official quant thread for cat08

[dare2]

Originally Posted by naiquevin

because minimum no of usable shoes is asked.
sorry i could not make my point properly here.
take another case...

size----left-----right------total
8' -------25----- 175------- 200
9' ------175------ 25 ------- 200
10' -----100 -----100------- 200
---------------------------------
total----300 ---- 300

so minimum usable ones here are 25
Likewise there can be many more cases

here again the usuable shoes are 25+25+100=150 much more than your last solution..

[MaskedMenace]

Originally Posted by naiquevin

d) none of these.

size----left-----right------total
8' -------0------ 200------- 200
9' ------200------ 0 ------- 200
10' -----100 -----100------- 200
---------------------------------
total----300 ---- 300
so minimum no of usable shhoes is 0....... or am i missing something?

comfortably u r missing...

u can't wear only left or only right ...ryt?

so 8' n 9' are unusable.....

10' can b used..since u r having equal pairs...

[KillFRenzy]

Originally Posted by naiquevin

because minimum no of usable shoes is asked.
sorry i could not make my point properly here.
take another case...

size----left-----right------total
8' -------25----- 175------- 200 means 50 usable(25 L ,25 R)
9' ------175------ 25 ------- 200 means 50 usable(25 L ,25 R)
10' -----100 -----100------- 200 means 200 usable(100 L ,100 R)
---------------------------------
total----300 ---- 300 total = 300

so minimum usable ones here are 25
Likewise there can be many more cases

We are talking about minimum...
hence 200 is the answer... (earlier post by naiquevin )
PS: we are not talking bout pairs...
its about shoes..(hence 100 is not the answer)

[naiquevin]

Originally Posted by dare2

here again the usuable shoes are 25+25+100=150 much more than your last solution..

okok!!!...........i realise my mistake..
Thanks

[shivam_01]

Originally Posted by samkris

Puys
A simple one

if z=5-5^(2/3)-5^(1/3)
find z^3+z^2+10z+60

I dont know the final answer ..... pls help me out with this

see take the z=5-5^(2/3)-5^(1/3)
and write it as
if 5-z = (if z=5-5^(2/3)-5^(1/3)
cubing both the sides na then substituting the value of 5^(2/3)-5^(1/3) as 5-z

[jha16june]

I think this is good one:
The positive integers m, n and k are such that m*n=1024 and k = (1024!)/[(m!)^n]. If k has the maximum possible number of zeros at the end, what is the min value that n can assume?
1. 256 2. 192 3. 128. 4. 64. 5. None of these

[shivam_01]

Lets practice some more
two questions from my side don not have explanation just answers see what others have to say
If a is a positive integer such that aa - (a - 1)a + 1 = 13, then the number of positive divisors of a are
20 15 10 5 3


The series 1,2,4,5,7,9,10,12,14,16,17... has one odd number followed by two evens, then three odds, followed by four evens, and so on. What number is the 2003th term?
3943 3845 3844 3945 None of these

[shivam_01]

Originally Posted by hi.shivani

Answer for that v5=7 one question is d) cant be determined..

New question:
A warehouse contains 600 shoes,300 of right leg and 300 of left leg.
There are three sizes 8,9,10 with number of shoes of each size be 200.What is the minimum number of usable shoes?
a) 50 b) 100 c) 200 d) None of these

Apply pigeon hole principle and the answer is 100
Nice article Rmbt
thanks

[samkris]

Originally Posted by shivam_01

see take the z=5-5^(2/3)-5^(1/3)
and write it as
if 5-z = (if z=5-5^(2/3)-5^(1/3)
cubing both the sides na then substituting the value of 5^(2/3)-5^(1/3) as 5-z

can u pls explain the steps and the final answer ???

[MaskedMenace]

Originally Posted by jha16june

I think this is good one:
The positive integers m, n and k are such that m*n=1024 and k = (1024!)/[(m!)^n]. If k has the maximum possible number of zeros at the end, what is the min value that n can assume?
1. 256 2. 192 3. 128. 4. 64. 5. None of these

is d ans 256 ?