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## Arithmetic: Averages Test-1

Congratulations - you have completed Arithmetic: Averages Test-1. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 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:
 A 0.6% B 1.36% C 1.34% D 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
 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?
 A 46 B 42 C 27 D 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
 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?
 A 94 B 91 C 89 D 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?
 A 1677 B 1517 C 1665 D 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
 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
 A 20 B 18 C 15 D 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
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