Please wait...

SBI Clerk Prelims 2018 Aptitude Test 1
Menu grid icon
Result Result point icon
SBI Clerk Prelims 2018 Aptitude Test 1
  • Goals icon

    /

    Score
  • Trophy icon

    -

    Rank
White alarm icon Time Taken: -
Result frame illustration
  • Question 1/35
    1 / -0

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

    Que. 1 1 2 6 ? 120

    Solutions

    0! = 1

    1! = 1

    2! = 2

    3! = 6

    4! = 24

    5! = 120

     

  • Question 2/35
    1 / -0

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

    Que. 4 13 17 ? 30 39

    Solutions

    4 + 9 = 13

    13 + 4 = 17

    17 + 9 = 26

    26 + 4 = 30

    30 + 9 = 39

     

  • Question 3/35
    1 / -0

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

    Que. 982 977 952 827 822 ?

    Solutions

    982−(5)1=977

    977−(5)2=952

    952−(5)3=827

    827−(5)1=822

    822−(5)2=797

     

  • Question 4/35
    1 / -0

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

    Que. 41472 5184 576 72 8 ?

    Solutions

    41472 ÷ 8 = 5184

    5184 ÷ 9 = 576

    576 ÷ 8 = 72

    72 ÷ 9 = 8

    8 ÷ 8 = 1

     

  • Question 5/35
    1 / -0

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

    Que. 7 33 137 553 ? 8873

    Solutions

    7 × 4 + 5 = 28 + 5 = 33

    33 × 4 + 5 = 132 + 5 = 137

    137 × 4 + 5 = 548 + 5 = 553

    553 × 4 + 5 = 2212 + 5 = 2217

    2217 × 4 + 5 = 8868 + 5 = 8873

     

  • Question 6/35
    1 / -0

    Directions: Study the table given below and answer the questions.

    Que. In which department did the number of employees (approximately) remain the same during the years 2006 to 2011?

    Solutions

    From the table, it is clear that in the department ‘Research’ the number of employees remain approximately the same from the year 2006 to 2011

     

  • Question 7/35
    1 / -0

    Directions: Study the table given below and answer the questions.

    Que. In how many years was the number of employees in the production department more than 50% of the total number of employees?

    Solutions

    From the above table, it is clear that there were five such years, i.e., 2007, 2008, 2009, 2010 and 2011 in which number of employees in production were more than 50% of the total employees.

     

  • Question 8/35
    1 / -0

    Directions: Study the table given below and answer the questions.

    Que. In which years did the departments have a greater number of employees than they had in the immediately preceding year?

    Solutions

    It is clear from the table that in the year 2010, each department had a larger number of employees than it had in the immediately preceding year.

     

  • Question 9/35
    1 / -0

    Directions: Study the table given below and answer the questions.

    Que. Which department had less than 10% of the employees through all the years shown in the table?

    Solutions

    10% of the total number of employees in the year 2006 = 10% of 690 = 69

    2007 = 10% of 884 = 8.4

    2008 = 10% of 1416 = 141.6

    2009 = 10% of 1500 = 150

    2010 = 10% of 1642 = 164.2 2011 = 10% of 1642 = 164.2

    From the above calculation and the table, it is clear that in Marketing, the number of employees through all the years is less than 10% of the total number of

    employees in the corresponding years. 

     

  • Question 10/35
    1 / -0

    Directions: Study the table given below and answer the questions.

    Que. In which year did the total number of employees reach (approximately) twice the total number of employees that the factory had in the year 2006?

    Solutions

    Number of employees in the factory in the year 2006 = 300 + 100 + 90 + 150 = 690

    2007 = 450 + 80 + 90 + 124 + 140 = 884

    2008 = 900 + 130 + 60 +180 + 146 = 1416

    2009 = 940 + 146 + 64 + 210 + 140 = 1500

    2010 = 1000 + 160 + 70 + 264 + 148 = 1642

    2011 = 1010 + 150 + 72 + 260 + 150 = 1642

    Thus, it is clear that the number of employees in the year 2008 is approximately double that it had in the year 2006.

     

  • Question 11/35
    1 / -0

    In each question, what is the approximate value of the number which should replace the question-mark (?)

    Que. 38.5 + 12.0025*4.5 = ?

    Solutions

    In the given question, 12.0025 ≈ 12

    So, the approximate value of the given expression equals 38.5 + 12*4.5 = 38.5 + 54 = 92.5

     

  • Question 12/35
    1 / -0

    In each question, what is the approximate value of the number which should replace the question-mark (?)

    Que. (1050+1470+1340+1240)÷856=?

    Solutions

    (1050+1470+1340+1240)÷856=5100÷856=(850×6)÷856≈6

     

  • Question 13/35
    1 / -0

    In each question, what is the approximate value of the number which should replace the question-mark (?)

    Que. 1.22×4.82÷8.1=?

    Solutions

     

  • Question 14/35
    1 / -0

    In each question, what is the approximate value of the number which should replace the question-mark (?)

    Que. 1.892−1.092=?

    Solutions

    The given expression is 1.892−1.092≈1.92−1.12=(1.9+1.1)∗(1.9−1.1)=3∗0.8=2.4

     

  • Question 15/35
    1 / -0

    In each question, what is the approximate value of the number which should replace the question-mark (?)

    Que. 

    Solutions

     

  • Question 16/35
    1 / -0

    In the following questions, what will come in the place of the question mark (?) 

    Que. 256/16+16∗2−16=?

    Solutions

    BEDMAS rule applies. So, Brackets first, exponent next, division next, multiplication next, addition next, subtraction last.

    256/16 = 16

    16*2 = 32

    So, the answer is 16 + 32 - 16 = 32

     

  • Question 17/35
    1 / -0

    In the following questions, what will come in the place of the question mark (?) 

    Que. 

    Solutions

     

  • Question 18/35
    1 / -0

    In the following questions, what will come in the place of the question mark (?) 

    Que. 

    Solutions

    The expression becomes 24/11 * 11/2 - 11/3 * 5/11 = 12 - 5/3 = 31/3

     

  • Question 19/35
    1 / -0

    In the following questions, what will come in the place of the question mark (?) 

    Que. 

    Solutions

    BODMAS rule applies. So, Brackets are solved first followed by of, next division, next multiplication, next addition and finally subtraction.

    Numerator = 24 * 5 + 28 = 120 + 28 = 148

    Denominator = 6*12 - 8 = 72 - 8 = 64

    So, fraction = 148/64 = 37/16

     

  • Question 20/35
    1 / -0

    In the following questions, what will come in the place of the question mark (?) 

    Que. 24/12+(6−3)∗22=?

    Solutions

    BEDMAS rule applies. So, Brackets first, exponent next, division next, multiplication next, addition next, subtraction last

    24/12 = 2

    6-3 = 3

    22=4

    So, the expression is 2+3∗4 = 14

     

     

  • Question 21/35
    1 / -0

    What is the area of the triangle whose vertices are (0,0), (5,0) and (0,6)?

    Solutions

    This is a right-angled triangle with the perpendicular sides being 5 and 6. So, area = ½ * 5 * 6 = 15 sq units

     

  • Question 22/35
    1 / -0

    Hitesh is 40 years old and Ronnie is 60 years old. How many years ago was the ratio of their ages 3:5?

    Solutions

    Suppose, the ratio was 3 :5, x years ago.

    Then, (40-x)/(60-x) = 3/5 ⇔5(40-x)= 3(60-x)⟺2x=20 ⇔x=10 years

     

  • Question 23/35
    1 / -0

    A train 600 m long passes a pole in 9 seconds. What is the speed of the train in km/hr?

    Solutions

    Speed = Distance/Time =600/9 m/sec

    =600/9×18/5 km/hr = 240 km/hr [∵ 1 m/sec = 18/5 km/hr]

     

  • Question 24/35
    1 / -0

    The sum of three consecutive natural numbers each divisible by 3 is 72. What is the largest among them?

    Solutions

    3x + (3x + 3) + (3x + 6) = 72

    9x + 9 = 72 => 9x = 72 - 9

    Or x=63/9=7

    The largest of them is 27.

     

  • Question 25/35
    1 / -0

    A and B together can complete a piece of work in 35 days while A alone can complete the same work in 60 days. In how many days, B alone will be able to complete the same work?

    Solutions

    B’s one day work = 1/35-1/60=(12-7)/60=5/420=1/84

    ∴ Required number of days =84

     

  • Question 26/35
    1 / -0

    The third proportional to 0.36 and 0.48, is:

    Solutions

    Let the third proportional to 0.36 and 0.48 be x.

    Then, 0.36: 0.48 :: 0.48 : x

    ⇔x=((0.48×0.48)/0.36)=0.64

     

  • Question 27/35
    1 / -0

    A rectangular field of dimensions 30 m x 20 m is surrounded by a footpath of uniform width. If the area of the footpath is 216 m2, what is the width of the footpath?

    Solutions

    Let the width of the footpath be w.

    So, area of the footpath = (30+2w)*(20+2w) - 20*30 = 216

    From the options, w = 2 satisfies the equation. So, option a) is the answer.

     

  • Question 28/35
    1 / -0

    55% of a number is more than one-third of that number by 52. What is two-fifth of that number?

    Solutions

    Let the number be x.

    55/100 x=1/3 x+52

    13/60 x=52

    => x=240

    2/5 x=2/5 x 240 = 96

     

  • Question 29/35
    1 / -0

    The price of a book is reduced by 25%, what is the ratio of change in price to the old price?

    Solutions

    Let the old price be Rs. 100.

    ∴ (change in price/(old price=25/100=1/4

     

  • Question 30/35
    1 / -0

    Find the compound interest on Rs. 9375 at 8% per annum for 2 yr.

    Solutions

    A = P(1+R/100)t= Rs. [9375×(1+8/100)2] = Rs.[9375×27/25×27/25] = Rs. 10935

    CI = 10935-9375 = Rs. 1560

     

  • Question 31/35
    1 / -0

    A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

    Solutions

    S.I. for 1 year = Rs. (854 - 815) = 39.

    S.I. for 3 years = Rs. (39 × 3) = Rs. 117.

    ∴ Principal =Rs. (815 - 117) = Rs. 698.

     

  • Question 32/35
    1 / -0

    Find the average of the first 100 natural numbers

    Solutions

    Sum of the first n natural numbers =n(n+1)/2

    Sum of the first 100 natural numbers = (100×(100+1))/2 = 5050

    Average of first 100 natural numbers = 5050/100= 50.5

     

  • Question 33/35
    1 / -0

    Find the average weight of 8 students 49 kg, 47 kg, 46 kg, 42 kg, 39 kg, 48 kg, 50 kg and 43 kg.

    Solutions

    Average weight = (49+47+46+42+39+48+50+43)/8

    = 364/8

    = 45.5 kg

     

  • Question 34/35
    1 / -0

    Find the LCM of 23×34×52,22×33×7,52×72

    Solutions

    LCM of given numbers =23×34×52×72

    (Take the greatest power of each term)

     

  • Question 35/35
    1 / -0

    If CP = Rs. 20, % loss = 25%, then SP =?

    Solutions

    SP= ((100-loss%)/100)× CP = 75/100×20 = Rs. 15

     

Close button icon
User Profile
-

Correct (-)

Wrong (-)

Skipped (-)

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11
  • 12
  • 13
  • 14
  • 15
  • 16
  • 17
  • 18
  • 19
  • 20
  • 21
  • 22
  • 23
  • 24
  • 25
  • 26
  • 27
  • 28
  • 29
  • 30
  • 31
  • 32
  • 33
  • 34
  • 35
Mockers logo Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Arrow pointer icon
Click on Allow to receive notifications
Notification bell icon ×
Open Now