Problems on Average - Solved Examples
 1. 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
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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 = $\dfrac{250}{40} = 6.25$

 2. 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
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Answer: Option B

Explanation:

Let the sale in the sixth month $=x$

Then $\dfrac{6435+6927+6855+7230+6562+x}{6}$ $=6500$

=> $6435+6927+6855+7230+6562+x$ $=6×6500$

=> $34009 + x = 39000$
=> $x = 39000 - 34009 = 4991$

 3. 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
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Answer: Option D

Explanation:

Average of 20 numbers = 0

=> $\dfrac{\text{Sum of 20 numbers}}{20} = 0$

=> 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)

 4. 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
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Answer: Option A

Explanation:

Number of members in the team = 11

Let the average age of of the team = $x$

=> $\dfrac{\text{Sum of ages of all 11 members}}{11} = x$

=> Sum of the ages of all 11 members = $11x$

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
$= \dfrac{11x - 55}{9}$

Given that $\dfrac{11x - 55}{9} = (x - 1)$

$\Rightarrow 11x - 55 = 9(x - 1)\\ \Rightarrow 11x - 55 = 9x - 9\\ \Rightarrow 2x = 46\\ \Rightarrow x = \dfrac{46}{2} = 23\text{ years}$

 5. 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
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Answer: Option C

Explanation:

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

 6. 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
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Answer: Option B

Explanation:

Total Cost = 4000 × 3

Total diesel used = $\dfrac{4000}{7.5}+\dfrac{4000}{8}+\dfrac{4000}{8.5}$

Average cost per litre of diesel
$=\dfrac{4000 \times 3}{\left(\dfrac{4000}{7.5} + \dfrac{4000}{8} + \dfrac{4000}{8.5}\right)}\\ = \dfrac{3}{\left(\dfrac{1}{7.5} + \dfrac{1}{8} + \dfrac{1}{8.5}\right)}$

It is important how you proceed from this stage. Remember time is very important and if we solve this in the normal method, it may take lot of time. Instead, we can find out the approximate value easily and select the right answer from the given choices.

In this case, answer
$=\dfrac{3}{\left(\dfrac{1}{7.5} + \dfrac{1}{8} + \dfrac{1}{8.5}\right)}\\ \approx \dfrac{3}{\left(\dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{8}\right)} \approx \dfrac{3}{\left(\dfrac{3}{8}\right)} \approx 8$

We got that answer as approximately equal to 8. From the given choices, the answer can be 8 or 7.98 or 8.1 . But which one from these?

It is easy to figure out. We approximated the denominator,$\left(\dfrac{1}{7.5}+\dfrac{1}{8}+\dfrac{1}{8.5}\right)$ to $\dfrac{3}{8}$. However
$\dfrac{1}{7.5}+\dfrac{1}{8.5}\\=\dfrac{1}{8-0.5}+\dfrac{1}{8+0.5}\\=\dfrac{8+0.5+8-0.5}{(8-0.5)(8+0.5)}\\ =\dfrac{16}{\left(8^2-0.5^2\right)} ~~\small{\left[∵ (a-b)(a+b)=a^2 - b^2\right]}\\ =\dfrac{16}{64-0.25}$

i.e., $\dfrac{1}{7.5} + \dfrac{1}{8.5} = \dfrac{16}{\left(64-0.25\right)}$
We know that $\dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}=\dfrac{16}{64}$

=> $\dfrac{1}{7.5}+\dfrac{1}{8.5} > \dfrac{1}{8} + \dfrac{1}{8}$

Early we had approximated the denominator to $\dfrac{3}{8}$. However, from the above mentioned equations, now we know that actual denominator is slightly greater than $\dfrac{3}{8}$. It means answer is slightly lesser than 8. Hence we can pick the choice 7.98 as the answer

Try to remember such relations between numbers which can save lot of time in calculations.

 7. 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
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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, x $\leq$ 68 ----(equation 3)

Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,we can write as

$65 < x \leq 68$

i.e., x = 66 or 67 or 68

Average of different probable weights of Kiran = $\dfrac{66+67+68}{3}=67$

 8. 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
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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 = $\dfrac{\left( 50.25 \times 16 \right)+\left( 45.15 \times 8\right)}{24}$
$= \dfrac{(50.25 \times 2)+(45.15 \times 1)}{3}\\=(16.75 \times 2)+ 15.05\\= 33.5 + 15.05\\=48.55$

 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
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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
$=\dfrac{(510 \times 5)+(240 \times 25)}{30}\\ =\dfrac{(51 \times 5)+(24 \times 25)}{3}\\=(17\times 5)+(8\times 25)\\=85+200\\=285$

 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 $\dfrac{1}{2}$. What is the number of students in the class? A. 45 B. 40 C. 35 D. 30
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Answer: Option B

Explanation:

Let the total number of students = $x$

The average marks increased by $\dfrac{1}{2}$ due to an increase of 83 - 63 = 20 marks.

But total increase in the marks = $\dfrac{1}{2}\times x = \dfrac{x}{2}$

Hence we can write as
$\dfrac{x}{2}=20\\ \Rightarrow x = 20 \times 2 = 40$

 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. $32 \dfrac{2}{7}$ years B. $31 \dfrac{5}{7}$ years C. $28\dfrac{1}{7}$ years D. $30 \dfrac{5}{7}$ years
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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
$=\dfrac{(67 \times 2)+(35 \times 2) + (6 \times 3)}{7}\\ =\dfrac{134+70+18}{7}\\=\dfrac{222}{7}\\= 31\dfrac{5}{7}$

 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. $28 \dfrac{1}{2}$ kg C. $32$ kg D. $30 \dfrac{1}{2}$ kg
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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 = 45
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 Kg

 13. 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
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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
$=\dfrac{(55 \times 50 )+ (60 \times 55)+(45 \times 60)}{160}\\ =\dfrac{275+ 330+ 270}{16}\\= \dfrac{875}{16}\\= 54.68$

 14. 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
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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 and their 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

i.e., Present age of the husband = 40

 15. 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. 75 Kg B. 50 Kg C. 85 Kg D. 80 Kg
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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

 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
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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
$=\dfrac{(36 \times 40)+(44 \times 35)}{80}\\ =\dfrac{(9 \times 40)+(11\times 35)}{20}\\ =\dfrac{(9 \times 8)+(11\times 7)}{4}\\ =\dfrac{72+77}{4}\\ =\dfrac{149}{4}\\ =37.25$

 17. 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
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Answer: Option A

Explanation:

Let the average after 17 innings = x
Total runs scored in 17 innings = 17x

Average after 16 innings = (x-3)
Total runs scored in 16 innings = 16(x-3)

Total runs scored in 16 innings + 87 = Total runs scored in 17 innings
=> 16(x-3) + 87 = 17x
=> 16x - 48 + 87 = 17x
=> x = 39

 18. 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
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Answer: Option C

Explanation:

$\dfrac{\text{3+11+7+9+15+13+8+19+17+21+14+}x}{12}$ $= 12$

$\Rightarrow \dfrac{137 + x}{12} = 12\\~\\ \Rightarrow 137 + x = 144\\~\\ \Rightarrow x = 144 - 137 = 7$

 19. 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
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Answer: Option D

Explanation:

Average mark = $\dfrac{76 + 65 + 82 + 67 + 85}{5}=\dfrac{375}{5}=75$

 20. 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
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Answer: Option C

Explanation:

-------------------------------------------
Solution 1
-------------------------------------------
If a car covers a certain distance at $x$ kmph and an equal distance at $y$ kmph. Then,
average speed of the whole journey = $\dfrac{2xy}{x+y}$ kmph.

By using the same formula, we can find out the average speed quickly.

Average speed
$=\dfrac{2 \times 84 \times 56}{84 + 56}=\dfrac{2\times84 \times 56}{140}\\ =\dfrac{2 \times 21 \times 56}{35}=\dfrac{2 \times 3 \times 56}{5}\\ =\dfrac{336}{5}=67.2$

-------------------------------------------
Solution 2
-------------------------------------------
Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics.

Let the distance between A and B = $x$

Train travels from A to B at 84 km per hour
Total time taken for traveling from A to B = $\dfrac{\text{distance}}{\text{speed}}=\dfrac{x}{84}$

Train travels from B to A at 56 km per hour
Total time taken for traveling from B to A = $\dfrac{\text{distance}}{\text{speed}}=\dfrac{x}{56}$

Total distance travelled = $x+x=2x$
Total time taken = $\dfrac{x}{84}+\dfrac{x}{56}$

Average speed $=\dfrac{\text{Total distance traveled}}{\text{Total time taken}}$

$=\dfrac{2x}{\dfrac{x}{84}+\dfrac{x}{56}}=\dfrac{2}{\dfrac{1}{84}+\dfrac{1}{56}}\\~\\ =\dfrac{2 \times 84 \times 56}{56 + 84}=\dfrac{2\times84 \times 56}{140}\\~\\ =\dfrac{2 \times 21 \times 56}{35}=\dfrac{2 \times 3 \times 56}{5}\\ =\dfrac{336}{5}= 67.2$

 21. The average age of boys in a class is 16 years and that of the girls is 15 years. What is the average age for the whole class? A. 15 B. 16 C. 15.5 D. Insufficient Data
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Answer: Option D

Explanation:

We do not have the number of boys and girls. Hence we cannot find out the answer.

 22. The average age of 36 students in a group is 14 years. When teacher's age is included to 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
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Answer: Option A

Explanation:

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 to 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

 23. 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
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Answer: Option D

Explanation:

Sum of 5 numbers = 5 × 27
Sum of 4 numbers after excluding one number = 4 × 25

Excluded number
= 5 × 27 - 4 × 25
= 135 - 100 = 35

 24. 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
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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 $=\dfrac{4}{2}$ = 2

Highest score = x + 172 = 2 + 172 = 174

 25. 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
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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 = $\dfrac{10 \times 38.9 - 6 \times 42}{4}$
$=\dfrac{389 - 252}{4}=\dfrac{137}{4}= 34.25$

 26. 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
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answer with explanation

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

 27. 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
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answer with explanation

Answer: Option C

Explanation:

-------------------------------------------
Solution 1
--------------------------------------------
If a car covers a certain distance at x kmph and an equal distance at y kmph. Then,
average speed of the whole journey = $\dfrac{2xy}{x+y}$ kmph.

Therefore, average speed
$=\dfrac{2 \times 50\times 30}{50 + 30} = \dfrac{2\times 50 \times 30}{80} \\ = \dfrac{2 \times 50 \times 3}{8}= \dfrac{50 \times 3}{4}\\ = \dfrac{25\times 3}{2} = \dfrac{75}{2} = 37.5$
-------------------------------------------
Solution 2
--------------------------------------------
Though it is a good idea to solve the problems quickly using formulas, you should know the fundamentals too. Let's see how we can solve this problems using basics.

Total time taken for traveling one side = $\dfrac{\text{distance}}{\text{speed}}=\dfrac{150}{50}$
Total time taken for return journey = $\dfrac{\text{distance}}{\text{speed}}=\dfrac{150}{30}$
Total time taken = $\dfrac{150}{50}+\dfrac{150}{30}$

Total distance travelled $= 150 + 150 = 2 × 150$

Average speed = $\dfrac{\text{Total distance traveled}}{\text{Total time taken}}$

$=\dfrac{2 \times 150}{\dfrac{150}{50}+\dfrac{150}{30}}=\dfrac{2}{\dfrac{1}{50}+\dfrac{1}{30}}\\~\\ =\dfrac{2 \times 50 \times 30}{30+ 50}=\dfrac{2\times 50 \times 30}{80}\\ =\dfrac{2 \times 50 \times 3}{8}=\dfrac{50 \times 3}{4}\\ =\dfrac{25\times 3}{2} = \dfrac{75}{2} = 37.5$

 28. 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
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answer with explanation

Answer: Option D

Explanation:

Total age of husband and wife (at the time of their marriage) = 2 × 23 = 46

Total age of husband and wife after 5 years + Age of the 1 year old child
= 46 + 5 + 5 + 1 = 57

Average age of the family = $\dfrac{57}{3}$ = 19

 29. 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
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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 = additional marks of the student = 20 + 2 = 22
=> 2x = 22
=> x = $\dfrac{22}{2}$ = 11

 30. 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
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answer with explanation

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

Count 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 = $\dfrac{42}{2}$ = 21

Comments(71) Sign in (optional)
showing 1-10 of 71 comments,   sorted newest to the oldest
Swati
2015-03-22 05:29:32
A batsman in his 20th inning makes a score of 110 and thereby increases his average by 4. What is his average before 20th inning?
(0) (0) Reply
Pankaj Maheshwari
2015-04-24 06:28:12
20*4 = 80.....
110-80 = 30
thats all
(0) (0) Reply
Dev
2015-04-05 17:32:24
Let the initial average be x
Then the total score in the first 19 innings = 19x

He scores 110 runs in the 20th inning.
Total score becomes 19x+110

average of the 20 innings is (x+4).
Total score of the 20 innings = 20(x+4)

19x+110 = 20(x+4)
x = 30
his average before 20th inning = 30
(0) (0) Reply
mahender
2015-03-17 05:11:48
A pool is there. In order to fill that pool, it takes 28 days.how many days does it take to fill half of the pool.day by day it will goes like twice
(0) (0) Reply
shyam
2015-09-25 11:05:47
since everyday pool it fills up twice, so it will fill to half in 27 days.
(0) (0) Reply
yavanika sharma
2015-03-12 07:11:49
At the end of a soccer season every player had a prime number of goals and the average of the 11 players was also a prime number. No player's individual tally was the same as of anyone else's or as the average. nobody has scored more than 45 goals.

Q1 what was the average of their goals scored?
Q2 what was the maximum goals scored by a single person?
Q3 what was the minimum number of goals scored by a single person?
(0) (0) Reply
Raj
2015-03-14 08:52:12
Consider all prime numbers less than 45 which is a possible super set of the 11 unique scores
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43}

Total of these numbers = 281
if we remove 3,5,7 (lowest three),  average will slightly above 24
if we remove
37,41,43, average will be slightly above 14
So, Possible values of the average are 17, 19 and 23 (because average is also a prime)

possible set is                                {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43}
reminder (when divided by by 11)   {2, 3, 5, 7,  0,    2,  6,   8,   1,  7,   9,   4,   8,  10
}

if the average is 17,
possible set is                                  {2, 3, 5, 7, 11, 13, 19, 23, 29, 31, 37, 41, 43}

Total of these 13 numbers = 264 .
Total of the actual 11 numbers = 17 * 11 = 187 (because average is 17)

Remainder of
264 when divided by 11 is 0.  So the two numbers which are subtracted from 264 should have the remainder sum of 11 (ie, sum of the numbers as 11, 22 etc) so that the net sum will be divisible by 11.

Even when we subtract any two numbers from 264 whose reminder sum is 11, we will not get 187. So average cannot be 17.

if the average is 19,
possible set is                                  {2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37, 41, 43}

Total of these 13 numbers = 262
Total of the actual 11 numbers = 19 * 11 = 209

Remainder of 262 when divided by 11 is 9. so the two numbers which are subtracted from 262 should have the remainder sum of 2
(ie, sum of the numbers as 2, 13, 24 etc) so that the net sum will be divisible by 11)

Even when we subtract any two numbers from 262 whose reminder sum is 2, we will not get 209 and hence, average cannot be 19

if the average is 23,
possible set is                                  {2, 3, 5, 7, 11, 13, 17, 19, 29, 31, 37, 41, 43}
remainder (when divided by by 11)    {2, 3, 5, 7,  0,    2,  6,  8,  7,   9,   4,   8,  10
}

Total of these 13 numbers = 258(remainder when divided by 11 is 5)
Total of the actual 11 numbers = 23 * 11 = 243
we can subtract 2 and 3 (
whose reminder sum is 5) such that is sum becomes 253 which is divisible by 11

So the average is 23
The numbers are
{5, 7, 11, 13, 17, 19, 29, 31, 37, 41, 43}

Now the questions can be easily answered
1. average of their goals scored : 23
2.
maximum goals scored by a single person : 43
3. minimum number of goals scored by a single person : 5
(0) (0) Reply
hyma
2015-03-11 10:48:18
there are 27 students in a class whose average weight is 36kg, among these 27 students 15 are boys and 12 are girls, the average weight of 15 boys is 6.75kg more than the average weight of 12 girls. what is the average weight of 12 girls(in kg)?
(0) (0) Reply
Dev
2015-03-11 17:41:36
Total weight of all students = 27*36 = 972

Let average weight of the 12 girls be x kg.
Then, the average weight of 15 boys = (x+6.75)

Total weight of the 12 girls = 12x
Total weight of the 15 boys = 15(x+6.75)

12x+15(x+6.75)=972
x = 32.25 kg
average weight of 12 girls = 32.25 kg
(0) (0) Reply
ishan
2015-01-20 15:15:13
the length of a rectangular plot is $4\dfrac{1}{2}$ times that of its breadth if the area of plot is 200 square metres then, then what is its length ?
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