Quant Test for IBPS Clerk 2018 Prelim Exam Set – 17

Directions(1-5): Find the relation between x and y, and choose a correct option.

  1. X^2 – 14x + 40 = 0
    y^2 – 25y + 144 = 0

    y>=x
    x>y
    x>=y
    y>x
    x=y or relation cannot be established.
    Option E
    X^2 – 4x – 10x + 40 = 0
    =>x(x-4)- 10(x-4) = 0
    => x = 4,10
    Y^2-25y+144 = 0
    =>y^2 – 16y – 9y + 144 = 0 => y(y-16)-9(y-16)= 0
    => y = 16,9
    x=y or relation cannot be established.

     

  2. 12x^2 + 13x – 35 = 0
    10y^2= 21y-9

    x>y
    x>=y
    y>=x
    x=y or relation cannot be established.
    y>x
    Option D
    12x^2 + 13x – 35 = 0
    =>12x^2 + 28x -15x – 35 = 0
    =>4x(3x+7)-5(3x+7)=0
    =>x = -7/3,5/4
    10y^2= 21y-9
    => 10y^2 -15y -6y+ 9 = 0
    => 5y(2y-3)-3(2y-3)=0
    =>y =3/2,3/5
    x=y or relation cannot be established.

     

  3. X= (1089)^1/2 – (14^2 – 190)
    y = 13^3 – (45^2+146)

    y>=x
    x>y
    x=y or relation cannot be established.
    y>x
    x>=y
    Option B
    X= (1089)^1/2 – (14^2 – 190)
    => x = 33 – (196-190)
    => x = 27
    y = 13^3 – (45^2+146)
    =>y =2917-(2025+146)
    =>y = 26
    x>y

     

  4. 3x+y = 20
    4x+3y = 30

    y>x
    x=y or relation cannot be established.
    y>=x
    x>y
    x>=y
    Option D
    X = 6
    y = 2
    x>y

     

  5. X^2 -36x + 323 = 0
    y^2-29y+210 = 0

    x>=y
    x=y or relation cannot be established.
    y>x
    x>y
    y>=x
    Option D
    X^2 -36x + 323 = 0
    => x^2 – 17x – 19x + 323 = 0
    => x(x-17)-19(x-17) = 0
    => x = 17,19
    y^2-29y+210 = 0
    =>y^2 – 15y – 14y + 210 = 0
    => y(y-15)-14(y-15) = 0
    => y = 15,14
    x>y

     

  6. The length and the breadth of a rectangular plot are in the ratio 5:4. If the cost incurred for fencing the boundary of the plot at Rs. 15 per metre is Rs. 2970, the find the area of the plot.
    2550 m^2
    2250 m^2
    2000 m^2
    2420 m^2
    2500 m^2
    Option D
    Perimeter = 2970/15 = 198 m
    Now, 2(5x+4x) = 198
    => x = 11 metres
    Length = 55 m
    Breadth = 44 m
    Required area = 2420 m^2

     

  7. A boy cycles from his home to his school with a speed of 12 km/hr. and reaches his school in 45 minutes. If he returns home at a speed of 3 km/hr. more than the speed of going to his school. Find his average speed for the whole journey.
    11.11 km
    12.25 km
    13.33 km
    10.11 km
    9.52 km
    Option C
    Distance between his home and his school = 45/60*12
    = 9 km
    Speed of boy while returning = 12+3 = 15km/hr.
    Time taken to cover the distance between home and school = 9/15 = 0.6 hour
    Required average speed = (9+9)/(0.6+0.75) = 13.33 km

     

  8. A man sold an article at 15% discount and earned 36% profit. The discount given on the article was Rs. 576. What should be the selling price if the shopkeeper wanted to earn 42% profit.
    Rs. 3500
    Rs. 4150
    Rs. 3408
    Rs. 3333
    Rs. 3000
    Option C
    Let MP be x. 15% of x = 576 x = (576*100)/15
    => x = Rs. 3840
    SP = 3840-576 = RS.3264
    CP = (3264*100)/136 = Rs.2400
    Required SP = 142% of 2400 = Rs. 3408

     

  9. Pipe A fills (1/4)th of the tank in 5 minutes and another pipe B fills (1/5)th of the tank in 6 minutes. Find the time taken to fill half of the tank if both the pipes work together.
    6 minutes
    7 minutes
    9 minutes
    5 minutes
    4 minutes
    Option A
    Time taken by pipe A = 5*4 = 20 minutes
    Time taken by pipe B = 6*5 = 30 minutes
    Let the total capacity of the tank LCM (20 and 30) = 60 units
    Number of units of water filled by A alone in one minute = 3 units
    Number of units of water filled by B alone in one minute = 2 units
    Number of units of water filled by both pipes in one minute = 5 units
    Required time taken by both pipes to fill the tank = 60/(2*5) = 6 minutes

     

  10. A bag contains 5 red balls, 3 black balls and 5 white balls. A man draws 3 balls at random from the bag. Find the probability that out of 3 balls at least 2 balls are white.
    40/143
    45/143
    33/143
    45/111
    41/134
    Option B
    Required Probability = [ (5C2*3C1)+(5C2*5C1)+5C3]/13C3
    = 45/143

     


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