Quant Boosters - Vikas Saini - Set 4



  • Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? - Yes
    Source -



  • Q1) If f(x) = 4^x / (4^x + 2) then find the value of f(1/100) + f(2/100) ... f(99/100).



  • f (x) = 4^x / (4^x + 2)
    f (1 – x) = 4^(1 – x) / [4^(1 – x ) + 2]
    = 4 / (4 + 2 * 4^x)
    = 2 / (2 + 4^x)

    f(x) + f(1 – x) = [4^x/(4^x + 2)] + [ 2/(4^x + 2)] = 1
    f(1/100) + f(1 – 1/100) = f(1) + f(99/100) = 1
    f (2/100) + f( 1 – 2/100) = f(2) + f(98/100) = 1
    Similarly f(49/100) + f(50/100) = 1
    Hence f(1/100) + f(2/100) + f(3/100) ... f(99/100) = 49



  • Q2) The minimum value of ax^2 + bx + c is 7/8 at x = 5/4. Find the value of expression at x = 5, if the value of the expression at x = 1 is 1



  • Let f(x) = ax^2 + bx + c
    f ’(x) = 2ax + b.
    2ax + b = 0.
    x = -b / 2a

    At x = 5/4
    b = -5, a = 2.
    At x = 1,
    a + b + c = 1.
    2 – 5 + c = 1.
    c = 4.

    f(x) = 2x^2 - 5x + 4.
    f(5) = 2(5)^2 – 5(5) + 4 = 29



  • Q3) If f(x) = x^4 + x^3 + x^2 + x + 1, where x is a positive integer greater than 1. What will be the remainder if f(x^5) is divided by f(x) ?



  • f(x) = x^4 + x^3 + x^2 + x + 1.
    f(x^5) = x^20 + x^15 + x^10 + x^5 + 1.
    Let’s take x = 2.
    f (x) = f(2) = 31.
    f (x^5) = f(32) = 2^20 + 2^15 + 2^10 + 2^5 + 1
    f(32) mod f(2) = (2^20 + 2^15 + 2^10 + 2^5 + 1) mod 31
    = (2^5)^4 + (2^5)^3 + (2^5)^2 + (2^5) + 1 mod 31
    = 1 + 1 + 1 + 1 + 1
    = 5



  • Q4) If x is real, then find smallest value of the expression 3x^2 – 4x + 7



  • Suppose f(x) = 3x^2 - 4x + 7
    f ’(x) = 6x – 4
    6x – 4 = 0
    x = 2/3.
    f(2/3) = 17/3.



  • Q5) Find the least number which has the highest power of 7 as 52



  • We know highest power comes by this formula.
    Highest power = [n/p]+[n/p^2]+[n/p^3]……..
    52 = [n/7] + [n/7^2] + [n/7^3]
    The only things strike in our mind to split 52.
    1 + 7 + 49 (not equal to 52)
    6 + 42 (not equal)
    7 + 49 (not equal)
    6 + 42 < 52 < 7 + 49
    6 + 43 < 52
    6 + 46 = 52.
    [n/7] = 46
    n > 7 * 46.
    n > 322.
    [n/49] = 6.
    n > 49 * 6.
    n > 294.
    For least possible number n = 322.



  • Q6) Find the least number which has highest power of 13 as 52.



  • 52 = [n/13] + [n/13^2]
    52 = [ n/13] + [n/169]
    Let’s think , how 52 can be split.
    3 + 39 < 52 < 4 + 52
    3 + 49 = 52.
    [n/13 ] = 49.
    n > 13 * 49
    n > 637.
    [n/169] = 3
    n = 169 * 3
    n > 507.
    For least value, n = 637.



  • Q7) Find highest power of 72 in 100!



  • 72 = 2^3 x 3^2.
    Highest power of 2 = [100/2] + [100/2^2] + [100/2^3] + [100/2^4] + [100/2^5] + [100/2^6]
    = 50 + 25 + 12 + 6 + 3 + 1
    = 97.

    But power of 2 is 3 here, so 97/3 = 32.
    Highest power of 3 = [100/3] + [100/3^2] + [100/3^3] + [ 100/3^4]
    = 33 + 11 + 3 + 1
    = 48.

    But power of 3 is 2 here, so 48/2 = 24.
    24 < 32.
    Power of 72 in 100! Is 24.



  • Q8) Find the highest power of 24 in 50!



  • 24 = 2^3 x 3.
    Power of 2 in 50! = [50/2] + [50/2^2] + [50/2^3] +[50/2^4]+[50/2^5]
    = 25+12+6+3+1
    = 47.
    Power of 2 is 3 here.
    47 / 3 = 15.

    Power of 3 in 50!
    [50/3]+[50/3^2]+[3^3]
    = 16+5+1
    = 21.

    15 < 21.
    Power of 24 in 50! is 15.



  • Q9) Find the total no of divisors of 17!



  • Prime factors are 2,3,5,7,11,13,17.
    Highest power of 2 = [17/2] + [ 17/2^2] + [ 17/2^3] + [17/2^4]
    8+4+2+1 = 17.
    Highest power of 3 = [17/3] + [17/3^2]
    5+1 = 6.
    Highest power of 5 = [17/5] + [17/5^2] = 3.
    Highest power of 7 = [17/7] + [17/7^2] = 2.
    Highest power of 11 = [17/11] = 1.
    Highest power of 13 = [17/13] = 1.
    Highest power of 17 = [17/17] = 1.
    17! = 2^17 x 3^6 x 5^3 x 7^2 x 11 x 13 x 17
    Factors = (17+1)(6+1)(3+1)(2+1)(1+1)(1+1)(1+1)
    = 18 x 7 x 4 x 3 x 2 x 2 x 2
    = 126 x 96
    = 12096.



  • Q10) Find the no of zeroes at the end of 100 * 99^2 * 98^3 * 97^4 ... 1^100


 

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