All Courses
Get the App Signup / Login
Sign in Open App

All questions of Averages for Delhi Police Constable Exam

Clear all your Delhi Police Constable doubts with EduRev

There are 7 members in a family whose average age is 25 years. Ram who is 12 years old is the second youngest in the family. Find the average age of the family in years just before Ram was born?
  • a)
    15.167
  • b)
    18.2
  • c)
    13
  • d)
    Cannot be determined
Correct answer is option 'D'. Can you explain this answer?

Rajeev Kumar answered
In order to find the average age of the family before Ram was born, we need to know the age of the youngest member of the family. 
Since, we do not know the age of the youngest member, we can not calculate the total age of the family before Ram was born.
Hence, we can not calculate the answer with the given conditions.
Thus, D is the right choice.

Can you explain the answer of this question below:
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

The answer is D.

Ishita Das answered
Average of 20 numbers = 0.
 Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).

Can you explain the answer of this question below:
 A car owner buys diesel at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends Rs. 4000 each year?
  • A:Rs. 8
  • B:Rs. 7.98
  • C:Rs. 6.2
  • D:Rs. 8.1

The answer is B.

Ishani Rane answered
Average cost per litre of petrol = Total amount / Total quantity of petrol

Re. 4000 is spent each year, so total amount spent = 3 * Rs. 4000 = Rs. 12,000

Total quantity of petrol consumed in 3 years = (4000/7.50) + (4000/8) + (4000/8.50) litres

= 533.3 + 500 + 470.6 = 1505

Average cost = Total amount / Total quantity

= 12000/1504 = 7.98

To find quickly, you can take cube root of (7.50 * 8 * 8.50), and it will be slightly less than 8.

The correct option is B.

The captain of a cricket team of 11 members is 26 years old and the wicket keeper 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
Correct answer is option 'A'. Can you explain this answer?

Pallabi Deshpande answered
Number of members in the team = 11
Let the average age of 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 = 11 x
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

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
Correct answer is 'B'. Can you explain this answer?

Manoj Ghosh answered
Total marks of first batch(55) is= 55*50=2750
Total marks of second batch (60) is= 60*55=3300
Total marks of third batch(45)= 45*60=2700

Total marks of second batch= 2750+3300+2700 = 8750
Total number of Students= 55+60+45 =160
Avg= 8750/160= 54.68

The average number of runs scored by Virat Kohli in four matches is 48. In the fifth match, Kohli scores some runs such that his average now becomes 60. In the 6th innings he scores 12 runs more than his fifth innings and now the average of his last five innings becomes 78. How many runs did he score in his first innings? (He does not remain not out in any of the innings)
  • a)
    30
  • b)
    50
  • c)
    70
  • d)
    90
Correct answer is option 'A'. Can you explain this answer?

Quantronics answered
Runs scored by Kohli in first 4 innings = 48*4 = 192
Average of 5 innings is 60, so total runs scored after 5 innings = 60*5 = 300
Hence runs scored by Kohli in fifth inning = 300 – 192 = 108
It is given that in 6th innings he scores 12 runs more than this, so he must score 120 in the sixth inning. Hence total runs scored in 6 innings = 300+120 = 420
Now average of last five innings is 78, so runs scored in last innings = 390
Hence runs scored in first inning = 420 – 390 = 30.

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
Correct answer is option 'A'. Can you explain this answer?

Manoj Ghosh answered
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

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
Correct answer is option 'A'. Can you explain this answer?

Arya Roy answered
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child (20 x 2 + 5 x 2) years  = 50 years.
Husband's present age = (90 - 50) years = 40 years.

The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
  • a)
    3500
  • b)
    4000
  • c)
    4050
  • d)
    5000
Correct answer is 'B'. Can you explain this answer?

Arya Roy answered
Let monthly income of A = a
monthly income of B = b 
monthly income of C = c
a + b = 2 * 5050 .... (Equation 1)
b + c = 2 * 6250 .... (Equation 2)
a + c = 2 * 5200 .... (Equation 3)
(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
i.e., Monthly income of A = 4000

Practice Quiz or MCQ (Multiple Choice Questions) with solution are available for Practice, which would help you prepare for "Average" under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.
 
Q. 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
Correct answer is option 'A'. Can you explain this answer?

Arya Roy answered
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

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
Correct answer is 'D'. Can you explain this answer?

Priyanka Datta answered
Average of 20 numbers = 0.
 Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).

The average weight of a class of 10 students is increased by 2 kg when one student of 30kg left and another student joined. After a few months, this new student left and another student joined whose weight was 10 less than the student who left now. What is the difference between the final and initial averages?
  • a)
    11
  • b)
    1
  • c)
    111
  • d)
    121
Correct answer is option 'B'. Can you explain this answer?

Rajeev Kumar answered
Change in total weight of 10 students = difference in weight of the student who joined and the student
=> weigth of first student who left = 30 + (10×2) = 50
weight of the student who joined last = 50 – 10 = 40...
Thus change in average weight = (40 – 30)/10 = 1...
 

The average age of a family of 5 members is 20 years. If the age of the youngest member is 10 years, what was the average age of the family at the birth of the youngest member?
  • a)
    12.50
  • b)
    15.25
  • c)
    21.25 
  • d)
    18.75 
Correct answer is option 'D'. Can you explain this answer?

Rajeev Kumar answered
At present the total age of the family = 5 × 20 =100
The total age of the family at the time of the birth of the youngest member,
= 100 - 10 - (10 × 4)
= 50
Therefore, average age of the family at the time of birth of the youngest member,
= 50/4 =12.5

Consider a class of 40 students whose average weight is 40 kgs. m new students join this class whose average weight is n kgs. If it is known that m + n = 50, what is the maximum possible average weight of the class now?
  • a)
    40.18 Kgs
  • b)
    40.56 Kgs
  • c)
    40.67 Kgs
  • d)
    40.49 Kgs
Correct answer is option 'B'. Can you explain this answer?

Rajeev Kumar answered
If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40.
So, n > 40
Consequently, m has to be < 10 (as n + m = 50)
Working with the “differences"? approach, we know that the total additional weight added by “m"? students would be (n - 40) each, above the already existing average of 40. m(n - 40) is the total extra additional weight added, which is shared amongst 40 + m students.
So, m * (n−40)(m+40)(n−40)(m+40) has to be maximum for the overall average to be maximum.

At this point, use the trial and error approach (or else, go with the answer options) to arrive at the answer.
The maximum average occurs when m = 5, and n = 45

And the average is 40 + (45 – 40) * 545545 = 40 + 5959 = 40.56 kgs
The question is "what is the maximum possible average weight of the class now?"
Hence, the answer is "40.56 kgs".

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
Correct answer is option 'D'. Can you explain this answer?

Ishani Rane answered
Average of 20 numbers = 0.

 Sum of 20 numbers (0 x 20) = 0.

It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).

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)
    28 kg
  • c)
    32 kg
  • d)
    30 kg
Correct answer is option 'A'. Can you explain this answer?

Kavya Saxena answered
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
 B's weight = 31 kg.

A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
  • a)
    Rs. 8
  • b)
    Rs. 7.98
  • c)
    Rs. 6.2
  • d)
    Rs. 8.1
Correct answer is option 'B'. Can you explain this answer?

Arya Roy answered
Average cost per litre of petrol = Total amount / Total quantity of petrol

Re. 4000 is spent each year, so total amount spent = 3 * Rs. 4000 = Rs. 12,000

Total quantity of petrol consumed in 3 years = (4000/7.50) + (4000/8) + (4000/8.50) litres

= 533.3 + 500 + 470.6 = 1505

Average cost = Total amount / Total quantity

= 12000/1504 = 7.98

To find quickly, you can take cube root of (7.50 * 8 * 8.50), and it will be slightly less than 8.

 A car owner buys diesel at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends Rs. 4000 each year?
  • a)
    Rs. 8
  • b)
    Rs. 7.98
  • c)
    Rs. 6.2
  • d)
    Rs. 8.1
Correct answer is option 'B'. Can you explain this answer?

Ishani Rane answered
Total quantity of petrol consumed in 3 years =(4000/7.50+4000/8+4000/8.50)  liters
= 4000(2/15+1/8+2/17) liters                                                   
= 76700/51 liters
Total amount spent = Rs. (3 x 4000) = Rs. 12000.
Average cost = Rs. (12000*51/76700) = Rs. 7.98.

The average weight of a class is 54 kg. A student, whose weight is 145 kg, joined the class and the average weight of the class now becomes a prime number less than 72. Find the total number of students in the class now.
  • a)
    7
  • b)
    13
  • c)
    15
  • d)
    Cannot be determined
Correct answer is option 'D'. Can you explain this answer?

Rajeev Kumar answered
Let the original number of students in the class be N.
Total weight of the class = 54N
New total weight of the class = 54N + 145
New average weight of the class = (54N + 145)/(N+1) = (54N + 54)/(N+1) + 91/(N+1) = 54 + 91/(N+1).
Since the new average is an integer, (N+1) should be a factor of 91.
If N+1 = 7, the new average becomes 54 + 91/7 = 54 + 13 = 67
and if N+1 = 13, then the new average becomes 54 + 91/13 = 54 + 7 = 61
Both 67 and 61 are prime numbers less than 72. So, we cannot uniquely determine the number of students in the class.

There are n weights in a bag measuring 1kg, 2kg and so on till n kg. These weights are divided into 3 parts. The first part contains the weights 1kg, 4kg, 7kg, and so on. The second part contains the weights 2kg, 5kg, 8kg and so on. The third part contains the remaining weights. The average weights any two of the three parts is equal to the weight present in those parts but the average weight of the remaining one part is not equal to the weight present in that part. Which of the following can be a possible value of n?
  • a)
    90
  • b)
    93
  • c)
    96
  • d)
    88
Correct answer is option 'D'. Can you explain this answer?

Rajeev Kumar answered
We know that if in an AP the number of terms in a series is odd, the average of the terms of the series is equal to the middle term of the series. However if the number of terms in the series is even, the average of all the terms of the series is not equal to one of the terms of the series. Hence the three part contain terms 2x+1, 2x+1, 2x or 2x-1, 2x-1, 2x
Hence the total number of parts = 2x+1+2x+1+2x or 2x-1+2x-1+2x = 6x+2 or 6x-2
Among the options, the only number of the form 6x+2 or 6x-2 is 88. Hence 88 can be the required value of n.

The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
  • a)
    3500
  • b)
    4000
  • c)
    4050
  • d)
    5000
Correct answer is option 'B'. Can you explain this answer?

Sameer Rane answered
Let monthly income of A = a
monthly income of B = b 
monthly income of C = c

a + b = 2 x 5050 .... (Equation 1)
b + c = 2 x 6250 .... (Equation 2)
a + c = 2 x 5200 .... (Equation 3)

(Equation 1) + (Equation 3) - (Equation 2)
=> a + b + a + c - (b + c) = (2 x 5050) + (2 x 5200) - (2 x 6250)
=> 2a = 2(5050 + 5200 - 6250)
=> a = 4000

i.e., Monthly income of A = 4000

Chapter doubts & questions for Averages - Quantitative Aptitude for Competitive Exams 2025 is part of Delhi Police Constable exam preparation. The chapters have been prepared according to the Delhi Police Constable exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Delhi Police Constable 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Averages - Quantitative Aptitude for Competitive Exams in English & Hindi are available as part of Delhi Police Constable exam. Download more important topics, notes, lectures and mock test series for Delhi Police Constable Exam by signing up for free.

Quantitative Aptitude for Competitive Exams

204 videos|108 docs|112 tests

Signup to see your scores go up within 7 days!

Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
© EduRev
Education Revolution
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily