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## Arithmetic: Averages Test-1

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*Arithmetic: Averages Test-1*. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%% Your answers are highlighted below.

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 â€“ 40 +90 = 3650

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

Hence option b

Wrong entry = 40

Right entry = 90

Therefore the corrected score = 3600 â€“ 40 +90 = 3650

Therefore the percentage error is {(3650 â€“ 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 â€¦â€¦â€¦â€¦..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 =Â w +1.5

= 1.5n +2w = 97â€¦â€¦â€¦â€¦â€¦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 â€¦â€¦â€¦â€¦..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 =Â w +1.5

= 1.5n +2w = 97â€¦â€¦â€¦â€¦â€¦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 =Â 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 fatherâ€˜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)/10â€¦â€¦â€¦â€¦â€¦.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)/10â€¦â€¦â€¦â€¦â€¦.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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