Quant Test for NIACL AO Prelims Exam Set – 2

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

  1. 23.97% of 7624.89 + 9.988% of ? – 14.78% of 1779.89 = 64.78% of 2959.9
    1005
    1100
    1215
    1000
    1805
    Option E
    23.97% of 7624.89 + 9.988% of ? – 14.78% of 1779.89 = 64.78% of 2959.9 =>24% of 7625 + 10% of ? – 15% of 1780 = 65% of 2960
    => 1830 + 0.20* ? – 267 = 1924
    =>? = 1805

     

  2. (4912.7)^1/3 – (5329.10)^1/2 + ?^2 = (7920.99)^1/2 + (13823.78)^1/3
    5
    10
    8
    13
    6
    Option D
    (4912.7)^1/3 – (5329.10)^1/2 + ?^2 = (7920.99)^1/2 + (13823.78)^1/3
    => (4912)^1/3 – (5329)^1/2 + ?^2 = (7921)^1/2 + (13824)^1/3
    => 17 – 73 + ?^2 = 89 + 24
    => ?^2 = 89 + 24 + 73 – 17
    => ? = 13

     

  3. (24.55)^5 / (8.42)^6 * (2.9)^3 = (2.9)^(? – 2)
    5
    4
    8
    7
    9
    Option C
    (24.55)^5 / (8.42)^6 * (2.9)^3 = (2.9)^(? – 2)
    => (2.9^3)^5 / (2.9^2)^6 * (2.9)^3 = (2.9)^(?-2)
    => ? = 8

     

  4. 17.98*7(1/7)% of 363.98 + 39.98% of 519.98 = ?^2
    26
    20
    25
    12
    19
    Option A
    17.98*7(1/7)% of 363.98 + 39.98% of 519.98 = ?^2
    => ?^2 = 18*1/14*364 + 0.4*520
    => ?^2 = 18*26 + 208
    => ? = 26

     

  5. 14(3/5) + 8(2/3) – 5(4/5) = (?)^1/2 + 6(7/15)
    92
    100
    121
    80
    111
    Option C
    14(3/5) + 8(2/3) – 5(4/5) = (?)^1/2 + 6(7/15)
    => (?)^1/2 = 73/5 + 26/3 – 29/5 – 97/15
    => (?)^1/2 = 44/5 + 26/3 – 97/15
    => (?)^1/2 = [132 + 130 – 97]/15
    => ? = 121

     

  6. A boat can cover a “r” km upstream in 7 hours. It can cover “r+16” km downstream in 5 hours. Find the time taken by boat to cover 66 km upstream and 70 km downstream if the speed of the boat in still water and speed of stream are in the ratio 4:1 resp.
    8 hours
    9 hours
    7 hours
    6 hours
    5 hours
    Option B
    Let the speed of the stream be x km/hr and speed of the still water be 4x km/hr.
    Upstream =4x – x = 3x kmhr.
    Downstream = 5x km/hr.
    r = 7*3x = 21x
    And r + 16 = 5*5x
    =>21x + 16 = 25x
    => x = 4
    Downstream speed = 20 km/hr.
    Upstream speed = 12 km/hr.
    Time taken to cover 66 km upstream and 70 km downstream speed
    = 66/12+ 70/20 = 9 hours

     

  7. Radius of the circular filed is “r” metre. Find the area of the circular field if the cost of fencing the field along its circumference at the rate of Rs. 18.5 per meter is Rs. 4652.
    44567 m^2
    41400 m^2
    40450 m^2
    49896 m^2
    44050 m^2
    Option D
    Circumference of the circular field = 2*22/7*r = 14652 /18.5
    => r = 126
    Area of the circular field = 22/7*126*126 = 49896 m^2

     

  8. A.B and C together can complete work in “x” days. B alone can complete the work in “3x” days. A is 20% efficient than B. C alone can complete the work in 30 days. In how many days A and C together can complete the work ?
    7 days
    9 days
    12 days
    10 days
    8 days
    Option C
    Time taken by A alone to complete the work = 3x*(100/120) = 5x/2 days
    Now, 1/3x + 2/5x + 1/30 = 1/x
    => [15-5-6]/15x = 1/30
    => x = 8
    A alone can complete the work = 5*8/2 = 20 days
    Time taken by A and C together to complete the work = 1/[1/20+1/30] = 12 days

     

  9. Present ages of Radha and Geeta are in the ratio 3:4 resp. Mohan is 6 years older than Geeta. 8 years ago the ratio of the ages of Mohan and Sohan was 11:24 resp. If the present age of Sohan is two more than thrice teh age of Radha then find the present average age of Radha,Geeta, Mohan and Sohan.
    28 years
    33 years
    32 years
    25 years
    30 years
    Option C
    Let the present ages of Radha and Geeta be 3x and 4x resp. and the present ages of Mohan and Sohan be 11y and 24y resp.
    Present age of Mohan and Sohan = 11y+8 years and 24y+8 years resp.
    Now, 4x+6 = 11y+8
    => x= [11y+2]/4
    Also, 24y + 8 = 3*3x+2
    => y = 2
    => x = 6
    Present ages of Radha, Geeta, Mohan and Sohan are 18 years, 24 years, 30 years and 56 years resp.
    Required average age = (18+24+30+56)/4 = 32 years

     

  10. A bag contains 10 red, 11 blue and “x” pink balls. Two balls are randomly drawn from the bag and the probability that both the balls are pink is 1/8. What is the probability that one ball is red and other ball is blue?
    7/20
    3/22
    5/24
    6/23
    5/21
    Option C
    Probability of both balls are pink = xC2/(10+11+x)C2 = 1/8
    =>xC2/(21+x)C2 = 1/8
    => x(x-1)/(21+x)(20+x) = 1/8
    => 7x^2 – 49x – 420 = 0
    => x = 12, -5
    Bag contains pink balls = 12
    Probability of one red and one blue ball = 10C1*11C1/33C2 = 5/24

     


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