official quant thread for cat08

[romy4allcse]

4) Let N=(3*5*7*...*99)/(2*4*6*...*100)
which one is true regarding the values of N?
a) 1/3<N<1/2
b) 1/5<N<1/4
c) 1/15<N<1/10
d) 1/10<N<1/5

soln: N = (1*3*5*7..........*97*99)/(2*4*6*...*100)
= (1/2)*(3/4)*(5/6)*............*(97/9*(99/100)

here we can see that lowest value is 0.5 and after that the value of the factors is increasing n reaches to 0.99
and the multiplication value gets lower and lower so it vl be less than 0.5 i.e 1/2 for sure
compute it for 3-4 factores
value goes down so as i m not able to find the exact soln
hence by seeing the variation i vl go with my "Tukka"
option(c)

wat is the ans.??

[masoom]

Originally Posted by hi.shivani

Masoom your schedule is good but there is one flaw in it..it has been planned on daily basis which means a &quot;great risk factor&quot;as most of us are working.
What I want to say is that suppose you have planned to have HM on 23rd and suddenly we have an urgent work at office which need our immediate response and which cannot be delayed,so obviously we wont be able to be active on PG and properly involve in that day's discussion which totally deprived us of the whole topic at one shot.

So I am just modifying your schedule after consulting with shivam.
Please let me know if everyone agrees with it or not.

As you wish....

but there is nothing mentioned against 12 sep..ie tomorrow..

btw i am preparing a questionaire on mixture and allegations for tom..

some might pitch in with other topics also..

[HYgoldking]

Originally Posted by dare2

one more with a bit higher difficulty level.

At a movie theater, the manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout a 365 day year, what position in line gives you the best chance of being the first duplicate birthday?

Answer: 366 ?

[poojadatta]

1) value of (0.16)^(log2.5(1/3+(1/3)^2+(1/3)^3.............infiniti))
1)3 2)4 3)0.4 4)none of this

2)Find the sum of 6+66+666.........n terms
1)( 6(10^(n+1)-10))/81
2)(6/81)*((10^(n+1))-9n)
3)(6/81)*((10^9)-9n-10)
4)(6/81)*((10^(n+1))-10-9n)

3) betwen 7/2 and -83/2,17 arithmetic means are inserted. Find the sixth aithmetic mean.
1) -23/2 2) -47/3 3)-45/2 4)none of these

Answer are in bold... please post ur approach.

@shivani, masoom
Good scedule...Keep up the good work
A query but what if someone like me has a query on topic other than scedule topic of the day?

[hi.shivani]

Originally Posted by romy4allcse

4) Let N=(3*5*7*...*99)/(2*4*6*...*100)
which one is true regarding the values of N?
a) 1/3<N<1/2
b) 1/5<N<1/4
c) 1/15<N<1/10
d) 1/10<N<1/5

soln: N = (1*3*5*7..........*97*99)/(2*4*6*...*100)
= (1/2)*(3/4)*(5/6)*............*(97/9*(99/100)

here we can see that lowest value is 0.5 and after that the value of the factors is increasing n reaches to 0.99
and the multiplication value gets lower and lower so it vl be less than 0.5 i.e 1/2 for sure
compute it for 3-4 factores
value goes down so as i m not able to find the exact soln
hence by seeing the variation i vl go with my "Tukka"
option(c)

wat is the ans.??

The answer is (c) itself..

[hi.shivani]

Originally Posted by masoom

As you wish....

but there is nothing mentioned against 12 sep..ie tomorrow..

btw i am preparing a questionaire on mixture and allegations for tom..

some might pitch in with other topics also..

12th ie,2morrow we keep it for Mixture and Alligation and then continue from the nxt day as per the schedule..

Now is it okay with you?? This will be followed with everyone's consent only..i m not insisting anybody to follow it...it was just a proposed schedule..

[dare2]

Originally Posted by romy4allcse

4) Let N=(3*5*7*...*99)/(2*4*6*...*100)
which one is true regarding the values of N?
a) 1/3<N<1/2
b) 1/5<N<1/4
c) 1/15<N<1/10
d) 1/10<N<1/5

soln: N = (1*3*5*7..........*97*99)/(2*4*6*...*100)
= (1/2)*(3/4)*(5/6)*............*(97/9*(99/100)

here we can see that lowest value is 0.5 and after that the value of the factors is increasing n reaches to 0.99
and the multiplication value gets lower and lower so it vl be less than 0.5 i.e 1/2 for sure
compute it for 3-4 factores
value goes down so as i m not able to find the exact soln
hence by seeing the variation i vl go with my "Tukka"
option(c)

wat is the ans.??

@shivam.
post the answer with the solution.

[hi.shivani]

Originally Posted by poojadatta

1) value of (0.16)^(log2.5(1/3+(1/3)^2+(1/3)^3.............infiniti))
1)3 2)4 3)0.4 4)none of this

2)Find the sum of 6+66+666.........n terms
1)( 6(10^(n+1)-10))/81
2)(6/81)*((10^(n+1))-9n)
3)(6/81)*((10^9)-9n-10)
4)(6/81)*((10^(n+1))-10-9n)

3) betwen 7/2 and -83/2,17 arithmetic means are inserted. Find the sixth aithmetic mean.
1) -23/2 2) -47/3 3)-45/2 4)none of these

Answer are in bold... please post ur approach.

@shivani, masoom
Good scedule...Keep up the good work
A query but what if someone like me has a query on topic other than scedule topic of the day?

that will be always entertain..

[deepakraam]

Here is huv I worked this

N = (1*3*5......*99)/(2*4*6*...*100)

Taking common 2 out frm the Denominator,we get

N= (1*3*5....*99) / 2 (1*2*3....*50).---(1)

All the odd terms in the denominator will get cancelled with the numerator.
So we get

N = (51*53*....*99) / 2 (2*4*6*...*50) ---(2)

Taking 2 again from the denominator,we get

N= (51*53*...*99) / 4(1*2*3*...*25)

So this effectively translates to

N= 0.25 (51*53...*99) / (1*2*3....*25) --(3)

N > 0.25 since the terms will yield value more than 1.So N >0.25.Looking at the options I will go with option 1)1/3<N<1/2

-Deepak.

[raghav507]

Originally Posted by poojadatta

2)Find the sum of 6+66+666.........n terms
1)( 6(10^(n+1)-10))/81
2)(6/81)*((10^(n+1))-9n)
3)(6/81)*((10^9)-9n-10)
4)(6/81)*((10^(n+1))-10-9n)

hi,
this can be approached in this way...
S = 6+66+666+6666+....+n terms...
= 6(1+11+111+1111+...)
= 6/9(9+99+999+9999+....)
= 6/9(10^1 - 1 + 10^2 - 1 + 10^3 - 1 + .....)
= 6/9 [(10+10^2+10^3+10^4+....)-(n)]

summing this GP we can get the answer as shown in bold...