Quant Test for IBPS PO 2018 Prelim Exam Set – 7

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

  1. 66, 91,120, 153, 190, ?, 276
    200
    231
    209
    198
    212
    Option B
    66 + 25 = 91
    91 + 29 = 120
    120 + 33 = 153
    153 + 37 = 190
    190 + 41 = 231
    231 + 45 = 276

     

  2. 13.76, 14.91, 17.21, 20.66, ?, 31.01
    30.11
    25.26
    28.30
    22.12
    20.21
    Option B
    13.76 + 1 × 1.15 = 14.91
    14.91 + 2 × 1.15 = 14 + 2.30 = 17.21
    17.21 + 3 × 1.15 = 17.21 + 3.45 = 20.66
    20.66 + 4 × 1.15 = 20.66 + 4.60 = 25.26
    25.26 + 5 × 1.15 = 25.26 + 5.75 = 31.01

     

  3. ? ,16, 28, 58, 114, 204
    10
    12
    14
    20
    17
    Option C
    14 + 1 × 2 = 16
    16 + 3 × 4 = 16 + 12 = 28
    28 + 5 × 6 = 28 + 30 = 58
    58 + 7 × 8 = 58 + 56 = 114
    114 + 9 × 10 = 114 + 90 = 204

     

  4. 21, 10.5, ?, 15.75, 31.5, 78.75
    10.5
    7.4
    11.8
    9.2
    8.5
    Option A
    21 × 0.5 = 10.5
    10.5 × 1 = 10.5
    10.5 × 1.5 = 15.75
    15.75 × 2 = 31.50
    31.50 × 2.5 = 78.75

     

  5. 96, 94, 373, 3353, ?, 1341069
    53415
    53643
    48600
    53773
    50544
    Option B
    (× 1 − 2), (× 4 − 3), (× 9 − 4), (× 16 − 5), (× 25 − 6)
    = 3353 × 16 − 5 = 53643

     

  6. A train travelling at 36 kmph completely crosses another train having half its length and travelling in the opposite direction at 54 kmph, in 12 seconds. If it also passes a railway platform in 1(1/2) minutes, find the length of the platform.
    750 m
    720 m
    685 m
    700 m
    660 m
    Option D
    Let the length of slower train be x metres and the length of faster train be x/2 meters.
    Relative speed = (36+54)km/hr. = (90×5/18) = 25 m/sec.
    3x/(2×25) = 12
    => 3x = 600
    => x = 200
    Length of slower train = 200 m
    Let the length of platform be y metres
    Then, (200+y)/36+5/18=90
    => 200 + y=900
    y = 700 m
    Length of platform = 700 m

     

  7. Vidal borrows Rs. 50,000 from a bank at 5% p.a. simple interest and clears the debt in five years. If the installments paid at the end of the first, second, third and fourth years to clear the debt are Rs. 5,000, Rs. 10,000, Rs. 15,000 and Rs. 20,000 respectively, what amount should be paid at the end of the fifth year to clear the debt?
    6800
    7700
    8000
    7520
    7500
    Option E
    Simple interest for 1st yr.= P*R*T / 100
    = 50000×1×5/ 100 = 2500
    Amount after 1st instalment = 50000- 5000=45000
    Simple interest for 3rd yr.= P*R*T / 100 = 35000×1×5/ 100 = 1750
    Amount after 3rd instalment = 35000- 15000 = 20000
    Simple interest for 4th yr.=P*R*T/ 100 = 20000×1×5/ 100 = 1000
    Amount after 4th instalment = 20000- 20000 = 0
    Balance of debt for the fifth year = 2500+2250+1750+1000 =7500

     

  8. A field is 90 metre long and 50 metre broad. A 25 metre long, 20 metre broad and 4 metre deep tank dug in the field and the earth taken out is spread evenly over the remaining field. How much the level of field will rise?
    1
    3
    2.5
    0.5
    2
    Option D
    Area of field = 90 x 50 = 4500 m^2
    Area of field dug out = 25 x 20 = 500 m^2
    Therefore, Area of remaining field = 4500 – 500 = 4000 m^2
    Volume of the earth dug out = 25 x 20 x 4 = 2000 m^3
    Therefore, Field will rise by 2000/4000 = 0.5 metre

     

  9. A cistern has a leak which would empty the cistern in 20 minutes. A tap is turned on which admits 4 liters a minute into the cistern, and it is emptied in 24 minutes. How many liters does the cistern hold ?
    420 litres
    380 litres
    480 litres
    390 litres
    400 litres
    Option C
    1/x – 1/20 = -1/24 x = 120
    120 x 4 = 480 litres

     

  10. There are 2 people who are going to take part in race. The probability that the first one will win is 2/7 and that of other winning is 3/5. What is the probability that one of them will win?
    11/34
    18/35
    17/36
    19/35
    11/35
    Option D
    Prob. of 1st winning = 2/7
    Not winning = 1 – 2/7 = 5/7
    Prob. of 2nd winning = 3/5
    Not winning = 1 – 3/5 = 2/5
    Required prob. = 2/7 * 2/5 + 3/5 * 5/7
    = 19/35

     


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