Quant Test for IBPS Clerk 2018 Main Exam Set – 11

Directions(1-5): Find the missing term of the following series.

  1. 192,96,?,360,1260,5670
    150
    100
    125
    144
    130
    Option D
    *0.5,*1.5,*2.5,*3.5,*4.5
    ? = 144

     

  2. 37,60,89,?,157,198
    107
    120
    115
    98
    90
    Option B
    +23,+29,+31,+37,+41
    ? = 120

     

  3. 25,46,88,?,235,340
    144
    160
    151
    142
    155
    Option C
    +21*1
    +21*2
    +21*3
    +21*4
    +21*5
    ? = 151

     

  4. 17,28,?,116,193,292
    48
    50
    30
    53
    61
    Option E
    +11*1
    +11*3
    +11*5
    +11*7
    +11*9
    ? = 61

     

  5. 39,40,49,74,123,?
    210
    204
    190
    200
    198
    Option B
    +1^2,+3^2,+5^2,+7^2,+9^2 = 204

     

  6. Directions(6-10): In the following questions you have two equations and solve both individually to establish a relation between them and choose a correct option.

  7. Quantity I: A shopkeeper marks up an item by 50% above the cost price and sold it after offering a discount of 20%. The profit earned by shopkeeper is what percentage less than the profit earned by him, if he marks up the item by 75% above the cost price and offers a discount of 20%.
    Quantity II: The area of the square built on the side of a square is what percentage less than the area of square built on the diagonal of a square?

    Quantity I =< Quantity II
    Quantity I < Quantity II
    Quantity I > Quantity II
    Quantity I >= Quantity II
    Quantity I = Quantity II
    Option
    From I: case 1: Let the CP of the item be Rs.x.
    SP = 1.5x*0.8 = Rs. 1.2x
    Profit = 1.2x – x = Rs. 0.2x
    case 2: Let the CP of the item be Rs.x.
    SP = 1.75x*0.8 = Rs.1.4x
    Required% = {(0.4x-0.2x)/0.4x}*100 = 50%
    From II: Let side of the square be a unit.
    Length of the diagonal of the square = (2)^1/2a
    Area of the square = a^2 units
    Area of the square built on the diagonal = [(2)^1/2*a]^2 = 2a^2 unit
    Required% = (2a^2-a^2)/2a^2*100 = 50%
    Quantity I = Quantity II

     

  8. >Quantity I: X and Y together started a business with an investment of Rs. 2400 and Rs. 3600 resp. After x months, Z joined them with an investment of Rs. 3000. If after a year, Y received Rs. 2700 out of total profit of Rs. 6000 find the value of x.
    Quantity II: A and B together started a business with total investments of Rs. 4500 in the ratio of investment 5:x resp. If after a year, B received Rs. 3200 as profit out of a total profit of Rs. 7200 find the value of x.

    Quantity I > Quantity II
    Quantity I < Quantity II
    Quantity I = Quantity II
    Quantity I =< Quantity II
    Quantity I >= Quantity II
    Option C
    From I: Ratio of investments of X:Y:Z = 4:6:5
    Ratio of profit share of X:Y:Z = 48:72:5(12-x) 72/[48+72+5(12-x)] = 9/20
    => x = 4
    From II: Ratio of profit share of A:B = 5:x x/(5+x) =4/9
    => x= 4
    Quantity I = Quantity II

     

  9. Quantity I: Radius of the base and the height of the circular cylinder are in the ratio is 3:7 resp. Find the diameter of the base, if its volume is 1584 cm^2.
    Quantity II: Radius of the base and the height of the circular cone are in the ratio 6:7 resp. Find the radius of the base if its volume is 2112 cm^3.

    Quantity I =< Quantity II
    Quantity I > Quantity II
    Quantity I >= Quantity II
    Quantity I = Quantity II
    Quantity I < Quantity II
    Option D
    From I:Volume = 22/7*(3x)^2*7x = 1584
    => x = 2
    Diameter of the base of the cylinder = 2*3*2 = 12 cm
    From II: Volume = 1/3*22/7*(6x)^2*7x =2112
    => x = 2
    Quantity I = Quantity II

     

  10. Quantity I: Two positive numbers are in the ratio of 3:5 resp. The difference between the squares of the two numbers is 1024. Find the larger of the two numbers. Quantity II: A number is 25% more than the another number. Average of these two numbers is 36. Find the largest number.
    Quantity I = Quantity II
    Quantity I >= Quantity II
    Quantity I =< Quantity II
    Quantity I < Quantity II
    Quantity I > Quantity II
    Option A
    From I: Let the numbers be 3x and 5x.
    (5x)^2 – (3x)^2 = 1024
    => x = -8,8
    Numbers are 24,40.
    Larger number = 40
    From II: First number be x and second number be 1.25x.
    x+1.25x = 2*36
    => x= 32
    Largest number = 1.25*32 = 40
    Quantity I = Quantity II

     

  11. Quantity I: x^2 + 17x + 72 =0
    Quantity II: y^2 + 15y + 56 =0

    Quantity I =< Quantity II
    Quantity I = Quantity II
    Quantity I < Quantity II
    Quantity I > Quantity II
    Quantity I >= Quantity II
    Option A
    From I:x^2 + 17x + 72 =0
    =>x^2 +8x + 9x + 72 =0
    => (x+8)(x+9) = 0
    => x = -9,-8
    From II:y^2 + 15y + 56 =0
    =>y^2 + 8y + 7y + 56 = 0
    => (y+8)(y+7) =0
    => y = -7,-8
    Quantity I =< Quantity II

     


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