each face of a cube is painted either white or black. in hw many diffrnt ways the cube can be painted.

guys the answr is 10. can u expalin hw??:

1..In a triangle ABC, right angled at B, the median through C bisects the angle between CA and the bisector of < C. Which among the following best describes the range of AC/CB?
(1) ((5 -1)/2, 2) (2) (2, 5/2) (3) (2+1, 23) (4) (5/2, 3) (5) (5, 4)

Solution:
Let E be the mid-point of AB such that AE = EB = c/2.
CD bisects < C => BD = ac/(a+b). Since, CE bisects < ACD, CD/CA = DE/EA.

CD^2 = a^2 + a^2c^2/(a+b)^2, DE^2 = (c/2 - ac/(a+b))^2
Simplifying, we get a^2{(a+b)^2 + c^2} = b^2(b-a)^2. Using b^2 = a^2 + c^2 and eliminating c we get b^3 - 2ab^2 -a^2b - 2a^3 = 0. Put p = b/a, we have p^3 - 2p^2 - p - 2 = 0.

Let f(p) = p^3 - 2p^2 - p - 2. The 1st derivative of f(p) is 3p^2 - 4p - 1 > 0 for p > 2 and thus a increasing function for p > 2. Also f(2).f(3) < 0 => p lies in (2, 3).

Looking at the options, we have to find the narrowest interval which fits the range of p.
f(2.5) < 0, and thus this works. p = b/a lies in (5/2, 3).

=> Choice (4) is the right answer

Im a new member of QQAD 08.Could any1 pls provide with the link to Q1-120 along with the answer key?

Q=118
Has anyone solved ..
In a triangle ABC, right angled at B, the median through C bisects the angle between CA and the bisector of < C. Which among the following best describes the range of AC/CB?

(1) ((5 -1)/2, 2) (2) (2, 5/2) (3) (2+1, 23) (4) (5/2, 3) (5) (5, 4)

hi ,

how to find the last non zero digit of n! ?

A lock has a code consisting of 5 alphabets from among the first 11 letters of the English alphabets. The lock will be opened, if any of the three letters of the code match, occurring not necessarily in the same positions. What is the minimum number of five letters codes one should try to be sure that the lock is opened.

Choices are:

i) 3
ii) 4
iii) 5
iv) 6

User Radhika_cool Banned for repeated spamming & advertising !
Plz be advised, this ain't a place for you to leech !

radhika_cool Says

radhika_cool Says

Had the highlighted part been a female ... Ur ad wud have been a 100% guranted success...
Waise I think ki... Rahul and Gaurav dono ki dosti Jai aur Viru ki dosti type hai... ya phir... Airtel or Hutch or whatever it is... they hv run out of new SIMs... Dono ke numbers ek hee jo hai...

radhika_cool Says

dudette (dude ?? ) stop spamming ...... u hv the audacity to go around lying and posting ths crap .... u thnk aftr ths ppl will actually attend ths so calld 'Workshop' ..... ystrday u were posting under some other id ... and now "female" ....... stop it ...

No advertising allowed ! User Banned !

Hi all puys & Quant gurus.
I've come across this question
But unable to solve it.
So give it a try.

Q. 100 cards are numbered 1 to 100 (each card different) and placed in 3 boxes (at least one card in each box). How many ways can this be done so that if two boxes are selected and a card is taken from each, then the knowledge of their sum alone is always sufficient to identify the third box?
1) 12 2) 14 3) 10 4) 20 5) 18

Is prashan ka jawab 12 hai kya harsh bhaiya.......
humko pata hai.....hum ganit mein thoda TEJ houn......aakhir BIHARI ENGG hein...chaliye....thoda dimag pe jor daaliye or sochiye ...kaahe 12 hoga iska jawab !!!!




Bihar BIHAR BIHAR........baaaki sab BIMAR:grab:

Harsh bhaiya ko namaskar !!!!
aapka ee sawal to hume humre IIT coaching ki yaad dilata hai....jab vo budbak Abhay kumar singh or Das Gupta Humra jaan khaat tha ......!!!!

thoda waqt dijiya....SK GOYAL SE REFER KAR KE AAPKA JAWAB ABHI PESH KARTE HEIN:grab: Meri bhens ko danda kyoun maara .......!!!!!

Maari Laathi Mar gayi bhens ....Chaara huvi gaya saara baste

Hi all puys & Quant gurus.
I've come across this question
But unable to solve it.
So give it a try.

Q. 100 cards are numbered 1 to 100 (each card different) and placed in 3 boxes (at least one card in each box). How many ways can this be done so that if two boxes are selected and a card is taken from each, then the knowledge of their sum alone is always sufficient to identify the third box?
1) 12 2) 14 3) 10 4) 20 5) 18

can any body solve this
aarav plz help me

1.Bilbo bought a triangular piece of land area 150m2.Bilbo took a
piece of rope and measured the two sides of the plot and found the
largest side to be 50 m and another side to be 10 m. What is the exact length of the third side?
2.Let a=3-5x
b=2x-4
c=1-4x
y=min(a,b,c) for different values of x.
The maximum value of y is?
:grab: :grab: :grab:

maximum value of y is (-7/3).plz confirm

can any body solve this
aarav plz help me

1.Bilbo bought a triangular piece of land area 150m2.Bilbo took a
piece of rope and measured the two sides of the plot and found the
largest side to be 50 m and another side to be 10 m. What is the exact length of the third side?
2.Let a=3-5x
b=2x-4
c=1-4x
y=min(a,b,c) for different values of x.
The maximum value of y is?
:grab: :grab: :grab:

My try,
Now, this follows a pattern of 8 no. repeated.
But since total no. 100 is not divisible by 8 and leave remainder 4. its not the pattern of 8 but 4 and hence the answer.
If this is in linear there wont be unique solution given the same conditions.

PPl,
I am not getting how the solution would differ if the numbers were arranged in a linear pattern i.e. a straight line.
Pls help me here.

http://pagalguy.com/newsletter/archive.php?ArchiveID=MjA%3D 1 to 46.
Please search the last 8 pages of this thread to get the rest.

aarav,
How wud i get archive of previos 150 quant quetions with its solutions?
sumit

Question:
a, b, c are positive integers forming an increasing geometric sequence, b-a is a square, and (loga + logb + logc)/log6 = 6. Then a + b + c is
(a) 54 (b) 78 (c) 111 (d) none of the foregoing
My Stake on this is C):111
Explanation: (loga + logb + logc)=Log(a.b.c)
So Log(a.b.c)=6.Log6=log(6^6)
So abc=6^6=46656....(i)
Now as a,b,c are in geometric progression then:
B^2=a.c
So put this value in (i)
We have b^3=46656 so b=36
Now as b-a has to be square,the only possible values for a ,b and c can be
A=27,b=36 and c=48
So a+b+c=111 and also a.b.c=46656

here u have assumed that (b-a) as 25
but why not 16 or 9 or 4 ??
please solve my doubt

regarding q. no. 148

aarav

the three eqns given are symmetrical in nature..
then doesn't it automatically means that x=y=z
also, if they were distinct.. they wont satisfy the give set of eqns.
please clear my doubt..

hi i am new to pagal guy.how to get the questions before ones posted to our mails

Does anyone know a easy method to calculate dates e.g. what say is 31 dec 2020, 19 May 1982, 20 Mar 1983 etc

Regards,
-S

http://www.pagalguy.com/forum/quantitative-questions-and-answers/17263-quantitative-questions-day-151-onwards.html

THIS IS THE NEW THREAD FOR QNS NO 151 onwards .... post thy solns there !!

soumyajyoti Says

Where is the new thread ...can anyone give the link???

Where is the new thread ...can anyone give the link???ME too

Where is the new thread ...can anyone give the link???

Please use the new thread opened for the discussion from 151 onwards. It is lying somewhere in the middle right now, if you can catch some moderator then request him to make 151 onwards thread sticky. I spoke to Rohit, but I guess he forgot.

x, y, z are distinct reals such that y = x(4-x), z = y(4-y), x = z(4-z). The possible value of x+y+z is
(a) 4 (b) 6 (c) 7 (d) 9

can i try my shout.. for this question...
i say add the three equations..
take x+y+z =p ..
so now we get x^2+y^2+z^2 = 3p..
now x+y+z =p so we get squaring..
xy+yz+zx =(p^2-3p) /2
well i don't know what .. but i remember ..
that.. x^2+y^2+z^2 >= xy+yz+zx
so 3p>= p^2-3p/2.. pr
9p >=p^2....between 0and 9. .. i am in a fix..:neutral:
but we have no idea about distinct solution as mentioned.. in the question.. but we know is a nondistinct solution..
now i think what to do.. well .. its all bit hazy now..
which choices to mark 7 looks a certain for distinct .. now. i try with 2,2,2 .. i don't get a solution.. 3,3,0 again i don't .. so i mark.. 6 as well..well u can try putting.. x=4, y =0 ,z=2.. so x+y+z =6 is a solution..:satisfie:
kya kare for x+y+z =4.. is it a solution.. or not.. how to decide..somebody help.. .. i need help..here. i think it is.. cause.. x,y,z are distinctfor.. except for x=y=z=0 and x=y=z=3...

can somebody help here..

I Got it ........

cycle will be a1-a8,a9-a16,......a87-a96,,,, a96-a4, a5-a12........ a93-a100,,,, Then it will repeat again. .... (1)
So, a1+a2+a3+a4=a9+a10+a11+a12 & a5+a6+a7+a8=a13+a14+a15+a16 do for the series..... (2)
Using data from Question & (2):
a9th = a1 = -1/2
a19th = a3 = 3/4
a20th = a4 = 2

a50th = a2 = ??????
Using data from Question & (2):
a1 + a2 + a3 + a4 = a97 + a98 + a99 + a100= (25/2)/2
a2 = 4

Hence option (a) 4

Hey Folks ....

The Answer am getting is 4 . The approach i have used is as follows .

Given the sum of every 8th consecutive number is 25/2 . And this follows a cyclic pattern (given that any 8 consecutive numbers ) . So we know that we will need 4 numbers for the sum of 8 consecutive numbers to be the same .For this we have 9th number as -1/2 , 19th number as 3/4 and the 20th number as 2 .So in all we have 3 numbers . Also the sum of those 4 numbers should be 25/4 . So the unknown number or the 50th number here is (25/4 - (-1/2 +2+3/4)) . Which gives the answer as 4 ..........

Hey Folks ....

The Answer am getting is 4 . The approach i have used is as follows .

Given the sum of every 8th consecutive number is 25/2 . And this follows a cyclic pattern (given that any 8 consecutive numbers ) . So we know that we will need 4 numbers for the sum of 8 consecutive numbers to be the same .For this we have 9th number as -1/2 , 19th number as 3/4 and the 20th number as 2 .So in all we have 3 numbers . Also the sum of those 4 numbers should be 25/4 . So the unknown number or the 50th number here is (25/4 - (-1/2 +2+3/4)) . Which gives the answer as 4 ..........

Hello
This Is My First Posting
I Really Liked Too Much Yours Section
How Can I Have Access To The Earlier Questions

answer is option a

my approach...
let the numbers be a1, a2,....a100

a1 = a9 = a97
a2 = a10 = a98
a3 = a11 = a99

hence a19 = a11 = a3
a20 = a12 = a4
a9 = a1
a2 = a50

now = 12 hence 96 numbers

the remaining 4 nos thats a97 to a100 are left
so a97+a98 + a99 +a100 +a1+a2+a3+a4 = 25/2
hence 2(a1+a2+a3+a4) = 25/2
hence a2 = 4

same here... bit easy this time

I am visiting this thread for the first time today. It's a gr8 one though, and i regret having missed out on all the previous questions. so, is there ne link where i could find all the auestions along with the answers at one go? please let me know.....and keep up the gr8 work..........PG ROCKS

Even after my mail server started functining once again, I didn't receive the question.

Thanks Superstar and Deepu, I have got it at this thread.
It is an easy question (less than a minute) while sitting in the comfort of home, but can test your nerves in the examination hall.

The answer as well as the method are already discussed. I went by the same approach and arrived at (4) as my answer.

I too go with choice (a) 4 like all others.

My try,
Now, this follows a pattern of 8 no. repeated.
But since total no. 100 is not divisible by 8 and leave remainder 4. its not the pattern of 8 but 4 and hence the answer.
If this is in linear there wont be unique solution given the same conditions.

ok..answer is now history....such a talented group of guys in pg....then let me repeat history..answer is option (A) 4 and method as used by fellow puys..

sum (x93+...+x100)=25/2
sum(x96+....x4)=25/2

the above equations implies sum(x93+..+x96)=sum(x1+...x4)=25/4
given x1,x3 and x4.....now do the math......

Me getting 4

no the numbers are not consecutive...just chck this series...
I am saying this series has a constant sum for every 4 consecutive inumbers of the series...

15641564156415641564......

Have you got the point???

Hey folks,
I am not understanding the question. I interpreted "consecutive numbers" as "numbers placed in immediate succession" in the circle. The question never says that all 100 numbers are in sequence i.e. 2,3,4...etc. Please explain.

This is my first post... and I have been enjoying each and every question/solution posted on this site.

Gr8 job guys...

: