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Thread: Quant Thread For CAT 2011

  1. #201
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    Quote Originally Posted by quixotic_me View Post
    A number of 6-digit numbers that can be made with the digits 0,1,2,3,4and 5 so that even digits occupy odd places,is
    a)24
    b)36
    c)48
    d)none of these
    24 is the answer....

    Even digits are 0,2 and 4. 0 cannot come into the first place , so there are two options for first digit that is 2 and 4 , so 2 ways , similarly for 3rd digit 2 ways and for 5th digit only 1 way.

    So 2*2*1= 4 ways

    now remaining can be filled in 3*2*1 = 6

    total 4*6 =24 ways

    so option (a) is the asnwer
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    Quote Originally Posted by quixotic_me View Post
    Let A be the set of 4-digit numbers a1a2a3a4 where a1>a2>a3>a4 then how many values of A are possible?
    a)126
    b)84
    c)210
    d)none of these

    There are 10 digits {0,1,2,...,9}
    From these any 4 can be selected in 10C4 ways.

    Now as in every selection of 4 digits all are distinct,they can be arranged in descending order from R to L in only 1 way.

    So 10C4 * 1 = 210

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    anjalig (16-Apr-11), maddy9021 (17-Jul-11), quixotic_me (16-Apr-11), sumit2goody (18-Apr-11)

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    Default Q24

    The total number of integral solutions for (x,y,z) such that xyz=24 is
    a)36
    b)90
    c)120
    d)none of these

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    Default Topic surds

    Hello Friends ...could u pls help me ...i have a problem related to surds & the query is:

    If x=2+۟√3+√5 then find the value of:

    x4-8x3+8x2+32x...?

    what the shortest trick to solve this query??

    pls give ur advice.....

    thanx 4 being with me...

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    Quote Originally Posted by quixotic_me View Post
    The total number of integral solutions for (x,y,z) such that xyz=24 is
    a)36
    b)90
    c)120
    d)none of these

    Now for x * y *z = 24

    We can have the abs values of (x,y,z) such that

    Case1 : 2 are same and 3rd one distinct
    1 * 1* 24
    2 * 2 * 6
    Arranged in 3!/2! = 3 ways each
    So there are 3*2=6 ways.

    Case2 : all 3 distinct
    1*2*12
    1*3*8
    1*4*6
    2*3*4
    As each can be arranged in 3! ways, there are 4*3! = 24 ways

    But as for all cases we can have any 2 out of x,y,z as -ve and all 3 +ve, each case has (3C2 +1 ) = 4 ways

    So total = (6+24)*4 = 120
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    Quote Originally Posted by Rajasekaran View Post
    Q21:There are 1 to 500 questions..
    In round 1,a student attempts Q1.
    In round 2,he attempts(or checks) Q2,Q4,Q6.....
    In round 3,he attempts(or checks) Q3,Q6,Q9....
    Similarly for all the rounds.
    What will be the no. of unattempted questions after round 90?

    a)0
    b)1
    c)4
    d)6
    option c) 4. is it correct?

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    Default Q 26

    Let S be the set of five-digit numbers formed by digits 1,2,3,4 and 5 using each digit exactly once such that exactly two odd positions are occupied by odd digits.what is the sum of the digits in the rightmost position of the numbers in S?
    a)228
    b)216
    c)294
    d)192

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    Quote Originally Posted by quixotic_me View Post
    Let S be the set of five-digit numbers formed by digits 1,2,3,4 and 5 using each digit exactly once such that exactly two odd positions are occupied by odd digits.what is the sum of the digits in the rightmost position of the numbers in S?
    a)228
    b)216
    c)294
    d)192
    op b)216
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    Quote Originally Posted by shaunak_87 View Post
    op b)216
    As always.. u r right ! how did u get it ?

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    _ _ _ _ _
    A B C D E

    Case 1: E has a even no. and A and C have odd

    E can be filled in 2 ways(2 or 4)
    A & C have to be filled by odd nos.{1,3,5} in 3P2 = 6 ways.
    remaining B & D can be filled in 2*1 ways

    So there are 6*2*1 = 12 nos. each with unit's digit 2 or 4

    So sum of unit place digit is 12*(2+4)=72


    Case 2: E has odd no. and A has odd no. and C has even

    E can have 1/3/5
    For each selection of E, A can each be filled in 2 ways(with 2 remaining odd nos.) and thus C has to be an even no.(2 or 4) in 2 ways.
    Remaining places can be filled in 2*1 ways
    So there are 2*2*2*1 = 8 nos. having 1/3/5 at unit place

    As another case can be E has odd no. and A has odd no. and C has even,the above case repeats
    and so sum of unit place is 2*8*(1+3+5) = 144

    So total sum = 72+144 = 216
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    Q27)

    One of the most concept clearing question for number system , want the newbies to try this ..so shaunak plz dont post solution first

    find the remainder when 128^1000 is divided by 153?
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    Quote Originally Posted by sumit2goody View Post
    Q27)

    One of the most concept clearing question for number system , want the newbies to try this ..so shaunak plz dont post solution first

    find the remainder when 128^1000 is divided by 153?
    Ok Sumit,

    I had PMed u....Please do check.

    And kindly get the DILR thread up and running, also the Quant thread

    I request Sumit, Tushar to take up initiative and speed up the QA/DI/VA threads even with qstns from past 2 yrs threads....plz, i request this to both of u....maybe this will be the only means of preps.....
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    Quote Originally Posted by shaunak_87 View Post
    Ok Sumit,

    I had PMed u....Please do check.

    And kindly get the DILR thread up and running, also the Quant thread

    I request Sumit, Tushar to take up initiative and speed up the QA/DI/VA threads even with qstns from past 2 yrs threads....plz, i request this to both of u....maybe this will be the only means of preps.....
    will try to do so..difficult to access net while in internships and will reply to ur PM
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    Quote Originally Posted by sumit2goody View Post
    Q27)

    One of the most concept clearing question for number system , want the newbies to try this ..so shaunak plz dont post solution first

    find the remainder when 128^1000 is divided by 153?
    The remainder is 52

  18. #215
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    Quote Originally Posted by anjalig View Post
    The remainder is 52
    method please
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    Quote Originally Posted by anjalig View Post
    The remainder is 52
    Yes, Please do share some good probs, i request u.....please...
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    Quote Originally Posted by shaunak_87 View Post
    Yes, Please do share some good probs, i request u.....please...
    ya sure ....

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    Ques.28 Find the highest natural number N,less than 400,such that N can be written as sum of consecutive natural numbers in 11 ways but can't be written as sum of 11 consecutive natural numbers.

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    Ques 29. P and Q are two natural numbers,both greater than 1 such that
    P * Q = 5! . If Q can be represented as xy where (x+y) is minimum and P is not a multiple of x, what is the value of P ?

    A. 10
    B. 30
    C. 40
    D. 60

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    Ques. 30 In how many zeroes does 2002 ! / (1001 !)^2 end in ?
    A. 0
    B. 1
    C. 2
    D. 3

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