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## Arithmetic: Averages Test-1

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Question 1 |

While calculating the average of a batsman in 100 matches, which came out to be 36, one of the scores of 90 was incorrectly noted as 40. The percentage error for calculating his average is:

0.6% | |

1.36% | |

1.34% | |

1.21% |

Question 1 Explanation:

Total score = 100 x 36 = 3600

Wrong entry = 40

Right entry = 90

Therefore the corrected score = 3600 – 40 +90 = 3650

Therefore the percentage error is {(3650 – 3600)/3650 } x 100 = 1.36%

Hence option b

Wrong entry = 40

Right entry = 90

Therefore the corrected score = 3600 – 40 +90 = 3650

Therefore the percentage error is {(3650 – 3600)/3650 } x 100 = 1.36%

Hence option b

Question 2 |

In a class with a certain number of students, if one new student weighing 50 kg is added then the average weight of the class is increased by 1 kg .If one more student weighing 50 kg is added, then the average weight of the class increases by 1.5 kg over the original average . What is the original average weight of the class?

46 | |

42 | |

27 | |

47 |

Question 2 Explanation:

let the number of students in the class = n

Let the average weight = w

Now the given conditions are if one new student weighing 50 kg is added then the

average weight of the class is increased by 1 kg

(nw +50)/n+1 = w +1

On simplifying n + w = 49 …………..1

And If one more student weighing 50 kg is added, then the average weight of the class

increases by 1.5 kg over the original average

(nw +50 + 50)/n+2 = w +1.5

= 1.5n +2w = 97……………2

On solving both the equations

W= 47

Hence option d

Let the average weight = w

Now the given conditions are if one new student weighing 50 kg is added then the

average weight of the class is increased by 1 kg

(nw +50)/n+1 = w +1

On simplifying n + w = 49 …………..1

And If one more student weighing 50 kg is added, then the average weight of the class

increases by 1.5 kg over the original average

(nw +50 + 50)/n+2 = w +1.5

= 1.5n +2w = 97……………2

On solving both the equations

W= 47

Hence option d

Question 3 |

The average marks of a student in 8 subjects are 87. Of these, the highest score is 2 more than the second best score. If these two subjects are eliminated, the average marks of the remaining subject are 85. What is the highest mark obtained by him?

94 | |

91 | |

89 | |

96 |

Question 3 Explanation:

Total marks in 8 subjects = 8 x 87 = 696
Total marks in 6 subjects = 6 x 85 = 510
Remaining marks = 186
Let the highest mark = a
And the second highest marks = a - 2
Therefore these two numbers would be = 186
a + a - 2 = 186
2a = 188
a = 94
So the highest marks = 94

Question 4 |

The average of 5 consecutive numbers A, B, C, D, E is 41. What is the product of A and E?

1677 | |

1517 | |

1665 | |

1591 |

Question 4 Explanation:

Since there are consecutive numbers so let us consider the first number as a

So the next will be a+1and so up to a+4

So we are given by that the average of these is 41

So we can say that

5a + 10 = 41 x 5

5a = 195

a = 39

so the product of A and E = a (a+4)

39 x 43 = 1677

So the next will be a+1and so up to a+4

So we are given by that the average of these is 41

So we can say that

5a + 10 = 41 x 5

5a = 195

a = 39

so the product of A and E = a (a+4)

39 x 43 = 1677

Question 5 |

The average age of three children in a family is 20% of the average of the age of father and the eldest child .The total age of the mother and the younger child is 39 yrs. If the father‘s age is 26 yrs, what is the age of the second child

20 | |

18 | |

15 | |

cannot be determined |

Question 5 Explanation:

We are here given by the average age of three children in a family is 20% of the average of the father and the eldest child

Therefore let us suppose the age of the three children be a,b, c

And the average would be

= (a+b+c)/3 = 20/100{(26 + c)/2}

= (a + b+c )/3

= (26 + c)/10…………….1

Let suppose that the age of mother is d

So given condition that the d+a = 39

But from this we cannot determined the value of b so option d is the best answer

Therefore let us suppose the age of the three children be a,b, c

And the average would be

= (a+b+c)/3 = 20/100{(26 + c)/2}

= (a + b+c )/3

= (26 + c)/10…………….1

Let suppose that the age of mother is d

So given condition that the d+a = 39

But from this we cannot determined the value of b so option d is the best answer

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