hi........:grin:

Ya I'm aware of that formula for GP..thanks :)actually I was confused coz I assumed that there are 6 terms in GP including 4 GM....Any luck in solving Quant q 98 that I had posted earlier...

MaskedMenace Says

yeah...for a gp common ratio is term2/term1 ....nd it remain same through out d series...tht is wht i hav done....if not clear just revert

this is regd following q :
Four Geometrical means are inserted between 1/8 and 128.find the 3rd Geometrical mean.


hey just one query....regd the way u calculated common ratio..
I thnk it shuld be...calculated using nth term of GP formula..In that putting first term as 1/8 and nth term as 128 and n =6we get common ratio as : 4....In your Method what I found dubious was that equaiton that you have formed....can you elaborate it a bit....coz assumin " x" is common ratio I thnk that equation can't be formed....

yeah...for a gp common ratio is term2/term1 ....nd it remain same through out d series...tht is wht i hav done....if not clear just revert

this is regd following q :
Four Geometrical means are inserted between 1/8 and 128.find the 3rd Geometrical mean.


hey just one query....regd the way u calculated common ratio..
I thnk it shuld be...calculated using nth term of GP formula..In that putting first term as 1/8 and nth term as 128 and n =6we get common ratio as : 4....In your Method what I found dubious was that equaiton that you have formed....can you elaborate it a bit....coz assumin " x" is common ratio I thnk that equation can't be formed....

Just find the common ratio..n keep multiplying it ...

this is how it can b done...

1/8, _ , _ , _ , _ ,128


Now,128/x = 8x == >x =4 (common ratio)

so, 1/8,1/2,2,8,32,128 ....3rd gm is 8..

hello all,

quant Question no 98 :
------------------------------------------------------
Quantitative Question # 098
------------------------------------------------------

Consider two distinct positive integers x and y having integer arithmetic, geometric and harmonic means. The minimum value of x - y is

(1) 15 (2) 20 (3) 30 (4) 40 (5) 50



And my logic for solving this is described in my post above...but as per that I'm not able to get an answer...

Need help .....

Hello all,

I'm Deepa..
just had this query regd the Quant Q 98 :It says AM,GM,HM all should be integral..so that means the no's x & y should be an even perfect square ...x & y can be both positive or both negative....coz otherwise we can't get an intergral square root...
now goin by all this....lets consider even perfect squares like 4,16,36,64....and x&y; can be (2,2) ; (4,4) ; (-2,-2)....like wise and if we go by this method we can't get an answer....:...can sum1 sugg a way to solve it......

hoping for quick replies ;)..

i cldnt able to find q.no 98...please post tht ques...

Four Geometrical means are inserted between 1/8 and 128.find the 3rd Geometrical mean.

I kno its a simple question but just cant find the answer!

Just find the common ratio..n keep multiplying it ...

this is how it can b done...

1/8, _ , _ , _ , _ ,128


Now,128/x = 8x == >x =4 (common ratio)

so, 1/8,1/2,2,8,32,128 ....3rd gm is 8..

Hello all,

I'm Deepa..
just had this query regd the Quant Q 98 :It says AM,GM,HM all should be integral..so that means the no's x & y should be an even perfect square ...x & y can be both positive or both negative....coz otherwise we can't get an intergral square root...
now goin by all this....lets consider even perfect squares like 4,16,36,64....and x&y; can be (2,2) ; (4,4) ; (-2,-2)....like wise and if we go by this method we can't get an answer....:...can sum1 sugg a way to solve it......

hoping for quick replies ;)..

Bhaiyon Aur Behno,

Main QQAD subscribe karne ka param ichhuk hun. Meri iccha hai ki ek prashna pratidin pratah kal mere mail-box mein aaye. Maine subscribe karne ki link ki khoj ki lekin meri khoj asaphal rahi. Kya koi meri madad karna chahega?

Dhanyawad,

QQAD ka subscription pane ka param icchuk.

Four Geometrical means are inserted between 1/8 and 128.find the 3rd Geometrical mean.

I kno its a simple question but just cant find the answer!

------------------------------------------------------
Quantitative Question # 065
------------------------------------------------------

If all palindromes (positive integers which is unchanged if you reverse the order of its digits) are written in increasing order, what is the possible number of prime values can the difference between successive palindromes take?



(1) 0 (2) 1 (3) 2 (4) 3 (5) none of these

The palindromes in increasing order are

1
2
3
...
9
11
22
33
...
99
101
111
121
131
...
191
202
212
222
...

We see that the prime difference can be either 2 or 11 - hence the answer is (3)

hei...I can't understand the concept of no: of steps taken in an escalator..plz help me wid dis,
(speed of a person/speed of escalator) = (no: of steps on moving escalator / no: of steps of the person when escalator is off)

cld any one plz explain the concept..?

..i too got 20/3..

Question 101:
At his normal speed, Shirish can travel 18 km downstream in a fast flowing stream in 9 hours less than what he takes to travel the same distance upstream. The downstream trip would take 1 hour less than what the upstream trip would take provided he doubles his rate of rowing. What is the speed of the stream in km/hr?

(a) 20/3 (b) 8 (c) 33/4 (d) none of the foregoing


I am getting speed of stream as either zero(0) or negative(-20/3).
9=18/(x+v)-18/(x-v)
and
1=18/(2x+v)-18/(2x-v)

where x and v are speeds of Sirish and stream, respectively.
on calculating first one the relation is: x^2=v^2-4v

and then putting in second and solving.
Am i wrong somewhere??

yup... its comin out to be...20/3
:

HI all

I am new to PG can some body mail me the qns from 50 to 100. Or id any link is available for archives can some one post it. Thanks in advance.

My mail id is pradeep.gorai@cgi.com

Hi I joined in late ,so if somebody can post/mail me all the questions from 1-100 of AQQAD with its respective solution,I will be highly gr8ful
Regards,
Akrant
akrant@gmail.com

oye me not getting ANY quant question. are the questionssupposed to be mailed everyday (as the name of the thread suggests) to the people who have enrolled for the same? if true, there's some problem somewhere to be solved. i registered a loooong time back.

ne suggestions.....fellow PGites?

hi guys i m new to this site i joined this forum and i recieve questions by mail but the main cause of worry for me is that i m not able to crack this questions can u tell me is this the level of questions appearing in cat

bhai apna bhi 20/3 ans hai
i too joined the majority!!!!!!!!!!!!!
anybody having sectional test 01 questions and ans ???????????????
Pls mail me at ................
:wow: Mein aapka bahut abhari rahunga

Quagmire Says

Shouldnt ur first eqn be 9 = (18/x-v) - (18/x+v)? Because x-v is going to be smaller than x+v, so 18/x-v > 18/x+v

Thanks a lot yaar, i did a very stupid error.Vaise we are on the wrong thread:) see you on the other one!!

hi guys ..this is my first post at PG ...
My answer is also 20/3

me also getting as option a - 20/3

Question 101:
At his normal speed, Shirish can travel 18 km downstream in a fast flowing stream in 9 hours less than what he takes to travel the same distance upstream. The downstream trip would take 1 hour less than what the upstream trip would take provided he doubles his rate of rowing. What is the speed of the stream in km/hr?

(a) 20/3 (b) 8 (c) 33/4 (d) none of the foregoing


I am getting speed of stream as either zero(0) or negative(-20/3).
9=18/(x+v)-18/(x-v)
and
1=18/(2x+v)-18/(2x-v)
where x and v are speeds of Sirish and stream, respectively.
on calculating first one the relation is: x^2=v^2-4v
and then putting in second and solving.
Am i wrong somewhere??

Shouldnt ur first eqn be 9 = (18/x-v) - (18/x+v)? Because x-v is going to be smaller than x+v, so 18/x-v > 18/x+v

18/(x-y)-18/(x+y) =9 =>2y/x^2-y^2=1/2(x spped of rowin,y speed of stream)
18/(2x-y)-18/(2x+y)=1=>2y/4x^2-y^2=1/18

hence 2y=(x^2-y^2)/2 =(4x^2-y^2)/18

from (x^2-y^2)/2 =(4x^2-y^2)/18 we get 5x^2=8Y^2 =>5(x^2-y^2)=3y^2=>x^2-y^2=3/5*y^2
put ths vaue in eqn 2y/x^2-y^2=1/2 u get 10y/3y^2=1/2 =>y=20/3

Question 101:
At his normal speed, Shirish can travel 18 km downstream in a fast flowing stream in 9 hours less than what he takes to travel the same distance upstream. The downstream trip would take 1 hour less than what the upstream trip would take provided he doubles his rate of rowing. What is the speed of the stream in km/hr?

(a) 20/3 (b) 8 (c) 33/4 (d) none of the foregoing


I am getting speed of stream as either zero(0) or negative(-20/3).
9=18/(x+v)-18/(x-v)
and
1=18/(2x+v)-18/(2x-v)

where x and v are speeds of Sirish and stream, respectively.
on calculating first one the relation is: x^2=v^2-4v

and then putting in second and solving.
Am i wrong somewhere??

hi me too getting the same answer, i approached with some steps like taking p=q=r=1 and then from the options was able to get the option 3). One help guys quote]

i think th question says there exists three DISTINCT real nos p,q & r.
whereas u took all of then as 1....??:

ONE HELP

guys pls could anyone forward the quatz quetion a day mails before question #89...it will be of gr8 help to me... thnx in advance..

U can get some of the Qs here....

http://pagalguy.com/newsletter/archive.php?ArchiveID=MTE%3D

amar_kashyap Says

Abhi mil gaya na answer, m=-n^2...But, its lil long..let's see if we can find some different approach...

ONE HELP

guys pls could anyone forward the quatz quetion a day mails before question #89...it will be of gr8 help to me... thnx in advance..

mpq+nq=1
mqr+nr=1
mpr+np=1
i took value of q from 1st eq..
put it in 2nd .
den d value of r frm 2nd eq n put it in 3rd one..

it was a long method.. n made mistakes too...

Abhi mil gaya na answer, m=-n^2...But, its lil long..let's see if we can find some different approach...

Aarav Says

The answer is not what you are saying, please check your answer again. Write you solution also :-)

mpq+nq=1
mqr+nr=1
mpr+np=1
i took value of q from 1st eq..
put it in 2nd .
den d value of r frm 2nd eq n put it in 3rd one..

it was a long method.. n made mistakes too...

Hey Mansoor...
Ur solution is fine...
but dont u think that the last term shud be...
(m+n^2)( mp^2 +np-1) = 0
That's wat i m gettin...
Though ultimately both leads to...
m=-n^2

yeah ur right...
Whatever the 2nd term be, it isnt zero.....
actually stopped solving when i saw (m+n)

f(p)=q, f(q)=r, f(r)=p => f(f(f(p)))=p
f(p)=1/(mp+n)
f(f(p))=1/(m(f(p)+n)=(mp+n)/(m+n+mnp)
similarly,
f(f(f(p)))=(m+mnp+n)/((m+n)p+n(2m+n))=p
=> (m+n)p+n(m+n)p+(m+n)=0
=>(m+n)(p+pn+1)=0
(p+pn+1) !=0 since similar relations will hold for q, r making p, q, r not distinct.
=> m=-n
choice (3)

hi me too getting the same answer, i approached with some steps like taking p=q=r=1 and then from the options was able to get the option 3). One help guys please..I joined late in this forum so was able to get questions only from #89..If anyone could forward the previous questions to this id:jpsbala@gmail.com i will be very grateful.thanx in advance.

Hey Mansoor...
Ur solution is fine...
but dont u think that the last term shud be...
(m+n^2)( mp^2 +np-1) = 0
That's wat i m gettin...
Though ultimately both leads to...
m=-n^2

@koolbuddy
u r right.... even im getting the it as (m+n^2)( mp^2 +np-1) = 0

f(p)=q, f(q)=r, f(r)=p => f(f(f(p)))=p
f(p)=1/(mp+n)
f(f(p))=1/(m(f(p)+n)=(mp+n)/(m+n+mnp)
similarly,
f(f(f(p)))=(m+mnp+n)/((m+n)p+n(2m+n))=p
=> (m+n)p+n(m+n)p+(m+n)=0
=>(m+n)(p+pn+1)=0
(p+pn+1) !=0 since similar relations will hold for q, r making p, q, r not distinct.
=> m=-n
choice (3)

Hey Mansoor...
Ur solution is fine...
but dont u think that the last term shud be...
(m+n^2)( mp^2 +np-1) = 0
That's wat i m gettin...
Though ultimately both leads to...
m=-n^2

got it Mansoor Bond !!

f(p)=q, f(q)=r, f(r)=p => f(f(f(p)))=p
f(p)=1/(mp+n)
f(f(p))=1/(m(f(p)+n)=(mp+n)/(m+n+mnp)
similarly,
f(f(f(p)))=(m+mnp+n)/((m+n)p+n(2m+n))=p
=> (m+n)p+n(m+n)p+(m+n)=0
=>(m+n)(p+pn+1)=0
(p+pn+1) !=0 since similar relations will hold for q, r making p, q, r not distinct.
=> m=-n

choice (3)

VAIBHAV_MBA Says

I suppose the answer is c. After working on equations i checked from options which relation can be possible.

Even I did that ... going by answer options after getting an eqn in terms of p, m and n.
But could not get LHS = RHS by any of the answer options......

Aarav Says

The answer is not what you are saying, please check your answer again. Write you solution also :-)

I suppose the answer is c. After working on equations i checked from options which relation can be possible.

Quanta Bhai
I m revolving in a loop of 3 eqns for the past 20 mins. ...
Cannot break this loop of three Eqns....

Richa , can u post ur method please....

Same here, not able to reach a solution.