Quant Boosters - Maneesh - Set 4


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    n[2] = 500-200/2 +1 = 151
    n[5] = 61
    n[7] = 43
    n[10] = 31
    n[14] = 21
    n[35] = 9
    n[70] = 5

    151 + 61 + 43 - 31 - 21 - 9 + 5 = 260 - 61 = 199

    coprime to 70 --> 301 - 199 = 102


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    Q26) Find the number of trailing zeros in 80C16


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    80!/64! * 16!
    power of 2 = 2^78/2^63*2^15 => 0
    Hence 0 trailing zeros


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    Q27) 4003 students line up in a row from left to right. Starting from the left, every 5th student is given a piece of quant book ; starting from the right, every 6th student is given a va book.
    a) how many students get only quant book ?
    b) how many students get only VA book ?
    c) How many students get both the book ?
    d) how many student got none of the book ?


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    solution by sagar gupta

    5,10,15............4000 : 5a+5
    2,8,14..............3998 : 6a +2

    quant books : 3995/5 + 1 = 800
    verbal books : 3996/6 + 1 = 667

    5a+5 = 6b+2
    5a = 6b - 3
    a = b + ( b-3)/5
    b=3 : 20
    b=8 : 50
    30c+20 form : both

    20,50.....4000
    3980/30 + 1= 133

    only quant = 800 - 133 =667
    only verbal = 534
    both = 133
    none = 2669


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    Q28) 17 numbers --- 2^0,2^1,2^2,.......2^16 were written on the board.You repeatedly take two numbers on the blackboard, subtract one from the other, erase them both, and write the result of the subtraction on the blackboard. What is the largest possible number that can remain on the blackboard when there is only one number left?


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    (solution by nitin gupta sir)
    Let’s take 4 numbers , 2^0, 2^1,2^2,2^3 which are basically 1,2,4,8
    Now, step 1 : 1-2 = -1,
    Now the three numbers are -1 ,4,8
    Step 2 : -1-4 = -5
    Now the two numbers are -5 & 8
    Step 3 : 8 – (-5) = 13
    So the largest number is 13
    Similarly for 5 numbers , 2^0, 2^1,2^2,2^3 ,2^4 which are basically 1,2,4,8,16
    Step 1: (1-2), 4,8,16
    Step 2: (1-2 -4), 8,16
    Step 3: (1-2 -4-8),16
    Step 4: 16 – (1-2-4-8) = 29
    So for 2^0,2^1,2^2,.......2^16
    Ans : 2^16 – (2^0 – 2^1 – 2^2 – 2^3 -2^4 -……..2^15)
    Rearranging this : 2^16 +2^15 + 2^14 + ……+ 2^1 - 2^0
    = 2^17 – 3 = 131069


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    Q29) Consider the series : 0, 0, 2, 8, 22, 52, 114 ....
    Find the sum of tenth and eleventh term of this series. (A10 + A11)


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    General term = 2^x - 2x
    A(10) = 2^10 - 20 = 1004
    A(11) = 2^11 - 22 = 2026
    Sum = 1004 + 2026 = 3030


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    Q30) The quantity of water (in ml) needed to reduce 9 ml saving lotion containing 50% alcohol to a lotion containing 30% alcohol is


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    4.5/(9+x) = 3/10
    27 + 3x = 45
    x = 6ml



  • Here
    nf(x)= f(x^n)

    F(343sqrt3)= 3f(7)+1/2f(3)
    F(7) and f(3) can be found using given values.


 

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