Quant Boosters - Swetabh Kumar - Set 3



  • 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



  • Q26) 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



  • Q27) 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



  • Q28) 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)



  • Q29) Choti has 11 different toys and Bade has 8 different toys.Find the number of ways in which they can exchange their toys so that each keeps their initial number of toys?



  • Goods have to be "exchanged". 19C8 or 19C11 is the total possibilities, but since they have to be "exchanged" so it is 19C8-1.
    (original configuration not to be counted)



  • Q30) Find the number of positive solutions for 2x + 3y > 60 and 2y + 3x < 60



  • Draw 3x+2y = 60 and 2x+3y = 60
    find the common region as per question
    a triangle will be formed with cordinates (12,12 ) ( 0,20 ) ( 0,30 )
    area of this triangle = 60
    use picks theorem : A = i + b/2 - 1
    61 = i + b/2
    calculate the number of boundary pts of triangle boundary , total 20 boundary pts will be there
    9 on the vertical line between 0,20 and 0,30
    3 on the line between 0,20 and 12,12
    5 on the line between 0,30 and 12,12
    3 vertices
    total 20 boundary points
    61 - 20/2 = 51 integral points



  • @swetabh_kumar

    shouldn't it be 89.7 mins?
    You might have missed 18 seconds both men have moved while from the moment the train meets the second man to the moment it passes him.
    so reqd time = {600(T+v2)-600(v1+v2)-18(v1+v2)}/(v1+v2)
    which gives 5400-18 seconds or 89.7 mins



  • We can do it another way too:

    Put x= y,
    F(2x) = f(x)^2
    Put y=2x
    F(3x)= f(x)^3
    =>f(nx)=f(x)^n

    F(-8)=F(-2×4)=f(4)^-2=1/9


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