Home | About Indiatry | Login | Register

    • ( 1 ) How many words can be formed from the letters of the word "AFTER", so that the vowels never comes together.

    • 1) 48
    • 2) 52
    • 3) 72
    • 4) 120
    • Discussion in forum
      Answer : 3) 72
      Solution : We need to find the ways that vowels NEVER come together.
      Vowels are A, E
      Let the word be FTR(AE) having 4 words.
      Total ways = 4! = 24
      Vowels can have total ways 2! = 2
      Number of ways having vowel together = 48
      Total number of words using all letter = 5! = 120 Number of words having vowels never together = 120-48 = 72








      discussion


      Answer : 3) 72

    • ( 2 ) Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side.

    • 1) 864
    • 2) 863
    • 3) 865
    • 4) 1728
    • Discussion in forum
      Answer : 4) 1728
      Solution : Required number of ways = 3C2 * 4! * 4! = 1728








      discussion


      Answer : 4) 1728

    • ( 3 ) There are 12 yes or no questions. How many ways can these be answered?

    • 1) 4096
    • 2) 2048
    • 3) 1024
    • 4) 144
    • Discussion in forum
      Answer : 1) 4096
      Solution : Each of the questions can be answered in 2 ways (yes or no)
      Therefore, no. of ways of answering 12 questions =212 = 4096 ways.








      discussion


      Answer : 1) 4096

    • ( 4 ) Four dice are rolled simultaneously. What is the number of possible outcomes in which at least one of the die shows 6?

    • 1) 6! / 4!
    • 2) 625
    • 3) 671
    • 4) 1296
    • Discussion in forum
      Answer : 3) 671
      Solution : When 4 dice are rolled simultaneously, there are 64 = 1296 outcomes. The converse of what is asked in the question is that none of the dice show '6'. That is all four dice show any of the other 5 numbers. That is possible in 54 = 625 outcomes.
      Therefore, in 1296 - 625 = 671 outcomes at least one of the dice will show 6.








      discussion


      Answer : 3) 671

    • ( 5 ) How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?

    • 1) 26
    • 2) 50
    • 3) 100
    • 4) 1000
    • Discussion in forum
      Answer : 3) 100
      Solution : 1 million distinct 3 digit initials are needed.
      Let the number of required alphabets in the language be 'n'.
      Therefore, using 'n' alphabets we can form n * n * n = n3 distinct 3 digit initials.
      Note distinct initials is different from initials where the digits are different.
      For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
      This n3 different initials = 1 million
      i.e. n3 = 106 (1 million = 106)
      => n3 = (102)3
      => n = 102 = 100
      Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.








      discussion


      Answer : 3) 100

    • ( 6 ) There are three places P, Q and R such that 3 roads connects P and Q and 4 roads connects Q and R. In how many ways can one travel from P to R?

    • 1) 8
    • 2) 10
    • 3) 12
    • 4) 14
    • Discussion in forum
      Answer : 3) 12
      Solution : The number of ways in which one can travel from P to R = 3 * 4 = 12








      discussion


      Answer : 3) 12

    • ( 7 ) Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?

    • 1) 210
    • 2) 1050
    • 3) 25200
    • 4) 21400
    • Discussion in forum
      Answer : 3) 25200
      Solution : Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
      = (7C3 * 4C2) = (7*6*5/3*2*1*4*3/2*1) = 210.
      Number of groups, each having 3 consonants and 2 vowels = 210.
      Each group contains 5 letters.
      Number of ways of arranging 5 letters among themselves
      = 5 ! = (5 * 4 * 3 * 2 * 1) = 120.
      ∴ Required number of words = (210 * 120) = 25200.








      discussion


      Answer : 3) 25200

    • ( 8 ) Find out the number of ways in which 6 rings of different types can be worn in 3 fingers?

    • 1) 120
    • 2) 720
    • 3) 125
    • 4) 729
    • Discussion in forum
      Answer : 4) 729
      Solution : The first ring can be worn in any of the 3 fingers
      => There are 3 ways of wearing the first ring
      Similarly each of the remaining 5 rings also can be worn in 3 ways
      Hence total number of ways
      = 3 * 3 * 3 * 3 * 3 * 6 = 36 =729








      discussion


      Answer : 4) 729

    • ( 9 ) 25 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?

    • 1) None of these
    • 2) 600
    • 3) 576
    • 4) 625
    • Discussion in forum
      Answer : 2) 600
      Solution : He can go in any bus out of the 25 buses.
      Hence He can go in 25 ways.
      Since he can not come back in the same bus that he used for travelling,
      He can return in 24 ways.
      Total number of ways = 25 * 24 = 600








      discussion


      Answer : 2) 600

    • ( 10 ) An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?

    • 1) 100
    • 2) 80
    • 3) 110
    • 4) 64
    • Discussion in forum
      Answer : 2) 80
      Solution : He has has 10 patterns of chairs and 8 patterns of tables
      Hence, A chair can be arranged in 10 ways and A table can be arranged in 8 ways
      Hence one chair and one table can be arranged in 10 * 8 ways = 80 ways








      discussion


      Answer : 2) 80





Top