*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).

Answer is 152.8

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.

Answer is 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.

Answer 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.

Answer is 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.

Answer is 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.

Answer is 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.

Answer is 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.

Answer is 570 Rs.

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

Answer is 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

Answer is 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.

Answer is 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

Answer is Rs.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.

Answer is 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.

Answer is 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.

Answer is 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.

Answer is 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.

Answer is 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.

Answer is 71 Kgs.

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.

Answer is 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

Answer is 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.

Answer is 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

Answer is 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.

Answer is 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

Answer is 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.

Answer is 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 12.4x26/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.

Answer is 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

Answer is 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

Answer is 24,12,6.

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.

Answer is 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.

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