Home » Quantitative Aptitude » Arithmetic » Average Level 1 » SBI PO Pre » Average Quiz 12

Direction : Read the following questions carefully and choose the right answer.
1
In an exam of 100 marks, the average marks of a class of 40 students are 76. If the top 3 scorers of the class leave, the average score falls “down by 1. If the other two toppers except “the highest topper scored not more than 85. “then what is the minimum score the topper can score?

A
86
» Explain it
C
Total score of 40 students = (40 × 76) = 3040
 
Total score of top 3 scorers = 3040 – (37 × 75) = 265
 
To minimize the score of the top scorer, we assume the other two top scorers score the maximum they can = 85 marks each.
 
So, the top scorer scored = 265 – 170 = 95 marks.
 
Hence, option C is correct.
2
Average weight of three friends Amar, Visera and Daman is 70 kg. Another person Vishal joins the group and now the average is 66 kg. If another person Tahir whose weight is 6 kg more than Vishal, joins the group replacing Amar, then average weight of Visera, Daman, Vishal and Tahir becomes 75 kg. What is the weight of Amar (in kg)?

A
18
» Explain it
D
Total weight of Amar, Visera and Daman = 70 × 3 = 210 kg
 
Again, Amar + Visera + Daman + Vishal = 66 × 4 = 264 kg   ......... (i)
 
∴ Weight of Vishal = 264 – 210 = 54 kg
 
∴ Weight of Tahir = 54 + 6 = 60 kg
 
Now, as per the question
 
Visera + Daman + Vishal + Tahir = 75 × 4 = 300 kg.  ................... (ii)
 
Subtracting (i) from (ii), we get
 
Tahir – Amar = 60 – Amar = 300 – 264 = 36
 
Therefore, weight of Amar = 60 – 36 = 24 kg
 
Hence, option D is correct.
3
The average salary of a company increases by 100 when the salary of the manager, which is Rs. 9500, is included. If the number of employees excluding the manager is the smallest cube divisible by 16, what is the final average of the company?
» Explain it
D
The smallest cube divisible by 16 is 64.
 
Lets assume the average salary before the manager's salary is included is x
 
After addition of Manager's salary the average increases by 100
 
We can write down the above information in form of  an equation as:
 
64x + 9500 = 65 × (x + 100)
 
Solving for x, we get x = 3000
 
The final average is 3000 + 100 = Rs. 3100
 
Hence, option D is correct.
4
An exam was conducted in a state over 222 centres. The average number of applicants per centre was found to be 1560. However, it was later realized that in one centre, the number of applicants was counted as 1857 instead of 1747. What was the correct average number of applicants per centre (upto two decimals)?
» Explain it
D
Number of applicants that have been counted extra = 1857 − 1747 = 110
 
Hence, decrease in average = 110/222 = 0.495 
 
∴ Correct average = 1560 − 0.495 = 1559.505 = 1559.51 
 
Hence, option D is correct.
5
In Champions league, Rohit scored an average of 120 runs per match in the first 3 match and an average of 140 runs per match in the last four match. What is Rohit’s average runs for the first match and the last two match if his average runs per match for all the five match is 122 and total number of matches are 5? 
» Explain it
A
Rohit’s average score in the first 3 exams = 120

Let the scores in the 5 exams be denoted by M1, M2, M3, M4, and M5

M1 + M2 + M3 = 120 × 3 = 360     .......(i)

Average of last 4 match = 140

⇒   M2 + M3 + M4 + M5  = 140
4

⇒ M2 + M3 + M4 + M5 = 560        ......(ii)

Average of all the exams
⇒   M1 + M2 + M3 + M4 + M5  = 122
5

∴   M1 + M2 + M3 + M4 + M5 = 122 × 5 = 610                    .......(iii)

From solving above equation, we get M1 + M4 + M5 = 300
 
Required average runs =  300  = 100
3

Hence correct option (A) is correct.