Quant Test for IBPS Clerk 2018 Prelim Exam Set – 19

Banya

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

  1. (1575/63*24+48)*3 = ?^2*6^3
    4
    5
    6
    8
    3
    Option E
    (1575/63*24+48)*3 = ?^2*6^3
    => (25*24+48)*3 = ?^2*216
    =>?^2 = (648*3)/216 = 9
    => ? = 3

     

  2. 1610*(17.64)^1/2/49 = ?^2 – (961)^1/2
    18
    13
    10
    16
    14
    Option B
    1610*(17.64)^1/2/49 = ?^2 – (961)^1/2
    => 1610*4.2/49 = ?^2 – 31
    => ?^2 = 138 – 31
    => ?^2 = 169
    => ? = 13

     

  3. (4095.8)^1/3 + 106.97% of 1599.8 = ?^3
    8
    15
    10
    12
    13
    Option D
    (4095.8)^1/3 + 106.97% of 1599.8 = ?^3
    => (4095)^1/3 + 107% of 1600 = ?^3
    => 16+1.07*1600 = ?^3
    => ?^3 = 1728
    => ? = 12

     

  4. 81^3 /27^5 * 9^7 = 3^?
    9
    13
    11
    14
    17
    Option C
    81^3 /27^5 * 9^7 = 3^?
    => 3^12 / 3^15*3^14 = 3^?
    => 3^11 = 3^?
    => ? = 11

     

  5. 32*45+16*35 + ? = 48*55+14*65
    1440
    1550
    1300
    1250
    1050
    Option B
    32*45+16*35 + ? = 48*55+14*65
    => 1440+560+? = 2640+910
    => ? = 1550

     

  6. A boat goes 90 km upstream and comes back downstream to its original place in overall 16 hours. The speed of boat in still water is x km/hr. Find the value of x if the speed of stream is 3 km/hr.
    14
    10
    15
    12
    16
    Option D
    Let speed of boat in still water be x km/hr. 90/[x+3]+ 90/[x-3] = 16 => 4x^2 -45x – 36 = 0
    => 4x(x-12)+3(x-12) = 0
    => x = 12,-3/4
    Hence, the value of x = 12 km/hr.

     

  7. A cricket player has an average of x runs after 10 matches. He scored 90 runs in his 11th match due to which the average of his runs becomes (x+7) in the 11 matches. Find his average runs in the 11 matches.
    18
    20
    15
    22
    24
    Option B
    His average runs after 10 matches be x.
    10x+90 = 11(x+7)
    => x= 13
    Average of his runs after 11th match = 13+7 = 20

     

  8. Akash bought a laptop at Rs. x and marked it 40% above the cost price and sold it to Rajan after giving 15% discount thus made a profit of Rs. 9880. What would have been his amount of profit or loss if he had sold it at 30% discount.
    1040
    1250
    1440
    1200
    1000
    Option A
    Let CP be x. MP = 1.40*x = Rs. 1.40x
    SP = 0.85*1.40x= Rs. 1.19x
    Profit = 1.19x – 1.40x = 9880
    => x = Rs. 52000
    New SP = 0.70*1.40x = Rs. 0.98x
    Loss = x – 0.98x = 0.02x = 0.02*52000 = Rs. 1040

     

  9. The present ages of Gaurav and Ashwani are in the ratio 11:13 resp. And the ratio of their age 12 years ago was 4:5 resp. If after Y years the ratio of their age would be 13:15 resp. The find the value of Y.
    9
    10
    6
    8
    5
    Option D
    Now , [11x-12]/[13x-12] = 4/5
    => x = 4
    Gaurav = 44 years and Ahswani = 52 years
    After Y years, 44+Y/52+Y = 13/15
    =>Y = 8 years

     

  10. A bag contains 30 oranges out of which 8 are rotten. 3 oranges are drawn randomly and the probability that at most one orange is rotten is “p” . Find the value of “1-p”.
    24/141
    23/145
    22/143
    20/141
    24/145
    Option E
    At most one rotten orange means either no rotten orange or one rotten orange.
    Probability = [22C3+22C2*8C1]/30C3 = 121/145
    1-P = 1 – 121/145 = 24/145

     


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