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good morning ,

sir i have doubts in PRACTICE QUESTIONS of AVERAGE lesson and questions number are as follows:

2,12,15,30

Sameer,

I have moved your query to this thread, please continue here for proper identification of the sheets.

Manish

sir i have doubts in PRACTICE QUESTIONS of AVERAGE lesson and questions number are as follows:

2,12,15,30

Sameer,

I have moved your query to this thread, please continue here for proper identification of the sheets.

Manish

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

These questions were asked as doubts of Average Class sheet - Practice Questions

Class SHEET - Average - Practice Question- Question-1

Q1. With an average speed of 40 kmph, the train reaches on time. With a speed of 35 kmph, the train is 15 minutes late , what is the distance of the journey ?

The best way to look at this question is:

A............................B

d

The distance AB is covered at two different speeds, therefore, the time required at two different speeds must have been different.

OUR basic equation becomes -

ts - tf = difference of time ( ts = time at slow speed; tf = time required at faster speed)

ts - tf = 15/60

d/35 - d/40 = 15/60 => d/280 = 1/4 => d = 70 km option b

Class SHEET - Average - Practice Question- Question-1

Q1. With an average speed of 40 kmph, the train reaches on time. With a speed of 35 kmph, the train is 15 minutes late , what is the distance of the journey ?

The best way to look at this question is:

A............................B

d

The distance AB is covered at two different speeds, therefore, the time required at two different speeds must have been different.

OUR basic equation becomes -

ts - tf = difference of time ( ts = time at slow speed; tf = time required at faster speed)

ts - tf = 15/60

d/35 - d/40 = 15/60 => d/280 = 1/4 => d = 70 km option b

Last edited by subish on Thu Jul 13, 2017 1:33 am, edited 1 time in total.

- subish
**Posts:**4**Joined:**Sun Sep 08, 2013 6:49 am

Class Sheet- Average- Practice Question- Q2

Typist A = 6 minutes ; Typist B= 7 minutes ; Typist C = 9 minutes

To find the average number of sheets typed by all the three we need to find the number of sheets typed by each of them in one hour.

Number of sheets typed by A in one hour = 60 /6 = 10

Number of sheets typed by B in one hour = 60 /7

Number of sheets typed by c in one hour = 60 /9

Total sheets typed by the three typists in one hour = 10 + 60/7 + 60 /9 = 1590 /63

Average of the three = 1590 /(63 x 3) = 530 /63 Option b

Typist A = 6 minutes ; Typist B= 7 minutes ; Typist C = 9 minutes

To find the average number of sheets typed by all the three we need to find the number of sheets typed by each of them in one hour.

Number of sheets typed by A in one hour = 60 /6 = 10

Number of sheets typed by B in one hour = 60 /7

Number of sheets typed by c in one hour = 60 /9

Total sheets typed by the three typists in one hour = 10 + 60/7 + 60 /9 = 1590 /63

Average of the three = 1590 /(63 x 3) = 530 /63 Option b

- subish
**Posts:**4**Joined:**Sun Sep 08, 2013 6:49 am

Class Sheet - Average-1

Practice Questions- Q3

Solution

Number of rooms on the first floor = 30

Number of rooms on the 2nd floor = 40

Number of rooms on the 3rd floor = 40

Total rooms occupied on the first floor = % occupancy x Number of rooms = 60 % x 30 = 18

Total rooms occupied on the 2nd floor = 40 % x 40 = 16

Total rooms occupied on the 3rd floor = 75 % x 40 = 30

Total collection = 200 x 18 + 100 x 16 + 150 x 30 = Rs 9700

Average per room = Total collection/ Total rooms occupied = 9700 /(18 + 16 + 30 ) = 151.5 Rs

Practice Questions- Q3

Solution

Number of rooms on the first floor = 30

Number of rooms on the 2nd floor = 40

Number of rooms on the 3rd floor = 40

Total rooms occupied on the first floor = % occupancy x Number of rooms = 60 % x 30 = 18

Total rooms occupied on the 2nd floor = 40 % x 40 = 16

Total rooms occupied on the 3rd floor = 75 % x 40 = 30

Total collection = 200 x 18 + 100 x 16 + 150 x 30 = Rs 9700

Average per room = Total collection/ Total rooms occupied = 9700 /(18 + 16 + 30 ) = 151.5 Rs

- subish
**Posts:**4**Joined:**Sun Sep 08, 2013 6:49 am

CLASS SHEET - AVERAGE

Practice Question - Q8

solution

Assume total number of candidates to be n

total marks = n * x

because the average marks of 94 candidates came down from 84 to 64, therefore we subtract the wrongly entered total marks for 94 candidates and add the corrected total marks for them.

Corrected total marks = n * x - 94 * 84 + 94 * 64 ................(1)

Because the average for n students came down by 18.8 marks , the total marks = n ( x -18.8 ) ...................(2)

Equating 1 and 2 we have

n*x - 94 *84 + 94* 64 = n*(x-18.8 )

n *18.8 = 94 (84-64) => n = (94 *20)/18.8 => n = 100 option a

Practice Question - Q8

solution

Assume total number of candidates to be n

total marks = n * x

because the average marks of 94 candidates came down from 84 to 64, therefore we subtract the wrongly entered total marks for 94 candidates and add the corrected total marks for them.

Corrected total marks = n * x - 94 * 84 + 94 * 64 ................(1)

Because the average for n students came down by 18.8 marks , the total marks = n ( x -18.8 ) ...................(2)

Equating 1 and 2 we have

n*x - 94 *84 + 94* 64 = n*(x-18.8 )

n *18.8 = 94 (84-64) => n = (94 *20)/18.8 => n = 100 option a

- subish
**Posts:**4**Joined:**Sun Sep 08, 2013 6:49 am

QCS-6 - Average -

Q8. let the fourth number = x

Therefore, the average of the first three numbers = 2x

sum of the first three numbers = 3 (2x) = 6x

Average of the four numbers = 28

sum of all the four numbers = 28 x 4 = 6x + x

( Note that 6x is the sum of the first three numbers, while x is the the fourth number itself)

7x = 112 or x = 16

Which is the fourth number

Q8. let the fourth number = x

Therefore, the average of the first three numbers = 2x

sum of the first three numbers = 3 (2x) = 6x

Average of the four numbers = 28

sum of all the four numbers = 28 x 4 = 6x + x

( Note that 6x is the sum of the first three numbers, while x is the the fourth number itself)

7x = 112 or x = 16

Which is the fourth number

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

QCS- 6 Average

Q9

The average of the team = 84 , provided the best marksman scored 92. ( Acutally the best marksman has not scored 92)

Total score assuming that the best marksman scored 92 = 84 x 8 = 672

From the above total we must subtract, the score of the best marksman that was wrong ( i.e. 92) and must add back the actual score of the best marksman

Actual total scored = 672 -92 + 85 = 665

Q9

The average of the team = 84 , provided the best marksman scored 92. ( Acutally the best marksman has not scored 92)

Total score assuming that the best marksman scored 92 = 84 x 8 = 672

From the above total we must subtract, the score of the best marksman that was wrong ( i.e. 92) and must add back the actual score of the best marksman

Actual total scored = 672 -92 + 85 = 665

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

QCS-6

Q10

Total number of men and women = 12000

If the number of men = m , then the number of women can be taken as 12000-m

to solve the above question- let us write a basic equation

The total height of all men + the total height of all women = Total height of the entire population

Total height of all men = Average height of all men x the number of men = m ( 175

Total height of all women = Average height of all women x number of women = (12000-m) 165

Substituting the above values in our basic equation we get :

175 m + (12000-m) 165 = 120000 ( 172.5)

=> 175m + 165 x 12000 - 165m = 172.5 x 120000

=> 10 m = 12000 ( 172.5 -165) or m = 1200 x 7.5 or m = 9000

Therefore the number of women = 12000 -9000= 3000

Q10

Total number of men and women = 12000

If the number of men = m , then the number of women can be taken as 12000-m

to solve the above question- let us write a basic equation

The total height of all men + the total height of all women = Total height of the entire population

Total height of all men = Average height of all men x the number of men = m ( 175

Total height of all women = Average height of all women x number of women = (12000-m) 165

Substituting the above values in our basic equation we get :

175 m + (12000-m) 165 = 120000 ( 172.5)

=> 175m + 165 x 12000 - 165m = 172.5 x 120000

=> 10 m = 12000 ( 172.5 -165) or m = 1200 x 7.5 or m = 9000

Therefore the number of women = 12000 -9000= 3000

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

QCS-6

12

To solve such questions - one has to logically form an equation

Initial total weight - the weight of the boy who goes away = Final total weight of the group

Since the average weight is not given - we should assume the average weight to be x

=> Initial total weight of all boys = 40 (x)

=> 40(x) - 40 = 39 ( x + 0.1)

x = 40 + 3.9 or x = 43.9

12

To solve such questions - one has to logically form an equation

Initial total weight - the weight of the boy who goes away = Final total weight of the group

Since the average weight is not given - we should assume the average weight to be x

=> Initial total weight of all boys = 40 (x)

=> 40(x) - 40 = 39 ( x + 0.1)

- Note the new average weight of all 40 boys is 100g (which is 0.1 kg) less than the initial average ; therefore, the new average = x -0.1

Also, one boy leaves, therefore, the number of boys in the group is 39 now

x = 40 + 3.9 or x = 43.9

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

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