Quant Test for SBI PO 2018 Prelim Exam Set – 46

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 will come in place of question mark ‘?’ in the given expression?

  1. 534.596 + 61.472 – 496.708 = ? + 27.271
    66
    80
    72
    93
    104
    Option C
    534.596 + 61.472 – 496.708 = ? + 27.271
    => ? = 27.271 – 534.596 – 61.472 + 496.708
    => ? = 72.089 ==72

     

  2. 56% of 958 + 67% of 1008 = ?% of 2000
    50
    33
    45
    61
    77
    Option D
    56% of 958 + 67% of 1008 = ?% of 2000
    =>536.48 + 675.36 = ?% of 2000
    => ? = 1211.84/20
    => ? = 60.592 == 61

     

  3. (?)² + (23)² = (106)² – (46)² – 133
    66
    111
    75
    87
    92
    Option E
    (?)² + (23)² = (106)² – (46)² – 133
    =>?^2 = 11236 – 2116 – 133 – 529
    => ?^2 = 8458
    => ? = 92

     

  4. 515.15 – 15.51 – 1.51 – 5.11 – 1.11 =?
    491.91
    371.11
    481.91
    391.11
    291.91
    Option A
    515.15 – 15.51 – 1.51 – 5.11 – 1.11 =?
    =>? = 491.91

     

  5. 811.81 + 88.11 + 0.88 + 1.88 + 8 = ?
    888
    911
    845
    900
    922
    Option B
    811.81 + 88.11 + 0.88 + 1.88 + 8 = ?
    => ? = 910.68 == 911

     

  6. Mixture A contains milk and water in the ratio of 3:2 . Another mixture B, containing milk and water in the ratio is 5:1 , is mixed with mixture A such that the ratio of the milk and water in the resultant mixture becomes 2:1 . If the quality of mixture B is 18 litres, then find the quantity of mixture A .
    40 l
    25 l
    30 l
    52 l
    45 l
    Option E
    Let the quantity of milk and water in mixture B be 5y and 1y litres resp.
    5y + y = 18
    => 6y = 18
    => y = 3
    The quantity of milk in B = 15 litres
    The quantity of water in B = 3 litres
    Now, [3x+15]/[2x+3] = 2/1 => x = 9litres
    Hence , the quantity of mixture A = 3x+2x = 27 + 18 = 45 litres

     

  7. A train running at a speed of 54 km/hr. crosses a man standing on the platform in 28 seconds. What would be the length of the platform which is 33.33% more than the length of the train?
    620 m
    500 m
    560 m
    660 m
    440 m
    Option C
    Length of the train = 15 * 28 = 420 m
    Length of the platform = 420 + 33.33% of 420
    = 560 m

     

  8. In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. A student attempted all the 200 questions and scored in all 200 marks. Find the number of questions, he answered correctly .
    100
    90
    70
    80
    60
    Option D
    Let the number of corrected answers be ‘x’.
    Number of wrong answers be 200−x
    4x− (200−x) =200
    =>5x=400
    =>x=80

     

  9. Mrs. Sharma invests 15% of her monthly salary, i.e., Rs. 4428 in Mutual Funds. Later she invests 18% of her monthly salary on Pension Policies also she invests another 9% of her salary on Insurance Policies. What is the total monthly amount invested by Mrs. Sharma?
    Rs.10390.3
    Rs.12398.4
    Rs.11300.5
    Rs.12250.4
    Rs.12520.5
    Option B
    15% → 4428
    1% →4428/15
    ( 15+18+9) =42%
    →4428/15×42 → Rs.12398.4

     

  10. A bag contains 10 coins of different values (Re.1, Rs.2 and Rs.5). If the probability of drawing a Rs.2 coin from the bag is 1/5 and the number of Re.1 coin is equal to the number of Rs.5 coins, then find the probability that each coin is of different value when three coins are drawn randomly.
    7/13
    2/11
    1/15
    4/15
    4/11
    Option D
    Number of 10 coins in the bag = 10
    Number of Rs.2 coins in the bag = 1/5*10 = 2
    Let the number of Re.1 coins be x.
    Then, x+x+2 = 10
    => 2x = 8
    => x = 4
    Number of Re.1 coin = 4
    Number of Rs.5 coin = 4
    Number of Rs.2 coin = 2
    Required probability = (2C1+4C1+4C1)/10C3 = 4/15

     


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