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 Example 1:

A man purchased 5 toys at the rate of Rs 200 each, 6 toys at the rate of Rs 250 each and 9 toys at the rate of Rs 300 each. Calculate the average cost of one toy.

Solutions:

Price of 5 toys = 200 x 5 = 1000

Price of 6 toys = 250 x 6 = 1500

Price of 9 toys = 300 x 9 = 2700

Aveage cost of one toy = (1000+1500+2700)/20= 5200/20= Rs. 260\-

Example 2:

In three numbers, the first is twice the second and thrice the third, if the average of these three numbers is 44, then the first number is:

Solution:

Let the three numbers be x,  y and z

Now,Therefore, x = 2y = 3z.
y=x/2 and z=x/3
Now, (x+x/2+x/3)/3=44
or 11 x/18=44 or x=72

Example 3:

The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers?

Solution:

Let the numbers be x, x + 2, x+ 4, x + 6 and x + 8

Then, (x+(x+2)+(x+4)+(x+6)+(x+8))/5= 61

or 5x + 20 = 305 or x = 57.

So, required difference = (57 + 8) – 57 = 8.

  • Average of a group consisting two different groups when their averages are known:

(a) Let Group A contains m quantities and their average is A and group B contains n quantities and their average is b, then average of group C containing a + b quantities

=(ma+mb)/(m+n)

Example 4:

             There are 30 student in a class. The average age of the first 10 students is 12.5 years. The average age of the next 20 students is 13.1 years. The average age of the whole class is:

Solution:

Total age of 10 students = 12.5×10 = 125 years

Total age of 20 students = 13.1×20 = 262 years

Average age of 30 students = (125+262)/30=12.9 years

Example 5:

The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girl is 15.4 years. The ration of the number of boys to the number of girls in the class is

Solution:

Let the number of boys in a class be x.

Let the number of girls in a class be y.

∴ Sum of the ages of the boys = 16.4 x

Sum of the ages of the girls = 15.4 y

∴15.8(x+y)=16.4x+15.4y
⇒0.6x=0.4y⇒x/y=2/3
∴ Required ratio = 2:3

(b) It average of m quantities is a  and the average of n quantities out of them is b then the average of remaining group rest of quantities is==(ma+nb)/(m+n).

Example 6:

Average salary of all the 50 employees increasing 5 officers of a company is Rs 850.  If the average salary of the officers is Rs 2500. Find the average salary of the remaining staff of the company.

(a) 560                         (b) 660

(c) 667                         (d) 670

Solution:

Here, m = 50, n =5, a = 850, b = 2500

 Average salary of remaining staff = (ma-nb)/(m-n)
=(50×850-5×2500)/(50-5)
=(42500-12500)/45
=667 (approx)

WEIGHTED AVERAGE

If we have two or more groups of members whose individual averages are know, then combined average of all the members of all the groups is known as weighted average. Thus if there are k groups having member of number n1, n2, n3… nk with averages A1, A2, A3, ……. Ak respectively then weighted average.

Example 7:

The average monthly expenditure of a family was Rs 2200 during the first 3 months; Rs 2250 during the next 4 months and Rs 3120 during the last 5 months of a year. If the total saving during the year were Rs 1260, then the average monthly income was

Solution:

Total annual income
=3×2200+4×2250+5×3120+1260
=6600+9000+15600+1260=32460
∴ Average monthly income
=32460/12=Rs 2705
If X is the average x_1+x_2+x_3+⋯+x_n then
The average of x_1+〖a,x〗_2+a,x_3+a,…x_n+a is X+a
The average of x_1-〖a,x〗_2-a,x_3-a,…x_n-a is X-a
The average of ax_1,ax_2,…ax_n is aX, provide a≠0
The average of x_1/a,x_2/a,x_3/a,…x_n/a is x/a provided a ≠0

  • If, in a group, one or more new quantities are added or excluded, then the new quantity or sum of added or excluded quantities = [Change in no. of quantities x original average][Change in average x final no. of quantities]

Take +ve sing.if quantities added and

Take –ve sing if quantities removed.

Example 8:

The average weight of 29 students in  a class is 48 kg. If the weight of the teacher is included, the average weight rises by 500 g. Find the weight of the teacher.

Solution:

Here, weight of the teacher is added    and final average of the group increases.

 Change in average is (+)ve, using the formula

Sum of the quantities added

⇒ weight of teacher = (1×48)+(0.5×30)=63kg
∴ weight of teacher is 63 kg.

Example 9:

The average age of 40 students in class is 15 years. When 10 new students are admitted, the average is increased by 0.2 year. Find the average age of the new students.

Solution:

            Here, 10 new students are admitted.

           Change in average is +ve. Using the

Formula

⇒ Sum of the weight of 10 new students admitted
=(10×15)+(0.2×50)=160 kg
∴ Average age of 10 new students = s_a/n_a =160/10=16 kg

  Average age of 10 new students is 16 years.

Example 10:

             A train covers the first 160 km at a speed of 120 km/h, another 160 km at 140 km/h, and the last 160 km at 80 km/h, Find the average speed of the train for the entire journey.

Solution:

Average speed =3xyz/(xy+yz+zx)
=(3×120×140×80)/(120×140+140×80+80×120)
=(360×140×80)/(1600+11200+9600)=4032000/37600
= 107 11/(47 ) km/h

  • If a person cover A km at a speed of x km/h,B km at a speed of y km/hr and c km at a speed of z km/h, the average speed during the entire journey is ((A+B+C)/(A/x+B/Y;+C/z)) km/h

= 107   km/h 

Example 11:

             A person cover 9 km at a speed of 3 km/h,25 km at a speed of 5 km/h and 30 kmat a speed of 10 km/h. Find the average speed for the entire journey. The average speed

 = ((A+B+C)/(A/x+B/y+C/z))
=((9+25+30)/(9/3+25/5+30/10))
=64/11=59/11 km/h

  • If a person covers At part of the distance at x km/h. Bth part of the distance at y km/h and the remaining Cth part at z km/h, then the average speed during the entire  journey is  (1/(A/x+B/y+C/z))km/h.

Example 12:

A train covers 50% of the journey at 30 km/h. 25% of the journey at 25 km/h and the remaining at 20 km/h. Find the average speed of the train during the entire journey.

Solution:

            The average speed

  ( Here, A=50,B=25 and C=25)

= ((A+B+C)/(A/x+B/y+C/z))
=((9+25+30)/(9/3+25/5+30/10))
=64/11=59/11 km/h

  • If a certain distance is covered at  a mph and  an equal  distance at b kmph, then the average speed during whole

 Journey =2ab/( a+b)kmph

Example 13:

A motorist travels to a place 150 km away at an average speed of 50km/hr and returns at 30 km/hr. His average speed for the whole journey in km/hr is

Solution:

 Average speed
= 2ab/(a+b) = km/hr
=((2×50×30)/(50+30))km/hr
=37.5km/hr.

GEOMETRIC MEAN OR GEOMETRIC AVERAGE

Geometric mean of x_1,x_2,….x_n is denoted by
G. M = √(x_1×x_2×…×x_n )

Example 14:

            The production of a company for three successive years has increased by 10% 20 % and 40% respectively what is the average increase of production.

Solution: G. M. = (10×20×10)^(1/3)=20%

Example 15:

                The mean of the marks secured by 25 students of section A of class X is 4,  that of 35 students of section B is 51 and that 30 students of section C is 53. Find the combined mean of the marks of students of three sections of class X.

Solution:

            Mean of the marks of 25 students of XA = 47

             Sum of the marks of 25 students

                          =25×47=1175 … (i)

            Mean of the marks of 35 students of XB = 51

             Sum of the marks of 35 students

                               =35×51=1785 … (ii)

            Mean of the marks of 30 students of XC = 53

             Sum of the marks of 30 students =

                              =30×53=1590 … (iii)

            Adding (i), (ii), and (iii)

            Sum of the marks of (25+35+30) i.e. 90

            student

            = 1175 + 1785 + 1590 = 4550

            Thus the combined mean of the marks of students of three sections =4550/90=50.56.

Example 16:

            Find the A.M. of the sequence 1, 2, 3, …., 100

Solution:

We have sum of first n natural numbers =n/2 (n+1)
here n =100
⇒Sum=100/2×101=101×50
⇒AM=Sum/100=(101×50)/100=50.5

Example 17:

            A sequence of seven consecutive integers is given. The average of the first five given integers is n. Find the average of all the seven integers.

Solution:

Let the seven consecutive integers be x,x+1,x+2,….x+6
The sum of the first five is x+x+1+x+2+x+3+x+4=5x+10
The average of these five is (5x+10)/5=x+2=n
The average of the seven will be (5x+10+x+5+x+6)/7=(7x+21)/7=x+3
As x + 2 = n, so x + 3 = x + 2 + 1 = n + 1

Example 18:

            The average of 11 results is 50. If the average of first six results is 49 and that of last six results is 52, find the sixth result.

Solution:

Average of 11 results

1 2 3 4 5 6 7 8 9 10 11

Average of last 6 results = 52

Average of last 6 results = 49

It is quite obvious that the sixth result is included twice, once in the first six results and second in the last six results.

 Value of the sixth result = (Sum of first six results) + (Sum of last six results) – Sum of 11 results

= 6×49+6×52-11×50=56

Example 19:

Typist A can type a sheet in 5 minutes, typist B is 6 minutes and typist C in 8 minutes. The average number of sheets typed per hour per typist is ……. sheets.

Solution:

            A types 12 sheets in 1 hour

            B types 10 sheets in 1 hour

            C types 7.5 sheets in 1 hour

            Average number of sheets types per hour per typist

= (12+10+7.5)/3=29.5/3=9.83

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