Quant Boosters - Swetabh Kumar - Set 9


  • BITS Pilani | CAT 2016 - 99.93 percentile in Quant


    Q1. A faulty clock gains 10 minutes every hour. If the time is set correctly at 12 Noon on 1st Jan 2010,then how many times will its minute-hand and hour-hand meet in the next 24 hours ?
    (a) 22
    (b) 26
    (c) 24
    (d) 25

    10 minutes every hour, so in 24 hours, clock gains 240 minutes.
    minute and hour hands meet every 65 5/11 minutes = 720/11 mins
    so number of extra meetings: 240*11/720 = 3.xx so 3 more meetings than usual.
    so 22+3=25 times

    Q2. If (x−a)(x−b)=1 and (a−b)+5=0, then (x−a)^3−1/(x−a)^3 is
    A. 100
    B. 140
    C. 200
    D. 280

    b=a+5
    (x-a)(x-a-5)=1 let (x-a)=t
    t (t-5) = 1
    t^2-5t =1
    divide by t
    t-1/t= 5
    cube both sides.
    t^3-1/t^3 -3*5 = 5^3
    t^3-1/t^3 = 125+15=140

    Q3. (1 x 2)/6 + (3 x 4)/6^2 + (5 x 6)/6^3 + (7 x 8 )/6^4 + ( 9 x 10)/6^5 + …
    a) 24/25
    b) 125/116
    c) 125/108
    d) 116/125
    e) 108/125

    S= 2/6 + 12/6^2 + 30/216 + 56/6^4 + 90/6^5....
    S/6 = 2/6^2 + 12/216 + 30/6^4 + 56/6^5...
    5S/6 = 2/6 + 10/6^2 + 18/6^3 + 26/6^4...
    5S/36 = 2/6^2 + 10/6^3+ 18/6^4....
    25S/36 = 2/6 + 8 (1/6^2 + 1/6^3....)
    25S/36 = 2/6 + 8 ( 1/30) = 18/30 = 3/5
    S= 108/125

    Q4. If y = (40/x) + x and x is less than 0, then what is the greatest possible integral value of y?

    Differentiate, -40/x^2 + 1 = 0
    x^2 = 40 x < 0 so x = -rt (40)
    so y = 40/(-rt40) - rt 40 = -2 rt 40= -rt (160) = bw -12 and -13 so max is -13

    Q5. Semi perimeter of a right angled triangle is 154 cm and smallest median is 72 cm. Area of Triangle= ??

    smallest median is to hypotenuse, which is equidistant from all vertices. so hypotenuse= 2 * 72=144
    Let the perpendicular legs be x,y
    so x^2+y^2=144^2
    and x+y+144=308 so x+y=164
    so xy = 1/2{(x+y)^2-(x^2+y^2)} = 1/2 {164^2-144^2}=1/2 * 308 * 20=3080
    so area = 1/2 * xy = 1/2 * 3080 = 1540.

    Q6. A polynomial p(x) leaves remainder 75 and 15 respectively, when divided by (x-1) and (x+2). Then the remainder when f(x) is divided by (x-1)(x+2) is
    a)5(4x+11)
    b) 5(4x-11)
    c) 5(3x+11)
    d) 5(3x-11)

    p(x)= (x-1)(x+2)Q(x) + ax+b
    Put x=1
    a+b = 75
    put x=-2
    -2a+b = 15
    From both, a=20, b=55
    so ax+b = 20x+55 = 5(4x+11)

    Q7) Gopal went to a fruit market with certain amount of money. With this money he can buy either 50 oranges or 40 mangoes. He retains 10% of the money for taxi fare. If he buys 20 mangoes, then the number of oranges he can buy is
    (a) 25
    (b) 18
    (c) 20
    (d) None of these

    1 orange= 4 Rs, 1 mango=5 Rs, say
    so He has Rs 200. and then 180 to spend.
    20 mangoes means he spent 100.
    80 left, so oranges= 80/4=20

    Q8) In Sivakasi, each boy’s quota of match sticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?
    (a) 200
    (b) 150
    (c) 125
    (d) 175

    NB = (N-25)(B+3)
    3N-25B = 75
    N=50, B=3 satisfies, so NB=50 * 3=150.

    Q9) Find the slope of the line which passes through the point (3,5) and cuts off the least area from the first quadrant.

    Any such line: y-5 = m(x-3)
    x intercept: -5/m + 3
    y intercept: -3m+5
    area = 1/2*(3-5/m)(5-3m)
    =1/2 (30-25/m-9m)
    Diff.
    =1/2 (25/m^2-9) = 0
    m= +-5/3 since its first quadrant so negative slope, so -5/3

    Q10) Straight lines m^2 x + 4y + 9 = 0, x + y = 1 and mx + 2y = 3 are concurrent for exactly how many values of 'm'

    Putting y=1-x in other two,
    m^2 x-4x+13 and mx-2x-1=0
    from 2nd, x= 1/(m-2)
    putting in 1st,
    m^2/(m-2) -4/(m-2) +13=0
    m^2+13m-30=0
    (m+15)(m-2) = 0
    m = 2, m = -15
    The first two equations are inconsistent for m = 2.
    So one possible value.



  • for question 10)
    for lines to be concurrent,
    X + Y = 1......eq1
    mX + 2Y = 3.....eq2
    solve eq1 nd eq2
    X = 1 / ( m - 2) and Y = (m - 3) / ( m - 2)
    put this in m^X + 4Y + 9 = 0
    m^2 / ( (m - 2) + 4 (m - 3) / (m - 2) + 9 = 0
    m^2 + 13m - 30 = 0
    so, m = (-13 +- 17 ) / 2
    m = -13 - 17 / 2 = -15
    or, m = -13 + 17 / 2 = 2


  • NMIMS, Mumbai (Gold Medallist) | Asia-Pacific Champion – CFA Global Research Challenge


    @deep OA seems to be one value.
    m = 2 is not possible.


  • Being MBAtious!


    @deep for m = 2, 1st equation becomes x + y = -9/4 which will contradict 2nd equation, x + y = 1



  • Yeah Guys you are correct I was lazy in cross checking values at the end .
    Thanks Zabeer Sir, harris sir for the correction !


  • Being MBAtious!


    @deep Neat explanations Deepak :slight_smile:


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