Quant Boosters by Hemant Malhotra - Set 6


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    A circus troupe was performing in a village. One evening all the 120 members of the village went to the circus and Rs.120 were collected in all. The charges were Rs.5 per male, Rs.2 per female & 10 paisa, per child. How many ladies were there?
    A-13
    B-17
    C-30
    D-19

    a+b+c=120
    5a+2b+0.1c=120
    50a+20b+c=1200
    so 49a+19b=1080
    so 19b=1080-49a)
    so b=((1080-49a)/19
    a=17 and b=13

    if N=(5 * sqrt5 + 11)^7 find N * f where f is fractional part of N .

    Note – X is conjugate to R

    1. in R = (a + b*rtc)^k
      if k is odd then f = X
      fractional part will be f=(a-b * rtc)^n
    2. and if k is even then f+X=1
      f = 1 - (a-b * rtc)^k

    Here odd
    so f = (5 sqrt5 - 11)^7
    so N * F = (5sqrt5 + 11) * (5sqrt5-11)))^7 = 4^7

    (5 * rt5 + 11)^2n+1 find fractional part of this number.

    let f is fractional part of this f = R - [R]
    let a number X= (5 * rt5 - 11)^2n+1
    X will lie in 0 < X < 1
    so R-X=(5 * rt5+11)^2n+1-(5 * rt5-11)^2n+1
    so R-X=2 * (2n+1c1 * (5 * rt5)^2n * 11+2n+1c3 * (5rt5)^2n-2 * (11)^3......
    so R-X=even number
    R=[R]+f
    so [R]+f-X=even
    so f-X=even-[R]
    so f-X=an integer (odd or even both possible)
    so f-X=0
    bcz 0< f < 1
    0 < X < 1
    so f-X=0
    so f = X = (5rt5-11)^2n+1
    so fractional part is (5rt5-11)^2n+1

    Find the fractional part of (5+2 * rt6)^n

    R=(5+2 * rt6)^n=[R]+f
    let X=(5-2rt6)^n
    so[R}+f+X=((5+2 * rt6)^n+(5-2rt6)
    so R+f+G=2 * (nc0 * 5^n+nc2 * 5^n-2 * (2rt6)^2....
    so [R]+f+X=even
    so f+X=integer so f+X=1
    so f=1-x so 1-(5-2rt6)^n

    Let F(x) = max (5 − x, x + 2). The smallest possible value of F(x) is
    A] 4.0
    B] 4.5
    C] 1.5
    D] None of these.

    5-x=x+2
    so 2x=3
    so at x=1.5 it will give smallest value
    max((5-1.5,1.5+2)
    =3.5 OA

    A function f is even if f(t) = f(–t), and it is odd if f(t) = –f(–t). Let f(x) = g(x) + h(x) and f(–x) = g(x) – h(–x). Which of the following is definitely correct?
    a) f is an even function and g is an odd function.
    b) Both f and g are even functions.
    c ) f is an even function and h is an odd function.
    d ) g is an even function and h is an odd function.

    G(x) is an even function et f(x)=g(x)+h(x) (1)
    f(-x)=g(x)-h(-x) (2)
    F(x)+f(-x)=2(g(x)
    means g(x)=(f(x)+f(-x))/2
    g(-x)=f(-x)+f(x)/2
    so g(x)=g(-x) so even

    For how many positive integer values of ‘x’ is ||||x – 1| – 2| – 3| – 4| < 5

    ||||x – 1| – 2| – 3| – 4| < 5
    so |x-1| < 5 + 4 + 3 + 2
    so |x-1| < 14
    so -14 < x-1 < 14
    so -13 < x < 15
    so max 14 and min=-12

    Let f be a function such that
    f(a,b) = f(a-b , b) if a=>b
    = a if a < b.
    Now, if f(n,5)= 3 and f(n,6)= 5, then n

    f(a,b)=f(a-b,b) if a>=b
    =a if a < b
    f(n,5) =3

    now f(n,6)=5
    If f(n,5)=3 and we need 3
    so n will be in form of 5k+3 for 3
    and n should be also in form of 6k+5
    now 5k+3=6k+5
    so first number will be 23 and So on so OA=CBD

    f(x, y) is a polynomial function such that f(2, 3) = 30, f(3, 4) = 84 and f(4, 5) = 180. Find f(5, 6).
    a) 310
    b) 330
    c) 340
    d) 350

    (3+2) * 3 * 2 = 30
    (3+4) * 4 * 3 = 84
    (4+5) * 4 * 5 = 180
    f(5,6) = (5 + 6) * 5 * 6 = 330

    To measure the weight between 1 to 300 how many minimum weighs are required so that all weight can be measured?

    for any weight we have 3 possibilities:-

    1. Place it on the weight pan
    2. Place it on other pan, along with something whose weight we need to find
    3. We don't use it

    So, it is similar to base 3, hence by taking weight of form 3^n, we can ensure that we are using least number of weights and when we just have two possibilities. so 3^n >=300 so n=6


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