Quant Test for NIACL AO Prelims Exam Set – 7

Banya

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

  1. 29.98% of 1780.21 + ? + 19.72% of 699.23 = 64.83% of 2359.69
    785
    800
    755
    820
    860
    Option E
    29.98% of 1780.21 + ? + 19.72% of 699.23 = 64.83% of 2359.69
    => 30% of 1780 + ? + 20% of 700 = 65% of 2360
    => 534 + ? + 140 = 1534
    =>? = 860

     

  2. (20.99)^2 – (13.01)^2 + (291.12 + 76.92)/? = 55.92% of 499.88
    40
    35
    46
    39
    30
    Option C
    (20.99)^2 – (13.01)^2 + (291.12 + 76.92)/? = 55.92% of 499.88
    => 21^2 – 13^2 + (291+77)/? = 56% of 500
    => 441 – 169 + 368/? = 280
    => ? = 46

     

  3. (2210)^1/2 + 979.90 / 13.927 = ?^2 – 5.099% of 1579.99
    14
    12
    10
    15
    16
    Option A
    (2210)^1/2 + 979.90 / 13.927 = ?^2 – 5.099% of 1579.99
    => (2210)^1/2 + 979/14 = ?^2 – 5% of 1580
    => 47 + 70 = ?^2 – 79
    => ? = 14

     

  4. 11.92% of 1199.98 – ?* 13.9*20.99 / (196)^1/2 = 43.21 – 297.78
    22
    19
    20
    14
    16
    Option B
    11.92% of 1199.98 – ?* 13.9*20.99 / (196)^1/2 = 43.21 – 297.78
    => 12% of 1200 – ?*14*221/(196)^1/2 = 43 – 298
    => 144 – ?*14*21/(196)^1/2 = – 255
    => 21 * ? = 399
    => ? = 19

     

  5. (2.197)^4 / (1.69)^2 *(1.3)^2 = (1.3)^(? – 2)
    11
    8
    7
    10
    12
    Option E
    (2.197)^4 / (1.69)^2 *(1.3)^2 = (1.3)^(? – 2)
    => (1.3)^12/(1.3)^4 *(1.3)^2 = (1.3)^(?-2)
    => (1.3)^10 = (1.3)^(?-2)
    => ? = 12

     

    6. Directions(6-10): The table shows the marks scored by five students in five different subjects.

  6. Marks obtained by D in all the six subjects together are approximately how much percent more than the marks scored by A in all the six subjects together?
    7%
    3%
    4%
    6%
    5%
    Option D
    Marks obtained by A in all subjects = 485
    Marks obtained by D in all subjects = 513
    Difference = 513 – 485 = 28
    Required% = 28/485*100 = 5.77% == 6%

     

  7. Marks obtained by B,C and D in Science together are approximately how much percent less than marks obtained by A,B and E in G.K. together.
    22.11%
    31.52%
    18.41%
    32.23%
    30.04%
    Option B
    Marks obtained by B,C and D in Science together = 120+105+90 = 315
    Marks obtained by A,B and D in G.K. together = 180+120+160 = 460
    Required% = (460-315)/460*100 = 31.52%

     

  8. Ram scored 20% more marks than A in all the subjects together and Shyam scored 25% less than C in all the subjects together. Find the difference between the total marks of Ram and Shyam.
    185
    200
    220
    231
    193
    Option D
    Marks obtained by A = 485
    Marks obtained by Ram = 1.2*485 = 582
    Marks obtained by C = 468
    Marks obtained by Shyam = 0.75*468 = 351
    Difference = 582 – 351 = 231

     

  9. How many number of students scored more than 65% of maximum marks.
    6
    3
    4
    5
    2
    Option C
    Maximum marks = 100+80+150+120+75+200 = 725
    65% of maximum marks = 0.65*725 = 471.75
    Marks obtained by A = 485
    Marks obtained by B = 480
    Marks obtained by C = 468
    Marks obtained by D = 513
    Marks obtained by E = 545
    Only 4 students = A,B,D and E.

     

  10. Average percentage of marks per student scored by the five students in which subject are the highest?
    Science
    English
    Hindi
    G.K.
    Maths
    Option E
    Average percentage per student for Maths
    = (70+72+73+89+91)/5*100*100 = 79%

     


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