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Re: [Ascent CAT] Doubt in Ratio & Proporti

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  • raju dictator
    ... so
    Message 1 of 21 , Mar 5, 2007
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      On 3/2/07, rajib samanta <rajibsam@...> wrote:

      Can anyone please solve the question given below?

      A certain city with a population of 132,000 is to be
      divided into 11 voting districts, and no district is
      to have a population that is more than 10% greater
      than the population of any other district. what is the
      minimum possible population that the least populated
      district could have?

      Answer choices are:

      10,700
      10,800
      10,900
      11,000
      11,100

      the answer is 11000 because when we take equal populations for all districts the population comes out to be 12000.so,now if you consider the option given only 11000 leaves us with a population of 121000 which has to be divided into 10 districts.so,if we take equal distribution the population of remaining district comes out to be 12100 which is exactly 10% greater than trhe min and hence just acceptable.any lower option wont satisfy this criteria                                                                                           


       


      so


       


    • rajib samanta
      a1, a2, a3, ...., a15 In the sequence shown, a[n] = a[n-1] + k, where 2
      Message 2 of 21 , Mar 14, 2007
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        a1, a2, a3, ...., a15

        In the sequence shown, a[n] = a[n-1] + k, where
        2<=n<=15 and k is a non-zero constant. How many of the
        term in the sequence are greater than 10?

        Given: a[8] = 10

        Please solve.

        Thanks,
        Rajib




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      • Arghya Basu
        On Tue, 06 Mar 2007 07:49:55 +0530, raju dictator
        Message 3 of 21 , Mar 22, 2007
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          On Tue, 06 Mar 2007 07:49:55 +0530, raju dictator <rajudictator@...>
          wrote:

          > On 3/2/07, rajib samanta <rajibsam@...> wrote:
          >>
          >> Can anyone please solve the question given below?
          >>
          >> A certain city with a population of 132,000 is to be
          >> divided into 11 voting districts, and no district is
          >> to have a population that is more than 10% greater
          >> than the population of any other district. what is the
          >> minimum possible population that the least populated
          >> district could have?
          >>
          >> Answer choices are:
          >>
          >> 10,700
          >> 10,800
          >> 10,900
          >> 11,000
          >> 11,100
          >>
          >> the answer is 11000 because when we take equal populations for all
          >> districts the population comes out to be 12000.so,now if you consider
          >> the
          >> option given only 11000 leaves us with a population of 121000 which has
          >> to
          >> be divided into 10 districts.so,if we take equal distribution the
          >> population of remaining district comes out to be 12100 which is exactly
          >> 10%
          >> greater than trhe min and hence just acceptable.any lower option wont
          >> satisfy this
          >> criteria
          >>
          >>
          >
          >
          >
          >
          > so
          >
          >
          >
          >
          >
          >>
        • srikanth viswanathan
          Since k can be either +ve or -ve we can say that 1.if k 0 ,then a9,a10,.....a15 10 2.if k 10 Either case,7 numbers are greater than
          Message 4 of 21 , Apr 9, 2007
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            Since k can be either +ve or -ve
            we can say that
            1.if k > 0 ,then a9,a10,.....a15>10
            2.if k < 0 ,then a7,a6,a5....a1>10
            Either case,7 numbers are greater than 10

            Answer : Seven


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          • basudev saha
            Solution: Given series is: a1,a2,a3,...........a15 and, a[n] = a[n-1] + k which clearly indicates that the given series is in A.P. Also given that, a8=10 so,
            Message 5 of 21 , Apr 10, 2007
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              Solution:
               
              Given series is: a1,a2,a3,...........a15
               and, a[n] = a[n-1] + k
              which clearly indicates that the given series is in A.P.
              Also given that, a8=10
              so, a8=a7+ k
              i.e  10 = a7+k ......(1)
               
              also, a9=a8+k
              i.e.  a9=10 +k........(2)
               
              Solving these two equations we get, a7 - a9 = 20
              which means the geiven series is of decreasing A.P. type
               
              From above we can figure out that vaule of k would be= -10
              and a9 = 0 , a7= 20
               
              a6=30, a5=40, a4=50, a3=60, a2=70, a1=80
               
              Hence, 7 values in the given series will be greater than 10.
               


              rajib samanta <rajibsam@...> wrote:
              a1, a2, a3, ...., a15

              In the sequence shown, a[n] = a[n-1] + k, where
              2<=n<=15 and k is a non-zero constant. How many of the
              term in the sequence are greater than 10?

              Given: a[8] = 10

              Please solve.

              Thanks,
              Rajib

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            • rajneesh sinha
              Dear Rajib, Ans - 7 Explaination - a(8) = a(7) + k or, 10 = a(7) +K 10-k = a(7) Like wise, following are value for a1 to a7 a1 = 10-7k a2 = 10- 6k....
              Message 6 of 21 , Apr 10, 2007
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                Dear Rajib,
                Ans - 7
                Explaination -
                 
                a(8) = a(7) + k
                or, 10 = a(7) +K
                10-k = a(7)
                 
                 
                Like wise, following are value for a1 to a7
                a1 = 10-7k
                a2 = 10- 6k....                       Total 7 elements a(1) to a(7)
                ......
                a(7) = 10-k
                 
                Values for a(9) to a(15) are followings -
                a(9) = 10+k
                a(10) = 10+2k...
                .....                                        Total 7 elements a(9) to a(15) 
                a(15) = 10 +7k

                 
                 
                 
                CASE 1 -Now if k is positive a(9) to a(15) will be more than 10.
                CASE 2 - If K is -ve a(1) to a(7) will be more than 10
                 
                In both cases it is 7 elements.
                 
                So for sure there are 7 elements in series more than 10.
                rajib samanta <rajibsam@...> wrote:
                a1, a2, a3, ...., a15

                In the sequence shown, a[n] = a[n-1] + k, where
                2<=n<=15 and k is a non-zero constant. How many of the
                term in the sequence are greater than 10?

                Given: a[8] = 10

                Please solve.

                Thanks,
                Rajib

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              • sathish vijayaraghavan
                Hi All I have got one question in prob solving which am facing difficulty in getting it solved. Plz help me. Problem Statement: If xy = 1 then what is the
                Message 7 of 21 , Apr 29, 2007
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                  Hi All
                   
                  I have got one question in prob solving which am facing difficulty in getting it solved. Plz help me.
                   
                  Problem Statement:
                  If xy = 1 then what is the value of
                   
                  {2* (x+y)^2} / {2* (x-y)^2}
                   
                  The answer options are:  a) 2, b) 4, c) 8, d) 16, e) 32
                   
                  Thanks,
                  Sathish
                   


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                • sathish vijayaraghavan
                  Hi All This is yet another question which also bothers me on solving it, For every positive even integer n, the function h(n) is defined to be the product of
                  Message 8 of 21 , Apr 29, 2007
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                    Hi All
                     
                     This is yet another question which also bothers me on solving it,
                     
                    For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
                     
                    a) between 2 and 10
                    b) between 10 and 20
                    c) between 20 and 30
                    d) between 30 and 40
                    e) greater than 40
                     
                    Thanks,
                    Sathish


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                  • raj kumar
                    The Function h(n) is given by h(n)+1=2^(n/2) x (n/2)!+1 So by substituting various even integer n, we can get consecutive Values .So by checking for prime
                    Message 9 of 21 , May 6, 2007
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                      The Function h(n) is given by
                                     h(n)+1=2^(n/2) x (n/2)!+1
                       So by substituting various even integer n, we can get consecutive Values .So by checking for prime numbers. I got the answer to be
                            Ans: Greater than 40
                          For the term h(100)+1 = (2^50 + 50! )+1
                       Iam not sure with the answer
                       
                      Rajkumar

                      sathish vijayaraghavan <sathish_vijayaraghavan@...> wrote:
                      Hi All
                       
                       This is yet another question which also bothers me on solving it,
                       
                      For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
                       
                      a) between 2 and 10
                      b) between 10 and 20
                      c) between 20 and 30
                      d) between 30 and 40
                      e) greater than 40
                       
                      Thanks,
                      Sathish

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                    • uday bhaskar
                      Hi All, The number in context here is 2^50*50!+1 let 2^50*50! be y so we want the divisibility of y+1 now we can clearly see that y is divisible be all the
                      Message 10 of 21 , May 9, 2007
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                        Hi All,
                        The number in context here is
                        2^50*50!+1
                        let 2^50*50! be y
                        so we want the divisibility of y+1 now we can clearly see that y is divisible be all the numbers till 50 including the prime numbers so obviously y+1 will not be divisibile by those numbers. Hence any number less than 50 will not divide y+1 so the smallest prime number has to be greater than 50, thus the option greater than 40 (e) is the answer.
                        Regards
                        uday

                        Funda: y and y+1 are coprimes, 100 n 101

                        sathish vijayaraghavan <sathish_vijayaraghavan@...> wrote:
                        Hi All
                         
                         This is yet another question which also bothers me on solving it,
                         
                        For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
                         
                        a) between 2 and 10
                        b) between 10 and 20
                        c) between 20 and 30
                        d) between 30 and 40
                        e) greater than 40
                         
                        Thanks,
                        Sathish

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                      • Kaushik Bar
                        h(100)+1 = 2*4*6*8*...*100 + 1 = 2 * (1*2*3*...*50) + 1 Look at this number carefully. The quantity inside the brackets is divisible by each of the numbers
                        Message 11 of 21 , May 14, 2007
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                          h(100)+1 = 2*4*6*8*...*100 + 1 = 2 * (1*2*3*...*50) + 1

                          Look at this number carefully. The quantity inside the brackets is divisible by each of the numbers from 2 to 50. So when you multiply it by 2, the quantity obtained [2 * (1*2*3*...*50)] will also be ivisible by each of the numbers from 2 to 50. So, the next integer, i.e., 2 * (1*2*3*...*50) + 1, cannot be divisible by any of the numbers from 2 to 50.

                          Therefore, the option e) must be the correct one.

                          Regards,
                          Kaushik

                          On 4/30/07, sathish vijayaraghavan <sathish_vijayaraghavan@...> wrote:

                          Hi All
                           
                           This is yet another question which also bothers me on solving it,
                           
                          For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
                           
                          a) between 2 and 10
                          b) between 10 and 20
                          c) between 20 and 30
                          d) between 30 and 40
                          e) greater than 40
                           
                          Thanks,
                          Sathish


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                        • rishabh jain
                          Hi, h(n) = 2.4.6.....100 + 1 = 2(50!)+1 I think the answer should be (e). Expl : any no can not divide the number obtained by adding 1 to the multiple of that
                          Message 12 of 21 , May 27, 2007
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                            Hi,

                            h(n) = 2.4.6.....100 + 1
                                  = 2(50!)+1

                            I think the answer should be (e).

                            Expl : any no can not divide the number obtained by adding 1 to the multiple of that number. e.g. since 25 is divisible by 5 so 25+1 is not divisible by 5.

                            since 2(50!) is divisible by all the no from 2 to 50 including prime no. so 2(50!)+1 can not be divisible by any prime no between 2 to 50. so the first prime no which can divide h(100) should be more than 50.



                            raj kumar <dv_rajkumar@...> wrote:
                            The Function h(n) is given by
                                           h(n)+1=2^(n/ 2) x (n/2)!+1
                             So by substituting various even integer n, we can get consecutive Values .So by checking for prime numbers. I got the answer to be
                                  Ans: Greater than 40
                                For the term h(100)+1 = (2^50 + 50! )+1
                             Iam not sure with the answer
                             
                            Rajkumar

                            sathish vijayaraghavan <sathish_vijayaragha van@yahoo. com> wrote:
                            Hi All
                             
                             This is yet another question which also bothers me on solving it,
                             
                            For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
                             
                            a) between 2 and 10
                            b) between 10 and 20
                            c) between 20 and 30
                            d) between 30 and 40
                            e) greater than 40
                             
                            Thanks,
                            Sathish

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                          • sheetal_feb
                            Dear All, Sheetal here .I want suggestion as to wat to do regarding verbal cutoff it worries me like a night mare. All possible suggestion and advices are
                            Message 13 of 21 , Jun 4, 2007
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                              Dear All,

                              Sheetal here .I want suggestion as to wat to do regarding verbal
                              cutoff it worries me like a night mare.
                              All possible suggestion and advices are welcome.
                              Completely lost .No confidence at all.

                              Regards
                              Sheetal
                            • alok mohanty
                              This is not addition that u will be taking common out here. On taking 2 common from each term it becomes 2^50. then u can follow rest of ur steps. Kaushik Bar
                              Message 14 of 21 , Jun 7, 2007
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                                This is not addition that u will be taking common out here.
                                On taking 2 common from each term it becomes 2^50.
                                then u can follow rest of ur steps.

                                Kaushik Bar <kaushikbar@...> wrote:
                                h(100)+1 = 2*4*6*8*...* 100 + 1 = 2 * (1*2*3*...*50) + 1

                                Look at this number carefully. The quantity inside the brackets is divisible by each of the numbers from 2 to 50. So when you multiply it by 2, the quantity obtained [2 * (1*2*3*...*50) ] will also be ivisible by each of the numbers from 2 to 50. So, the next integer, i.e., 2 * (1*2*3*...*50) + 1, cannot be divisible by any of the numbers from 2 to 50.

                                Therefore, the option e) must be the correct one.

                                Regards,
                                Kaushik

                                On 4/30/07, sathish vijayaraghavan <sathish_vijayaragha van@yahoo. com> wrote:
                                Hi All
                                 
                                 This is yet another question which also bothers me on solving it,
                                 
                                For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
                                 
                                a) between 2 and 10
                                b) between 10 and 20
                                c) between 20 and 30
                                d) between 30 and 40
                                e) greater than 40
                                 
                                Thanks,
                                Sathish

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                              • Tarun Aggrawal
                                Hey Sheetal 3 things can help you... 1. Start reading A LOT. It sounds cliched... and maybe even a little too late. But it should help. Make sure that
                                Message 15 of 21 , Jul 17 10:55 PM
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                                  Hey Sheetal

                                  3 things can help you...
                                  1. Start reading A LOT. It sounds cliched... and maybe even a little too late. But it should help. Make sure that everything you read, you do give it a deep introspection. Like what one line could summarize the whole idea, whats the tone of the author etc. You might want to have someone to verify all that you make out of the paragraph, but even if you don't you'll be just fine.
                                  2. Practice. Take loads of "reputed" (to avoid wrong answers) material.... and keep practicing. You'll get better with time. Many a times the problem with the verbal section is that you never read the stuff properly owing to the "time running away" factor. Take your time. Read properly. Solve. And then check your answers. Dont expect this to give you a leap in your marks. You might fair just as bad. But then when you'll analyze you would understand better. And fair better in the following tests. Do not forget to learn all the new words you encounter. Surprisingly, contrary to what many of us believe, they ARE very common words.
                                  3. Improve your reading speed. Its a byproduct of the first point so you dont really have to make a special effort for this. Just make sure theres a small sense of urgency while reading. It will help in subduing the negative effect of the "time running away" factor.
                                  Just try. And be patient. Above all, be confident. Your mindset should be - If I am not getting to solve this question, probably there's no one else either.

                                  All the best.
                                  Tarun

                                  On 6/4/07, sheetal_feb <sheetal_feb@...> wrote:

                                  Dear All,

                                  Sheetal here .I want suggestion as to wat to do regarding verbal
                                  cutoff it worries me like a night mare.
                                  All possible suggestion and advices are welcome.
                                  Completely lost .No confidence at all.

                                  Regards
                                  Sheetal


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