Please solve these quants questions

[MrAnderson]

One qn from P&C
A student has to answer 9 qns out of 12 qns in a qn paper such that he has to answer atleast 3 qns from the first 4 qns . In how many ways can he do this ?
My Soln...
------------
He can choose the 3 qns out of the first 4 in 4C3 ways ...
and the qns left out now are 9 and he has to answer any 6 of those in 9C6 ways...
Therefore number of ways = 4C3 * 9C6
= 4 *84
= 336
Soln Given ...
-------------
Case 1 : He attempts 3 qns out of the first 4 and the other 6 qns from the rest
No of ways = 4C3 * 8C6

Case 2: He attempts all the 4 qns out of the first 4 qns and the other 5 from the rest
No of ways = 4C4 * 8C5

Total no of ways = 4C3*8C6 + 4C4*8C5
= 4*28 + 56
= 168
Which soln is correct and why ?? What's the diff between the both of them ??
Thanks in advance

hi,
lemme try...
this one was really interesting one..puzzled me for some time
tried out a simplified version of the problem to find the PROBLEM..
the second soln. given is right.
and the root cause of the mistake in ur soln. is that in it
the selections are not exclusive ...i.e, there are repetitons...
thats why ur getting a higher value for the answer..
for eg..
supose in the prob ...say a,b,c,d,e,f,g,h,i,j,k,l are the 12 answers..
now u select 3 out of the first 4...
ie abc,bcd,acd or abd ...now from the remaining nine ur selecting 6...
let the initial selection be abc ...
from the remaining u r selcting defghi ..so ur full selection becomes..
abcdefghi...
now for another initial selection
say abd ...ur remaining selction cud be cefghi =>totally abcdefghi
both selections are the same...but as per ur soln,these will get get counted as 2.hence the error...

if still not clear try the same problem as
3 out of 4 qs to be answered and atleast 1 from first 2...ull see wht im tying to xplain..
hope my xplanation made some sense...
MrAndy

[newguy]

Originally Posted by MrAnderson @ Wed Nov 10, 2004 2:11 pm

hi all,
need help ..
plz tell me how to solve probs of this kind..
1.Find no of common roots btw 2 eqns.
a)x^3+3x^2+4x+5=0
b)x^3+2x^2+7x+3=0

thnx in advance
MrAnderson.

Subtract both of them :

x^2-3x+2=0
(x-1)(x-2)=0
Roots are 1 and 2

This means that the two plots cut each other at x value of 1 and 2. They may or may not be the roots of the cubic equation . Now, put these values of x the cubic equations given...Whichever value satisfies both the equation at the same time is the common root.

In your case, no root is common...

I hope I made myself clear..In case of any doubt, feel free...

[rohit_anand]

just got a small query...
we always take roots in pairs
as in if we know that 2+root(3) is a root then we assume that 2-root(3) will be a root.
but isn't this true only for equations with rational coefficients?
say if i have an equation
(x-3)(x-2+root(3))=0
then 2+root(3) is not a root even tho 2- root(3) is one...
so what shud be the right approach?...
isnt the assumption that coefficients will be rational wrong?
plz reply
rohit

[MrAnderson]

thnx new guy,
understood the method to do the problem..
will go home and think bout how it works..(which is still a bit unclear to me )
MrAndy

[raghav_krishna]

Hi,


what is perfect number


how to solve the probem: ( CL MOCK 9)
1, 8 , 15, 22, 29........ 701 ; How many numbers are perfect squares in this series.

Thanks
krishna

[THUSSU]

Originally Posted by MrAnderson @ Wed Nov 10, 2004 2:11 pm

hi all,
need help ..
plz tell me how to solve probs of this kind..
1.Find no of common roots btw 2 eqns.
a)x^3+3x^2+4x+5=0
b)x^3+2x^2+7x+3=0

thnx in advance
MrAnderson.

equating the two,
we have X^2-3X+2=0 or (X-1)(X-2)=0

Neither of these values work in the eq. above;

so none!

[methuselah]

Originally Posted by MrAnderson @ Wed Nov 10, 2004 5:10 pm

hi,
lemme try...
this one was really interesting one..puzzled me for some time
tried out a simplified version of the problem to find the PROBLEM..
the second soln. given is right.
and the root cause of the mistake in ur soln. is that in it
the selections are not exclusive ...i.e, there are repetitons...
thats why ur getting a higher value for the answer..
for eg..
supose in the prob ...say a,b,c,d,e,f,g,h,i,j,k,l are the 12 answers..
now u select 3 out of the first 4...
ie abc,bcd,acd or abd ...now from the remaining nine ur selecting 6...
let the initial selection be abc ...
from the remaining u r selcting defghi ..so ur full selection becomes..
abcdefghi...
now for another initial selection
say abd ...ur remaining selction *cud be *cefghi =>totally abcdefghi
both selections are the same...but as per ur soln,these will get get counted as 2.hence the error...

if still not clear try the same problem as
3 out of 4 qs to be answered and atleast 1 from first 2...ull see wht im tying to xplain..
hope my xplanation made some sense...
MrAndy

Thanks a lot ....cleared my dbts

One karma for u

Regards

Methuselah

[whocarez]

Originally Posted by raghav_krishna @ Thu Nov 11, 2004 12:35 am

Hi,


what is perfect number

A number n is said to be perfect if it is equal to the sum of its proper divisors (or, if the sum of all its divisors is equal to its double 2n). The first obvious example is 6 = 1+2+3.

how to solve the probem: ( CL MOCK 9)
1, 8 , 15, 22, 29........ 701 ; How many numbers are perfect squares in this series.

Thanks
krishna

Max sq less than 701 is 26 square so we have to check till 26 square....
Now the given series is of the form 7x+1 (where x=0-100)
Therefore any square of the form 7k+1 would be part of the answer. For the square to be 7k+1 the number is of the form (7n+1) or (7n-1).
so numbers whose squares are part of the series are
1,6,8,13,15,20,22
Thus 7 numbers are possible!
I know the sol. doesn't look elegant but i hope u get the drift

[Rahul Srivatsa]

Hi All..

can anyone pls solve the foll problem for me???
Q. The product of two nos. is 'y' and the HCF of two nos. is 'x' with y=kx^2..For which of the foll values of 'k' will there b a unique difference of the nos.?
(1) 42 (2) 32 (3) 44 (4) 24

The key to this problem has given option (2) as the ans..but no clue as 2 how to solve it..pls help!!

[GauravShah]

Originally Posted by Rahul Srivatsa @ Fri Nov 12, 2004 10:50 pm

Hi All..

can anyone pls solve the foll problem for me???
Q. The product of two nos. is 'y' and the HCF of two nos. is 'x' with y=kx^2..For which of the foll values of 'k' will there b a unique difference of the nos.?
(1) 42 (2) 32 (3) 44 (4) 24

The key to this problem has given option (2) as the ans..but no clue as 2 how to solve it..pls help!!

I figured out the answer coz the answer option was given .

Hope this part is understood that the two terms are (Ax) and (Bx), where k = AB.
y = Ax*Bx = (AB)*(x^2)....

Now option (1) for k

AB = 42, Here we can have (A=42, B=1), (A=6, B=7) and more sets to get HCF still as "x"

But for option (2) we can AB = 32...

The only option we can is (A=32 and B=1) or vice versa keeping the differnce same...

for other options like A = 4, B=8.... HCF will be '4x'

Hope it clarifies...

Gaurav.