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INET Numerical Ability & Average Test 541
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INET Numerical Ability & Average Test 541

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  • Question 1/5
    1 / -0.33
    The average of first 10 natural numbers is how much less than the average of first 10 prime numbers?
    Solutions

    Sum of first n natural number =

    Sum of first 10 natural numbers =  =  =  = 55

    Average of first 10 natural numbers =  = 5.5

    Sum of first 10 prime numbers = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129

    Average of first 10 prime numbers =  = 12.9

    Required Difference = 12.9 – 5.5 = 7.4

  • Question 2/5
    1 / -0.33
    How does average of a sample change if two quantities with the same average are added to the sample?
    Solutions

    Average of the sample will remain same if two samples with equal average are added.

  • Question 3/5
    1 / -0.33
    The average height of 45 students of a class is 120 cm. If 5 among them whose average height is 125cm left the class and 10 new boys of average height 130 cm are included in the class, then what will be the new average height of the students of the class?
    Solutions

    The total height of 45 students = 120 ×45 =5400 cm

    Total decrease in the height when 5 students left the class = 125×5=625 cm

    Total increase in the height when 10 students included in the class = 130×10=1300 cm

    Now, total height of 50 students = 5400 – 625 + 1300 = 6075 cm

    New average = Sum of height of 50 students / total number of students =6075/50 = 121.5

  • Question 4/5
    1 / -0.33
    The average of ten consecutive even numbers is k. If the next five even number are also included, the average of all numbers will be:
    Solutions
    Let ten consecutive even numbers are = x , x+2 ,x+4 , x+6 , x+8, x+10 , x +12 , x+14 , x+16 , x + 18

    We know that , Average =

    Hence, average of five consecutive even numbers =

    Given that average of five consecutive even numbers is m

    ⇒ k = x+9

    ⇒ x =k - 9

    Now, Next five consecutive even numbers are = x+20, x+22 ,x+24 , x+26 and x+28

    Sum of next five consecutive even numbers =

    Sum of all numbers = 10x+90+5x+120 = 15x + 210

    We know that , Average =

    ⇒ Average = k – 9 + 14 = k + 5

  • Question 5/5
    1 / -0.33
    Let a, b, c, d, e, f, g be consecutive even numbers and j, k, l, m, n be consecutive odd numbers. What is the average of all the numbers?
    Solutions
    As, a, b, c, d, e, f, g are consecutive even numbers then numbers are
     d-6,d-4,d-2,d,d+2,d+4,d+6
    Total = d-6+d-4+d-2+d+d+2+d+4+d+6=7d  
    Also, when j, k, l, m, n are consecutive odd numbers then numbers be

    Total =
    Average =
    Hence option (d)
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