Average For Competitive Exams

An average, or more accurately arithmetic mean is, in crude terms, the sum of n different data divided by n.
For example, if a batsman scores 35,45 and 37 runs in first, second and third innings respectively, then his average runs in 3 innings is equal to 35+45+37/3 = 39 runs.
Therefore, the two mostly used formula in this chapter are :

Average = Sum of all observation number of observation

Sum of all observation= Average x number of observation

There is many type of average questions :

Type1 Problem

Q1. The average age of 30 boys of a class is equal to 14 yrs. When the age of the class teacher is included the average becomes 15 yrs. Find the age of the class teacher?
Solution 1
Total age of 30 boys = 14×30 = 420 yrs.
Total ages when class teacher is included = 15 x 31 = 465 yrs.
Thus, age of class teacher = 465 – 420 = 45 yrs.
Solution 2
By the direct formula :
Age of new entrant = new average + no. of old x increase in average
= 15 + 30(15 – 14) = 45 yrs.


Q2. The average weight of 4 men is increased by 3 kg when one of them who weighs 120 kg. is replaced by another man. What is the weight of the new man?
Solution:
Quicker approach: if the average is increased by 3 kg, then the sum of weight is increases by 3×4 = 12 kg.
And this increase in weight is due to the extra weight included due to the inclusion of new person.
Thus, weight of new man = 120+12 = 132 kg.
By the direct formula :
Weight of new person = weight of removed person + no. of persons x increase in average
= 120 + 4 x 3 = 132 kg.

Type 2 Problem

Q1. The average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of the failed candidates is 15, what is the number of candidates who passed the examination?
Solution:
Let the number of passed candidates be X.
Then total marks = 120 x 35 = 39X + (120 – X) x 15
Or, 4200 = 39X + 1800 – 15X
Or, 24X = 2400
Thus, X = 100
Number of passed candidates = 100


Q2. The average of 11 results is 50. If the average of first six results is 49 and that of last six is 52, find the sixth result?
Solution:
The total of 11 results = 11 x 50 = 550
The total of first 6 results = 6 x 49 = 294.
The total of last 6 results = 6 x 52 = 312.
The 6th result is common in both ;
Thus, sixth result = 294 + 312 – 550 = 56
By the direct formula:
6th result = 50 + 6 [(52 – 50) + (49 – 50)] = 50 + 6 {2 – 1} = 56

Type 3 Problem

Q1. The average age of 8 persons in a committee is increased by 2 years when two men aged 35 yrs and 45 yrs are substituted by two women. Find the average age of these two women.
Solution:
By the direct formula :
Weight of new person = weight of removed person + no. of persons x increase in average
Then, the total age of two women = 2 x 8 + (35 + 45) = 16+80 = 96 yrs
Thus, average age of two women = 96/2 = 48 yrs.


Q2. The average age of a family of 6 members is 22 yrs. If the age of the youngest member be 7 yrs, then what was the average age of the family at the birth of the youngest member?
Solution:
Total ages of all the family members = 6×22 = 132 yrs.
7 yrs ago, total sum of ages = 132 – (6 x 7) = 90 yrs.
But at that time there were 5 members in the family.
Thus, the average at that time = 90/5 = 18 yrs.

Type 4 Problem

Q1. A batsman in his 17th innings makes a score of 85, and thereby increase his average by 3. What is his average after 17th innings?
Solution:
Let the average after 16th innings be X, then 16X + 85
= 17(X + 3) = total score after 17th innings.
Thus, X = 85 – 51 = 34
Thus, average after 17th innings = X + 3 = 34 + 3 = 37.
By the direct formula :
Average after 16th innings = 85 – 3 x 17 = 34.
Average after 17 innings = 85 – 3(17 -1) = 37


Q2. A cricketer has completed 10 innings and his average 21.5 runs. How many runs must he make in his next innings so as to raise his average to 24?
Solution:
Total of 10 innings = 21.5 x 10 = 215
Suppose he needs a score of X in 11th innings; then average in 11 innings = 215+X/11 = 24
Or, X = 264 – 215 = 49
By the direct formula :
Required score = 11 x 24 – 21.5 x 10 = 49

Type 5 Problem

Q1. In a one-day cricket match, Viru the captain of Indian Cricket team scored 94 runs more than the average runs scored by the remaining ten batsmen of the team. If the total runs scored by all the batsmen of the team was 358, then how many runs did Viru score?
Solution:
Let the average of the runs made by the other 10 batsmen be X.
Thus, Runs made by the captain = X + 94
Now, X + 94 + 10x = 358
or, 11X = 264
X = 264/11 = 24
Thus, Runs scored by Viru = 24 + 94 = 118


Q2. The average runs of 65 players in heroCUP season 10 is 90. If highest five players are removed, the average drops by 5 runs. If the runs of highest five scorers are in consecutive decreasing integers by order, then find the run of highest scorer?
Solution:
addition of the total runs before removal of highest 5 run scorers = 65 x 90 = 5850
addition of the total runs after removal of top 5 run scorers = 60 x 85 = 5100
Therefore, the runs scored by highest 5 players = 5850 – 5100 = 750
Let the runs scored by highest run beaver be x.
Therefore,
addition of the runs scored by highest 5 scorers
= x + (x – 1) + (x – 2) + (x – 3) + (x – 4) = 750
or, 5x = 750 + 10
Thus, x = 152

Type 6 Problem

Q1. The average age of boys in the class is 15 years. The average age of girls in the class also added, then the average becomes 18. If there are 18 girls in the class and the average age of girls in the class is 20, then find how many boys in the class?
Solution:
Let the total number of boys in the class be X,
15X + (20 x 18) = (X + 18) x 18
15X + 360 = 18X+324
36 = 3X
X = 12
Total boys in the class is 12


Q2. The average salary of the whole employees in a firm is Rs. 300 per day. The average salary of probationary officers is Rs. 800 per day and that of guards is Rs. 240 per day. If the number of probationary officers is 30, then find the number of guards in the firm?
Solution:
Let the number of guards in the firm be X,
800 x 30 + 240 x X = 300(30 + X)
24000 + 240X = 9000 + 300X
15000 = 60X
X = 15000/60 = 250
Total number of guard in the firm = 250

Type 7 Problem

Q1. A train travels from A to B at the rate of 20 km per hour and from B to A at the rate of #0 km/h. what is the average rate for the whole journey?
Solution:
Here use the average speed formula = 2xy/x+y km/hr.
This formula is apply on the speed during up and down journey.
By the formula of average speed = 2 x 20 x 30/ 20 + 30 = 24 km/hr.
Q2. A person divides his total route of journey into three equal parts and decides to travel the three parts with speeds of 40, 30 and 15 km/hr respectively. Find his average speed during the whole journey?
Solution:
Here use the average speed during the whole journey :
Formula is = 3xyzxy + yz + xz km/hr.
Hence: average speed = 3 x 40 x 30 x 1540 x 30 + 30 x 15 + 40 x 15
3 x 40 x 30 x 152250
= 24 km/hr.


Q3. The average salary of the entire staff in an office is 120 rs. Per month. The average salary of officers is 460 rs. And that of non-officers is 110 rs. If the number of officers is 15, then find the number of non-officers in the office?
Solution:
Let the required number of non-officers = X
Then, 110X + 40 x 15 = 120(15 + X)
Or, 120X – 110X = 460 x 15 – 120 x 15 = 15(460 – 120)
Or, 10X = 15 x 340;
Thus, X = 15 x 34 = 510.
By the direct formula:
No. of non-officers
= no. of officers x ([average salary of officers – mean average]/[mean average – average salary of non-officers])

= 15([460 – 120]/[120 – 110]) = 510

I hope, this article will help you a lot to understand the Average | Formulas | Properties | Examples. If you still have any doubts and problems with any topic of mathematics you can ask your problem in the Ask Question section. You will get a reply shortly.

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