Categories
civil Maths

Maths : Problems on Average (Solved in Steps)

1. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs.5200. What is the monthly income of A?

A. 2000

B. 3000

C. 4000

D. 5000

Answer : Option C

Explanation :

Let the monthly income of A = a
monthly income of B = b
monthly income of C = a
a +b=2×5050————(Equation1)
b + c=2×6250————(Equation2)
a +c=2×5200————(Equation3)
(Equation 1) + (Equation 3) – (Equation 2)
=> ( a +b +a +c) “ (b +c)
=(2×5050)+(2×5200)”(2×6250)
=> 2a = 2(5050 + 5200 – 6250)
=> a = 4000
=> Monthly income of A = 4000


2. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?

A. 6.25,

B. 5.5,

C. 7.4 ,

D. 5

Answer : Option A

Explanation :

Runs scored in the first 10 overs = 10×3.2=32
Total runs = 282
remaining runs to be scored = 282 – 32 = 250
remaining overs = 40
Run rate needed = 250/40=6.25


3. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs.6500?

A. 4800

B. 4991

C. 5004

D. 5000

Answer : Option B

Explanation :

Let the sale in the sixth month = x
Then( 6435+6927+6855+7230+6562+x)/6=6500
= > 6435 + 6927 + 6855+ 7230 + 6562 + x
= 6 × 6500 = 39000
= > 34009 + x = 39000
= > x = 39000 – 34009 = 4991


4. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?

A. 1

B. 20

C. 0

D. 19

Answer : Option D

Explanation :

Average of 20 numbers = 0
=>Sum of 20 numbers/20=0
=> Sum of 20 numbers = 0
Hence at the most, there can be 19 positive numbers.
(Such that if the sum of these 19 positive numbers is x, 20th number will be -x)


5. The captain of a cricket team of 11 members is 26 years old and the wicketkeeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team.

A. 23 years

B. 20 years

C. 24 years

D. 21 years

Answer : Option A

Explanation :

Number of members in the team = 11
Let the average age of the team = x
=>(Sum of the ages of all the 11 members of the team)/11=x
=> Sum of the ages of all the 11 members of the team = 11x
Age of the captain = 26
Age of the wicket keeper = 26 + 3 = 29
Sum of the ages of 9 members of the team excluding captain and wicket keeper = 11x – 26 – 29 = 11x – 55
Average age of 9 members of the team excluding captain and wicket keeper =(11x”55)/9
Given that( 11x”55)/9=(x”1)
=> 11x – 55 = 9(x – 1)
=> 11x – 55 = 9x – 9
=> 2x = 46
=>x=46/2=23 years


6. In Kiran’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran?

A. 70 kg

B. 69 kg

C. 61 kg

D. 67 kg

Answer : Option D

Explanation :

Let Kiran’s weight = x. Then
According to Kiran, 65 < x < 72 ——————(equation 1)
According to brother, 60 < x < 70 ——————(equation 2)
According to mother, xd”68 ——————(equation3)
Given that equation 1, equation 2 and equation 3 are correct. By combining these equations, we can write as 65<xd”68
That is x = 66 or 67 or 68
the average of different probable weights of Kiran = (66+67+68)/3=67


7. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

A. 48.55

B. 42.25

C. 50

D. 51.25

Answer : Option A

Explanation :

Average weight of 16 boys = 50.25
Total Weight of 16 boys = 50.25×16
Average weight of remaining 8 boys = 45.15
Total Weight of remaining 8 boys = 45.15×8
Total weight of all boys in the class = (50.25×16)+(45.15×8)
Total boys = 16+8=24
Average weight of all the boys = (50.25×16)+(45.15×8)/24
= (50.25×2)+(45.15×1)/3
= (16.75×2) + 15.05
=33.5+15.05=48.55


8. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, what is the average marks of all the students?

A. 53.23

B. 54.68

C. 51.33

D. 50

Answer : Option B

Explanation :

Average marks of batch1 = 50
Students in batch1 = 55
Total marks of batch1=55×50
Average marks of batch2 = 55
Students in batch2 = 60
Total marks of batch2=60×55
Average marks of batch3 = 60
Students in batch3 = 45
Total marks of batch3=45×60
Total students = 55 + 60 + 45 = 160
Average marks of all the students
= {(55×50)+(60×55)+(45×60)}/160
= (275+330+270)/16
= 875/16=54.68


9. A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday?

A. 290

B. 304

C. 285

D. 270

Answer : Option C

Explanation :

in a month of 30 days beginning with a Sunday, there will be 4 complete weeks and another two days which will be Sunday and Monday Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday
Average visitors on Sundays = 510
Total visitors of 5 Sundays = 510×5
Average visitors on other days = 240
Total visitors of other 25 days = 240×25
Total visitors = (510×5)+(240×25)
Total days= 30
Average number of visitors per day ={ (510×5) + (240×25)}/30
= { (51×5)+(24×25) }/3
= (17×5) + (8×25)
=85+200=285


10. A student’s mark was wrongly entered as 83 instead of 63. Due to that, the average marks for the class got increased by half 1/2. What is the number of students in the class?

A. 45

B. 40

C. 35

D. 30

Answer : Option B

Explanation :

Let the total number of students = x
The average marks increased by 1/2 due to an increase of 83-63=20 marks.
But total increase in the marks = 1/2×x=x/2
Hence we can write asx/2=20•Ëx=20×2=40


11. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. The average age of the family is

A. 32.27 years

B. 31.71 years

C. 28.17 years

D. 30.57 years

Answer : Option B

Explanation :

Total age of the grandparents = 67×2
Total age of the parents = 35×2
Total age of the grandchildren = 6×3
Average age of the family ={(67×2)+(35×2)+(6×3)}/7=(134+70+18)/ 7=222/7=31.71


12. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, what is the weight of B?

A. 31 kg

B. 2812 kg

C. 32 kg

D. 3012 kg

Answer : Option A

Explanation :

Let the weight of A, B and C are a,b and c respectively.
Average weight of A,B and C = a+b+c=45×3=135—equation(1)
average weight of A and B = 40 a+b=40×2=80—equation(2)
average weight of B and C = 43 b+c=43×2=86 —equation (3)
equation(2) +equation(3)- equation(1)
=> a+ b+ b +c “(a +b +c)=80+86”135
=>b=80+86″135=166″135=31
=> weight of B = 31


13. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. What is the present age of the husband?

A. 40

B. 32

C. 28

D. 30

Answer : Option A

Explanation :

let the present age of the husband = h
present age of the wife = w
present age of the child = c
3 years ago, average age of husband, wife andtheir child = 27
= >Sum of age of husband, wife and their child before 3 years = 3×27=81
= > (h-3) + (w-3) + (c-3) = 81
= > h + w + c = 81+9 = 90 ————————— equation(1)
5 years ago, average age of wife and child = 20
= >Sum of age of wife and child before 5 years = 2×20=40
= > (w-5) + (c-5) = 40
= > w + c = 40+10 = 50 —————————equation(2)
Substituting equation(2) in equation(1)
= > h + 50 = 90
= > h = 90 – 50 = 40
= >present age of the husband = 40


14. The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person?

A. 70

B. 75

C. 85

D 80

Answer : Option C

Explanation :

Total increase in weight = 8×2.5=20
If x is the weight of the new person, total increase in weight = x – 65
=> 20 = x – 65
=> x = 20 + 65 = 85


15. A batsman makes a score of 87 runs in the 17th inning and thus increases his averages by 3. What is his average after 17th inning?

A. 39

B. 35

C. 42

D. 40.5

Answer : Option A

Explanation :

Let the average after 17 innings = x
Total runs scored in 17 innings = 17x
then average after 16 innings = (x-3)
Total runs scored in 16 innings = 16(x-3)
We know that Total runs scored in 16 innings + 87
= Total runs scored in 17 innings
= > 16(x-3) + 87 = 17x
= > 16x – 48 + 87 = 17x
= > x = 39


16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If the average weight of divisions A is 40 kg and that of division b is 35 kg. What is the average weight of the whole class?

A. 38.25

B. 37.25

C. 38.5

D. 37

Answer : Option B

Explanation:

Total weight of students in division A = 36×40
Total weight of students in division B = 44×35
Total students = 36+44=80
Average weight of the whole class= {(36×40)+(44×35)}/80
= {(9×40)+(11×35)}/20
= {(9×8)+(11×7)}/4
= (72+77)/4
= 149/4=37.25


17. A student needed to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What is the value of x?

A. 12

B. 5

C. 7

D. 9

Answer : Option C

Explanation :

(3+11+7+9+15+13+8+19+17+21+14+x)/12=12
= > (137+x)/12=12
= > 137 + x = 144
= > x = 144 – 137 = 7


18. Arun obtained 76, 65, 82, 67 and 85 marks (out in 100) in English, Mathematics, Chemistry, Biology and Physics. What is his average mark?

A. 53

B. 54

C. 72

D. 75

Answer : Option D

Explanation :

Average mark = (76+65+82+67+85)/5=375#5=75


19. The distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of the train during the whole journey?

A. 69.0 km /hr

B. 69.2 km /hr

C. 67.2 km /hr

D. 67.0 km /hr

Answer : Option C

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph.
Then, the average speed of the whole journey = 2xy/x+y kmph.By using the same formula, we can find out the average speed quickly average speed = 2×84×56# (84+56) = (2×84×56)#140==67.2


20. The average age of 36 students in a group is 14 years. When teacher’s age is included in it, the average increases by one. Find out the teacher’s
age in years?

A. 51 years,

B. 49 years

C. 53 years

D. 50 years

Answer : Option A

Explanation :

the average age of 36 students in a group is 14
= > Sum of the ages of 36 students = 36×14
When teacher’s age is included in it, the average increases by one => average = 15
= > Sum of the ages of 36 students and the teacher =37×15
Hence teachers age = 37×15″36×14=37×15″14(37″1)=37×15″37×14+14=37(15″14)+14=37+14=51


21. The average of five numbers id 27. If one number is excluded, the average becomes 25. What is the excluded number?

A. 30,

B. 40,

C. 32.5,

D. 35

Answer : Option D

Explanation :

Sum of 5 numbers = 5×27Sum of 4 numbers after excluding one number = 4×25Excluded number = 5×27″4×25=135″100=35


22. The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then

A. None of these

B. x = y + z

C. 2x = y + z

D. x = 2y + 2z

Answer : Option C

Explanation :

Average of 6 numbers = x
=> Sum of 6 numbers = 6x
Average of the 3 numbers = y
=> Sum of these 3 numbers = 3y
Average of the remaining 3 numbers = z
=> Sum of the remaining 3 numbers = 3z
Now we know that 6x = 3y + 3z
=> 2x = y + z


23. The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Find out the highest score of the player.

A. 150

B. 174

C. 180

D. 166

Answer : Option B

Explanation :

Total runs scored by the player in 40 innings = 40×50
Total runs scored by the player in 38 innings after excluding two innings= 38×48
Sum of the scores of the excluded innings = 40×50″38×48=2000″1824=176
Given that the scores of the excluded innings differ by 172. Hence let’s take the highest score as x + 172 and lowest score as x
Now x + 172 + x = 176
= > 2x = 4
= >x=4# 2=2
highest score as x + 172 = 2 + 172 = 174


24. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, what is the average for the last four matches?

A. 34.25

B. 36.4

C. 40.2

D. 32.25

Answer : Option A

Explanation :

Total runs scored in 10 matches = 10×38.9
Total runs scored in first 6 matches = 6×42
Total runs scored in the last 4 matches = 10×38.9″6×42
Average of the runs scored in the last 4 matches
= (10×38.9″6×42)#4
= (389″252)#4
= 137#4=34.25


25. Suresh drives his car to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/hr. What is his average speed for the whole journey?

A. 32.5 km/hr.

B. 35 km/hr

C. 37.5 km/hr

D. 40 km/hr

Answer : Option C

Explanation :

If a car covers a certain distance at x kmph and an equal distance at y kmph. Then, the average speed of the whole journey = 2xy/x+y kmph By using the same formula, we can find out the average speed quickly average speed = ( 2×50×30)#(50+30)
= (2×50×30)#80
= (2×50×3) # 8
= (50×3)#4
=(25×3)# 2=75#2=37.5


26. The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one-year-old child. What is the average age of the family?

A. 21 years

B. 20 years

C. 18 years

D. 19 years

Answer : Option D

Explanation :

Total of the age of husband and wife = 2×23=46
Total of the age of husband and wife after 5 years
+ Age of the 1-year-old child =46+5+5+1=57
The average age of the family=57#3=19


27. In an examination, a student’s average marks were 63. If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65. How many subjects were there in the examination?

A. 12

B. 11

C. 13

D. 14

Answer : Option B

Explanation :

Let the number of subjects = x
Then, Total marks he scored for all subjects = 63x
If he had obtained 20 more marks for his Geography and 2 more marks for his history, his average would have been 65
=> Total marks he would have scored for all subjects = 65x
Now we can form the equation as 65x – 63x = the
additional marks of the student = 20 + 2 = 22
=> 2x = 22
=> x= 22#2
=11


28. The average salary of all the workers in a workshop is Rs.8000. The average salary of 7 technicians is Rs.12000 and the average salary of the rest is Rs.6000. How many workers are there in the workshop?

A. 21

B. 22

C. 23

D.24

Answer: Option A

Explanation:

Let the number of workers = x
Given that average
salary of all the workers = Rs.8000
then total salary of all workers = 8000x
Given that average salary of 7 technicians is Rs.12000
=>
Total salary of 7 technicians = 7×12000=84000
Number of the rest of the employees = (x – 7)
Average salary of the rest of the employees = Rs.6000
Total salary of the rest of the employees = (x – 7)(6000) 8000x = 84000 + (x – 7)(6000)
= > 8x = 84 + (x – 7)(6)
= > 8x = 84 + 6x – 42
= > 2x = 42
= >x=21