### APTITUDE- Problems on Averages

Averages

Formula:

1.Average=Sum of quantities/Number of quantities.

2.Suppose a man covers a certain distance at x kmph
and an equal distance at y kmph ,then the average speed
during the whole journey is (2xy/x+y) kmph.

Examples:

1.Find the average of all these numbers.142,147,153,165,157.
Solution:
142 147 153 165 157
Here consider the least number i.e, 142
comparing with others,
142 147 153 165 157
+5 +11 +23 +15
Now add 5+11+23+15 = 52/5 = 10.8
Now add 10.8 to 142 we get 152.8
(Average of all these numbers).

2.Find the average of all these numbers.4,10,16,22,28

Solution:
4,10,16,22,28
As the difference of number is 6
Then the average of these numbers is central one i.e, 16.

3.Find the average of all these numbers.4,10,16,22,28,34.
Solution:
Here also difference is 6.
Then middle numbers 16,22 take average of these
two numbers 16+22/2=19
Therefore the average of these numbers is 19.

4.The average marks of a marks of a student in 4 Examination is 40.If he got 80 marks in 5th Exam then what is his new average.
Solution:
4*40+80=240
Then average means 240/5=48.

5.In a group the average income of 6 men is 500 and that of 5 women is 280, then what is average income of the group.

Solution:
6*500+5*280=4400
then average is 4400/11=400.
Another Method: here consider for 6 men
6 men â€“ each 500.
so 5th women is 280.
then 500-280=220.
then 220*6/11=120.
therefore 120+280=400.

6.The average weight of a class of 30 students is 40 kgs if the
teacher weight is included then average increases by 2 kgs then
find the weight of the teacher?

Solution:
30 students average weight is 40 kgs.
So,when teacher weight is added it increases by 2 kgs
so total 31 persons ,therefore 31*2=62.
Now add the average weight of all student to it
we get teachers weight i.e, 62+40=102 kgs.

7.The average age of Mr and Mrs Sharma 4 years ago is 28 years .
If the present average age of Mr and Mrs Sharma and their son
is 22 years. What is the age of their son.

Solution:
4 years ago their average age is 28 years.
So their present average age is 32 years.
32 years for Mr and Mrs Sharma then 32*2=64 years.
Then present age including their son is 22 years.
So 22*3 =66 years.
Therefore son age will be 66-64 = 2 years.

8.The average price of 10 books is increased by 17 Rupees when
one of them whose value is Rs.400 is replaced by a new book.
What is the price of new book?

Solution:
10 books Average increases by 17 Rupees
so 10*17= 170.
so the new book cost is more and by adding its cost average
increase,therefore the cost of new book is 400+170=570Rs.

9.The average marks of girls in a class is 62.5. The average marks
of 4 girls among them is 60.The average marks of remaining girls
is 63,then what is the number of girls in the class?

Solution:
Total number of girls be x+4.
Average marks of 4 girls is 60.
therefore 62.5-60=2.5
then 4*2.5 =10.
the average of remaining girls is 63
here 0.5 difference therefore 0.5*x=10(since we got from 4 girls)
(this is taken becoz both should be equal)
x=10/0.5
x=20.
This clear says that remaining are 20 girls
therefore total is x+4=20+4=24 girls

10.Find the average of first 50 natural numbers.
Solution:
Sum of the Natural Numbers is n(n+1)/2
therefore for 50 Natural numbers 50*51/2=775.
the average is 775/50=15.5

11.The average of the first nine prime number is?
Solution:
Prime numbers are 2,3,5,7,11,13,17,19,23
therefore 2+3+5+7+11+13+17+19+23=100
then the average 100/9= 11 1/9.

12.The average of 2,7,6 and x is 5 and the average of and the
average of 18,1,6,x and y is 10 .what is the value of y?

Solution:
2+7+6+x/4=5
=>15+x=20
=>x=5.
18+1+6+x+y/5=10
=>25+5+y=50
=>y=20.

13.The average of a non-zero number and its square is 5 times the
number.The number is

Solution:
The number be x
then x+x2/2=5x
=>x2-9x=0
=>x(x-9)=0
therefore x=0 or x=9.
The number is 9.

14.Nine persons went to a hotel for taking their meals . Eight of
them spent Rs.12 each on their meals and the ninth spent Rs.8 then
the average expenditure of all the nine. What was the total money
spent by them?

Solution:
The average expenditure be x.
then 8*12+(x+8)=9x
=>96+x+8=9x.
=>8x=104
=>x=13
Total money spent =9x=>9*13=117

15.The average weight of A.B.C is 45 Kgs.If the average weight of
A and B be 40 Kgs and that of Band C be 43 Kgs. Find the weight of B?

Solution:
The weight of A,B,Care 45*3=135 Kgs.
The weight of A,B are 40*2=80 Kgs.
The weight of B,C are 43*2=86 Kgs.
To get the Weight of B.
(A+B)+(B+C)-(A+B+C)=80+86-135
B=31 kgs.

16.The sum of three consecutive odd number is 48 more than the average of these number .What is the first of these numbers?
Solution:
let the three consecutive odd numbers are x, x+2, x+4.
By adding them we get x+x+2+x+4=3x+6.
Then 3x+6-(3x+6)/3=38(given)
=>2(3x+6)=38*3.
=>6x+12=114
=>6x=102
=>x=17.

17.A family consists of grandparents,parents and three grandchildren.
The average age of the grandparents is 67 years,that of parents is 35
years and that of the grand children is 6 years . What is the average
age of the family?

Solution:
grandparents age is 67*2=134
parents age is 35*2=70
grandchildren age is 6*3=18
therefore age of family is 134+70+18=222
average is 222/7=31 5/7 years.

18.A library has an average of 510 visitors on Sundays and 240 on
other days .The average number of visitors per day in a month 30
days beginning with a Sunday is?
Solution:
Here specified that month starts with Sunday
so, in a month there are 5 Sundays.
Therefore remaining days will be 25 days.
510*5+240*25=2550+6000
=8550 visitors.
The average visitors are 8550/30=285.

19.The average age of a class of 39 students is 15 years .If the age of the teacher be included ,then average increases by 3 months. Find the age of the teacher.
Solution: Total age for 39 persons is 39*15=585 years.
Now 40 persons is 40* 61/4=610 years
(since 15 years 3 months=15 3/12=61/4)
Age of the teacher =610-585 years
=>25 years.

20.The average weight of a 10 oarsmen in a boat is increases by 1.8 Kgs .When one of the crew ,who weighs 53 Kgs is replaced by new man. Find the weight of the new man.
Solution: Weight of 10 oars men is increases by 1.8 Kgs
so, 10*1.8=18 Kgs
therefore 53+18=71 Kgs will be the weight of the man.

21.A bats man makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find the average after 17th inning.

Solution: Average after 17 th inning =x
then for 16th inning is x-3.
Therefore 16(x-3)+87 =17x
=>x=87-48
=>x=39.

22.The average age of a class is 15.8 years .The average age of boys in the class is 16.4 years while that of the girls is 15.4 years .What is the ratio of boys to girls in the class.
Solution: Ratio be k:1 then
k*16.4 + 1*15.4 = (k+1)*15.8
=>(16.4-15.8)k=15.8-15.4
=>k=0.4/0.6
=>k=2/3
therefore 2/3:1=>2:3

23.In a cricket eleven ,the average of eleven players is 28 years .Out of these ,the average ages of three groups of players each are 25 years,28 years, and 30 years respectively. If in these groups ,the captain and the youngest player are not included and the captain is
eleven years older than the youngest players , what is the age of the captain?

Solution: let the age of youngest player be x
then ,age of the captain =(x+11)
therefore 3*25 + 3*28 + 3*30 + x + x+11=11*28
=>75+84+90+2x+11=308
=>2x=48
=>x=24.
Therefore age of the captain =(x+11)= 24+11= 35 years.

24.The average age of the boys in the class is twice the number of girls in the class .If the ratio of boys and girls in the class of 36 be 5:1, what is the total of the age (in years) of the boys in the class?
Solution: Number of boys=36*5/6=30
Number of girls =6
Average age of boys =2*6=12 years
Total age of the boys=30*12=360 years

25.Five years ago, the average age of P and Q was 15 years ,average age of P,Q, and R today is 20 years,how old will R be after 10 years?

Solution: Age of P and Q are 15*2=30 years
Present age of P and Q is 30+5*2=40 years.
Age of P Q and R is 20*3= 60 years.
R ,present age is 60-40=20 years
After 10 years =20+10=30 years.

26.The average weight of 3 men A,B and C is 84 Kgs. Another man D joins the group and the average now becomes 80 Kgs.If another man E whose weight is 3 Kgs more than that of D ,replaces A then the
average weight B,C,D and E becomes 79 Kgs. The weight of A is.

Solution:Total weight of A, B and C is 84 * 3 =252 Kgs.
Total weight of A,B,C and Dis 80*4=320 Kgs
Therefore D=320-252=68 Kgs.
E weight (68+3)=71 kgs
Total weight of B,C,D and E = 79*4=316 Kgs
(A+B+C+D)-(B+C+D+E)=320-316 =4Kgs
A-E=4Kgs
A-71=4 kgs
A=75 Kgs

27.A team of 8 persons joins in a shooting competition.The best marksman scored 85 points.If he had scored 92 points ,the average score for the team would have been 84.The team scored was.

Solution: Here consider the total score be x.
therefore x+92-85/8=84
=>x+7=672
=>x=665.

28.A man whose bowling average is 12.4,takes 5 wickets for 26 runs and there by decrease his average by 0.4. The number of wickets,taken by him before his last match is:
Solution: Number of wickets taken before last match be x.
therefore 12.4×26/x+5=12(since average decrease by 0.4
therefore 12.4-0.4=12)
=>12.4x+2612x+60
=>0.4x=34
=>x=340/4
=>x=85.

29.The mean temperature of Monday to Wednesday was 37 degrees
and of Tuesday to Thursday was 34 degrees .If the temperature on Thursday was 4/5th that of Monday. The temperature on Thursday was:

Solution:
The total temperature recorded on Monday,Wednesday was 37*3=111.
The total temperature recorded on Tuesday,
Wednesday,Thursday was 34*3=102.
and also given that Th=4/5M
=>M=5/4Th
(M+T+W)-(T+W+Th)=111-102=9
M-Th=9
5/4Th-Th=9
Th(1/4)=9
=>Th=36 degrees.

30. 16 children are to be divided into two groups A and B of 10 and 6 children. The average percent marks obtained by the children of group A is 75 and the average percent marks of all the 16 children is 76. What is the average percent marks of children of groups B?
Solution: Here given average of group A and whole groups .
So,(76*16)-(75*10)/6
=>1216-750/6
=>466/6=233/3=77 2/3

31.Of the three numbers the first is twice the second and the second is twice the third .The average of the reciprocal of the numbers is 7/72,the number are.
Solution:Let the third number be x
Let the second number be 2x.
Let the first number be 4x.
Therefore average of the reciprocal means
1/x+1/2x+1/4x=(7/72*3)
7/4x=7/24
=>4x=24
x=6.
Therefore
First number is 4*6=24.
Second number is 2*6=12
Third number is 1*6=6

32.The average of 5 numbers is 7.When 3 new numbers are added the average of the eight numbers is 8.5. The average of the three new number is:
Solution: Sum of three new numbers=(8*8.5-5*7)=33
Their average =33/3=11.

33.The average temperature of the town in the first four days of a month was 58 degrees. The average for the second ,third,fourth and fifth days was 60 degree .If the temperature of the first and
fifth days were in the ratio 7:8 then what is the temperature on the fifth day?

Solution :
Sum of temperature on 1st 2nd 3rd
and 4th days =58*4=232 degrees.
Sum of temperature on 2nd 3rd 4th
and 5th days =60*4=240 degrees
Therefore 5th day temperature is 240-232=8 degrees.
The ratio given for 1st and 5th days be 7x and 8x degrees
then 8x-7x=8
=>x=8.
therefore temperature on the 5th day =8x=8*8=64 degrees.

1. Problems were very interestig, I Required lot of problems
on Average,Trains, especially on AGE……………

• Hi this is saif, Probles were very interesting, I required lot of problems on Average, Trains, espesially on AGE……………… Thanks

2. ravi says:

Problems are awesome and please i want some more problems please forward to my mail. my mail is eswarkumar55@gmail.com

3. priya says:

In an exam the average score of the class was 70 marks.
if the average score of the candidates who failed in the exam was 40 marks and that of the passed students is 80 marks. find the pass percentage of the class??????

• Jay says:

Total students = x
students passed = y
students failed = x-y
therefore
70x = 80y + 40(x-y)
=> 30x=40y
=>3x=4y
=>y/x=3/4

pass %
= (y/x)*100%
= (3/4)*100%
= 75%

4. sushree s. dash says:

i want some problems of time,speed and distance. please forward those to my mail

5. Aparna patil says:

please provide a quick method for solving the following question with an easy solution such that we can solve it verbally
Question:
In what ratio an alloy containing 8.8% copper be mixed with another alloy containing 11.5% copper so that the resultant alloy contains 9.7% copper?

• Jay says:

required ratio
= (11.5-9.7)/(9.7-8.8)
= 2:1

6. Aparna patil says:

want some examples of averages problems solved logically?

7. aparna says:

th given problems are very interesting.I want some more problems on averages.pls send on my e-mail.

8. neeraja says:

four persons a,b,c and d start a business.d invests Rs.5 lakhs more than c.b invests Rs.5lakhs more than a.c invests Rs.5 lakhs more than b.later ,e joins the business with a capital of Rs.40lakhs.now the average capital increases by Rs.4.5 lakhs.how much did c invest?

9. Yakshap sharma says:

Average weight of a class of 38 students is 63 kg. If 2 students weighing 58 and 77 kgs leaves the class. What will be the new average

10. Yakshap sharma says:

Runs scored by dhoni in first 4 innings are 38,68,54,76. How much he score in next innings to make his average 50

• bn says:

average of dhoni in 4 innigs(38+68+54+76)/4=59
in 5 inning the average is given as 50 .so,x/5=50
x=250
total number of runs in 4 innigs=38+68+54+76=236
runs to be got in 5 inning for a average of 50 is 250-236=14 runs

11. Yakshap sharma says:

The average age of a newly married couple is 20 years . After 5 years the average age of the couple and a child born during the period is 18 years. How old is the child

12. Yakshap sharma says:

Average age of newly married couple is 20 years. After 5 years the average age of the couple and a child born during the period is 18 years.HOW old is the child?

13. Rajalakshmi.R says:

Hi

Problems are very useful in this iste. Please forward some more problems and pdf file to my mail id