Quant by Arun Sharma

maskedmenace · October 20, 2008, 6:43pm

Hello all,who visit this forum often. I am trying to solve one problem but failed.The problem goes in the following fashion.

f{f(x,y),f(y,x)}=1-f(x,y);f(1,1)!=1;f(1,1)=?

is it 3 ??

rupam2u · October 21, 2008, 8:08am

No the answer is 1/2.Thanks for the try.

nayan_dawn · October 21, 2008, 9:12am

Rupam,

1/2 is not a possible answer.Please re confirm.This might be 0

regards,
nayan

.

aravind_raju · October 21, 2008, 9:38am

Hey puys,

Nice problem for you all......
What will be the remainder when
(2^2+22^2+222^2+2222^2+.....................+22... ..(2upto 49 times) )
is divided by 9?

Soln :

Rem(22^2/9) = Rem(22*22/9) = Rem(4*4/9) = Rem(4^2/9)

Similarly,
R(222^2/9) = R(222*222/9) = R(6*6/9) = R (6^2/9)
and R(2222^2/9) becomes R(8^2/9), R(22222^2/9) becomes R(10^2/9) and so on upto R(222..(49times)^2/9) becomes R(98^2/9)

So the qn is now reduced to

R ( (2^2 + 4^2 + 6^2 +.......+ 98^2)/9 )
= R( 4(1^2+2^2+......+49^2)/9 )
= R( /9 )
= R( (4*40425)/9 ) = R (24/9) = 6

rupam2u · October 21, 2008, 9:58am

I have not solved the problem myself.Actually it was sent to the mail box of one of my friend from this particular website.But they have not provided him with the solution but only the answer and it is indeed 1/2.

Thanks

divyaakanksha · October 23, 2008, 2:56pm

If a0 = 2, a1 = 1 and a n + 1 a n 1 = a n 1 a n + 1, then find the value of a1000.
(1) 1/1001
(2) 1/2002
(3) 1/2
(4) 2/1001
(5) 3/1003

divyaakanksha · October 23, 2008, 2:59pm

If a#0 = 2, a#1 = 1 and a #(n + 1) a #(n 1) = a #(n 1) a #(n + 1), then find the value of a#1000.
(1) 1/1001
(2) 1/2002
(3) 1/2
(4) 2/1001
(5) 3/1003

*digits and exp after # are subscripts

maskedmenace · October 23, 2008, 7:35pm

If a#0 = 2, a#1 = 1 and a #(n + 1) a #(n - 1) = a #(n - 1) - a #(n + 1), then find the value of a#1000.
(1) 1/1001
(2) 1/2002
(3) 1/2
(4) 2/1001
(5) 3/1003

*digits and exp after # are subscripts

For this type of questions ..just put values ...Like n =1, n=2 ....thn u vil observe some pattern to be following ....

1/a(n+1) - 1/ a(n-1) = 1

put n =1

1/a2 - 1/a0 = 1

=> 1/a2 = 3/2

put n =3

1/a4 - 1/a2 = 1

=> 1/a4 = 5/2

so, wht's da pattern ...1/an = /2

Like wise v get 1/a6 = 7/2 , 1/a8 = 9/2 ........1/a1000 = 1001/2

=> a1000 = 2/1001

nuisance · October 24, 2008, 6:35am

Hello all,who visit this forum often. I am trying to solve one problem but failed.The problem goes in the following fashion.

f{f(x,y),f(y,x)}=1-f(x,y);f(1,1)!=1;f(1,1)=?

I assumed f(x,y) =a+bx+cy
=> f(y,x) = a+by+cx
so,
f{f(x,y),f(y,x)} = a+b(a+bx+cy)+c(a+by+cx) =a(1+b+c) +(b^2+c^2)x+2bcy ---(1)

and 1-f(x,y) = (1-a) -bx -cy ---(2)

by equatin 1 and 2 I got
a=1/2,b=-1/2,c=1/2 and a=1,b=-1/2,c=-1/2

so f(x,y) =1/2(1-x+y) which gives f(1,1) =1/2
and f(x,y) =1-1/2(x+y) which gives f(1,1) =1/2 again

netsync07 · October 30, 2008, 4:18pm

Hi I need the explanation for solving this problem from progressions:
sum to n terms of the series:8 + 88 + 888 + ....
it will be really nice if someone can help.

say2sayeed · October 30, 2008, 4:41pm

Hi I need the explanation for solving this problem from progressions:
sum to n terms of the series:8 + 88 + 888 + ....
it will be really nice if someone can help.

8 + 88+ 888......
=8(1+1+111....)
=8/9(9+99+999.....)
=8/9(10-1+100-1+1000-1.....)
=8/9((10+100+1000+..)-(1+1+..))
=8/9((10(10^(n-1) -1)/(10-1))-n)

put n and get result

hope that helps

netsync07 · October 30, 2008, 7:09pm

hey thanks!!
got an another problem in progression:
sum of 1.2 + 2.2^2 +3.2^3+.......+100.2^100??

shashank_chaos · October 30, 2008, 8:19pm

Find the maximum value of n such that 157! is perfectly divisible by 18^n .

Answer is 37.

Please show the steps.

Thanks!

the_hate · October 31, 2008, 9:36am

hey thanks!!
got an another problem in progression:
sum of 1.2 + 2.2^2 +3.2^3+.......+100.2^100??

S = 1.2 + 2.2^2 +3.2^3.............100.2^100
2S = 1.2^2+2.2^3+...........+99.2^100 + 100.2^101

subtracting eqn 1 from 2=>

S = -(1.2 + 1.2^2 + ..............+2^100) + 100.2^101
= -1. 2(2^100 -1) + 100. 2^101 (G.P.)
= 99.2^101 + 2

there might be few calcln mistakes but i hope u get the method

bishweshwar1234 · November 1, 2008, 11:49am

Find the maximum value of n such that 157! is perfectly divisible by 18^n .

Answer is 37.

Please show the steps.

Thanks!

I'm getting ans. as 26.
From where you got 37.

shashank3012 · November 1, 2008, 7:42pm

Find the maximum value of n such that 157! is perfectly divisible by 18^n .

Answer is 37.

Please show the steps.

Thanks!

18 can be expressed as 3^2*2
so,the answer would be
157/3+157/9+157/27+157/81
52+17+5+1=75
but,power of 3 is 2
so,answer is 75/2=37

hope that helps.

netsync07 · November 2, 2008, 4:13pm

Please provide explanation for the below question:
A wall clock gains 2 min in 12 hrs while a table clock loses 2 mins in 36 hrs;both are set right at noon on tuesday.The correct time when they both show the same time next would be
a) 12:30 night b)12 noon c)1:30 night d) 12 night

shashank3012 · November 2, 2008, 9:59pm

After 36 hrs
Wall-12.06 night
Table-11.58 night
They will meet,when 12 hours are covered between them.
36 hrs=>8 mins
x hrs=>720 mins
x=3240 hrs
x=135 days
So,12 noon.

Is the answer correct?

ket_gmat · November 5, 2008, 1:51pm

Test has 80 questions
each wrong question -1/2
each not attempted question -1/4
final score is 34.5.
How many questions are right?

45
48
54
can not be determined.

shashank3012 · November 5, 2008, 4:07pm

Test has 80 questions
each wrong question -1/2
each not attempted question -1/4
final score is 34.5.
How many questions are right?

45
48
54
can not be determined.

let,questions right be x
questions wrong y
not attempted z
x+y+z=80
x-0.5y-0.25z=34.5
1.5y+1.25z=45.5
6y+5z=182----(1)
6x+z=298-----(2)
5x-y=218-----(3)
if x=45,y=7,z=28
if x=48,y=22,z=10
if x=54,y=52=>not possible
as you can see from the options too,2 cases are possible,and also,as there are 2 equations for 3 variables.i think,it cannot be determined.

what does the original solution say??