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Average Test 135
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Average Test 135

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  • Question 1/10
    1 / -0.25
    How much is the average of squares of first n natural numbers less than the average of cubes of first n natural numbers?
    Solutions

    Average of squares of first n natural numbers

    Average of cubes of first n natural numbers

    Difference



  • Question 2/10
    1 / -0.25
    A number is such that when it is multiplied by 11, it gives another number; which is as much above from 390 as the twice of original number is below 390. The average of the original number and the resultant number is.
    Solutions
    Let the number be .
    A.T.Q.
    390 - 2x = 11x - 390
    13x = 780
    x = 60

    Then,
    11x = 60 × 11 = 660
    x = 60

    Average of both numbers = (660+60)/2 = 360
  • Question 3/10
    1 / -0.25
    Three numbers are such that if the average of any two of them is added to the third number, the sums obtained are 168, 174 and 180 respectively. What is the average of the original three numbers?
    Solutions

    Let the numbers are x, y and z. Then A.T.Q,


    and 

    and 

    Adding (i), (ii) and (iii):



    ⇒ 2(x + y + z) = 522

    ⇒ x + y + z = 261

    So, Average = 

  • Question 4/10
    1 / -0.25
    There are three sections A, B and C of class 10th. If number of students in section A, B and C are 105, 63 and 42 respectively and average marks of section A, B, and C in an exam are 63, 67 and 58 respectively. What are the average marks of Class 10th?
    Solutions
    Given, Number of students in section A, B and C are 105, 63 and 42 respectively and average marks of section A, B, and C in an exam are 63, 67 and 58 respectively.

    ⇒ Sum of observations = Average × Total number of observations

    So, Sum of marks of all students of section A of class 10th =105 × 63 = 6615

    And Sum of marks of all students of section B of class 10th = 63 × 67= 4221

    And Sum of marks of all students of section C of class 10th = 42 × 58 = 2436

    Total marks scored by class 10th = 6615 + 4221 + 2436 = 13272

    Total number of students = 105 + 63 + 42 = 210

    So, Required average =  = 63.2

  • Question 5/10
    1 / -0.25
    Average of two numbers is 90 and one of the numbers is twice the other. Find both the numbers.
    Solutions

    Let one number be x and the other will be 2x.

    Average = (x + 2x)/2 = 90    (given)

    ⇒ 3x = 180

    ⇒ x = 60

    and, 2x = 120

    So, the numbers will be 60 and 120.

  • Question 6/10
    1 / -0.25
    The average income of P, Q and R is Rs. 14000 per month and the average income of Q, R and S is Rs. 18000 per month. If the salary of S is twice to that of P, then the avg. salary of Q and R is:
    Solutions
    P + Q + R = 14000 X 3
    Q + R + S = 18000 X 3
    ∴S – P = Rs.12000
    Since S = 2P
    ∴2P – P = 12000 ⇒ P = Rs.12000
    Required
    The avg. salary of (Q + R) is = (42000-12000)/2 
    = Rs. 15000
  • Question 7/10
    1 / -0.25
    There are 6 consecutive odd numbers in increasing order. If the sum of the squares of the first and the last element is 178, then the average value of all the six numbers is:
    Solutions

    Let us suppose that the consecutive odd numbers be (a-5) , (a-3) , (a-1) , (a+1) , (a+3) , (a+5)

    Then the average =

    Now again by next condition we have

    Average of all six numbers is 8.

  • Question 8/10
    1 / -0.25
    Average of 40 innings of a cricket player is 45. Difference between his highest and lowest score is 106. Average of remaining 38 innings is 40 then what is his highest score?
    Solutions
    Total runs scored by player = 40x45 = 1800
    Total runs scored by him in 38 innings = 38x40 = 1520
    Sum of highest and lowest score= 1800-1520 = 280
    Let highest score be x and lowert score be x-106
    According to question
  • Question 9/10
    1 / -0.25
    In a cricket match of 50 over’s the run rate of the first 35 over’s is 5.8 runs/over. Calculate the required run rate in the remaining over’s to reach the target of 314 runs?
    Solutions
    Total runs scored by team till 35 overs = 35×5.8 = 203
    Required runs in remaining 15 overs = 314 – 203 = 111
    Required run rate =
  • Question 10/10
    1 / -0.25
    The average of twelve numbers is 58. The average of the first five numbers is 56 and the average of the next four numbers is 60. The 10th number is 4 more than the 11th number and the 11th number is one less than the 12th number. What is the average of the 10th and 12th numbers?
    Solutions

    Total sum of 12 numbers = 58 × 12 = 696

    Total sum of first five number = 56 × 5 = 280

    Total sum of next four numbers = 4 × 60 = 240

    Let the 11th number = x

    10th number = x + 4

    12th number = x + 1

    So, x + x + 4 + x + 1 = 696 – 280 – 240

    ⇒ 3x + 5 = 176

    ⇒ x =

    ⇒ x = 57.

    Required, (61+58)/2 = 119/2 = 59.5

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