hone dete

Puys!!! Please continue here:

http://www.pagalguy.com/forum/cat-and-related-discussion/83384-official-quant-thread-cat-2012-a.html#post3577256


Mods I request you to close the thread...:)

could any of you help me with these ques :

1.DS Ques :
Find the unit digit of the number,
n= a^17+ b^13+c^21
a,b,c are natural numbers.
I. unit digit of the number a+b=4
II. unit digit of the number c=2

2.find the last two digits of 47^89.
3. find the last two digits of the number n=20025x20026x20027x20035

TIA :D

47^89
UNIT DIGIT =7; 47^88*47=(.......1)*47=47
tens digit= (47^4)^22*47=(.........81)^22*47
take product of 8 and 2i.e 16 =>(.........61)*47=..............67(direct method)
6 tens 7 unit digit

Quote:
Originally Posted by Kungfu Panda
(5^11^22^33)/9 ... puys can you please let me know the remainder .
-----------------------------------
E(9)= 6
(11^22^33)/6=> /6=> (-1)^even=1

=> (5^1)/9=> 5

tn/t(n+1) = n^2/(n-1)(n+1)

t5/t6 = 5^2/4*6

t6/t7 = 6^2/5*7

t7/t8 = 7^2/6*7

t8/t9 = 8^2/7*9
.
.
.
t24/t25 = 24^2/23*25

multiple all RHSs and LHSs

t5/t6 * t6/t7 * t7/t8 * . . . * t23/t24 * t24/t25 = 5^2/4*6 * 6^2/5*7 * 7^2/6*7 * . . . 23^2/22*24 * 24^2/23*25

= t5/t25 = 5/4 * 24/25 = 6/5

as t5 = 6 given

so t25 = 5

tn/t(n+1) = n^2/(n-1)*(n+1)

t5/t6 = 5^2/4*6

t6/t7 = 6^2/5*8

t7/t8 = 7^2/6*9

and so on till t24/t25 = 24^2/23*26

Multiply all the terms to get

t5/t25 = (5^2 * 6^2 * 7^2 * ...... * 24^2 ) /(4*6) * (5*8 ) * ...... * (23*26)

Calculate ho jayega ab ;)

I think 5 aayega :)

25/6?


tn+1 = /n^2

t6 = t5*(4*6)/5^2

t7 = t6*(5*7)/7^2 = t5*(4*6)/5^2 * (5*7)/7^2
...

t25 = t5*(4*6)/5^2 * (5*7)/7^2*...*(23*25)/24^2

All terms except t5*(4/5)(25/24) ll get cancelled*
on substituting t5 value,it becomes 25/6

5 it is

anyone got this?

In a series , the nth term tn=t(n+1)*n^2/(n^2-1) , for all n>=2.If t5=6,find t25

5?

tn+1 = /n^2

t6 = t5*(4*6)/5^2

t7 = t6*(5*7)/7^2 = t5*(4*6)/5^2 * (5*7)/7^2
...

t25 = t5*(4*6)/5^2 * (5*7)/7^2*...*(23*25)/24^2

All terms except t5*(4/5)(25/24) ll get cancelled

on substituting t5 value,it becomes 5

anyone got this?

In a series , the nth term tn=t(n+1)*n^2/(n^2-1) , for all n>=2.If t5=6,find t25

tn/t(n+1) = n^2/(n-1)*(n+1)

t5/t6 = 5^2/4*6

t6/t7 = 6^2/5*8

t7/t8 = 7^2/6*9

and so on till t24/t25 = 24^2/23*26

Multiply all the terms to get

t5/t25 = (5^2 * 6^2 * 7^2 * ...... * 24^2 ) /(4*6) * (5*8 ) * ...... * (23*26)

Calculate ho jayega ab ;)

I think 5 aayega

Q5) sanjay writes a letter to his frnd. it is known that one out 'n' letters that are posted does not reach its destination. If sanjay does not receive the reply to his letter. then what is the prob that kesari didnot recvd sanjay's letter?? It is certain that kesari will definately reply to sanjay's letter if he recves it???

p(1)=sanjay posted but not reached destination=1/n
p(2)=kesari received but his reply not recevied by sanjay =(n-1)/n * 1/n
now,
if sanjay doesn't recive reply then total sample cases =1/n+(n-1)/n^2
required prob = 1/n / = n/(2n-1)

is it correct?

====================================

Yesch ..its absolutely right...Thankx d Qsss.. bhai.

bhai its written

1 out of n is not received so can we say

for n only the Favourable probability is 1-1/n ??

plzz check??

yes .....1 out of n letters do not reach destination so it means (n-1) reach destination ....so favourable prob is (n-1)/n

anyone got this?

In a series , the nth term tn=t(n+1)*n^2/(n2-1) , for all n>=2.If t5=6,find t25

tn/t(n+1) = n^2/(n-1)(n+1)

t5/t6 = 5^2/4*6

t6/t7 = 6^2/5*7

t7/t8 = 7^2/6*7

t8/t9 = 8^2/7*9
.
.
.
t24/t25 = 24^2/23*25

multiple all RHSs and LHSs

t5/t6 * t6/t7 * t7/t8 * . . . * t23/t24 * t24/t25 = 5^2/4*6 * 6^2/5*7 * 7^2/6*7 * . . . 23^2/22*24 * 24^2/23*25

= t5/t25 = 5/4 * 24/25 = 6/5

as t5 = 6 given

so t25 = 5

Kungfu Panda Says

(5^11^22^33)/9 ... puys can you please let me know the remainder .

the other way is by cyclicity

5/9= 5
25/9= 7
7*5/9= 8
8*5/9= 4
4*5/9=2
2*5/9= 1

hence there is a cyclicity of 6 for 5/9

when divided by 11/6 = 5
5*11/6 = 1

hence 2 cyclity which is divisible to 12^13 hence it will leave 0

which in turn leave 11^0=1

and 5^1/9= 5

question ka denominator clear karo:(

P.S: at ashish 100 bhai tune Galti se culdip bhai Ki post pe Groan kar diya .. on 987 page pe

oops! sorry ..done:)

Q5) sanjay writes a letter to his frnd. it is known that one out 'n' letters that are posted does not reach its destination. If sanjay does not receive the reply to his letter. then what is the prob that kesari didnot recvd sanjay's letter?? It is certain that kesari will definately reply to sanjay's letter if he recves it???

p(1)=sanjay posted but not reached destination=1/n
p(2)=kesari received but his reply not recevied by sanjay =(n-1)/n * 1/n
now,
if sanjay doesn't recive reply then total sample cases =1/n+(n-1)/n^2
required prob = 1/n / = n/(2n-1)

is it correct?

bhai its written

1 out of n is not received so can we say

for n only the Favourable probability is 1-1/n ??

plzz check??

question ka denominator clear karo:(

P.S: at ashish 100 bhai tune Galti se culdip bhai Ki post pe Groan kar diya .. on 987 page pe

armaan bhai.....i have already removed my groan... galti se ho gaya tha...
nahi to culdip bhai ke posts par groan...na baba na

catgirl2012 Says

Come again please

madam ji!
kindly see my previous post...i have edited my solution
actually euler of 9 is 6 9*(1-1/3)=6
euler wala question kaafi time baad kiya , to concept bhul gaya tha

anyone got this?

In a series , the nth term tn=t(n+1)*n^2/(n2-1) , for all n>=2.If t5=6,find t25

question ka denominator clear karo:(

P.S: at ashish 100 bhai tune Galti se culdip bhai Ki post pe Groan kar diya .. on 987 page pe

=============================================
?????
bhai

1/6+5/6*5/6*1/6+ 5/6*5/6*5/6*5/6*1/6.... continue

these highlighted are the failure of previous steps of A and B???

Yesch and Brother i made a mistake in Calculation 36-25=not 9 it is 11 :((

so i think answer should be 6/11

euler of 9 is 8
check divisibility of 11^22^33 by 8
or 3^22^33 by 8
which leaves remainder 1
so expression is
5^11^22^33-1 *5
so OA is 5

Come again please

anyone got this?

In a series , the nth term tn=t(n+1)*n^2/(n^2-1) , for all n>=2.If t5=6,find t25

could any of you help me with these ques :

3. find the last two digits of the number n=20025x20026x20027x20035

TIA :D

n%25=0
n%4=1*2*3*3=18=2
25k=4l+2
k-2=4l
k=6
so last two digits=50

Kungfu Panda Says

(5^11^22^33)/9 ... puys can you please let me know the remainder .

euler of 9 is 6
check divisibility of 11^22^33 by 6
or -1^22^33 by 6
which leaves remainder 1
so expression is
5^11^22^33-1 *5
so OA is 5

A and B throw one dice for the stake of Rs 11. w/c is to be won by the player who first throws a six, the game ends when the stake is won by A or B. If has the first throw, what are their respective expectation?

it means till the winner is not declared it will continue

we have to find with respect to first


Now Case will be

first player will win on first throw
first player will win on 2nd throw ( that means both player loses or get sumthng else during there throw)
and continue

Favourable outcome only 6 = 1 out of 1 2 3 4 5 6 hence p(a)= 1/6

unfavourable = 1-1/6=5/6

hence

it will be

1/6+5/6*5/6*1/6+ 5/6*5/6*5/6*5/6*1/6.... continue


G.p Is there

so first term is 1/6 and C.R = 25/36

so sum= 1/6* ( 1/ 1-25/36)

=1/6*36/9= 6/9=2/3 :)

I hope you are clear with step :)

Dont hesitate to ask :)

=============================================
?????
bhai

1/6+5/6*5/6*1/6+ 5/6*5/6*5/6*5/6*1/6.... continue

these highlighted are the failure of previous steps of A and B???

could any of you help me with theseues :



2.find the last two digits of 47^89.

TIA :D

47^89%25
=47^4=-3^4=81=6
=6
47^89%4=-1
=3
25k+6=4l+3
25k+3=4l
l=k+3+(24k)%4
so k=1
ans. = 31

Kungfu Panda Says

(5^11^22^33)/9 ... puys can you please let me know the remainder .

my take 5

Q2) A and B throw one dice for the stake of Rs 11. w/c is to be won by the player who first throws a six, the game ends when the stake is won by A or B. If has the first throw, what are their respective expectation?

assuming A throws first :
P(A)=1/6 + 5/6*5/6*1/6 +5/6*5/6*5/6*5/6*1/6+....
P(A)=1/6(1-25/36)= 6/11

P(B)= 5/6*1/6 + 5/6*5/6*5/6*1/6 +....
= 5/11

(5^11^22^33)/9 ... puys can you please let me know the remainder .

Q5) sanjay writes a letter to his frnd. it is known that one out 'n' letters that are posted does not reach its destination. If sanjay does not receive the reply to his letter. then what is the prob that kesari didnot recvd sanjay's letter?? It is certain that kesari will definately reply to sanjay's letter if he recves it???

p(1)=sanjay posted but not reached destination=1/n
p(2)=kesari received but his reply not recevied by sanjay =(n-1)/n * 1/n
now,
if sanjay doesn't recive reply then total sample cases =1/n+(n-1)/n^2
required prob = 1/n / = n/(2n-1)

is it correct?

Q3)(i) The prob. of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. the least no. of bombs req so that the prob of the bridge being destroyed is greater than 0.9 is: ???


In this question remember too hits are sufficient to destroy the Bridge but probability should be >0.9


take 2 bombs = maximum probability = 3/4=.75
take 3 bombs = it will be 7/8=87.5
and for 4 it will = 15/16= 93.75 :)

which is > .9

hence least will be 4 :)

this is a Shortcut to remember and do such question

----------------------------------------------------------

Bhai ans : 7 bombs diya hai

P & C (confusions..) approach plz..

Q:1) 3 of the 6 vertices of regular hexagon are choosen at randomthe prob that the triangle with these vertices is equilateral?

getting two triangles so 2/6c3 =2/20 =1/10

Q3)(i) The prob. of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. the least no. of bombs req so that the prob of the bridge being destroyed is greater than 0.9 is: ???

suppose at least n bombs are needed for given condition...

then prob that no bomb hits + only 1 bomb hits < (1-0.9)

so 1/2^n + nc1*1/2*1/2^(n-1) < 0.1
=> (n+1)/2^n < 0.1

least value of n is 7

so at least 7 bombs are needed for given condition