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Average Questions Answers

Average Questions Answers (MCQ) listings with explanations includes calculating average in different type of questions a which are important for many competitive exams.

1) When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 stude

  • 1) 55
  • 2) 56
  • 3) 57
  • 4) 58
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Ans.   C
Explanation :
Let the average weight of the 59 students be A. So the total weight of the 59 of them will be 59*A. The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45 When this student is also included, the average weight decreases by 0.2 kgs \begin{aligned} \frac{59A + 45}{60} = A - 0.2 \end{aligned} => 59A + 45 = 60A - 12 => 45 + 12 = 60A - 59A => A = 57

2) Average of all prime numbers between 30 to

  • 1) 37
  • 2) 37.8
  • 3) 39
  • 4) 39.8
View Answer View Explanation
Ans.   D
Explanation :
Prime numbers between 30 and 50 are: 31, 37, 41, 43, 47 Average of prime numbers between 30 to 50 will be \begin{aligned} (\frac{31+37+41+43+47}{5}) = \frac{199}{5} = 39.8 \end{aligned}

3) Reeya obtained 65, 67, 76, 82 and 85 out of 100 in different subjects, What will be the averag

  • 1) 70
  • 2) 75
  • 3) 80
  • 4) 85
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Ans.   B
Explanation :
\begin{aligned} ( \frac {65+67+76+82+85}{5} ) = 75 \end{aligned}

4) Find the sum of first 30 natural numbe

  • 1) 470
  • 2) 468
  • 3) 465
  • 4) 463
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Ans.   C
Explanation :
Sum of n natural numbers \begin{aligned} = \frac{n(n+1)}{2} \end{aligned} \begin{aligned} = \frac{30(30+1)}{2} = \frac{30(31)}{2} = 465 \end{aligned}

5) Find the average of first 10 multiples of

  • 1) 35.5
  • 2) 37.5
  • 3) 38.5
  • 4) 40.5
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Ans.   C
Explanation :
\begin{aligned} = \frac {7(1+2+3+...+10)}{10} \end{aligned} \begin{aligned} = \frac {7(10(10+1))}{10 \times 2} \end{aligned} \begin{aligned} = \frac {7(110)}{10 \times 2} = 38.5 \end{aligned}

6) The average of four consecutive odd numbers is 24. Find the largest numbe

  • 1) 25
  • 2) 27
  • 3) 29
  • 4) 31
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Ans.   B
Explanation :
Let the numbers are x, x+2, x+4, x+6, then \begin{aligned} => \frac{x+(x+2)+(x+4)+(x+6)}{4} = 24 \end{aligned} \begin{aligned} => \frac{4x+12)}{4} = 24 \end{aligned} \begin{aligned} => x+3 = 24 => x = 21 \end{aligned} So largest number is 21 + 6 = 27

7) Average of 10 numbers is zero. At most how many numbers may be greater than ze

  • 1) 0
  • 2) 1
  • 3) 5
  • 4) 9
View Answer View Explanation
Ans.   D
Explanation :

8) Find the average of all numbers between 6 and 34 which are divisible by

  • 1) 15
  • 2) 20
  • 3) 25
  • 4) 30
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Ans.   B
Explanation :
\begin{aligned} Average = (\frac{10+15+20+25+30}{5}) = \frac{100}{5} =20 \end{aligned}

9) Average of 10 matches is 32, How many runs one should should score to increase his average by 4 ru

  • 1) 70
  • 2) 76
  • 3) 78
  • 4) 80
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Ans.   B
Explanation :
Average after 11 innings should be 36 So, Required score = (11 * 36) - (10 * 32) = 396 - 320 = 76

10) If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students

  • 1) 54.48
  • 2) 54.68
  • 3) 54.60
  • 4) 54.58
View Answer View Explanation
Ans.   B
Explanation :
\begin{aligned} \frac{(55 \times 50) +(60 \times 55) +(45 \times 60) }{55 + 60 + 45} \end{aligned} \begin{aligned} \frac{8750}{160} = 54.68 \end{aligned}
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