Quant Boosters - Swetabh Kumar - Set 10


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q1) If a, b, c are integers such that – 50 < a, b, c < 50 and a + b + c = 30, what is the maximum possible value of abc?

    Make one positive max so 49, so other two should add to -19. hence -9 and -10
    49 * (-9) * (-10) = 4410

    Q2) Find the range of x where ||x - 3| - 4| > 3.

    case 1: |x-3| - 4 > 3
    so |x - 3| > 7
    so x > 10 or x < -4
    so -(inf, -4) U (10,inf)
    case 2: |x - 3| - 4 < -3|x - 3| < 1
    so -1 < x - 3 < 1
    2 < x < 4
    take union of the three regions.

    Q3) a and b are roots of the equation x2 - px + 12 = 0. If the difference between the roots is at least 12, what is the range of values p can take?

    D= p^2 - 48
    roots = p + - rt (p^2-48) /2
    so given: rt (p^2-48) >= 12
    p^2 - 48 >= 144
    (p-8rt3) (p+8rt3) >= 0
    so (-inf, -8 rt 3] U [8 rt3, inf)

    Q4) (|x| - 3) (|y| + 4) = 12. How many pairs of integers (x, y) satisfy this equation?

    (a-3)(b+4) = 1 * 12, -1 * -12, 2 * 6, -2 * -6, 3 * 4, -3 * -4
    a >= 0, b >= 0
    so a = 4, b = 8 so 4 cases here (+- for mod)
    a=5, b=2 so 4 cases here.
    a=6, b=0 so 2 cases here.
    Total = 10

    Q5) [x] = greatest integer less than or equal to x. If x lies between 3 and 5, what is the approximate probability that [x^2] = [x]^2?

    between 3 and 4, [x]^2=9
    works till root 10, so from 3 to 3.16
    between 4 and 5, [x]^2=16
    works till root 17 = 4.12 so from 4 to 4.12
    so (0.16+0.12)/2 = 0.14

    Q6) f(x + y) = f(x)f(y) for all x, y.
    f(4) = 3 , what is f(–8)?

    f(0 + 4) = f(0)f(4) so f(0)=1
    f(4 - 4) = f(4)f(-4) so f(-4) = 1/3
    so f(-8)= f(-4) * f(-4)= 1/3 * 1/3 = 1/9

    Q7) An escalator is moving downwards at a speed of 4 steps/minute. Neerja takes 6 minutes less to reach the bottom from the top of the escalator, if he comes down on the moving escalator, as compared to when he does so on the stationary escalator. Gia takes 6 more min to reach the top from the bottom of the escalator if he goes up on the escalator moving downward as compared to when he does so on the stationary escalator. They start simultaneously from the top and the bottom of the escalator, moving downward, respectively and meet after 4 minutes. How many steps are there in the escalator?
    a. 60
    b. 56
    c. 48
    d. Cannot be determined

    total steps be T
    T/N -T/(N+4) =6
    3N^2+12N-2T=0 N=-12+rt (144+24T) /6 --(1)
    and T/(G-4)-T/G = 6
    3G^2 -12G-2T=0
    G= 12+rt (144+24T) /6 ---(2)
    relative speed = N+G
    so T= 4N+4G --(3)
    using 1 and 2,
    T = 4 * rt (144+24T)/3
    9T^2 = 16 (144+24T)
    9T^2 - 384T-2304 = 0
    T = (384 + 480)/18 = 48 steps

    Q8) Two cars P and Q are moving at uniform speeds, 50 km/hr and 25 km/hr respectively, on two straight roads intersecting at right angle to each other. P passes the intersecting point of the roads when Q has still to travel 50 km to reach it. What is the shortest distance between the cars during the journey?
    a. 20√5 km
    b. 50 km
    c. 25 km
    d. 25√2 km

    P's distance be Y, Q's be X
    Y/50 = (X-50)/25
    Y=2X-100
    say Y=100, X=100
    distance at any time=d
    d^2 = (100-50t)^2 + (100-25t)^2
    diff wrt t for minima
    (100-50t)*2 +(100-25t) = 0
    t=12/5
    d^2 = 20^2 + 40^2 = 2000 so d= 20 rt 5

    Q9) Two men are walking towards each other alongside a railway track. A freight train overtakes one of them in 20 seconds and exactly 10 minutes later meets the other man coming from the opposite direction. The train passes this man is 18 seconds. Assume the velocities are constant throughout. How long after the train has passed the second man will the two men meet?
    A. 89.7 minutes
    B. 90 minutes
    C. 90.3 seconds
    D. 91 seconds

    Train speed= T, men speed: v1, v2. length L.
    so L/(T-v1)=20 and L/(T+v2) = 18
    so 20T-20v1=18T+18v2
    T= 9v2+10v1 --(1)
    Ditsane between the men = 600 (T+v2)
    Distance covered in these 10 mins = 600(v1+v2)
    so reqd time = {600(T+v2)-600(v1+v2)}/(v1+v2)
    = (6000-600)= 5400 sec= 90 mins ( using 1)

    Q10) A and B started running from the same point and in opposite directions around a circular track of radius 24.5 m. A’s speed was twice that of B’s speed. They met each other aиer 14 seconds. What was A’s speed?

    A=2x, B=x
    length= 2pi*r = 49pi = 154 m.
    so given, 154/3x = 14 x=11/3 so 2x=22/3


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