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SBI PO Prelims 2018 Aptitude Test 1
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SBI PO Prelims 2018 Aptitude Test 1
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  • Question 1/20
    1 / -0

    DIRECTION: What should come in place of the question mark (?) in the following question?, 

    Solutions

     

  • Question 2/20
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    DIRECTIONS: In the given question two expressions numbered I and II are given. You have to solve both the expressions and give answer: , 

  • Question 3/20
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    DIRECTIONS: In the given question two expressions numbered I and II are given. You have to solve both the expressions and give answer: , 

    I. 1596 × 196 ÷ 49 – 484 , 

    II. 81 × 49 + (4)3 × 29

  • Question 4/20
    1 / -0

    DIRECTION: What should come in place of the question mark (?) in the following question? , 

    (5 x 7)% of (34 x 55) + 456.60 = 699.1 + ?

    Solutions

     

  • Question 5/20
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    DIRECTIONS: In the given question two expressions numbered I and II are given. You have to solve both the expressions and give answer:

    I. (243)÷ (27)× 32 ÷ 12

  • Question 6/20
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    DIRECTIONS: Find the missing term in the following series. , 

    11240, 9524, 8204, 7214, ?, 5990

    Solutions

    11240 – (1716) = 11240 – (1728 – 12) = 9524

    9524 – (1320) = 9524 – (1331 – 11) = 8204

    8204 – (990) = 8204 – (1000 – 10) = 7214

    7214 – (720) = 7214 – (729 – 9) = 6494

    6494 – (504) – (512 – 8) = 5990

     

  • Question 7/20
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    DIRECTIONS: Find the missing term in the following series. , 

    5111808, 159744, 9984, 1248, ?, 156

    Solutions

    5111808 ÷ 32 = 159744

    159744 ÷ 16 = 9984

    9984 ÷ 8 = 1248

    1248 ÷ 4 = 312

    312 ÷ 2 = 156

     

  • Question 8/20
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    DIRECTIONS: Find the missing term in the following series. , 

    363, ?, 431, 516, 635, 788

    Solutions

    363 + 17 × 1 = 380

    380 + 17 × 3 = 431

    431 + 17 × 5 = 516

    516 + 17 × 7 = 635

    635 + 17 × 9 = 788

     

  • Question 9/20
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    DIRECTIONS: Find the missing term in the following series. , 

    856, 874, 899, ?, 990, 1066

    Solutions

    856 + 18 = 874

    874 + (18 + 7) = 874 + 25 = 899

    899 + (25 + 7 + 5) = 899 + (25 + 12) = 899 + 37 = 936

    936 + (37 + 12 + 5) = 936 + (37 + 17) = 936 + 54 = 990

    990 + (54 + 17 + 5) = 990 + (54 + 22) = 990 + 76 = 1066

    1066 + (76 + 22 + 5) = 1066 + (76 + 27) = 1066 + 103 = 1169

     

  • Question 10/20
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    DIRECTIONS: Find the missing term in the following series. , 

    122, ?, 741, 2968, 14845, 89076

    Solutions

    122 × 2 + 2 = 246

    246 × 3 + 3 = 741

    741 × 4 + 4 = 2968

    2968 × 5 + 5 = 14845

    14845 × 6 + 6 = 89076

     

  • Question 11/20
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    A bag contains a total of 70 coins, in denominations of 50 paise coins and Rs.2 coins, totally worth Rs. 80. Find the number of 50 paise coins in the bag.

    Solutions

    Let the number of 50p coins and Rs.2 (200p) coins be a and b.

    Total value = Rs. 80 = 8000p

    Therefore, 50a + 200b = 8000

    a + 4b = 160---------(1)

    And a + b = 70--------(2)

    Subtract (2) from (1)

    3b = 90

    b = 30 and a = 40

     

  • Question 12/20
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    If the height of a cyllinder becomes 6 times and area of base becomes 1/9 times, then curved surface area of the cyllinder will become how many times?

    Solutions

     

  • Question 13/20
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    Two times Abhi’s age is thrice the age of Balu and four times Balu’s age is five times the age of Srikant. Also six times Srikant’s age is four times the age of David. If Abhi’s age is 30 years, what is David’s age?

    Solutions

    Let the ages of Abhi, Balu, Srikant and David are a, b, s and d respectively.

    Given, a = 30 years.

    2a = 3b; b = (2/3)a = (2/3) × 30 = 20 years.

    4b = 5s; s = (4/5)b = (4/5) × 20 = 16 years.

    6s = 4d; d = (6/4)s = (6/4) × 16 = 24 years.

     

  • Question 14/20
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    There are 50 balls numbered 1 to 50 in a tray. Every ball bearing a number which is a multiple of some other number ‘n’ (n > 1) is removed and put in a basket. What is the total number of balls removed?

    Solutions

    The balls having prime number are not removed.

    Also the ball with number 1 is not removed

    Number of prime numbers between 1 to 50 = 15

    Number of balls that are removed = 50 – (15 + 1) = 34

     

  • Question 15/20
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    Pawan and Raman started a business by investing Rs.18000 and Rs.25200 respectively. What is the share of Pawan in the profit of Rs.8640 earned at the end of the year?

    Solutions

    The ratio of the shares of Pawan and Raman in the profit = Rs.18000 : Rs.25200 = 5 : 7

    Therefore, share of Pawan in the profit = (5/12) x Rs.8640 = Rs.3600

     

  • Question 16/20
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    Anil marks up the price of an article by 50% and then offers a discount of 20% to Raman. Raman sells it for Rs.20 more than the price at which he purchased it. If Raman’s selling price is 30% more than the original cost price of the article, then Raman’s profit percentage is

    Solutions

    Let the cost price of Anil = Rs.a

    Then his marked price = Rs.1.5a

    and his selling price = Rs.1.5a(0.8) = Rs.1.2a

    Raman’s cost price = Rs. 1.2a

    Raman’s selling price = Rs.(1.2a + 20) = Rs.1.3a

    ⇒ a = 200

    ∴ Raman’s cost price = 1.2a = Rs. 1.2(200) = 240

    ∴ Raman’s profit % = (20/240) x 100% = 8.33%

     

  • Question 17/20
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    A train X running at a speed of 60 kmph crossed another train Y of length 540 m running in the opposite direction at a speed of 72 kmph in 33 seconds. What is the length of train X?

    Solutions

    let the length of train X = ‘a’ m

    Relative speed of train X with respect to train Y = (60 + 72) x 5/18 m/s = 110/3 m/s

    Therefore, (a + 540) = (110/3) x 33

    a + 540 = 1210

    a = 1210 – 540 = 670 m

     

  • Question 18/20
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    Variety A of tea costs Rs. 180 per kg. Variety B of tea costs Rs. 270 per kg. Variety C of tea was formed by mixing A and B in the ratio 3 : 2. Variety D of tea was formed by mixing A and B in the ratio 5 : 4. C and D were mixed in a certain ratio and the resulting mixture was sold at Rs. 261 per kg at 20% profit. Find the ratio in which C and D were mixed.

    Solutions

    Costs of tea of variety of C =  = Rs.216/kg

    Costs of tea of variety of D =  = Rs.220/kg

    S.P. of mixture of C and D = Rs.261/kg

    And profit = 20%

    C.P. of mixture of C and D =  = Rs.217.5/kg

    Let the required ratio = x : y, then

     = 217.5

    216x + 220y = 217.5x + 217.5y

    2.5y = 1.5x

    x : y = 5 : 3

     

  • Question 19/20
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    In a class of 300 students, 130 play football, 160 play hockey, 150 play cricket, 70 play hockey and cricket, 40 play football and cricket, 84 play football and hockey and 16 play all the three games. , 

    The number of students who do not play any of the three games is

    Solutions

    Number of students who play = 22 + 22 + 56 + 68 + 24 + 54+ 16 = 262

    Number of students who do not play any game = 300 – 262 = 38

     

  • Question 20/20
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    The ratio of copper and zinc in a 63 kg alloy is 4 : 3. Some amount of copper is extracted from the alloy and the ratio becomes 10 : 9. How much copper is extracted.

    Solutions

    Copper: Zinc

    4:3

    Copper = (4/7) × 63 = 36 kg

    Zinc = (3/7) × 63 = 27 kg

    Let x kg copper is extracted

    Remaining copper = 36 – x kg

    New ratio = (10/9)

     

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