Quant Test for SBI PO 2018 Prelim Exam Set – 38

Hello Aspirants

State Bank of India (SBI) is going to conduct examination for its recruitment for the post of Probationary Officers (SBI PO 2018) for a total of 2000 vacancies.

Click here to know the details of the Examination

The examination will be held in three phases i.e. Preliminary Examination, Main Examination and Group Exercise & Interview. The Preliminary Exam is scheduled on 1st, 7th & 8th of July 2018. Details of the exam are as under:

Practice the questions so as to familiarize yourself with the pattern of questions to be asked in the exam. 


 

Directions(1-5): What should come in place of the question mark (?) in the following number series?

  1. 57, 59 , 66 , 94, 157 , ?
    250
    222
    225
    283
    211
    Option D
    +1³+1, +2³ – 1, +3³ + 1, +4³ – 1…………

     

  2. 15, 18, 16, 19, 17, 20, ?
    13
    22
    18
    11
    8
    Option C
    15 + 3 = 18
    18 – 2 = 16
    16 + 3 = 19
    19 – 2 = 17
    17 + 3 = 20
    20 – 2 = 18

     

  3. 10, 14, 25, 55, 140,?
    316
    388
    302
    350
    374
    Option B
    14 – 10 = 4
    25 – 14 = 11 = 4 × 3 -1
    55 – 25 = 30 = 11 × 3 – 3
    140 – 55 = 85 = 30 × 3 – 5
    ? = 140 + 85 × 3 – 7 = 140 + 248 = 388

     

  4. 1200, 480, 192, 76.8, 30.72, 12.288, ?
    5.4100
    3.1122
    4.9152
    4.0845
    5.8102
    Option C
    1200 / 2.5 = 480
    480 / 2.5 = 192
    192 / 2.5 =76.8
    76.8 / 2.5 = 30.72
    30.72 / 2.5 = 12.288
    ? = 12.288 / 2.5 = 4.9152

     

  5. 963, 927, 855, 747, 603, 423, ?
    210
    215
    198
    200
    207
    Option E
    963 – 1 × 36 = 963 – 36 = 927
    927 – 2 × 36 = 927 – 72 = 855
    855 – 3 × 36 = 855 – 108 = 747
    747 – 4 × 36 = 747 – 144 = 603
    603 – 5 × 36 = 603 – 180 = 423
    ? = 423 – 6 × 36 = 423 – 216 = 207

     

  6. The height of a room is 40% of its semiperimeter. It cost Rs 260 to paper the walls of the room with paper 50 cm wide at the rate of Rs 2 per m allowing an area of 15 m2 for doors and windows. What is the height of the room ?
    10m
    6m
    4m
    12m
    9m
    Option C
    Let the length, breadth and height of the room be l, b and h respectively.
    Then, h = 0.4 ( l + b)
    Area of the four walls = 2 (l + b) h = 2 (l + b) × 0.4 (l + b) = 0.8 (l + b)^2
    Required area where paper has to be pasted = 4/5 (l + b)^2 – 15
    Now, area of paper = area of wall Length × breadth = 4/5 (l + b)^2 – 15
    Length = [4/5 (l + b)^2 – 15]/1/2
    Given, 2*2[4/5 (l + b)^2 – 15] = 260
    => (l + b)^2 = 320
    => (l + b)^2 = 100
    => (l + b) = 10
    =>h = 0.4 × 10 = 4 m

     

  7. Kamya purchased an item of Rs 46,000 and sold it at a loss of 12 percent. With that amount she purchased another item and sold it at a gain of 12 percent. What was her overall gain/loss ?
    Loss Rs.666.2
    Gain Rs.650.15
    Gain Rs.699.8
    Loss Rs.662.4
    Loss Rs.677.5
    Option D
    First S.P. = 46000 * 88/100 = Rs. 40480
    Second S.P. = 40480 * 112/100 = Rs. 45337.6
    Loss = Rs. (46000 – 45337.6) = Rs. 662.4

     

  8. A grocer purchased 2 kg. of rice at the rate of Rs. 15 per kg. and 3 kg. of rice at the rate of Rs. 13 per kg. At what price per kg should he sell the mixture to earn 33(1/3)% profit on the cost price ?
    Rs.17.4
    Rs.15.7
    Rs.20.2
    Rs.18.4
    Rs.10.5
    Option D
    2 kg of rice at Rs. 15/kg + 3 kg of rice at Rs. 13/kg Total weight = 2 + 3 = 5 kg
    Total cost price = (2 × 15) + (3 × 13) = 30 + 39 = Rs. 69
    Cost price per kg of the mixture = 69/5 = Rs.13.80
    Selling price to get 33(1/3) % profit = [10 + 33 (1/3)]/100× Rs 13.80
    = 400× Rs 13.80/3*100
    = 4/3 × Rs 13.80
    = Rs 18.40

     

  9. A bus started its journey from Ramgarh and reached Devgarh in 44 minutes with its average speed of 50 km/hour. If the average speed of the bus is increased by 5 km/hour, how much time will it take to cover the same distance ?
    60 mins.
    50 mins.
    70 mins.
    40 mins.
    20 mins.
    Option D
    Distance between Ramgarh and Devgarh = 50 * 44/60
    = 110/3 km
    New speed = 55 kmph = 55/60 km/minute
    Required time = Distance/Speed
    = 110/3 × 60/55 = 40 minutes

     

  10. In how many different ways can 4 blue beads and 3 red beads be arranged in a row such that all the blue beads are together and all the red beads are together.
    288
    243
    205
    255
    278
    Option A
    Required no. of ways = 4! x 3! x 2!

     


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