Quant Test for IBPS Clerk 2018 Prelim Exam Set – 10

Directions(1-5): What will come in place of question mark “?” in the following questions.

  1. ? = (84.97 *102.06)/(17.03)^2.01 + 11.9*4.9
    112
    100
    98
    90
    85
    Option D
    ? = (84.97 *102.06)/(17.03)^2.01 + 11.9*4.9
    => ? = (85*102)/(17*17)+12*5
    => ? = 30+60
    => ? = 90

     

  2. ?^2 = 17*23 + 131 – 9^2
    21
    32
    30
    25
    20
    Option A
    ?^2 = 17*23 + 131 – 9^2
    => ?^2 = 391 + 131 – 81
    => ?^2 = 441
    => ? = 21

     

  3. [18*(?^3 /5)]/3^2 + 18 = 68
    3
    7
    5
    8
    4
    Option C
    [18*(?^3 /5)]/3^2 + 18 = 68
    => 18*(?^3/5)/9 = 50
    => ?^3/5= 25 =>?^3 = 125
    => ? = 5

     

  4. ? = 47% of (14^2 + 4)+35% of (14*5*4)
    192
    180
    177
    185
    199
    Option A
    ? = 47% of (14^2 + 4)+35% of (14*5*4)
    => ? = 47% of (196+4)+ 35% of 280
    => ? = 0.47*200+ 0.35*280
    => ? = 94+98
    =>? = 192

     

  5. ? + 14 = (40% of 215 / (1369)^1/2)*3
    333
    300
    355
    398
    390
    Option C
    ? + 14 = (40% of 215 / (1369)^1/2)*3
    => ? + 14 = (86+37)*3
    => ? = 369 – 14
    => ? = 355

     

  6. If the total number of biscuit packets of cuboidal shape each of the dimension 2.5 cm*2cm *1cm is stored in a cuboidal carton of dimensions 25 cm * 7 cm *4 cm , then find the total cost of all the biscuit packet stored in the carton if each biscuit packet costs Rs. 10.
    Rs.1500
    Rs.1800
    Rs.1400
    Rs.1200
    Rs.1000
    Option C
    Volume of the cuboidal carton = 25*7*4 = 700 cm^3
    Volume of each biscuit packet = 2.5 *2*1 = 5 cm^3
    Required number of biscuit packet = 700/5 = 140
    Total cost of all the biscuit packet stored in the carton = 140*10 = Rs.1400

     

  7. If a two-digit number when increased by 20% then we get the reverse of the number. If the sum of the original number and its reverse is 99, then find the square of the original number.
    2500
    2025
    2121
    2030
    2000
    Option B
    Let the original number be x.
    Then, the reverse of the number = 1.2x x+1.2x = 99
    => x = 45
    So, x^2 = 45^2 = 2025

     

  8. The average age of 11 players of the Indian cricket team is decreased by 1 year when Ishant Sharma aged 31 years and Suresh Raina aged 28 years are replaced by two new players (Shikhar Dhawan and Bhubaneshwar Kumar). Find the average age of those new players.
    15 years
    24 years
    27 years
    18 years
    20 years
    Option B
    Decrease in the sum of the ages of all the 11 players on replacedment = 11*1 = 11 years
    Sum of the ages of the two new players = 31+28- 11 = 48 years
    Average age of new players = 48/2 = 24 years

     

  9. X deposited Rs. 800000 at 6% simple interest for 5 years whereas Y deposited Rs. 400000 at 4% compound interest for 2 years. What was the ratio of the accumulated amounts of X and Y ?
    125:57
    122:53
    121:50
    125:52
    132:51
    Option D
    Amount for X = 800000*(1+0.06*5)= Rs. 1040000
    Amount for Y = 400000*(1+0.04)^2 = Rs. 432640
    Required ratio = 1040000:432640 = 125:52

     

  10. A player gets 2 chances to roll a die in a game . He wins if the sum of the numbers obtained on the faces of the die in the two attempts is divisible by 6. What is the probability that the player wins the game?
    1/3
    1/5
    1/7
    1/6
    1/8
    Option D
    Favourable cases
    = (1,5),(5,1),(4,2),(2,4),(3,3),(6,6)
    Required Probability = 6/36 = 1/6

     


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