Quant Test for IBPS Clerk 2018 Main Exam Set – 1

  1. 6,17,?,105,182,281
    44
    42
    50
    52
    41
    Option C
    +11*1 +11*3 +11*5 +11*7 +11*9 ? = 50

     

  2. 16,10,12,20,42,?
    100
    111
    105
    107
    104
    Option D
    *0.5+2 *1+2 *1.5+2 *2+2 *2.5+2 ? = 107

     

  3. 1400,1399,1372,1247,?,175
    984
    858
    800
    850
    904
    Option E
    -1^3 -3^3 -5^3 -7^3 -9^3 ? = 904

     

  4. 40,65,90,115,?,165
    140
    130
    125
    155
    147
    Option A
    (10*1)+(15*2) = 40 (10*2)+(15*3) = 65 (10*3)+(15*4) = 90 (10*4)+(15*5) =115 (10*5)+(15*6) =140 (10*6)+(15*7) =165

     

  5. 70,195,370,595,870,?
    1050
    1115
    1220
    1195
    1333
    Option D
    10^2 – 30 = 70 15^2 – 30 = 195 20^2 – 30 = 370 25^2 – 30 = 595 30^2 – 30 = 870 35^2 – 30 = 1195

     

  6. A, B and C started a business by investing Rs. x, Rs. (x+20000) and Rs. (x+35000) resp. If it is known that the ratio of the initial capitals of A and C was 5:12 , then what was the difference between share of 1st year’s profit of B and C given that the total 1st year’s profit was Rs. 7800 and all of them invested for the whole year?
    Rs. 900
    Rs. 880
    Rs. 800
    Rs. 935
    Rs. 755
    Option A
    x/(x+35000) = 5/12 =>x = 25000 Profit at the end of the first year = 25000:45000:60000 = 5:9:12 B’s share =9/26*7800 =Rs. 2700 C’s share = 12/26*7800 = Rs. 3600 Required Difference = 3600 – 2700 = Rs. 900

     

  7. A hemisphere having radius greater than the radius of a cylinder is attached to one side of the cylinder. The radius of the hemisphere and the cylinder are x cm and (x-14) cm resp. The curved surface area of the hemisphere is 11088 cm^2 and the height of the cylinder is 20 cm. What is the total volume of the combined solid?
    211250 cm^3
    110453 cm^3
    154500 cm^3
    100500 cm^3
    204512 cm^3
    Option E
    11088 = 2*22/7*x^2 => x = 42 cm Radius of the cylinder = 42 – 14 = 28 cm Total volume of the figure = 2/3*22/7*x^3 + 22/7*(x-14)^2 *20 => 2/3*22/7*42^3+22/7*28^2*20 = 204512 cm^3

     

  8. In Kolkata , Any person consuming electricity has to pay a fixed charge and an additional amount propotional to the electricity consumed. But if one consumes more than 60 kwh per month. One has to pay the fixed charge, amount propotional to the electricity consumed and an additional amount to the square root of the energy consumed . If a person pays Rs. 400 for 50 kwh Rs. 360 for 30 kwh and Rs. 498 for 81 kwh. How much should he payfor 100 kwh?
    Rs. 540
    Rs. 650
    Rs. 500
    Rs. 566
    Rs. 420
    Option A
    Let Rs. x be fixed charge. X+ 50h = 400 —-(1) x + 30h = 360 —–(2) x+81h +(81)^1/2 i = 498 —–(3) Solving (1) and (2) , we get x = Rs. 300 and h = 2 and i = 4 Required amount to pay = 300 + 100*2+10*4 = Rs. 540

     

  9. A and B start from the same point but, in opposite direction towards each other and run around a rectangular park of area 1950 m^2 . It is known that the length of the park is 53 m more than the breadth of the park. A runs at 2x m/s and B runs at 3x m/s and they meet for the first time in (103/15) seconds. How long would it take them to meet each other if both of them ran at x m/s towards each other?
    101/5 seconds
    103/6 seconds
    101/2 seconds
    105/3 seconds
    101/6 seconds
    Option C
    L = B + 53 L*B = 1950 => (B+53)*B = 1950 =>B^2 +53B-1950 = 0 =>B = 25 m L = 25 + 53 = 78 m Perimeter of the park = 2*(78+25) = 206 m 206/(2x+3x) = 103/15 => x = 6 Time taken to meet each other = 206/x+2 = 103/6 seconds

     

  10. There are some brown, black and red kites in a packet out of which x are brown. 4 are black and y are red. y/2 black kites are added to the packet and the probability of selecting a black kite becomes 4/13. Also the average of number of brown and red kites in the packet is 9.What is the probability os selecting 2 brown kites before adding y/2 black kites to the packet?
    13/73
    10/71
    12/77
    13/71
    15/77
    Option E
    [4+y/2]/[x+y+(4+y/2)] = 4/13 => 8x – y = 72 —–(1) Also , (x+y)/2 = 9 => x + y = 18 —–(2) On solving (1) and (2) , we get x = 10 and y = 8 Required Probability = 10/22*9/21 =15/77

     


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