CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[rajeev_hts]

I got completely messed up by language of question.

I understood it now.
X-100*X/N >= 1
X>=N/N-100
N >= 27N/(N-100)
N >= 127
=> X <= 4.7

so N>=27*5 = 135

So min value of N=135. answer (e) None of these.

This was an eye wash for me, have to understand language very deeply.

[Aarav]

Originally Posted by rajeev_hts

I got completely messed up by language of question.

I understood it now.
So min value of N=135.

The answer is not this. As I told Slam earlier, you can better this number.

[slam]

Originally Posted by Aarav

The answer is not this. As I told Slam earlier, you can better this number.

Am working on it Sir, will post it as soon as I get a breakthrough.

Also, just so you know, I received the newsletter a few minutes ago.

[rajeev_hts]

Originally Posted by Aarav

The answer is not this. As I told Slam earlier, you can better this number.

It shall be 134 (c)

as for N=134

X>= 134/34 =~ 3.94

so we can take one candidate with 4 votes and 26 with 5 votes.

below than 134, X >= 4.XX

means we can not take value below 134.

[adarshiitm]

Here is my answer to #41

Suppose Ni is the number of votes ith candidate has got and the percentage of votes are Pi

We know that N1+N2+….+N27 = N
And P1 + P2+ .. + P27 = 100
And we also know that
Ni >= Pi + 1 for all i
=> N1+N2+….+N27 >=P1+P2+..+P27 + 27
=> N >= 127 ...... (1)

Again from:
Pi <=Ni-1
=>N1*100/N <=Ni-1
=>N>=(Ni*100)/(Ni-1)
Now Ni can not be 1 here
For Ni = 2, N>=200
For Ni = 3, N>=150
For Ni = 4, N>=133.33 => N >=134
For Ni = 5, N>=125 which doesn’t make any difference to our previous condition N>=127
So we can have 26 Nis equal to 5 and one Ni equal to 4 and N = 134

[zion1729]

The minimum value of n can be obtained by equating all the terms. Let each candidate get 'k' votes.

Hence,

100/27 <= k-1 => k>= 127/27 => min value of k = 5

Hence min candidates = 135.

Ans. (e) None of these

[sabsebadapaagal]

Hi Arav ,

I havent received NL today, please help me out in this.

[zion1729]

oops... I interpreted the question completely wrong.
It should be 100ki/sum(ki) <= ki-1, i=1,2,3..,27

adding, 127 >= sum(ki)

Let us assume sum(ki) = s

Since s > 100 , 400/s <= 3 => 500/s <= 4

Hence, we need to have 400/s <= 3

=> s >= 133.33...

The minimum integer s is 134 and holds for 26 5's and 1 4.

[vineet.nitd]

133/33>4=>Each candidate's vote is atleast 5 in count=>Least number of votes is 5*27=135 which is contradictory.

134/34>3.9=> Each candidate's vote is atleast 4 in count=>Least number of votes is 4*27=108 => 134 votes can be got, e.g. 26 people getting 5 votes and one person getting 4 or, 13 people getting 6 votes and 14 person getting 4

Hence the least value if N is 134

[vineet.nitd]

Many combinations can be generated by using 4a+5b+6c+7d+...=134 and a+b+c+d+...=27 => and there will be a uniques value of a coresponding to various non negative integer set(b,c,d,...)