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Aptitude Test 247
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Aptitude Test 247
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  • Question 1/10
    1 / -0.33

    Directions For Questions

    Directions : The following bar graph represents number red coloured balls decorated in five different halls.

    (Total number of balls in each hall = Number of red balls + number of blue balls)

    The following table shows the ratio of number of red balls and number of blue balls decorated in five different halls.

    ...view full instructions

    Find the total number of red balls in all the halls together.

    Solutions

    Number of red balls in Hall A = 70

    Number of red balls in Hall B = 64

    Number of red balls in Hall C = 52

    Number of red balls in Hall D = 80

    Number of red balls in Hall E = 58

    Required sum = 70 + 64 + 52 + 80 + 58 = 324

     

  • Question 2/10
    1 / -0.33

    What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.)

    24.98% of (?)2 = 93.02 + 692.91 ÷ 3.08

    Solutions

    25% of (?)2 = 93 + 693/3 = 93 + 231 = 324

    ?2 = 324 x 100/25 = 1296

    ? = 36

     

  • Question 3/10
    1 / -0.33

    Find the next term in the series.

    5, 6, 20, 87, 412, ?

    Solutions

    The pattern is as follows:

    (5 x 1) + 13 = 6

    (6 x 2) + 23 = 20

    (20 x 3) + 33 = 87

    (87 x 4) + 43 = 412

    (412 x 5) + 53 = 2185

     

  • Question 4/10
    1 / -0.33

    A shopkeeper marked an article at Rs.300 and after giving a discount of 30% he sold that article at 5% profit. If shopkeeper sold the article in order to earn a profit of 20%, then find the percentage increase in selling price of the article.

    Solutions

    MP of article = 300

    SP = 70% of 300 = 210

    profit% = 5%

    CP = 210 x 100/105 = 200

    New profit% = 20%

    New SP = 120% of 200 = 240

    Difference between selling price = 240 - 210 = 30

    Percentage increase = (30/210) x 100 = 14.28%

     

  • Question 5/10
    1 / -0.33

    Find the next term in the series.

    11, 18, 45, 164, 795, ?

    Solutions

    The pattern is as follows:

    (11 - 2) x 2 = 18

    (18 - 3) x 3 = 45

    (45 - 4) x 4 = 164

    (164 - 5) x 5 = 795

    (795 - 6) x 6 = 4734

     

     

  • Question 6/10
    1 / -0.33

    If certain principal invested at the rate of 15% per annum compounded annually becomes Rs.55545 in 2 years, then what will be final amount if same principal is invested at the rate of 20% per annum compound interest compounded annually in 3 years?

    Solutions

    Let principal be P

    Amount after 2 years if it is invested at the rate of 15% per annum compound interest compounded annually.

    = P x (1 + (15/100))2

    = P x (1 + (3/20))2

    = P x (23/20)2 = 55545

    P = Rs.42000

    Required final amount

    = 42000 x (1 + (20/100))3

    = 42000 x (6/5)3

    = Rs.72576

     

     

  • Question 7/10
    1 / -0.33

    Find the next term in the series.

    28, 57, 173, 868, 6080, ?

    Solutions

    The pattern multiplies with prime numbers to the previous term and adds an increasing constant as show below:

    28 x 2 + 1 = 57

    57 x 3 + 2 = 173

    173 x 5 + 3 = 868

    868 x 7 + 4 = 6080

    6080 x 11 + 5 = 66885

     

  • Question 8/10
    1 / -0.33

    What approximate value will come in place of question mark (?) in the following question?

    Solutions

    ? ≈ (800/8) + 5 x 11 - (140/2) + 41

    = 100 + 55 - 70 + 41

    =126

     

  • Question 9/10
    1 / -0.33

    What approximate value will come in place of question mark (?) in the following questions? (You are not expected to calculate the exact value.)

    640.05 ÷ 32.02 x 46.95 = ? + 29.03

    Solutions

    ? = 640 ÷ 32 x 47 - 29

    = 20 x 47 - 29

    = 940 - 29 = 911

     

  • Question 10/10
    1 / -0.33

    The average weight of a group of 12 persons was increased by 2.5 kg when a new person came in place of one of them weighing 44 kg. Again, the new person left the group. What had become the average weight of the remaining person after leaving the new person?

    Solutions

    Let the average weight of 12 person =a kg

    The sum = 12a kg

    After leaving one among them weighing 44 kg

    The sum = 12a - 44 kg

    Let the weight of new person = b kg

    Then, 12a - 44 + b = 12(a + 2.5)

    b = 44 + 30 = 74 kg

    Again, after leaving the new person

    The sum of the weight = 12a - 44 - 74 = 12a - 118 kg

    The average = (12a - 118)/11

    Without knowing the value of a, we cannot determine the average

     

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