Sponsored Ad:
See the top rated post in this thread. Click here

Page 3 of 5 FirstFirst 12345 LastLast
Results 21 to 30 of 41

Thread: challenge problems

  1. #21
    Trying to make mom and pop proud
    Join Date
    May 2009
    Posts
    9
    Rep Power
    12


    Good post? Yes | No
    Sponsored Ad:
    Quote Originally Posted by Md. Minuddin View Post
    In how many ways can the letters of the word “PROBLEM” be rearranged to make 7 letter words such that
    none of the letters repeat?

    A. 7

    B. 7C7
    C. 77
    D. 49
    E. None of these


    Answer should be E. Because the actual result should 7!

    The question says - 'none of the letters repeat'...this shuld be reg the positions(atleast from OG perspective).
    so, according to OG it is 7 instead of 7!

    1 -- PROBLEM
    2 -- ROBLEM P
    3 -- 0BLEM PR
    and so on.... which will stop after 7th combination.

    The Answer - A


    Not sure how excatly do we need to interpret this question...!!! Any inputs...?

    thank you.
    Last edited by givinggmat; 05-08-2009 at 03:00 PM.

  2. #22
    Within my grasp! Md. Minuddin's Avatar
    Join Date
    Mar 2009
    Location
    Bangladesh
    Posts
    258
    Rep Power
    13


    Good post? Yes | No
    Quote Originally Posted by givinggmat View Post
    The question says - 'none of the letters repeat'...this shuld be reg the positions(atleast from OG perspective).
    so, according to OG it is 7 instead of 7!

    1 -- PROBLEM
    2 -- ROBLEM P
    3 -- 0BLEM PR
    and so on.... which will stop after 7th combination.

    The Answer - A


    Not sure how excatly do we need to interpret this question...!!! Any inputs...?

    thank you.
    PROBLEM has 7 letter which is different from each other

    1----2---3----4---5----6---7

    Actually I think repeat means using same letter more then once. If so then for first position we have 7 choice. As we can't repeat same letter so now from the rest 6 letter we can choice second position in 6 way
    And so on,
    So total way 7*6*5*4*3*2*1=7!

  3. #23
    Within my grasp! abhishek_mumbai's Avatar
    Join Date
    Apr 2009
    Location
    Pune, India
    Posts
    202
    Rep Power
    12


    Good post? Yes | No
    How many different positive integers exist between 10^6and 10^7, the sum of whose digits is equal to 2?
    A. 6
    B. 7
    C. 5
    D. 8
    E. 18

    My answer is (A) 6
    110000
    101000
    100100
    100010
    100001
    200000

  4. #24
    Within my grasp! abhishek_mumbai's Avatar
    Join Date
    Apr 2009
    Location
    Pune, India
    Posts
    202
    Rep Power
    12


    Good post? Yes | No
    Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?
    A. 210
    B. 29
    C. 3*28
    D. 3*29
    E. None of these

    Answer should be 2^9..

  5. #25
    Within my grasp! abhishek_mumbai's Avatar
    Join Date
    Apr 2009
    Location
    Pune, India
    Posts
    202
    Rep Power
    12


    Good post? Yes | No
    A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit
    his target?
    A. 1
    B.1/256
    C.81/256
    D.175/256
    E.108/256

    1/4 * 1/4 * 1/4 * 1/4 = 1/256

  6. #26
    Within my grasp! abhishek_mumbai's Avatar
    Join Date
    Apr 2009
    Location
    Pune, India
    Posts
    202
    Rep Power
    12


    Good post? Yes | No
    What is the probability that the position in which the consonants appear remain unchanged when the letters of
    the word Math are re-arranged?
    A. 1/4
    B. 1/6
    C. 1/3
    D. 1/24
    E. 1/12

    E. 1/12

  7. #27
    Within my grasp! abhishek_mumbai's Avatar
    Join Date
    Apr 2009
    Location
    Pune, India
    Posts
    202
    Rep Power
    12


    Good post? Yes | No
    In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear
    together?
    A.6!/2!
    B. 3!*3!
    C.4!/2!
    D.4!*3!/2!
    E.3!*3!/2!

    Answer is D.4!*3!/2!

    (AAu)BCS => can be arranged in 4! ways..
    then AAU can be arranged in 3!/2! ways
    total 4!*3!/2!

  8. #28
    Within my grasp! abhishek_mumbai's Avatar
    Join Date
    Apr 2009
    Location
    Pune, India
    Posts
    202
    Rep Power
    12


    Good post? Yes | No
    How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?
    A. 59
    B.11!/3!*2!*2!*2!
    C. 56
    D. 23
    E.11!/2!*2!*2!
    before using, the number of times alphabet appear is
    m=1; e=3; r=2; a=2; n=2; d=1; i=1; t=1

    We make the word E _ _ R so number of times E and R is reduced by 1
    so m=1; e=2; r=1; a=2; n=2; d=1; i=1; t=1

    total combinations = 11!/2!*2!*2!

  9. #29
    Within my grasp! Md. Minuddin's Avatar
    Join Date
    Mar 2009
    Location
    Bangladesh
    Posts
    258
    Rep Power
    13


    Good post? Yes | No
    Quote Originally Posted by abhishek_mumbai View Post
    How many different positive integers exist between 10^6and 10^7, the sum of whose digits is equal to 2?
    A. 6
    B. 7
    C. 5
    D. 8
    E. 18

    My answer is (A) 6
    110000
    101000
    100100
    100010
    100001
    200000
    106= 1000000
    107= 10000000

    There is
    1000001
    1000010
    1000100
    1001000
    1010000
    1100000
    2000000

  10. #30
    Within my grasp!
    Join Date
    May 2009
    Posts
    207
    Rep Power
    12


    Good post? Yes | No
    How many different positive integers exist between 10^6and 10^7, the sum of whose digits is equal to 2?
    A. 6
    B. 7
    C. 5
    D. 8
    E. 18

    10^6=1000000 , 10^7= 10000000
    any integer between these two values will have 7 digits
    for a 7 digit number to have 2 as sum of its digits, there are only two possibilities :
    1) one of the digits has to be 2 and the rest 0...for the number to remain a 7 digits number, that one digit has to be the left most one..hence it should be 2000000
    2)two of the digits have to be 1 and the rest 0..for the number to remain a 7 digit number, the leftmost digit has to be 1..and of the other six positions, 1 of them have to be 1 and the rest 0..there are 6 such possibilities (6C1)

    Hence answer : 1+6 =7

Page 3 of 5 FirstFirst 12345 LastLast

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Similar Threads

  1. CR challenge -18
    By abhasjha in forum GMAT Critical Reasoning
    Replies: 7
    Last Post: 03-30-2009, 06:41 AM
  2. CR challenge -17
    By abhasjha in forum GMAT Critical Reasoning
    Replies: 3
    Last Post: 03-27-2009, 07:34 AM
  3. Challenge Problems
    By espyn in forum GMAT Data Sufficiency
    Replies: 11
    Last Post: 11-23-2007, 08:46 PM
  4. Challenge Problems
    By espyn in forum GMAT Problem Solving
    Replies: 9
    Last Post: 11-22-2007, 12:48 AM
  5. Another Challenge
    By Chrissie in forum GMAT Problem Solving
    Replies: 1
    Last Post: 10-11-2003, 05:41 PM

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •