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Aptitude::Average

Home > Quantitative Aptitude > Average > Subjective Solved Examples

Example 1 / 9

The average weight of A, B and C is 50 kg. If the average weight of A and B be 40 kg and that of B and C be 40 kg, then the weight of B is:
Note: Before solving any average questions do remember to take variables and solve. It will be very easy. Whenever the average of any n quantities are given we should first find the total value of that n quantities.
=> If we approach this way half the question is solved. In this case let the individual weights be a,b,c so the average of the weights will be a+b+c/3 = 50
So the total weight of 3 persons would be a+b+c=3*50=150...........eq.1
Now coming to the second part of the question where the average weight of 2 persons is given so by following the above steps. a+b=40*2=80 .........eq. 2 and b+c=40*2=80 .................eq. 3
so by adding eq. 2 and eq. 3
we get a+2b+c=160 .............eq. 4 and a+c= 150-b from eqn1
=> 2b +150-b =160 => b=10
So, the weight of B is 10kg. Ans.
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Example 2 / 9

What is the average (arithmetic mean) of all the multiples of ten from 10 to 100 inclusive?
The best way to solve this question is to understand the question first as to what is being asked. The average and the arithmetic mean has the same meaning.
So, we have to consider multiples of 10 from 10-100 including both so they are->
(10+20+30+40+50+60+70+80+90+100)= 550
So, its' average will be 550/10 = 55

If 10 and 100 were excluded then it would be 20+30+40+50+60+70+80+90 = 440/8= 55 Ans.
The above addition of 10+20+30+40 and so on is bit lengthy so we can use the formula i.e n(n+1)/2 for the addition of n consecutive integers where n is the length.

In this case we would have taken 10 common and then it would have been 10 (1+2+3+4+5+6++7+8+9+10)
=> 10{n(n+1)/2} = 10*{(10*11)/2}= 550
And average would be 550/10= 55 Ans.
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Example 3 / 9

The average length of three tapes is 7500 feet. None of the tapes is less than 6400 feet. What is the greatest possible length of one of the other tapes?
Assuming the first and second tape is of equal length;(6400 feet each. As per question.)

Total length of the three tapes : 7500*3 = 22500 ft.
Greatest Possible length : (22500 - 12800) = 9700 ft. Ans.
Reading this question the first idea comes is that the minimum length of any tape will be 6400 feet. Now the confusion arises that all three can be of different lengths with their average being 7500 i.e a+b+c= 7500*3=22500 ...............eq. 1
Here only one equation is there and 3 variables so impossible to find the values. Now coming to the condition that the minimum length has to be 6400ft, it cant be less than that so the values of a,b,c will lie above 6400ft now considering one tape to be having the greatest length which is only possible when the other two tapes are having the minimum length and the minimum length can not cross below 6400ft so considering the 2 tapes of length 6400ft the third tape length would be 22500- 6400*2= 9700ft is the greatest possible length.
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(6400+6400+x)/3=7500
=> 12800+x = 22500
=> x = 22500 - 12800
=> x = 9700 Ans.
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Example 4 / 9

The average salary of 3 workers is 120 Rs. per week. If one earns Rs.100 and second earns Rs.80 how much is the salary of the 3rd worker.
Let the third workers salary = x Rs.

So, by following the average salary (As per questions average is 120 Rs.)

=>(100+80+x)/3 = 120

(180+x)/3 = 120

=> 180+x = 360

=> x = (360 - 180) = 180 Rs. Ans.
Before solving any average questions do remember to take variables and solve. It will be very easy. Whenever the average of any n quantities are given we should first find the total value of that n quantities.
By following the step (a+b+c)/3=120
So, a+b+c=360 now out of three variables we know the value of two variables. So, now its simple maths 100+80+c=360.
=> c = 180Rs. Ans.
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Let the third workers salary = x Rs.

x = (120*3 - 100 - 80)
=> x = 360 - 280 = 180 Rs.
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Example 5 / 9

When a student weighing 45 kgs. left a class, the average weight of the remaining 59 students decreased by 200g. What is the average weight of the remaining 59 students?
Lets assume the avg. weight of 59 students = x Kg.
So total weight of 59 student = 59*x
As per question:

(59x + 45)/60 = (x - 0.2)
=> (59x + 45) = 60x - 12
=> 60x-59x = 45+12
=> x = 57 Kg.

Note:convert grams to Kg. for easy calculation.
This type is the most typical type asked in almost all the exams where the average of a certain quantity or group gets altered by addition or removal of some quantities or persons. In such questions, we need to find the total value of n quantities and then add or subtract the altered value and then again find the average with a new altered value of n.
Prior to the leaving of the particular student there were total 60 students so n=60 now 1 student left so it became 59, let the average weight of the 59 students be x hence total weight of 59 students will be 59x, Now according to the question there is a decrease in average weight by 200g or 0.2kg hence =>
(59x+45)/60= (x-0.2), Solving for x we will get x=57 Hence now the average weight of 59 students is 57kg.
Ans.
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Let the avg. weight of 59 students = x Kg.

(59x + 45) / 60 = (x - 0.2)
=> (59x + 45) = 60x - 12
x = 57

Note:convert grams to Kgs for easy calculation.
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Example 6 / 9

A library has an average of 500 visitors on Sundays and 250 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
if a month starts from Sunday then there are going to be a maximum of 5 Sundays in that month, 1st, 8th,15th, 23rd,30th all these dates will be Sundays.

So now to calculate the total number of visitors on Sundays= 500*5=2500
And on rest 25 days of the month = 25*250 = 6250
So average = (6250+2500)/30= 291.6 ~ 291 visitors. Ans.
The tip is to read the question carefully, and get to know how many exactly Sundays are there and the remaining days. Rest all calculations are easy. So firstly if a month starts from Sunday then there are going to be a maximum of 5 Sundays in that month, 1st, 8th,15th, 23rd,30th all these dates will be Sundays.
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Example 7 / 9

The average of four consecutive odd numbers is 28. Find the largest number.
Let the first number be x so next consecutive odd/even number will be x+2 and then x+4 and then x+6 and so on.

Hence calculating average of 4 numbers = (x+x+2+x+4+x+6)/4 = 28
Hence 4x+12 = 112
=> 4x= 100
=> x = 25
So, the largest number will number be 25+6 = 31 Ans.
Note : This is also a very typical question asked everywhere (IT Companies Written Exam, Bank, and other exams.). It can be for odd numbers, even numbers or divisible by a particular number. In all such cases, the approach is similar.
Let the first number be x so next consecutive odd/even number will be x+2 and then x+4 and then x+6 and so on.
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Example 8 / 9

The number of students in the three section of a class is in the ratio 2:3:4. The average marks scored in each of these sections is in the ratio 4:3:1. By what percent is the average mark of the second section more than the class average?
Note: To approach this question you must have a good understanding of the ratios. We don't know the number of students so let them be 2x, 3x and 4x.
Similarly, the average marks ratio will be 4y,3y,y Now calculating the total marks in each section=> 8xy,9xy,4xy
Let's now find the average marks of the class= total marks/total number of students=> 21xy/9x= 7y/3
Now coming to the question, average marks of the second section=3y
So the average mark of the second section is greater than the class average by 3y-7y/3=>2y/3
Percentage difference= (2y/3 )/(7y/3) hence 0.2857 or 28.57%Ans.
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Example 9 / 9

I drove 100 km at 25 kmph and then an additional 100 km at 50 kmph. Compute my average speed over my 120 km.
Note:The best way to find the solution is to use the formula [Average speed = Total distance /Total Time taken] .

So we know the total distance = 200km
T1= 100/25= 4hr and
T2= 100/50= 2hr

Hence, the Average speed = 200/6 = 33.33km/hr. Ans.

Keep in mind the unit's i.e m/sec or km/hr.
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