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Re: [Ascent CAT] Doubt in Ratio & Proportio

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  • Ruli Roy
    x,y,z are in continued proportion implies........ x/y=y/z *hence*, xz=y^2 (cross multiplication) xz/x^2=y^2/x^2 (dividing both sides of equation
    Message 1 of 21 , Feb 19, 2007
      x,y,z are in continued proportion implies........
                                x/y=y/z
       
      hence, xz=y^2        (cross multiplication)
       
      xz/x^2=y^2/x^2        (dividing both sides of equation by x^2)
       
      z/x=y^2/x^2
       
      thatz it,
                 z:x=y^2:x^2
      answer is a)z:x=y^2:x^2

      THANX..

       



       
      On 2/13/07, Ganesan MahendraBabu <babumathi246@...> wrote:

      Hi all,

      I have a doubt in the follow question. can anyone solve this equation and tell me the logic

      1. There are 3 unequal quantities x,y&z in continued proportion. Which of the following equals   z:x?

      ans a)y^2:x^2  b)(z^2-y^2)/(y^2-x^2)  c) z^2:y^2  d) all of these

                        THANKS IN ADVANCE

      Regards
      Ganesan m


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    • rajib samanta
      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
      Message 2 of 21 , Mar 1, 2007
        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

        Thanks,
        Rajib

        --- raju dictator <rajudictator@...> wrote:

        > have a doubt in the follow question. can anyone
        > solve this equation and
        > tell me the logic
        >
        > 1. There are 3 unequal quantities x,y&z in continued
        > proportion. Which of
        > the following equals z:x?
        > ans a)y^2:x^2 b)(z^2-y^2)/(y^2-x^2) c) z^2:y^2 d)
        > all of these
        >
        >
        > well the ans to this question is quite easy
        > now,we know that x,y and z are i cont.prop
        > so,x/y=y/z
        > now,z:x=z/x=(z/y)\(x/y)=(y/x)\(x/y)=y^2/x^2=z^2/y^2
        > also as y^2/x^2=z^2/y^2=z^2-y^2)/(y^2-x^2) (refer
        > any standard book on
        > ratios for the proof of this prop.)
        > so,ans is (D)
        >




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      • nitin sharma
        let x = minimum possible population other districts max possible population = 1.1x so 10 other ditricts pop = 10*1.1x = 11x total population = x+11x = 132000
        Message 3 of 21 , Mar 5, 2007
          let x = minimum possible population
           
          other districts max possible population  = 1.1x
          so 10 other ditricts pop = 10*1.1x = 11x
          total population = x+11x = 132000
          x= 11000

          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

          Thanks,
          Rajib

          --- raju dictator wrote:

          > have a doubt in the follow question. can anyone
          > solve this equation and
          > tell me the logic
          >
          > 1. There are 3 unequal quantities x,y&z in continued
          > proportion. Which of
          > the following equals z:x?
          > ans a)y^2:x^2 b)(z^2-y^2)/(y^2-x^2) c) z^2:y^2 d)
          > all of these
          >
          >
          > well the ans to this question is quite easy
          > now,we know that x,y and z are i cont.prop
          > so,x/y=y/z
          > now,z:x=z/x=(z/y)\(x/y)=(y/x)\(x/y)=y^2/x^2=z^2/y^2
          > also as y^2/x^2=z^2/y^2=z^2-y^2)/(y^2-x^2) (refer
          > any standard book on
          > ratios for the proof of this prop.)
          > so,ans is (D)
          >




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        • pvvd29
          ... Solution: answer is 11,000 say least possible population is x, which is possible when all other districts hav max. possible population, that is, just 10%
          Message 4 of 21 , Mar 5, 2007
            --- In ascent4cat@yahoogroups.com, 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
            >
            > Thanks,
            > Rajib
            >

            Solution:
            answer is 11,000

            say least possible population is x,
            which is possible when all other districts hav max. possible
            population, that is, just 10% more thn x, i.e 1.1x
            so, total population= 1.1x*10+x = 132,000
            =>12*x = 132,000;
            =>x = 11,000
            which is the answer.
          • raju dictator
            ... so
            Message 5 of 21 , Mar 5, 2007
              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 6 of 21 , Mar 14, 2007
                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 7 of 21 , Mar 22, 2007
                  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 8 of 21 , Apr 9, 2007
                    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 9 of 21 , Apr 10, 2007
                      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 10 of 21 , Apr 10, 2007
                        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 11 of 21 , Apr 29, 2007
                          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 12 of 21 , Apr 29, 2007
                            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 13 of 21 , May 6, 2007
                              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 14 of 21 , May 9, 2007
                                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 15 of 21 , May 14, 2007
                                  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 16 of 21 , May 27, 2007
                                    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 17 of 21 , Jun 4, 2007
                                      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 18 of 21 , Jun 7, 2007
                                        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 19 of 21 , Jul 17, 2007
                                          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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