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  • in reply to: parcels. #3050
    administration
    Keymaster

      The sum of lowest 2 weight = 94 and the sum of highest 2 weights =104

      Taking Option a) 40
      It has to be the lowest weight.
      so,  2nd lowest = 54
      but sum of highest two = 104

      So,  One of them will be less than or equal to 52 (not possible as both of them should be greater than 54….greater than the 2nd lowest)
      b) 45
      It has to be the lowest weight.
      2nd lowest = 49
      Highest = 101 – 49 = 52
      => 3rd lowest = 104 – 52 = 52

      c) 48
      It has to be the 2nd lowest
      => Lowest = 46
      => Highest = 101 – 48 = 53
      => 3rd lowest = 104 – 53 = 51
      But 51 + 48 = 99 (not possible, as only possibilities are 94, 97, 101, 104)

      d) 53
      We will get the above values (so not possible)

      So, 45 is the only possibility

      in reply to: possible values #3041
      administration
      Keymaster

        (x2 + 4x + 5) / (x + 2) = (x2 + 4x +4 + 1) / (x + 2) = [(x+2)2 +1]/(x+2) = (X+2) +1/(x+2)

         

        Since y + 1/y >=2 or <=-2 for y=/= 0

        so (x+2) +1/(x+2)  >=2 or <=-2 for x=/= -2 Hence -1 is the correct answer

        in reply to: Q1™ – The official Quant Race #3036
        administration
        Keymaster

          The correct answers are … Q1)  b i.e. 500 Q2) d i.e. 3360 Q3) c i.e. 8400 Q4) d i.e. 34 🙂 @cattack

          in reply to: Q1™ – The official Quant Race #3029
          administration
          Keymaster

            Yup @magnumopus22605 shadab… this one is also correct … 🙂 Ans is 34… 2 more to go 🙂

            in reply to: Q1™ – The official Quant Race #3026
            administration
            Keymaster

              ok right answer @magnumopus22605 shadab :)…. carry on guys/gals Q2, Q3, Q4 … go , go , go …

              in reply to: Q1™ – The official Quant Race #3023
              administration
              Keymaster

                So let me give a green signal to the Race with these 4 questions 🙂

                 

                Q1. A candidate must secure 40% of the total aggregate to pass an examination in which each question has equal marks. He attempts only 8 questions out of every 10 questions. On 75% of his answers, he secures only 26% of the marks and on the remaining of the answers, 100% of the marks. Thus he gets 22 marks, too few to pass. The maximum number of marks is
                a) 425  b) 500  c) 380  d) 620

                 

                Q2.   In how many ways we can form committee of 7 members out of 6 boys nd 4 girls if at least one of the president and the secretary is female?

                a) 768   b) 812 c) 1680 d) 3360  e) None

                 

                Q3. One plans to prepare schedule for 7 day period in which he shall play exactly one sport out of A,B,C,D on each day,number of ways in which a schedule can be prepared such that each sport is played on atleast one day in the 7 day period is ?
                a) 4200  b) 6300   c) 8400   d) 53760   e) None

                 

                Q4.  All 4 digit no. using digits 1,2,3,4,5 once are taken and their product is found How many zeroz at the end will such a product have ?
                a)18  b) 24  c) 30 d)34

                in reply to: Find the value #3022
                administration
                Keymaster

                  a4– 62a2 + 1 = 0

                  Dividing by a2 we get
                  => a2 + 1/a2 = 62
                  => (a + 1/a) = 8

                  a3 + 1/a3 = (a + 1/a)(a2 – 1 + 1/a2) = 8x(62-1) = 488 Answer

                   

                  in reply to: Locomotive speed #3020
                  administration
                  Keymaster

                    We need to minimize E/s

                    E/s = s^2 – 20s + 124 = (s – 10)^2 + 24
                    So, s = 10

                    => E/s = 24

                    => d = 72*10^5/24 = 300 km

                     

                    soln by Hemant Yadav

                    in reply to: shortage of tubelight, bulb and fan #3019
                    administration
                    Keymaster

                      Let there are 100 houses
                      so number of items required = 300
                      number of item missing in whole village = 67+83+73=223
                      19 houses not have any 1 of these so
                      ->100-19=81 houses should not have 2 or 3 item in there house
                      -> 223-19=204 item missing in this 81 house
                      let X number of house have 2 item missing so 81-X houses have 3 item missing
                      so-> 2X + 3(81-X)=204
                      X=39
                      81-x= 42 houses not have any of these item ..and 39 have 2 items missing

                      in reply to: function problem #3018
                      administration
                      Keymaster

                        g(k) = g(k-1)+f(k) = [g(k-2)+f(k-1)] +f(k)= [g(k-3)+f(k-2)]+f(k-1)+f(k) = [g(4)+f(5)] + f(6)+….+f(k)

                         

                        so g(k)- g(4) = f(k)+ …+f(5);

                        g(10)-g(8) = f(10)+f(9)

                        f(1)=1, f(2)=2,

                        f(3)=6, f(4)=10,

                        f(5)=15, f(6)=21,

                        f(7)=28,f(8)=36,

                        f(9)=45, f(10)=55.

                        so g(10)-g(8) = 100,

                         

                        now f(5)+f(6) +… f(k)= 100 so k =8.

                        in reply to: last two non-zero digit #3017
                        administration
                        Keymaster

                          yeah acc to options it shud be 64 !! going by last 2-digits will be very lengthy

                          see 36! have 8 zeros and 24! will have 4

                          so wen we subtract fr last 2 non zerodigit we’ll have
                          xxx00- last 2 non zero digit of 24!

                          now last non zerodigit of 24! is 6
                          so ans has to be in form of x4 using options 64!! (but yeah risky!! as none of dese is dere) u cn find last two digits also bt a bit lengthy

                          answer by Yatish Jain

                          in reply to: Quadratic equation #3016
                          administration
                          Keymaster

                            putting x=3
                            9a+3b+c=0
                            from f(5)=-3f(2)
                            25a+5b+c=-12a-6b-3c
                            –>37a+11b+4c=0
                            als o 36a+12b+4c=0 from f(3)=0 multiplied by 4
                            –>a-b=0
                            –>from f(3)=0 we have 12a=-c
                            –>ax^2+ax-12a=0
                            –>x^2+x-12=0
                            –>2nd root=-4

                            in reply to: sum of terms #3015
                            administration
                            Keymaster

                              x4 + 1/x4 = [x2+1/x2]^2 – 2 (x2)(1/x2) = [(x-1/x)2 +2x(1/x)]2 -2 = [22 +2]2 -2 = 34

                              in reply to: arithematic progression #3014
                              administration
                              Keymaster

                                For an AP, Tn=a+(n-1)d
                                so its clear that, Tn=a(mod d)
                                Considering above question, Tn=a(mod 9)
                                So if we find  the remainder of  (Tn/9), we get the first term
                                66541=4(mod 9)
                                48768=(43)2922 x 42=1×16(mod 9)=7(mod 9)
                                Hence a = 7 Ans.

                                in reply to: grandmother’s age #3013
                                administration
                                Keymaster

                                  Let grandmother has x sons

                                  so each son has x-1 brothers and hence x-1 grandsons.

                                  so total number of grandsons = x(x-1)

                                   

                                  also since grandmother’s age is in between 50-60 .

                                  so 50< x(x-1) <60 which is satisfied by x=8  and no other natural number
                                  Hence grandmother’s age = 8×7 = 56

                                Viewing 15 posts - 31 through 45 (of 53 total)